![Section 1.1 Definitions and Examples 3
symbol C denotes the set of complex numbers.) Other fields can be employed
in place of JR and C, but they are rarely useful in applied mathematics. The
field elements are often termed scalars, and the elements of X are often called
vectors.
Let X be a vector space. A norm on X is a real-valued function, denoted
by II II, that fulfills three axioms:
(i) Ilxll > 0 for each nonzero element in X.
(ii) IIAxl1 = IAlllxl1 for each Ain JR and each x in X.
(iii) Ilx + YII ~ Ilxll + IIYII for all x, YE X. (Triangle Inequality)
A vector space in which a norm has been introduced is called a normed linear
space. Here are eleven examples.
Example 1.
function.
Let X = JR, and define Ilxll = lxi, the familiar absolute value
•
Example 2. Let X = C, where the scalar field is also C. Use Ilxll = lxi, where
Ixl has its usual meaning for a complex number x. Thus if x = a + ib (where a
and b are real), then Ixl = v'a2 + b2 . •
Example 3. Let X = C, and take the scalar field to be lR. The terminology
we have adopted requires that this be called a real vector space, since the scalar
field is lR. •
Example 4. Let X = JRn . Here the elements of X are n-tuples of real numbers
that we can display in the form x = [x(l), x(2), ... ,x(n)] or x = [Xl, X2, . .. ,xn ].
A useful norm is defined by the equation
IIxlioo = max Ix(i)1
l,;;;.';;;n
Note that an n-tuple is a function on the set {l, 2, ... ,n}, and so the notation
x(i) is consistent with that interpretation. (This is the "sup" norm.) •
Example 5. Let X = JRn , and define a norm by the equation Ilxll =
L~l Ix(i)l· Observe that in Examples 4 and 5 we have two distinct normed
linear spaces, although each involves the same linear space. This shows the ad-
vantage of being more formal in the definition and saying that a normed linear
space is a pair (X, II II) etc. etc., but we refrain from doing this unless it is
necessary. •
Example 6. Let X be the set of all real-valued continuous functions defined
on a fixed compact interval [a, b]. The norm usually employed here is
(The notation maxa~s';;;b Ix(s)1 denotes the maximum of the expression Ix(s)1 as
s runs over the interval [a, b].) The space X described here is often denoted
by C[a, b]. Sticklers would insist on C([a, b]), because C(S) will be used for
the continuous functions on some general domain S. (This again is the "sup"
norm.) •](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-16-320.jpg)
![Section 1.1 Definitions and Examples 5
(Linear independence is not a property of a point; it is a property of a set
of points. Because of this, the usage "the vectors... are independent" is mis-
leading.) The reader probably recalls how this notion enters into the theory
of nth-order ordinary differential equations: A general solution must involve a
linearly independent set of n solutions.
Some other basic concepts to recall from linear algebra are mentioned here.
The span of a set S in a vector space X is denoted by span(S), and consists
of all vectors in X that are expressible as linear combinations of vectors in S.
Remember that linear combinations are always finite expressions of the form
L~=l AiXi' We say that "S spans X" when X = span(S). A base or basis
for a vector space X is any set that is linearly independent and spans X. Both
properties are essential. Any set that is linearly independent is contained in a
basis, and any set that spans the space contains a basis. A vector space is said
to be finite dimensional if it has a finite basis. An important theorem states
that if a space is finite dimensional, then every basis for that space has the same
number of elements. This common number is then called the dimension of the
space. (There is an infinite-dimensional version of this theorem as well.)
The material of this chapter is accessible in many textbooks and treatises,
such as: [Au], [Av], [BN], [Ban], [Bea], rep], [Day], [Dies], [Dieu], [DS], [Edw],
[Frie2], [Fried], [GP], [Gre], [Gri], [HS], [HP], [Hoi], [Horv], [Jam], [KA], [Kee],
[KF], [Kre], [LanI], [Lo], [Moo], [NaSn], rOD], [Ped], [Red], [RS], [RN], [Roy],
[Rul], [Sim], [Tay2], [Yo], and [Ze].
Problems 1.1
Here is a Chinese proverb that is pertinent to the problems: I hear, I forget; I see, I
remember; I do, I understand!
1. Let X be a linear space over the complex field. Let XT be the space obtained from X by
restricting the scalars to the real field. Prove that XT is a real linear space. Show by an
example that not every real linear space is of the form XT for some complex linear space
X. Caution: When we say that a linear space is a real linear space, this has nothing to
do with the elements of the space. It means only that the scalar field is IR and not IC.
2. Prove the norm axioms for Examples 4-7.
3. Prove that in any normed linear space,
11011 =0 and !llxll - Ilyll! ~ Ilx - yll
4. Denote the norms in Examples 4 and 5 by II IL", and II Ill' respectively. Find the best
constants in the inequality
Prove that your constants are the best. (The "constants" a and (3 will depend on n but
not x.)
5. In Examples 4, 5, 6, and 7 find the precise conditions under which we have Ilx +yll =
IIxll + Ilyll·
6. Prove that in any normed linear space, if x # 0, then x/llxli is a vector of norm 1.
7. The Euclidean norm on IRn is defined in Example 11. Find the best constants in the
inequality ollxlloc ~ IIxl12 ~ (3llxllx'](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-18-320.jpg)
![6 Chapter 1 Normed Linear Spaces
8. What theorems in elementary analysis are needed to prove the closure axioms for Example
6?
9. What is the connection between the normed linear spaces f and II defined in Examples
8 and 1O?
10. For any t in the open interval (0,1), let t be the sequence [t, t2 , t3 , .. .J. Notice that
t E foe. Prove that the set {t: 0 < t < I} is linearly independent.
11. In the space II we define special elements called monomials. They are given by xn(t) =
tn where n =0, 1,2, ... Prove that {Xn : n =0, 1,2,3 ...} is linearly independent.
12. Let T be a set of real numbers. We say that T is bounded above if there is an M
in ]R such that t ~ M for all t in T. We say that M is an upper bound of T. The
completeness axiom for ]R asserts that if a set T is bounded above, then the set of
all its upper bounds is an interval of the form [b,oo). The number b is the least upper
bound, or supremum of T, written b = l.u.b.(T) = sup(T). Prove that if x < b, then
(x, oo)nT is nonempty. Give examples to show that [b, oo)nT can be empty or nonempty.
There are corresponding concepts of bounded below, lower bound, greatest lower
bound, and infimum.
13. Which of these expressions define norms on ]R2? Explain.
(a) max{lx(l)l, Ix(l) + x(2)1}
(b) Ix(2) - x(l)1
(c) Ix(l)1 + Ix(2) - x(l)1 + Ix(2)1
14. Prove that in any normed linear space the conditions Ilxll = 1 and IIx - yll < £ < 1 imply
that Ilx - Y/llylill < 2£.
15. Prove that if NI and N2 are norms on a linear space, then so are olNI + 02N2 (when
Or > 0 and 02 > 0) and (N'f + Ni)I/2.
16. Is the following set of axioms for a norm equivalent to the set given in the text? (a) Ilxll ¥
oif x ¥ 0, (b) IIAXII = -Allxll if A ~ 0, (c) IIx + yll ~ IIxll + lIyll·
17. Prove that in a normed linear space, if Ilx+yll = Ilxll+llyll, then Ilox+,Byll = lIoxll+ll,Byll
for all nonnegative 0 and ,B.
18. Why is the word "distinct" essential in our definition of linear independence on page 4?
19. Is the set of functions J;(x) = Ix - ii, where i = 1,2 ..., linearly independent?
20. One example of an "exotic" vector space is described as follows. Let X be the set
of positive real numbers. We define an "addition", Ell, by x Ell y = xy and a "scalar
multiplication" by a 0 x = xa. Prove that (X, Ell, 0) is a vector space.
21. In Example 10, two norms (say NI and N2) were suggested. Do there exist constants
such that NI ~ ON2 or N2 ~ ,BNI?
22. In Examples 4 and 5, let n = 2, and draw sketches of the sets {x E]R2 : Ilxll = I}.
(Symmetries can be exploited.)
1.2 Convexity, Convergence, Compactness, Completeness
A subset K in a linear space is said to be convex if it contains every line segment
connecting two of its elements. Formally, convexity is expressed as follows:
[XEK & yEK & O~"~l] ~ ..x+(l-..)YEK
The notion of convexity arises frequently in optimization problems. For example,
the theory of linear programming (optimization of linear functions) is based on](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-19-320.jpg)
![Section 1.2 Convexity, Convergence, Compactness, Completeness 7
the fact that a linear function on a convex polyhedral set must attain its extrema
at the vertices of the set. Thus, to locate the maxima of a linear function
over a convex polyhedral set, one need only test the vertices. The central idea
of Dantzig's famous simplex method is to move from vertex to vertex, always
improving the value of the objective function.
Another application of convexity occurs in studying deformations of a physi-
cal body. The "yield surface" of an object is generally convex. This is the surface
in 6-dimensional space that gives the stresses at which an object will fail struc-
turally. Six dimensions are needed to account for all the variables. See [Mar],
pages 100-104.
Among examples of convex sets in a linear space X we have:
(i) the space X itself;
(ii) any set consisting of a single point;
(iii) the empty set;
(iv) any linear subspace of X;
(v) any line segment; i.e. a set of the following form in which a and bare
fixed:
{Aa+(I-A)b: O~A~I}
In a normed linear space, another important convex set is the unit cell or unit
ball:
{x EX: Ilxll ~ I}
In order to see that the unit ball is convex, let Ilxll ~ 1, IIYII ~ 1, and 0 ~ A~ 1.
Then, with Jl = 1 - A,
If we let n = 2 in Examples 4 and 5 of Section 1.1, then we can draw pictures
of the unit balls. They are shown in Figures 1.1 and 1.2.
1
-1 1 -1
-1 -1
Figures 1.1 and 1.2. Unit balls
There is a family of norms on ]Rn, known as the t'p-norms, of which the norms
in Examples 4 and 5 are special cases. The general formula, for 1 ~ p < 00, is
(
n ) lip
Ilxllp = ~ Ix(i}IP](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-20-320.jpg)
![8 Chapter 1 Normed Linear Spaces
The case p = 00 is special; for it we use the formula
IIXlloo = max Ix(i)1
l~t~n
It can be shown (Problem 1) that limp
--+oo Ilxllp = Ilxlloo. (This explains the
notation.) The unit balls (in ]R2) for II lip are shown for p = 1, 2, and 7, in
Figure 1.3.
0.5
·0.5 0.5
-0.5
Figure 1.3. The unit balls in fp, for p = 1, 2, and 7.
In any normed linear space there exists a metric (and its corresponding
topology) that arises by defining the distance between two points as
d(x, y) = Ilx - YII
All the topological notions from the theory of metric spaces then become avail-
able in a normed linear space. (See Problem 23.) In Chapter 7, Section 6,
the theory of general topological spaces is broached. But we shall discuss here
topological concepts restricted to metric spaces or to normed linear spaces. A
sequence Xl, X2, ... in a normed linear space is said to converge to a point X
(and we write Xn --+ x) if
lim Ilxn - xii = 0
n--+oo
For example, in the space of continuous functions on [0,1J furnished with the
max-norm (as in Example 6 of Section 1, page 3), the sequence of functions
xn(t) = sin(t/n) converges to 0, since
IIXn - 011 = sup Isin(t/n)1 = sin(1/n) --+ 0
O~t~l
The notion of convergence is often needed in applied mathematics. For example,
the solution to a problem may be a function that is difficult to find but can be
approached by a suitable sequence of functions that are easier to obtain. (Maybe
they can be explicitly calculated.) One then would need to know exactly in what
sense the sequence was approaching the actual solution to the problem.
A subset K in a normed space is said to be compact if each sequence
in K has a subsequence that converges to a point in K. (Caution: In general
topology, this concept would be called sequential compactness. Refer to Section
7.6.) A subsequence of a sequence Xl, X2, ... is of the form x nl ' x n2 ' ... , where
the integers ni satisfy nl < n2 < n3 < .... Our notation for a sequence is [xn J,
or [xn : n E NJ, or [Xl, X2, .. .J. With this meagre equipment we can already
prove some interesting results.](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-21-320.jpg)
![Section 1.2 Convexity, Convergence, Compactness, Completeness 9
Theorem 1. Let K be a compact set in a normed linear space X.
To each x in X there corresponds at least one point in K of minimum
distance from x.
Proof. Let x be any member of X. The distance from x to K is defined to be
the number
dist (x, K) = inf Ilx - zll
zEK
By the definition of an infimum (Problem 12 in Section 1.1, page 6), there exists
a sequence [Yn] in K such that Ilx - Ynll-+ dist (x, K). Since K is compact,
there is a subsequence converging to a point in K, say Yni -+ Y E K. Since
Ilx - YII ~ Ilx - Yni II + IIYni - YII
we have in the limit Ilx-YII ~ dist (x, K) ~ Ilx-YII. (The final inequality follows
from the definition of the distance function.) •
The preceding theorem can be useful in problems involving noisy measure-
ments. For example, suppose that a noisy measurement of a single entity x is
available. If a set K of admissible noise-free values for x is prescribed, then
the best noise-free estimate of x can be taken to be a point of K as close as
possible to x. Theorem 1 is also important in approximation theory, a branch
of analysis that provides the theoretical underpinning for many areas of applied
mathematics.
Example 1. On the real line, an open interval (a, b) is not compact, for we
can take a sequence in the interval that converges to the endpoint b, say. Then
every subsequence will also converge to b. Since b is not in the interval, the
interval cannot be compact. On the other hand, a closed and bounded interval,
say [a, b], is compact. This is a special case of the Heine-Borel theorem. See the
discussion before Lemma 1 in Section 1.4, page 20. •
Given a sequence [xn] in a normed linear space (or indeed in any metric
space), is it possible to determine, from the sequence alone, whether it con-
verges? This is certainly an important matter for practical purposes, since we
often use algorithms to generate sequences that should converge to a solution
of a given problem. The answer to the posed question is that we cannot infer
convergence, in general, solely from the sequence itself. If we confine ourselves to
the information contained in the sequence, we can construct the doubly indexed
sequence Cnm = Ilxn- xmll. If [cnm] does not converge to zero, then the given
sequence [xn] cannot converge, as is easily proved: For any x in the space, write
This shows that if Cnm does not converge to 0, then [xn] cannot converge. On
the other hand, if Cnm converges to zero, one intuitively thinks that the sequence
ought to converge, and if it does not, there must be a flaw in the space itself: The
limit of the sequence should exist, but the limiting point is somehow missing from
the space. Think of the rational numbers as an example. The missing ingredient
is completeness of the space, to which we now turn.](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-22-320.jpg)
![10 Chapter 1 Normed Linear Spaces
A sequence [xn] in a normed linear space X is said to have the Cauchy
property or to be a Cauchy sequence if
lim sup Ilxi - Xj II = °
n~oo i~n
j~n
If every Cauchy sequence in the space X is convergent (to a point of X, of
course), then the space X is said to be complete. A complete normed linear
space is termed a Banach space, in honor of Stefan Banach, who lived from 1892
to 1945. His book [Ban] stimulated the study of functional analysis for several
decades. Examples 1-7, 9, and 11, given previously, are all Banach spaces.
The real number field IR is complete, and so is the complex number field C.
The rational field <Q is not complete. These facts are established in elementary
analysis courses.
Completeness is important in constructing solutions to a problem by taking
the limit of successive approximations. One often wants information about the
limit (i.e., the solution). Does it have the same properties as the approximations?
For example, if all the approximating functions are continuous, must the limit
also be continuous? If all the approximating functions are bounded, is the limit
also bounded? The answers to such questions depend on the sense in which the
limit is achieved; in other words, they depend on the norm that has been chosen
and the function space that goes with it. Typically, one wants a norm that leads
to a complete normed linear space, i.e., a Banach space.
Here is an example of a normed linear space that is not a Banach space:
Example 2. Let the space be the one described in Example 8 of Section 1.1,
page 4. This is e, the space of "finitely-nonzero sequences," with the "sup norm"
Ilxll = maxi Ix(i)l· Define a sequence [Xk] in eby the equation
Xk = [1,~,~, ... ,~,0, 0, ...J
If m > n, then
Xm - Xn = [0, ... ,0, n : l' ... , ~,0, ...J
Since IIXm -Xnll = 1/(n+ 1), we conclude that the sequence [Xk] has the Cauchy
property. If the space were complete, we would have Xn -+ y, where y E e. The
point y would be finitely nonzero, say y(n) = 0 for n > N. Then for m > N, Xm
would have as its Nth term the value liN, while the Nth term of y is O. Thus
IIXm - YII ~ liN, and convergence cannot take place. •
Theorem 2.
Banach space.
The space C[a,b] with norm Ilxll = maxs Ix(s)1 is a
Proof. Let [xn] be a Cauchy sequence in C[a, b]. (This space is described in
Example 6, page 3.) Then for each s, [xn(s)] is a Cauchy sequence in R Since IR
is complete, this latter sequence converges to a real number that we may denote](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-23-320.jpg)
![Section 1.2 Convexity, Convergence, Compactness, Completeness 11
by x(s). The function x thus defined must now be shown to be continuous, and
we must also show that Ilxn - xii -+ o. Let t be fixed as the point at which
continuity is to be proved. We write
This inequality should suggest to the reader how the proof must proceed. Let
e > o. Select N so that Ilxn - xmll ~ e/3 whenever m ~ n ~ N (Cauchy
property). Then for m ~ n ~ N, Ixn(s) - xm(s)1 ~ e/3. By letting m -+ 00 we
get Ixn(s) - x(s)1 ~ e/3 for all s. This shows that Ilxn- xii ~ e/3 and that the
sequence Ilxn - xii converges to o. By the continuity of Xn there exists a 6 > 0
such that IXn(s) - xn(t)1 < e/3 whenever It - sl < 6. Inequality (1) now shows
that Ix(s) - x(t)1 < e when It - sl < 6. (This proof illustrates what is sometimes
called "an e/3 argument.") •
Remarks. Theorem 2 is due to Weierstrass. It remains valid if the interval
[a, b] is replaced by any compact Hausdorff space. (For topological notions, refer
to Section 7.6, starting on page 361.) The traditional formulation of this theorem
states that a uniformly convergent sequence of continuous functions on a closed
and bounded interval must have a continuous limit. A sequence of functions [In]
converges uniformly to I if
(2) Ve 3n Vk Vs [k>n ====> IIk(s)-I(s)l<e]
(In this succinct description, it is understood that e > 0, n E N, kEN, and s is
in the domain of the functions.) By contrast, pointwise convergence is defined
by
Vs Ve 3n Vk [k>n ====> 1!k(s)-I(s)l<e]
Our use of the austere and forbidding logical notation is to bring out clearly
and to emphasize the importance of the order of the quantifiers. Thus, in the
definition of uniform convergence, n does not (cannot) depend on s, while in
the definition of pointwise convergence, n may depend on s. Notice that by the
definition of the norm being used, (2) can be written
or simply as limn
.....oo IIIn - 11100 = O. The latter is conceptually rather simple, if
one is already comfortable with this norm (called the "supremum norm" or the
"maximum norm").
The (perhaps) simplest example of a sequence of continuous functions that
converges pointwise but not uniformly to a continuous function is the sequence
[In] described as follows. The value of In(x) is 1 everywhere except on the
interval [0,2/n], where its value is given by Inx - 11.
Problems 1.2
1. Prove that limp-tx Ilxlip = maxU:;i~n Ix(i)1 for every x in IRn.](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-24-320.jpg)
![12 Chapter 1 Normed Linear Spaces
2. Is this property of a sequence equivalent to the Cauchy property?
lim sup IIXk - xnll =0
n---too k~n
Answer the same question for this property: For every positive E there is a natural number
n such that Ilxm - xnll < E whenever m ~ n.
3. Prove that if a sequence [Xn] in a Banach space satisfies 2::'=1 Ilxnll < 00, then the
series 2::'=1 Xn converges.
4. Prove that Theorem 2 is not true for the norm JIx(t)1 dt.
5. Prove that the union of a finite number of compact sets is compact. Give an example to
show that the union of an infinite family of compact sets can fail to be compact.
6. Prove that II lip on IRn does not satisfy the triangle inequality if 0 < p < 1 and n ~ 2.
7. Prove that if Xn -+ x, then the set {X,X1, X2, ... } is compact.
8. A cluster point (or accumulation point) of a sequence is the limit of any convergent
subsequence. Prove that if a sequence lies in a compact set and has only one cluster
point, then it is convergent.
9. Prove that the convergence in Problem 1 above is monotone.
10. Give an example of a countable compact set in IR having infinitely many accumulation
points. If your example has more than a countable number of accumulation points, give
another example, having no more than a countable number.
11. Let Xo and Xl be any two points in a normed linear space. Define X2, X3, ... inductively
by putting
n =0,1,2, ...
Prove that the resulting sequence is a Cauchy sequence.
12. A particular Banach space of great importance is the space '-=(S), consisting of all
bounded real-valued functions on a given set S. For X E loo(8) we define
Ilxli oo =sup Ix(s)1
sES
Prove that this space is complete. Cultural note: The space loc (ll) is of special interest.
Every separable metric space can be embedded isometrically in it! You might enjoy
trying to prove this, but that is not part of problem 12.
13. Prove that in a normed linear space a sequence cannot converge to two different points.
14. How does a sequence [Xn : n E !I] differ from a countable set {Xn : n E !I}?
15. Is there a norm that makes the space of all real sequences a Banach space?
16. Let Co denote the space of all real sequences that converge to zero. Define Ilxll
sUPn Ix(n)l. Prove that Co is a Banach space.
17. If K is a convex set in a linear space, then these two sets are also convex:
u + K = {u + X : x E K} and AK = {AX: x E K}
18. Let A be a subset of a linear space. Put
AC
= {tAiai : n E!I , Ai ~ 0 , ai E A, t Ai =1}
Prove that A C AC Prove that AC is convex. Prove that AC is the smallest convex set
containing A. This latter assertion means that if A is contained in a convex set B, then
AC is also contained in B. The set AC is the convex hull of A.](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-25-320.jpg)
![Section 1.2 Convexity, Convergence, Compactness, Completeness 13
19. If A and B are convex sets, is their vector sum convex? The vector sum of these two sets
isA+B={a+b: aEA,bEB}.
20. Can a norm be recovered from its unit ball? Hint: If x E X, then X/A is in the unit
ball whenever IAI ~ Ilxli. (Prove this.) On the other hand, X/A is not in the unit ball if
IAI < Ilxll· (Prove this.)
21. What are necessary and sufficient conditions on a set 8 in a linear space X in order that
8 be the unit ball for some norm on X?
22. Prove that the intersection of a family of convex sets (all contained in one linear space)
is convex.
23. A metric space is a pair (X, d) in which X is a set and d is a function (called a metric)
from X x X to IR such that
(i) d(x, y) ~ 0
(ii) d(x, y) =0 if and only if X = y
(iii) d(x, y) =d(y, x)
(iv) d(x, y) ~ d(x, z) + d(z, y)
Prove that a normed linear space is a metric space if d(x, y) is defined as II x - y II.
24. For this problem only, we use the following notation for a line segment in a linear space:
(a, b) = {Aa+ (1-A)b: 0 ~ A ~ 1}
A polygonal path joining points a and b is any finite union of line segments
U~=l (ai, ai+l), where al =a and an+l =b. If the linear space has a norm, the length
of the polygonal path is L.:~=l Ila; - ai+lli. Give an example of a pair of points a, bin
a normed linear space and a polygonal path joining them such that the polygonal path
is not identical to (a, b) but has the same length. A path of length Iia - bll connecting a
and b is called a geodesic path. Prove that any geodesic polygonal path connecting a
and b is contained in the set {x: Ilx - all ~ lib - all}.
25. If Xn -+ x and if the Cesaro means are defined by an = (Xl +...+xn)/n, then an -+ x.
(This is to be proved in an arbitrary normed linear space.)
26. Prove that a Cauchy sequence that contains a convergent subsequence must converge.
27. A compact set in a normed linear space must be bounded; i.e., contained in some multiple
of the unit ball.
28. Prove that the equation f(x) = L.:;;"=o ak cos bkx defines a continuous function on IR,
provided that 0 ~ a < l. The parameter b can be any real number. You will find
useful Theorem 2 and Problem 3. Cultural Note: If 0 < a < 1 and if b is an odd
integer greater than a-I, then f is differentiable nowhere. This is the famous Weierstrass
nondifferentiable function. (See Section 7.8, page 374, for more information about this
function.)
29. Prove that a sequence [xn] in a normed linear space converges to a point x if and only if
every subsequence of [Xn] converges to x.
30. Prove that if ¢ is a strictly increasing function from N into N, then ¢(n) ~ n for all n.
3l. Let 8 be a subset of a linear space. Let 81 be the union of all line segments that join
pairs of points in 8. Is 81 necessarily convex?](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-26-320.jpg)
![14 Chapter 1 Normed Linear Spaces
32. (continuation) What happens if we repeat the process and construct 82 , 83 , ...? (Thus,
for example, 82 is the union of line segments joining points in 81.)
33. Let I be a compact interval in JR, I = [a, b]. Let X be a Banach space. The notation
C(I, X) denotes the linear space of all continuous maps I : I --t X. We norm C(I, X)
by putting 11111 = SUPtEI III(t)lI. Prove that C(I, X) is a Banach space.
34. Define In(x) =e-nx . Show that this sequence of functions converges pointwise on [0,1]
to the function 9 such that g(O) = 1 and g(t) = 0 for t -# O. Show that in the L2-norm
on [0,1], In converges to O. The L2-norm is defined by 11111 = {fo1 II(t)i2dt}1/2.
35. Let [Xn] be a sequence in a Banach space. Suppose that for every c > 0 there is a
convergent sequence [Yn] such that sUPn IIxn - Ynll < c. Prove that [xn] converges.
36. In any normed linear space, define K(x, r) = {y : IIx - yll ~ r}. Prove that if K(x, ~) c
K(O, 1) then 0 E K(x, ~).
37. Show that the closed unit ball in a normed linear space cannot contain a disjoint pair of
closed balls having radius ~.
38. (Converse of Problem 3) Prove that if every absolutely convergent series converges
in a normed linear space, then the space is complete. (A series 2:Xn is absolutely
convergent if 2: IIxnll < 00.)
39. Let X be a compact Hausdorff space, and let C(X) be the space of all real-valued
continuous functions on X, with norm 11111 =supII(x)l. Let [In] be a Cauchy sequence
in C(X). Prove that
lim lim In(x) = lim lim In(x)
X---+Xo n-+oo n-+<Xl x---t-XQ
Give examples to show why compactness, continuity, and the Cauchy property are needed.
40. The space £1 consists of all sequences x = [x(1),x(2), ... ] in which x(n) E JR and
2: Ix(n)1 < 00. The space £2 consists of sequences for which 2: Ix(n)12 < 00. Prove
that £1 C £2 by establishing the inequality 2: Ix(n)12 ~ (2: Ix(n)1)2.
41. Let X be a normed linear space, and 8 a dense subset of X. Prove that if each Cauchy
sequence in 8 has a limit in X, then X is complete. A set 8 is dense in X if each point
of X is the limit of some sequence in 8.
42. Give an example of a linearly independent sequence [xo, Xl, X2,"'] of vectors in loo such
that 2::'=0 Xn =O. Don't forget to prove that 2: Xn =O.
43. Prove, in a normed space, that if Xn --t X and Ilxn - Ynll --t 0, then Yn --t X. If Xn --t X
and IIxn - Yn II --t 1, what is lim Yn ?
44. Whenever we consider real-valued or complex-valued functions, there is a concept of
absolute value of a function. For example, if x E C[O, 1], we define Ixl by writing Ixl(t) =
/x(t)/. A norm on a space of functions is said to be monotone if IIxll ;;:: IIYII whenever
Ixl ;;:: IYI· Prove that the norms II 1100 and II lip are monotone norms.
45. (Continuation) Prove that there is no monotone norm on the space of all real-valued
sequences.
46. Why isn't the example of this section a counterexample to Theorem 2?](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-27-320.jpg)
![Section 1.3 Continuity, Open Sets, Closed Sets
Theorem 4. The inverse image of a closed set by a continuous map
is closed.
17
Preof. Recall that the inverse image of a set A by a map f is defined to be
f-' (A) = {x : f(x) E A}. Let f : X -t Y, where X and Yare normed spaces
and I is continuous. Let K be a closed set in Y. To show that f-1(K) is closed,
we start by letting [xn ] be a convergent sequence in f-1(K). Thus Xn -t x and
f(xn ) E K. By continuity, f(xn ) -t f(x). Since K is closed, f(x) E K. Hence
x E f-1(K). •
As an example, consider the unit ball in a normed space:
{x: Ilxll ~ I}
This is the inverse image of the closed interval [0,1] by the function x H Ilxll.
This function is continuous, as shown above. Hence, the unit ball is closed.
Likewise, each of the sets
{x: IIx-all ~r}
is closed.
{x: IIx-all ~r} {x: Q ~ IIx - all ~ ,6}
An open set is a set whose complement is closed. Thus, from the preceding
remarks, the so-called "open unit ball," i.e., the set
u = {x : Ilxll < I}
is open, because its complement is closed. Likewise, all of these sets are open:
{x: Ilxll > I} {x: IIx - all < r} {x: Q < Ilx - all <,6}
An alternative way of describing the open sets, closer to the spirit of general
topology, will now be discussed.
The open c-cell or c-ball about a point Xo is the set
B(xo,c) = {x : Ilx - xoll < c}
Sometimes this is called the c-neighborhood of Xo. A useful characterization of
open sets is the following: A subset U in X is open if and only if for each x E U
there is an c > 0 such that B(x, c) C U. The collection of open sets is called the
topology of X. One can verify easily that the topology T for a normed linear
space has these characteristic properties:
(1) the empty set, 0 , belongs to T;
(2) the space itself, X, belongs to T;
(3) the intersection of any two members of T belongs to T;
(4) the union of any subfamily of T belongs to T.
These are the axioms for any topology. One section of Chapter 7 provides an
introduction to general topology.
A series 2::%"=1 xk whose elements are in a normed linear space is conver-
gent if the sequence of partial sums Sn = 2::Z=l Xk converges. The given series
is said to be absolutely convergent if the series of real numbers 2::%"=1 Ilxkll
is convergent. That means simply that 2::%"=1 IIXkl1 < 00. Problem 3, page 13,
asks for a proof that absolute convergence implies convergence, provided that
the space is complete. See also Problem 38, page 14. The following theorem
gives another important property of absolutely convergent series.](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-30-320.jpg)
![Section 1.4 More about Compactness 19
2. Let U be an arbitrary subset of a normed space. Prove that the function x ...... dist(x, U)
is continuous. This function was defined in the proof of Theorem 1 in Section 1.2, page
9. Prove, in fact, that it is "nonexpansive":
Idist(x, U) - dist(y, U)I ::::; Ilx - yll
3. Let X be a normed space. We make X x X into a normed linear space by defining
lI(x,y)1I = IIxll + Ilyll· Show that the map (x,y) ...... x + y is continuous. Show that the
norm is continuous. Show that the map (A, x) ...... AX is continuous when lR x X is normed
by II(A,x)11 = IAI + IIxll·
4. Prove that the intersection of a family of closed sets is closed.
5. If x # 0, put x= x/llxll. This defines the mdial projection of x onto the surface of the
unit ball. Prove that if x and y are not zero, then
Ilx - ilil ::::; 211x - yll/llxli
6. Use Theorem 2 and Problem 2 in this section to give a brief proof of Theorem 1 in
Section 2, page 9.
7. Using the definition of an open set as given in this section, prove that a set U is open if
and only if for each x in U there is a positive c such that B(x, c) C U.
8. Prove that the inverse image of an open set by a continuous map is open.
9. The (algebraic) sum of two sets in a linear space is defined by A + B = {a + b: a E A,
b E B}. Is the sum of two closed sets (in a normed linear space) closed? (Cf. Problem
19, page 13.)
10. Prove that if the series L:':l Xi converges (in some normed linear space), then Xi --t 0.
11. A common misconception about metric spaces is that the closure of an open ball S = {x :
d(a, x) < r} is the closed ball S' = {x : d(a, x) ::::; r}. Investigate whether this is correct
in a discrete metric space (X,d), where d(x,y) = 1 if x # y. What is the situation in a
normed linear space? (Refer to Problem 23, page 13.)
12. Let L Xn and L Yn be two series of nonnegative terms. Prove that if one of these series
converges but the other does not, then the series L(Xn - Yn) diverges. Can you improve
this result by weakening the hypotheses?
13. Let L Xn be a convergent series of real numbers such that L IXn I = 00. Prove that the
series of positive terms extracted from the series L Xn diverges to 00. It may be helpful
to introduce Un = max(xn, 0) and Vn = min(xn,O). By using the partial sums of series,
one reduces the question to matters concerning the convergence of sequences.
14. Refer to Problem 12, page 12, for the space too(S). We write::::; to signify a pointwise
inequality between two members of this space. Let 9n and In be elements of this space,
for n = 1,2, ... Let 9n ~ 0, In-l - 9n-l ::::; In ::::; M, and L~ 9i ::::; M for all n. Prove
that the sequence [In] converges pointwise. Give an example to show that convergence
in norm may fail.
1.4 More About Compactness
We continue our study of compactness in normed linear spaces. The starting
point for the next group of theorems is the Heine-Borel theorem, which states
that every closed and bounded subset of the real line is compact, and conversely.
We assume that the reader is familiar with that theorem.](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-32-320.jpg)
![20 Chapter 1 Normed Linear Spaces
Our first goal in this section is to show that the Heine-Borel theorem is true
for a normed linear space if and only if the space is finite-dimensional. Since most
interesting function spaces are infinite-dimensional, verifying the compactness of
a set in these spaces requires information beyond the simple properties of being
bounded and closed. Many important theorems in functional analysis address
the question of identifying the compact sets in various normed linear spaces.
Examples of such theorems will appear in Chapter 7.
Lemma 1.
each ball {x
In the space IRn with norm Ilxlloo = maxl~i~n Ix(i)1
Ilxlloo ~ c} is compact.
Proof. Let [xkl be a sequence of points in IRn satisfying Ilxklloo ~ c. Then
the components obey the inequality -c ~ xk(i) ~ c. By the compactness of the
interval [-c, c], there exists an increasing sequence heN having the property
that lim [Xk(1) : k E Id exists. Next, there exists another increasing sequence
12 C h such that lim [xk(2) : k E 12l exists. Then lim [Xk(1) : k E 12l
exists also, because h C h. Continuing in this way, we obtain at the nth
step an increasing sequence In such that lim[xdi) : k E Inl exists for each
i = 1, ... , n. Denoting that limit by x'(i), we have defined a vector x' such that
Ilxk - x'iloo ---+ 0 as k runs through the sequence of integers In. •
Lemma 2. A closed subset of a compact set is compact.
Proof. If F is a closed subset of a compact set K, and if [xnl is a sequence in
F, then by the compactness of K a subsequence converges to a point of K. The
limit point must be in F, since F is closed. •
A subset S in a normed linear space is said to be bounded if there is a
constant c such that Ilxll ~ c for all XES. Expressed otherwise, sUPrES Ilxll <
00.
Theorem 1. In a finite-dimensional normed linear space, each
closed and bounded set is compact.
Proof. Let X be a finite-dimensional normed linear space. Select a basis for
X, say {Xl, ... ,xn}. Define a mapping T : IRn ---+ X by the equation
n
Ta = La(i)Xi a= (a(l), ... ,a(n)) EIRn
i=l
If we assign the norm II 1100 to IRn , then T is continuous because
IITa - Tbll = II t(a(i) - b(i))Xill ~ tla(i) - b(i)1 Ilxill
n n
~ mfx la(i) - b(i)I' L IIXjl1 = Iia - blloo L IIXj II
j=l j=l](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-33-320.jpg)
![Section 1.4 More about Compactness 21
Now let F be a closed and bounded set in X. Put M = T-1(F). Then M is
closed by Theorem 4 in Section 1.3, page 17. Since F = T(M), we can use
Theorem 1 in Section 1.3, page 15, to conclude that F is compact, provided that
M is compact. To show that M is compact, we can use Lemmas 1 and 2 above
if we can show that for some c,
In other words, we have only to prove that M is bounded. To this end, define
{3 = inf{ IITal1 :Iialloo = I}
This is the infimum of a continuous map on a compact set (prove that). Hence
the infimum is attained at some point b. Thus Ilblloo = 1 and
{3 = IITbl1 = II ~b(i)Xill
Since the points Xi constitute a linearly independent set, and since b i- 0, we
conclude that Tb i- 0 and that {3 > O. Since F is bounded, there is a constant
csuch that Ilxll ~ cfor all x E F. Now, if a E ]Rn and a i- 0, then ajllalloo is a
vector of norm 1; consequently, IIT(ajllalloo)11 ~ {3, or
This is obviously true for a = 0 also. For a E M we have Ta E F, and
{3llalloo ~ IITal1 ~ c, whence Iialloo ~ cj{3. Thus, M is indeed bounded. •
Corollary 1. Every finite-dimensional normed linear space is com-
plete.
Proof. Let [xnl be a Cauchy sequence in such a space. Let us prove that the
sequence is bounded. Select an index m such that Ilxi - Xjll < 1 whenever
i, j ~ m. Then we have
(i ~ m)
Hence for all i,
Since the ball of radius c is compact, our sequence must have a convergent
subsequence, say xni ~ x'. Given E: > 0, select N so that Ilxi - Xj II < E: when
i,j ~ N. Then IIXj - XnJ < E: when i,j ~ N, because ni > i. By taking the
limit as i ~ 00, we conclude that IIXj - x'il ~ E: when j ~ N. This shows that
Xj ~ x. •](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-34-320.jpg)
![22 Chapter 1 Normed Linear Spaces
Corollary 2. Every finite-dimensional subspace in a normed linear
space is closed.
Proof. Recall that a subset Y in a linear space is a subspace if it is a linear
space in its own right. (The only axioms that require verification are the ones
concerned with algebraic closure of Y under addition and scalar multiplication.)
Let Y be a finite-dimensional subspace in a normed space. To show that Y is
closed, let Yn E Y and Yn -+ y. We want to know that Y E Y. The preceding
corollary establishes this: The convergent sequence has the Cauchy property and
hence converges to a point in Y, because Y is complete. •
Riesz's Lemma. If U is a closed and proper subspace (U is neither
onor the entire space) in a normed linear space, and if0 < A < 1, then
there exists a point x such that 1 = Ilxll and dist(x, U) > A.
Proof. Since U is proper, there exists a point z EX" U. Since U is
closed, dist(z, U) > o. (See Problem 11.) By the definition of dist(z, U) there
is an element u in U satisfying the inequality liz - ull < A-I dist(z, U). Put
x = (z - u)/llz - ull. Obviously, Ilxll = 1. Also, with the help of Problem 7, we
have
dist(x, U) = dist(z - u, U)/llz - ull = dist(z, U)/llz - ull > A •
Theorem 2. If the unit ball in a normed linear space is compact,
then the space has finite dimension.
Proof. If the space is not finite dimensional, then a sequence [xn ] can be
defined inductively as follows. Let Xl be any point such that IlxI11 = 1. If
Xl, ... ,xn-l have been defined, let Un - l be the subspace that they span. By
Corollary 2, above, Un - l is closed. Use Riesz's Lemma to select Xn so that
IIXnl1 = 1 and dist(xn,Un-I) > 1· Then IIXn - xiii> 1whenever i < n. This
sequence cannot have any convergent subsequence. •
Putting Theorems 1 and 2 together, we have the following result.
Theorem 3. A normed linear space is finite dimensional ifand only
if its unit ball is compact.
In any normed linear space, a compact set is necessarily closed and bounded.
In a finite-dimensional space, these two conditions are also sufficient for compact-
ness. In any infinite-dimensional space, some additional hypothesis is required
to imply compactness. For many spaces, necessary and sufficient conditions for
compactness are known. These invariably involve some uniformity hypothesis.
See Section 7.4, page 347, for some examples, and [DS] (Section IV.14) for many
others.
Problems 1.4
1. A real-valued function f defined on a normed space is said to be lower semicontinuous
if each set {x : f{x) ~ >.} is closed (>. E JR). Prove that every continuous function is lower](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-35-320.jpg)
![Section 1.4 More about Compactness 23
semicontinuous. Prove that if f and -fare lower semicontinuous, then f is continuous.
Prove that a lower semicontinuous function attains its infimum on a compact set.
2. Prove that the collection of open sets (as we have defined them) in a normed linear space
fulfills the axioms for a topology.
3. Two norms, Nl and N2, on a vector space X are said to be equivalent if there exist
positive constants 0 and (3 such that ONl ~ N2 ~ (3Nl . Show that this is an equivalence
relation. Show that the topologies engendered by a pair of equivalent norms are identical.
4. Prove that a Cauchy sequence converges if and only if it has a convergent subsequence.
5. Let X be the linear subspace of all real sequences x = [x(l), x(2), ... J such that only a
finite number of terms are nonzero. Is there a norm for X such that (X, II II) is a Banach
space?
6. Using the notation in the proof of Theorem 1, prove in detail that F = T(M).
7. Prove these properties of the distance function dist(x, U) (defined in Section 1.2, page 9)
when U is a linear subspace in a normed linear space:
(a) dist(Ax, U) = IAI dist(x, U)
(b) dist(x - u, U) = dist(x, U) (u E U)
(c) dist(x + y, U) ~ dist(x, U) +dist(y, U)
8. Prove this version of Riesz's Lemma: If U is a finite-dimensional proper subspace in a
normed linear space X, then there exists a point x for which IIxll = dist(x, U) = 1.
9. Prove that if the unit ball in a normed linear space is complete, then the space is complete.
10. Let U be a finite-dimensional subspace in a normed linear space X. Show that for each
x E X there exists a u E U satisfying IIx - ull =dist(x, U).
11. Let U be a closed subspace in a normed space X. Prove that the distance functional has
the property that for x EX" U, dist(x,U) > O.
12. In any infinite-dimensional normed linear space, the open unit ball contains an infinite
disjoint family of open balls all having radius ~ (!!) (Prove it, of course. While you're at
it, try to improve the number ~.)
13. In the proof of Theorem 1, show that M is bounded as follows. If it is not bounded,
let ak E M and lIaklloc --+ 00. Put a~ = ak/llakiloc. Prove that the sequence [a~]
has a convergent subsequence whose limit is nonzero. By considering Ta~, obtain a
contradiction of the injective nature of T.
14. Prove that the sequence [xnJ constructed in the proof of Theorem 2 is linearly indepen-
dent.
15. Prove that in any infinite-dimensional normed linear space there is a sequence [xn] in
the unit ball such that IIxn - Xm II > 1 when n # m. If you don't succeed, prove the
same result with the weaker inequality IIxn - Xm II ~ 1. (Use the proof of Theorem 2 and
Problem 8 above.) Also prove that the unit ball in eoe contains a sequence satisfying
Ilxn - xmll =2 when n # m. Reference: [DiesJ.
16. Let S be a subset of a normed linear space such that IIx - yll ~ 1 when x and y are
different points in S. Prove that S is closed. Prove that if S is an infinite set then it
cannot be compact. Give an example of such a set that is bounded and infinite in the
space e[O, 1J.
17. Let A and B be nonempty closed sets in a normed linear space. Prove that if A + B is
compact, then so are A and B. Why do we assume that the sets are nonempty? Prove
that if A is compact, then A +B is closed.](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-36-320.jpg)
![Section 1.5 Linear Transformations
Theorem 1. A linear transformation acting between normed linear
spaces is continuous if and only if it is continuous at zero.
25
Proof. Let T : X -t Y be such a linear transformation. If it is continuous, then
of course it is continuous at O. For the converse, suppose that T is continuous
atO. For each € > 0 there is a 8 > 0 such that for all x,
jjxjj < 8 => jjTxjj < €
Hence
jjx - yjj < 8 => jjTx - Tyjj = jjT(x - y)jj < €
•
A linear transformation T acting between two normed linear spaces is said
to be bounded if it is bounded in the usual sense on the unit ball:
sup{ jjTxjj : jjxjj ~ I} < 00
Example 7. Let X = C 1[0,1]' the space of all continuously differentiable
functions on [0,1]. Give X the norm jjxjjoo = sup Ix(s)l. Let f be the linear
functional defined by f (x) == x'(1). This functional is not bounded, as is seen
by considering the vectors Xn (s) = sn. On the other hand, the functional in
Example 2 is bounded since If(x)1 ~ J; Ix(s)1 ds ~ jjxjjoo· •
Theorem 2. A linear transformation acting between normed linear
spaces is continuous if and only if it is bounded.
Proof. Let T : X -t Y be such a map. If it is continuous, then there is a 8 > 0
such that
jjxjj ~ 8 => jjTxjj ~ 1
If jjxjj ~ 1, then 8x is a vector of norm at most 8. Consequently, jjT(8x)jj ~ 1,
whence jjTxjj ~ 1/8. Conversely, if jjTxjj ~ M whenever jjxjj ~ 1, then
jjxll ~ :r => lI~xll ~ 1 => liT (~x) II ~ M => IITxl1 ~ €
This proves continuity at 0, which suffices, by the preceding theorem. •
If T : X -t Y is a bounded linear transformation, we define
IITII = sup{ IITxl1 : Ilxll ~ I}
It can be shown that this defines a norm on the family of all bounded linear
transformations from X into Y; this family is a vector space, and it now becomes
a normed linear space, denoted by £(X, Y).
The definition of jjTjjleads at once to the important inequality
jjTxjj ~ jjTjj jjxjj
To prove this, notice first that it is correct for x = 0, since TO = O. On the other
hand, if x#- 0, then x/jjxjj is a vector of norm 1. By the definition of jjTjj, we
have jjT(x/jjxjj)jj ~ jjTjj, which is equivalent to the inequality displayed above.
That inequality contains three distinct norms: the ones defined on X, Y, and
£(X, Y).](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-38-320.jpg)
![26 Chapter 1 Normed Linear Spaces
Theorem 3. A linear functional on a normed space is continuous if
and only if its kernel ("null space") is closed.
Proof. Let f : X -t lR be a linear functional. Its kernel is
ker(f) = {x : f(x) = O}
This is the same as f-1({O}). Thus if f is continuous, its kernel is closed, by
Theorem 4 in Section 1.3, page 17. Conversely, if f is discontinuous, then it is
not bounded. Let Ilxnll ::::;; 1 and f(xn ) -t 00. Take any x not in the kernel and
consider the points x - EnXn, where En = f(x)/ f(xn ). These points belong to
the kernel of f and converge to x, which is not in the kernel, so the latter is not
closed. •
Corollary 1. Every linear functional on a finite-dimensional normed
linear space is continuous.
Proof. If f is such a functional, its null space is a subspace, which, by Corollary
2 in Section 1.4, page 22, must be closed. Then Theorem 3 above implies that
f is continuous. •
Corollary 2. Every linear transformation from a finite-dimensional
normed space to another normed space is continuous.
Proof. Let T : X -t Y be such a transformation. Let {b1 , . .. , bn } be a basis
for X. Then each x E X has a unique expression as a linear combination of
basis elements. The coefficients depend on x, and so we write x = L~=1 Ai(X)bi.
These functionals Ai are in fact linear. Indeed, from the previous equation and
the equation u = L Ai(u)bi we conclude that
n
ax + j3u = Z)aAi(X) + j3Ai(U)] bi
i=1
Since we have also
n
ax + j3u = L Ai(ax + j3u)bi
i=1
we may conclude (by the uniqueness of the representations) that
Now use the preceding corollary to infer that the functionals Ai are continuous.
Getting back to T, we have
and this is obviously continuous.
•](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-39-320.jpg)
![Section 1.5 Linear Transformations 27
Corollary 3. All norms on a finite-dimensional vector space are
equivalent, as defined in Problem 3, page 23.
Proof. Let X be a finite-dimensional vector space having two norms 11111 and
11112· The identity map J from (X, II Ill) to (X, II IU is continuous by the
preceding result. Hence it is bounded. This implies that
By the symmetry in the hypotheses, there is a (3 such that IIxl11 ~ (3llxI12. •
Recall that if X and Yare two normed linear spaces, then the notation
£(X, Y) denotes the set of all bounded linear maps of X into Y. We have seen
that boundedness is equivalent to continuity for linear maps in this context. The
space £(X, Y) has, in a natural way, all the structure of a normed linear space.
Specifically, we define
(aA +(3B)(x) = a(Ax) +(3(Bx)
IIAII = sup{ IIAxlly :x EX, Ilxllx ~ l}
In these equations, A and B are elements of £(X, Y), and x is any member of
X.
Theorem 4. If X is a normed linear space and Y is a Banach space,
then £(X, Y) is a Banach space.
Proof. The only issue is the completeness of £(X, Y). Let [An] be a Cauchy
sequence in £(X, Y). For each x E X, we have
This shows that [Anx] is a Cauchy sequence in Y. By the completeness of Y we
can define Ax = lim Anx. The linearity of A follows by letting n -+ 00 in the
equation
The boundedness of A follows from the boundedness of the Cauchy sequence
[An]. If IIAnl1 ~ M then IIAnXl1 ~ Mllxll for all x, and in the limit we have
IIAxl1 ~ Mllxll· Finally, we have IIAn- AII-+ 0 because if IIAn- Am II ~ c when
m,n;;;: N, then for all x of norm 1 we have IIAnx - AmXl1 ~ c when m,n;;;: N.
Then we can let m -+ 00 to get IIAnx - Axil ~ c and IIAn - All ~ c. •
The composition of two linear mappings A and B is conventionally written
as AB rather than AoB. Thus, (AB)x = A(Bx). If AA is well-defined (Le., the
range of A is contained in its domain), then we write it as A2. All nonnegative
powers are then defined recursively by writing AD = J, An+1 = AAn.](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-40-320.jpg)
![28 Chapter 1 Normed Linear Spaces
Theorem 5. The Neumann Theorem. Let A be a bounded
linear operator on a Banach space X (and taking values in X). If
IIAll < 1, then I - A is invertible, and
00
(I - A)-I = L Ak
k=O
Proof. Put Bn = 2:~=o Ak. The sequence [Bn] has the Cauchy property, for
if n > m, then
IIBn - Bmll = 1 kf+1Akll ~ kf+IIIAkll ~ ~IIAllk
00
= IIAllmL IIAllk = IIAllm/(1-IIAII)
k=O
(In this calculation we used Problem 20.) Since the space of all bounded linear
operators on X into X is complete (Theorem 4), the sequence [Bn] converges to
a bounded linear operator B. We have
n n+1
(I - A)Bn = Bn - ABn = L Ak - L Ak = J - An+1
k=O k=1
Taking a limit, we obtain (J - A)B = I. Similarly, B(I - A) = I. Hence
B = (I - A)-I. •
The Neumann Theorem is a powerful tool, having applications to many
applied problems, such as integral equations and the solving of large systems of
linear equations. For examples, see Section 4.3, which is devoted to this theorem,
and Section 3.3, which has an example of a nonlinear integral equation.
Problems 1.5
1. Prove that the closure of a linear subspace in a normed linear space is also a subspace.
(The closure operation is defined on page 16.)
2. Prove that the operator norm defined here has the three properties required of a norm.
3. Prove that the kernel of a linear functional is either closed or dense. (A subset in a
topological space X is dense if its closure is X.)
4. Let {Xl, ... , Xk} be a linearly independent finite set in a normed linear space. Show that
there exists a 8 > 0 sum that the condition
max IIXi-Yill<8
l~i~k
implies that {Yl, ... ,Yk} is also linearly independent.
5. Prove directly that if T is an unbounded linear operator, then it is discontinuous at O.
(Start with a sequence [xnJ such that Ilxnll ~ 1 and IITxnll -+ 00.)](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-41-320.jpg)
![Section 1.5 Linear Transformations 29
6. Let A be an m x n matrix. Let X = IRn , with norm IIxllco =maxl ';:; i';:;n Ix(i)l. Let Y =
IRm , with norm lIyllco = maxl ';:;i';:;m ly(i)l· Define a linear transformation T from X to
Y by putting (Tx)(i) =2::7= 1 a;j x(j), 1 ~ i ~ m. Prove that IITII = max; 2::7= 1Ia;jl.
7. Prove that a linear map is injective (i.e., one-to-one) if and only if its kernel is the 0
subspace. (The kernel of a map T is {x : Tx = O}.)
8. Prove that the norm of a linear transformation is the infimum of all the numbers l'v! that
satisfy the inequality IITxl1 ~ Mllxll for all x.
9. Prove the (surprising) result that a linear transformation is continuous if and only if it
transforms every sequence converging to zero into a bounded sequence.
10. If f is a linear functional on X and N is its kernel, then there exists a one-dimensional
subspace Y such that X = Y EEl N. (For two sets in a linear space, we define U + V as
the set of all sums u + v when u ranges over U and v ranges over V. If U and V are
subspaces with only 0 in common we write this sum as U EEl V.)
11. The space eco(S) was defined in Problem 12 of Section 1.2, page 12. Let S = N, and
define T: eoc(N) -+ c[-!,!l by the equation (Tx)(s) = 2::;;"=1 x(k)sk. Prove that T is
linear and continuous.
12. Prove or disprove: A linear map from a normed linear space into a finite-dimensional
normed linear space must be continuous.
13. Addition of sets in a vector space is defined by A + B = {a + b : a E A , b E B} .
Better: A + B = {x : 3 a E A & 3 b E B such that x = a + b}. Scalar multiplication
is AA = {Aa : a E A}. Does the family of all subsets of a vector space X form a vector
space with these definitions?
14. Let Y be a closed subspace in a Banach space X. A "coset" is a set of the form x +Y =
{x +y : y E V}. Show that the family of all cosets is a normed linear space if we use the
norm III x + Y III = dist(x, V).
15. Refer to Problem 12 in the preceding section, page 23. Show that the assertion there is
not true if ~ is replaced by !.
16. Prove that for a bounded linear transformation T : X -+ Y
IITII = sup IITxll = sup IITxll/llxll
IIrll=1 #0
17. Prove that a bounded linear transformation maps Cauchy sequences into Cauchy se-
quences.
18. Prove that if a linear transformation maps some nonvoid open set of the domain space
to a bounded set in the range space, then it is continuous.
19. On the space C[O, 1] we define "point-evaluation functionals" by t*(x) = x(t). Here
t E (0, 1] and x E C(O, 1]. Prove that IIt'li = 1. Prove that if I/> = 2::7=1 A;t: , where
tl , t2, · .·,tn are distinct points in [0, 1], then 111/>11 = 2::7=1 IA; I·
20. In the proof of the Neumann Theorem we used the inequality IIAk II ~ IIAlik . Prove this.
21. Prove that if {<PI, ... ,qln} is a linearly independent set of linear functionals, then for
suitable Xj we have qI;(Xj) =8ij for 1 ~ i,j ~ n.](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-42-320.jpg)
![30 Chapter 1 Normed Linear Spaces
22. Prove that if a linear transformation is discontinuous at one point, then it is discontinuous
everywhere.
23. Linear transformations on infinite-dimensional spaces do not always behave like their
counterparts on finite-dimensional spaces. The space Co was defined in Problem 1.2.16
(page 12). On the space Co define
Ax =A[x(I), x(2), ...j = [x(2), x(3), ...j
Bx = B[x(I), x(2), ...j = [0, x(I), x(2), ...j
Prove that A is surjective but not invertible. Prove that B is injective but not invertible.
Determine whether right or left inverses exist for A and B.
24. What is meant by the assertion that the behavior of a linear map at any point of its
domain is exactly like its behavior at o?
25. Prove that every linear functional f on IRn has the form f(x) = 2:~=1 oix(i), where
x(I), x(2), ... ,x(n) are the coordinates of x. Let 0 = [01,02, ... , On] and show that the
relationship f ...... 0 is linear, injective, and surjective (hence, an isomorphism).
26. Is it true for linear operators in general that continuity follows from the null space being
closed?
27. Let <Po,(Pl, ... ,<Pn be linear functionals on a linear space. Prove that if the kernel of <Po
contains the kernels of all <Pi for 1 :(; i :(; n, then <Po is a linear combination of <PI,... ,<Pn.
28. If L is a bounded linear map from a normed space X to a Banach space Y, then L has a
unique continuous linear extension defined on the completion of X and taking values in
Y. (Refer to Problem 1.2.47, page 15.) Prove this assertion as well as the fact that the
norm of the extension equals the norm of the original L.
29. Let A be a continuous linear operator on a Banach space X. Prove that the series
2::'=0 An In! converges in £(X, X). The resulting sum can be denoted by eA. Is eA
invertible?
30. Investigate the continuity of the Laplace transform (in Example 5, page 24).
1.6 Zorn's Lemma, Hamel Bases, and the Hahn-Banach Theorem
This section is devoted to two results that require the Axiom of Choice for their
proofs. These are a theorem on existence of Hamel bases, and the Hahn-Banach
Theorem. The first of these extends to all vector spaces the notion of a base,
which is familiar in the finite-dimensional setting. The Hahn-Banach Theorem
is needed at first to guarantee that on a given normed linear space there can
be defined continuous maps into the scalar field. There are many situations
in applied mathematics where the Hahn-Banach Theorem plays a crucial role;
convex optimization theory is a prime example.
The Axiom of Choice is an axiom that most mathematicians use unre-
servedly, but is nonetheless controversial. Its status was clarified in 1940 by a
famous theorem of Godel [Go]. His theorem can be stated as follows.](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-43-320.jpg)
![Section 1.6 Zorn's Lemma, Hamel Bases, Hahn-Banach Theorem 31
Theorem 1. If a contradiction can be derived from the Zermelo-
Fraenkel axioms of set theory (which include the Axiom of Choice),
then a contradiction can be derived within the restricted set theory
based on the Zermelo-Fraenkel axioms without the Axiom of Choice.
In other words, the Axiom of Choice by itself cannot be responsible for intro-
ducing an inconsistency in set theory. That is why most mathematicians are
willing to accept it. In 1963, Paul Cohen [Coh] proved that the Axiom of Choice
is independent of the remaining axioms in the Zermelo-F'raenkel system. Thus
it cannot be proved from them. The statement of this axiom is as follows:
Axiom of Choice. If A is a set and / a function on A such that
/(a) is a nonvoid set for each a E A, then / has a "choice function."
That means a function c on A such that c(a) E /(a) for all a E A.
For example, suppose that A is a finite set: A = {a1, ... ,an}. For each i in
{1,2, ... ,n} a nonempty set /(ai) is given. In n steps, we can select "repre-
sentatives" Xl E /(a1), X2 E /(a2), etc. Having done so, define c(ad = Xi for
i = 1,2, ... , n. Attempting the same construction for an infinite set such as
A = JR, with accompanying infinite sets /(a), leads to an immediate difficulty.
To get around the difficulty, one might try to order the elements of each set /(a)
in such a way that there is always a "first" element in /(a). Then c(a) can be
defined to be the first element in /(a). But the proposed ordering will require
another axiom at least as strong as the Axiom of Choice! For a second example,
see Problem 45, page 40.
A number of other set-theoretic axioms are equivalent to the Axiom of
Choice. See [Kel] and [RR]. Among these equivalent axioms, we single out
Zorn's Lemma as being especially useful. First, we require some definitions.
Definition 1. A partially ordered set is a pair (X, -<) in which X is a set
and -< is a relation on X such that
(i) X -< X for all X
(ii) If x -< y and y -< Z, then x -< Z
Definition 2. A chain, or totally ordered set, is a partially ordered set in
which for any two elements x and y, either x -< y or y -< x.
Definition 3. In a partially ordered set X, an upper bound for a subset A
in X is any point x in X such that a -< x for all a E A.
Example 1. Let S be any set, and denote by 28 the family of all subsets of
S, including the empty set 0 and S itself. This is often called the "power set"
of S. Order 28 by the inclusion relation c. Then (28 , C) is a partially ordered
set. It is not totally ordered. An upper bound for any subset of 28 is S. •
Example 2. In JR2, define x -< y to mean that x(i) ~ y(i) for i = 1 and 2.
This is a partial ordering but not a total ordering. Which quadrants in JR2 have
upper bounds? •
Example 3. Let:F be a family of functions (whose ranges and domains need
not be specified). For / and g in :F we write / -< g if two conditions are fulfilled:](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-44-320.jpg)
![Section 1.6 Zorn's Lemma, Hamel Bases, Hahn-Banach Theorem 35
Proposition. The mapping u f--t 1>u is an isometric isomorphism
between £1 and co' Thus we can say that Co "is" £1.
Proof. Perhaps we had better give a name to this mapping. Let A : £1 -+
Co be defined by Au =: 1>u. It is to be shown that for each u, Au is linear
and continuous on co. Then it is to be shown that A is linear, surjective, and
isometric. Isometric means IIAul1 =: Ilulll. That 1>u is well-defined follows from
the absolute convergence of the series defining 1>u(x):
The linearity of 1>u is obvious:
1>u(ax +(3y) =: L u(n) [ax(n) +(3y(n)] =: a L u(n)x(n) +(3 L u(n)y(n)
=: a1>u(x) +(31)u(Y)
The continuity or boundedness of 1>u is easy:
By taking a supremum in this last inequality, considering only x for which
Ilxlloo ::;; 1, we get
On the other hand, if c > 0 is given, we can select N so that I:~N+l lu(n)1 < c.
Then we define x by putting x(n) =: sgn u(n) for n ::;; N, and by setting x(n) =: 0
for n > N. Clearly, x E Co and Ilxlloo =: 1. Hence
N N
II1>ull ~ 1>u(x) =: Lx(n)u(n) =: L lu(n)1 > Ilulll - c
n=l n=l
Since c was arbitrary, II1>ull ~ Ilulll. Hence we have proved
Next we show that A is surjective. Let '¢ E co. Let <5n be the element of Co
that has a 1 in the nth coordinate and zeros elsewhere. Then for any x,
00
x=: L x(n)<5n
n=l
Since '¢ is continuous and linear,](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-48-320.jpg)
![36 Chapter 1 Normed Linear Spaces
Consequently, if we put u(n) ='I/J(8n ), then 'I/J(x) =¢u(x) and 'I/J =¢u' To verify
that u E £1, we define (as above) x(n) = sgnu(n) for n ~ Nand x(n) = 0 for
n> N. Then
N
2:lu(n)1 =2: x(n)u(n) ='I/J(x) ~ 11'l/Jllllxll = 1I'l/J1l
n=l
Thus Ilull l ~ 1I'l/J1l.
Finally, the linearity of A follows from writing
¢O:U+{JV(x) = 2:(au +,Bv)(n)x(n) =a 2:u(n)x(n) +,B 2:v(n)x(n)
= (a¢u +,B¢v)(x) •
Corollary 4. For each x in a normed linear space X, we have
IIxll = max{I¢(x)1 : ¢ E X' , II¢II = 1}
Proof. If ¢ E X' and II¢II = 1, then
1¢(x)1 ~ 1I¢lIlIxll = IIxll
Therefore,
sup{I¢(x)1 : ¢ E X' , II¢II = 1} ~ IIxll
For the reverse inequality, note first that it is trivial if x = o. Otherwise, use
Corollary 3. Then there is a functional 'I/J E X' such that 'I/J(x) = IIxll and
1I'l/J1l = 1. Note that the supremum is attained. •
A subset Z in a normed space X is said to be fundamental if the set of
all linear combinations of elements in Z is dense in X. Expressed otherwise, for
each x E X and for each e > 0 there is a vector L~=l .Zi such that Zi E Z,
Ai E JR, and
IIx - 2:AiZili < e
We could also state that dist(x,span Z) = 0 for all x E X. As an example, the
vectors
81 = (1,0,0, ... )
82 = (0,1,0, ... )
etc.
form a fundamental set in the space Co.
Example 6. In the space C[a, b], with the usual supremum norm, an important
fundamental set is the sequence of monomials
uo(t) = 1 , U1(t) = t , U2(t) = t2 , ...
The Weierstrass Approximation Theorem asserts the fundamentality of this se-
quence. Thus, for any x E C[a, bJ and any e > 0 there is a polynomial u for
which IIx - ulloo < e. Of course, uis of the form L~o AiUi' •
Definition 5. If A is a subset of a normed linear space X, then the annihilator
of A is the set
AJ.. = {¢ E X' : ¢(a) = 0 for all a E A}](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-49-320.jpg)
![38 Chapter 1 Normed Linear Spaces
13. Let loo denote the normed linear space of all bounded real sequences, with norm given
by IIxIL", =sUPn Ix{n)l. Prove that loe is complete, and therefore a Banach space. Prove
that lj = loo, where the equality here really means isometrically isomorphic.
14. A hyperplane in a normed space is any translate of the null space of a continuous,
linear, nontrivial functional. Prove that a set is a hyperplane if and only if it is of the
form {x : ¢(x) = ,X}, where ¢ E X' "0 and ,X E lIt A translate of a set S in a vector
space is a set of the form v +S = {v +s : S E S}.
15. A half-space in a normed linear space X is any set of the form {x : ¢(x) ~ ,X}, where
¢ E X' "0 and ,X E JR. Prove that for every x satisfying IIxll = 1 there exists a half-space
such that x is on the boundary of the half-space and the unit ball is contained in the
half-space.
16. Prove that a linear functional ¢ is a linear combination of linear functionals ¢1, ... , ¢n if
and only if N(¢) :l n::"1 N(¢i). Here N(¢) denotes the null space of ¢. (Use induction
and trickery.)
17. Prove that a linear map transforms convex sets into convex sets.
18. Prove that in a normed linear space, the closure of a convex set is convex.
19. Let Y be a linear subspace in a normed linear space X. Prove that
dist(x, Y) =sup{¢(x) : ¢ E X' , ¢ 1. Y , II¢II = I}
Here the notation ¢ 1. Y means that ¢(y) =0 for all y E Y.
20. Let Y be a subset of a normed linear space X. Prove that Y -L is a closed linear subspace
in X'.
21. If Z is a linear subspace in X', where X is a normed linear space, we define
Z-L ={x EX: ¢(x) =0 for all ¢ E Z}
Prove that for any closed subspace Y in X, (Y -L h = Y. Generalize.
22. Let J(z) = L.::;;"=o anzn, where [an] is a sequence of complex numbers for which nan --t O.
Prove the famous theorem of Tauber that L.:: an converges if and only if limz--+1 J(z)
exists. (See [DS], page 78.)
23. Do the vectors tln defined just after Corollary 4 form a fundamental set in the space lX)
consisting of bounded sequences with norm Ilxlloo =maxn Ix(n)l?
THREE EXERCISES (24-26) ON SCHAUDER BASES (See [Sem] and [Sing].)
24. A Schauder base (or basis) for a Banach space X is a sequence [un] in X such that eacil
x in X has a unique representation
00
x = L,XnUn
n=l
This equation means, of course, that limN--+ocllx - L.::~=1 'xnunll = O. Show that one
Schauder base for Co is given by un(m) =tlnm (n, m = 1,2,3, ...).](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-51-320.jpg)
![Section 1.6 Zorn's Lemma, Hamel Bases, Hahn-Banach Theorem 39
25. Prove that the An in the preceding problem are functions of x and must be, in fact, linear
and continuous.
26. Prove that if the Banach space X possesses a Schauder base, then X must be separable.
That is, X must contain a countable dense set.
27. Prove that for any set A in a normed linear space all these sets are the same:
A-L, (closure A)-L, (span A)-L, [closure (span A)]-L,.
28. Prove that for x E co,
29. Use the Axiom of Choice to prove that for any set S having at least 2 points there is a
function f : S --t S that does not have a fixed point.
30. An interesting Banach space is the space C consisting of all convergent sequences. The
norm is IIxlloc = sUPn Jx(n)J. Obviously, we have these set inclusions among the examples
encountered so far:
Prove that Co is a hyperplane in c. Identify in concrete terms the conjugate space c'.
31. Prove that if H is a Hamel base for a normed linear space, then so is {h/llhil :h E H}.
32. Let X and Y be linear spaces. Let H be a Hamel base for X. Prove that a linear map
from X to Y is completely determined by its values on H, and that these values can be
arbitrarily-assigned elements of Y.
33. Prove that on every infinite-dimensional normed linear space there exist discontinuous
linear functionals. (The preceding two problems can be useful here.)
34. Using Problem 33 and Problem 1.5.3, page 28, prove that every infinite-dimensional
normed linear space is the union of a disjoint pair of dense convex sets.
35. Let two equivalent norms be defined on a single linear space. (See Problem 1.4.3, page
23.) Prove that if the space is complete with respect to one of the norms, then it is
complete with respect to the other. Prove that this result fails (in general) if we assume
only that one norm is less than or equal to a constant multiple of the other.
36. Let Y be a subspace of a normed space X. Prove that there is a norm-preserving injective
map J : Y' --t X' such that for each <P E Y', J<p is an extension of <p.
37. Let Y be a subspace of a normed space X. Prove that if Y -L =0, then Y is dense in X.
38. Let T be a bounued linear map of Co into Co. Show that T must have the form (Tx)(n) =
2.::':1 ani X
(i) for a suitable infinite matrix [ani]. Prove that sUPn 2.::':1 Jani J = IITII·
39. Prove that if #S =n, then #2s = 2n.
40. What implications exist among these four properties of a set S in a normed linear space
X? (a) S is fundamental in X; (b) S is linearly independent; (c) S is a Schauder base
for X; (d) S is a Hamel base for X.](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-52-320.jpg)
![Section 1.7 The Baire Theorem and Uniform Boundedness 41
Then take S~ C S2 n O2 , S~ C S3 n 0 3 , and so on. At the same time we can
insist that Tn .J.. o. Then for all n,
The points Xn form a Cauchy sequence because Xi, Xj E Sn if i,j > n, and so
Since X is complete, the sequence [xnl converges to some point X'. Since for
i > n,
Xi E S~+1 C S1 nOn
we can let i -+ 00 to conclude that x' E S~+1 C S1 n On. Since this is true for
all n, the set n:=1On does indeed intersect S1. •
Corollary. If a complete metric space is expressed as a countable
union of closed sets, then one of the closed sets must have a nonempty
interior.
Proof. Let Xbe a complete metric space, and suppose that X= U:=1 Fn ,
where each Fn is a closed set having empty interior. The sets On = X "Fn are
open and dense. Hence by Baire's Theorem, n:=1On is dense. In particular, it
is nonempty. If X E n:=1On, then X EX" U:=1 Fn, a contradiction. •
A subset in a metric space X (or indeed in any topological space) is said to
be nowhere dense in X if its closure has an empty interior. Thus the set of
irrational points on the horizontal axis in ]R2 is nowhere dense in 1R2 . A set that
is a countable union of nowhere dense sets is said to be of category I in X. A
set that is not of category I is said to be of category II in X.
Observe that all three of these notions are dependent on the space. Thus
one can have E C X C Z, where E is of category II in X and of category I in
Z. For a concrete example, the one in the preceding paragraph will serve.
The Corollary implies that if X is a complete metric space, then X is of the
second category in X.
Intuitively, we think of sets of the first category as being "thin," and those
of the second category as "fat." (See Problems 5, 6, 7, for example.)
Theorem 2. The Banach-Steinhaus Theorem. Let {Ao}
be a family of continuous linear transformations defined on a Banach
space X and taking values in a normed linear space. In order that
sUPo IIAol1 < 00, it is necessary and sufficient that the set {x EX:
sUPo IIAoxl1 < oo} be of the second category in X.
Proof. Assume first that c = sUPo IIAo II < 00. Then every x satisfies IIAoxl1 :::;;
cllxll, and every x belongs to the set F = {x : sUPo IIAoxl1 < oo}. Since F = X,
the preceding corollary implies that F is of the second category in X.
For the sufficiency, define
Fn = {x EX: sup IIAoxl1 :::;; n}
o](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-54-320.jpg)
![on the ends of small branchlets and often persisting for some
time.
The balsam fir (Abies balsamea).—Leaves narrow, less than one inch
long, borne singly, very numerous and standing out from the
branchlets in much the way of the spruce; cones about three
inches long, cylindrical, composed of thin scales, and standing
upright on the branches, or recurved; bark smooth, light green
with whitish tinge.
The arbor-vitæ (Thuya occidentalis).—Leaves very small, scale-like,
and over-lapping one another in four rows, adhering closely to
the branchlets; the cones oblong and small,—a half-inch or less
in length,—and composed of but few scales.
LEAFLET XXXIV.
THE CLOVERS AND THEIR KIN.[46]
By ANNA BOTSFORD COMSTOCK.
The pedigree of honey does not concern the bee,
A clover any time to him is aristocracy.
—Emily Dickinson.
here is a deep-seated prejudice that usefulness
and beauty do not belong together;—a
prejudice based obviously on human
selfishness, for if a thing is useful to us we
emphasize that quality so much that we forget
to look for its beauty. Thus it is that the clover](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-66-320.jpg)
![injury. The scientists became interested in the supposed disease, and
they finally ascertained that these nodules are filled with bacteria,
which are the underground partners of the clovers and other legumes.
These bacteria are able to fix the free nitrogen of the air, and make it
available for plant-food. As nitrogen is the most expensive of the
fertilizers, any agency which can extract it from the free air for the
use of plants is indeed a valuable aid to the farmer. Thus it is that in
the modern agriculture, clover or some other legume is put on the
land once in three or four years in the regular rotation of crops, and it
brings back to the soil the nitrogen which other crops have
exhausted. An interesting fact about the partnership between the root
bacteria and the clover-like plants is that the plants do not flourish
without this partnership, and investigators have devised a method by
which these bacteria may be scattered in the soil on which some kinds
of clover are to be planted, and thus aid in growing a crop. This
method is to-day being used for the introduction of alfalfa here in
New York State. But the use of clover as a fertilizer is not limited to its
root factory for capturing nitrogen; its leaves break down quickly and
readily yield the rich food material of which they are composed, so
that the farmer who plows under his second-crop clover instead of
harvesting it, adds greatly to the fertility of his farm.
The members of three distinct genera are popularly called clovers:
The True Clovers (Trifolium), of which six or seven species are found
in New York State, and more than sixty species are found in the
United States. The Medics (Medicago), of which four species are
found here. The Melilots (Melilotus), or sweet clovers, of which we
have two species.
The True Clovers. (Trifolium.)
The Red Clover (Fig. 245). (Trifolium pratense.[47])—This beautiful
dweller in our fields came to us from Europe, and it is also a native of
Asia. It is the clover most widely cultivated in New York State for
fodder, and is one of our most important crops. Clover hay often
being a standard of excellence by which other hay is measured. The](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-68-320.jpg)
![18. Describe a clover seed. Describe a seed of alfalfa.
19. What insects do you find visiting the red clover blossoms? The
white clover blossoms?
Alfalfa, or Lucerne.[48]
The alfalfa plant is just now coming into great prominence in New
York State. Every teacher, particularly in the rural schools, will need to
know the plant and to have some information about it.
What alfalfa is.—It is a clover-like plant. It is perennial. It has violet-
purple flowers. The leaves have three narrow leaflets. It sends up
many stiff stems, 2 to 3 feet high. The roots go straight down to great
depths.
Why it is important.—It is an excellent cattle food, and cattle-raising
for dairy purposes is the leading special agricultural industry in New
York State. In fact, New York leads all the States in the value of its
dairy products. Any plant that is more nutritious and more productive
of pasture and hay than the familiar clovers and grasses will add
immensely to the dairy industry, and therefore to the wealth of the
State. Alfalfa is such a plant. It gives three cuttings of hay year after
year in New York State, thereby yielding twice as much as clover
does. In the production of digestible nutrients per acre ranks above
clover as 24 ranks above 10. When once established it withstands
droughts, for the roots grow deep.
Alfalfa is South European. It was early introduced into North America.
It first came into prominence in the semi-arid West because of its
drought-resisting qualities, and now it has added millions of dollars to
the wealth of the nation. Gradually it is working its way into the East.
It is discussed in the agricultural press and before farmers' institutes.
Last year the College of Agriculture offered to send a small packet of
seeds to such school children in New York State as wanted to grow a
little garden plat of it. About 5,000 children were supplied. The
teacher must now learn what alfalfa is.](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-76-320.jpg)
![Alfalfa should be cut when it opens into flower. At this time the stems
and leaves contain their highest percentage of nutrients, the leaves
do not so easily fall off in curing, and the stems are not so woody.
Besides these reasons, if cutting be delayed until after flowering, the
plant may not spring quickly into subsequent growth.
Disease does not spare the alfalfa plant. Both leaves and roots are
attacked, the leaf spot being serious. The parasitic dodder is a serious
enemy in some parts of New York State.
LEAFLET XXXV
HOW PLANTS LIVE TOGETHER.[49]
By L. H. BAILEY.
o the general observer, plants seem to be distributed in
a promiscuous and haphazard way, without law or
order. This is because he does not see and consider.
The world is now full of plants. Every plant puts forth
its supreme effort to multiply its kind. The result is an
intense struggle for an opportunity to live.
Seeds are scattered in profusion, but only the few can grow. The
many do not find the proper conditions. They fall on stony ground. In
Fig. 250 this loss is shown. The trunk of an elm tree stands in the
background. The covering of the ground, except about the very base
of the tree, is a mat of elm seedlings. There are thousands of them in
the space shown in the picture, so many that they make a sod-like
covering which shows little detail in the photograph. Not one of these
thousands will ever make a tree.](https://image.slidesharecdn.com/21512891-250601214743-213e3621/85/Analysis-For-Applied-Mathematics-Ward-Cheney-80-320.jpg)
Analysis For Applied Mathematics Ward Cheney
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- 6. Graduate Texts in Mathematics 208 Editorial Board S. Axler F.w. Gehring K.A. Ribet Springer Science+Business Media, LLC
- 7. Graduate Texts in Mathematics TAKEUTIIZARING. Introduction to 35 ALEXANDERIWERMER. Several Complex Axiomatic Set Theory. 2nd ed. Variables and Banach Algebras. 3rd ed. 2 OXTOBY. Measure and Category. 2nd ed. 36 KELLEy/NAMIOKA et al. Linear 3 SCHAEFER. Topological Vector Spaces. Topological Spaces. 2nded. 37 MONK. Mathematical Logic. 4 HILTON/STAMMBACH. A Course in 38 GRAUERTIFRITZSCHE. Several Complex Homological Algebra. 2nd ed. Variables. 5 MAC LANE. Categories for the Working 39 ARVESON. An Invitation to C*-Algebras. Mathematician. 2nd ed. 40 KEMENY/SNELLIKNAPP. Denumerable 6 HUGHES/PIPER. Projective Planes. Markov Chains. 2nd ed. 7 SERRE. A Course in Arithmetic. 41 ApOSTOL. Modular Functions and Dirichlet 8 TAKEUTIIZARING. Axiomatic Set Theory. Series in Number Theory. 9 HUMPHREYS. Introduction to Lie Algebras 2nded. and Representation Theory. 42 SERRE. Linear Representations ofFinite 0 COHEN. A Course in Simple Homotopy Groups. Theory. 43 GILLMAN/JERISON. Rings ofContinuous 11 CONWAY. Functions ofOne Complex Functions. Variable I. 2nd ed. 44 KENDIG. Elementary Algebraic Geometry. 12 BEALS. Advanced Mathematical Analysis. 45 LoiNE. Probability Theory I. 4th ed. 13 ANDERSON/FuLLER. Rings and Categories 46 LOEVE. Probability Theory II. 4th ed. ofModules. 2nd ed. 47 MOISE. Geometric Topology in 14 GOLUBITSKy/GUILLEMIN. Stable Mappings Dimensions 2 and 3. and Their Singularities. 48 SACHSlWu. General Relativity for 15 BERBERIAN. Lectures in Functional Mathematicians. Analysis and Operator Theory. 49 GRUENBERGIWEIR. Linear Geometry. 16 WINTER. The Structure ofFields. 2nd ed. 17 ROSENBLATT. Random Processes. 2nd ed. 50 EDWARDS. Fermat's Last Theorem. 18 HALMOS. Measure Theory. 51 KLINGENBERG. A Course in Differential 19 HALMOS. A Hilbert Space Problem Book. Geometry. 2nd ed. 52 HARTSHORNE. Algebraic Geometry. 20 HUSEMOLLER. Fibre Bundles. 3rd ed. 53 MANIN. A Course in Mathematical Logic. 21 HUMPHREYS. Linear Algebraic Groups. 54 GRAVERIWATKINS. Combinatorics with 22 BARNES/MACK. An Algebraic Introduction Emphasis on the Theory of Graphs. to Mathematical Logic. 55 BROWN/PEARCY. Introduction to Operator 23 GREUB. Linear Algebra. 4th ed. Theory I: Elements ofFunctional 24 HOLMES. Geometric Functional Analysis Analysis. and Its Applications. 56 MASSEY. Algebraic Topology: An 25 HEWITT/STROMBERG. Real and Abstract Introduction. Analysis. 57 CROWELLlFox. Introduction to Knot 26 MANES. Algebraic Theories. Theory. 27 KELLEY. General Topology. 58 KOBLITZ. p-adic Numbers, p-adic Analysis, 28 ZARISKIiSAMUEL. Commutative Algebra. and Zeta-Functions. 2nd ed. Vol.I. 59 LANG. Cyclotomic Fields. 29 ZARISKIiSAMUEL. Commutative Algebra. 60 ARNOLD. Mathematical Methods in Vol.II. Classical Mechanics. 2nd ed. 30 JACOBSON. Lectures in Abstract Algebra I. 61 WHITEHEAD. Elements ofHomotopy Basic Concepts. Theory. 31 JACOBSON. Lectures in Abstract Algebra II. 62 KARGAPOLOv/MERLZJAKOV. Fundamentals Linear Algebra. ofthe Theory ofGroups. 32 JACOBSON. Lectures in Abstract Algebra 63 BOLLOBAS. Graph Theory. III. Theory ofFields and Galois Theory. 64 EDWARDS. Fourier Series. Vol. I. 2nd ed. 33 HIRSCH. Differential Topology. 65 WELLS. Differential Analysis on Complex 34 SPITZER. Principles of Random Walk. Manifolds. 2nd ed. 2nded. (continued after index)
- 8. Ward Cheney Analysis for Applied Mathematics With 27 Illustrations , Springer
- 9. Ward Cheney Department of Mathematics University of Texas at Austin Austin, TX 78712-1082 USA Editorial Board S. Axler Mathematics Department San Francisco State University San Francisco, CA 94132 USA F.W. Gehring Mathematics Department East Hall University of Michigan Ann Arbor, MI 48109 USA K.A. Ribet Mathematics Department University of California at Berkeley Berkeley, CA 94720-3840 USA Mathematics Subject Classification (2000): 46Bxx, 65L60, 32Wxx, 42B 10 Library of Congress Cataloging-in-Publication Data Cheney, E. W. (Elliott Ward), 1929- Analysis for applied mathematics / Ward Cheney. p. em. - (Graduate texts in mathematics; 208) Includes bibliographical references and index. ISBN 978-1-4419-2935-8 ISBN 978-1-4757-3559-8 (eBook) DOI 10.1007/978-1-4757-3559-8 1. Mathematical analysis. I. Title. II. Series. QA300.C4437 2001 515-dc21 2001-1020440 Printed on acid-free paper. © 2001 Springer Science+Business Media New York Originally published by Springer-Verlag New York, Inc in 2001. Softcover reprint ofthe hardcover 1st edition 2001 All rights reserved. This work may .not be translated or copi~d in whole or in part without the written permission of the publisher (Springer Science+Business Media, LLC ), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use of general descriptive names, trade names, trademarks, etc., in this publication, even if the former are not especially identified, is not to be taken as a sign that such names, as understood by the Trade Marks and Merchandise Marks Act, may accordingly be used freely by anyone. Production managed by Terry Kornak; manufacturing supervised by Jerome Basma. Photocomposed from the author's TeX files. 987 6 5 4 321 SPIN 10833405
- 10. Preface This book evolved from a course at our university for beginning graduate stu- dents in mathematics-particularly students who intended to specialize in ap- plied mathematics. The content of the course made it attractive to other math- ematics students and to graduate students from other disciplines such as en- gineering, physics, and computer science. Since the course was designed for two semesters duration, many topics could be included and dealt with in de- tail. Chapters 1 through 6 reflect roughly the actual nature of the course, as it was taught over a number of years. The content of the course was dictated by a syllabus governing our preliminary Ph.D. examinations in the subject of ap- plied mathematics. That syllabus, in turn, expressed a consensus of the faculty members involved in the applied mathematics program within our department. The text in its present manifestation is my interpretation of that syllabus: my colleagues are blameless for whatever flaws are present and for any inadvertent deviations from the syllabus. The book contains two additional chapters having important material not included in the course: Chapter 8, on measure and integration, is for the ben- efit of readers who want a concise presentation of that subject, and Chapter 7 contains some topics closely allied, but peripheral, to the principal thrust of the course. This arrangement of the material deserves some explanation. The ordering of chapters reflects our expectation of our students: If they are unacquainted with Lebesgue integration (for example), they can nevertheless understand the examples of Chapter 1 on a superficial level, and at the same time, they can begin to remedy any deficiencies in their knowledge by a little private study of Chapter 8. Similar remarks apply to other situations, such as where some point-set topology is involved; Section 7.6 will be helpful here. To summarize: We encourage students to wade boldly into the course, starting with Chapter 1, and, where necessary, fill in any gaps in their prior preparation. One advantage of this strategy is that they will see the necessity for topology, measure theory, and other topics - thus becoming better motivated to study them. In keeping with this philosophy, I have not hesitated to make forward references in some proofs to material coming later in the book. For example, the Banach contraction mapping theorem is needed at least once prior to the section in Chapter 4 where it is dealt with at length. Each of the book's six main topics could certainly be the subject of a year's course (or a lifetime of study), and many of our students indeed study functional analysis and other topics of the book in separate courses. Most of them eventu- ally or simultaneously take a year-long course in analysis that includes complex analysis and the theory of measure and integration. However, the applied math- ematics course is typically taken in the first year of graduate study. It seems to bridge the gap between the undergraduate and graduate curricula in a way that has been found helpful by many students. In particular, the course and the v
- 11. vi Preface book certainly do not presuppose a thorough knowledge of integration theory nor of topology. In our applied mathematics course, students usually enhance and reinforce their knowledge of undergraduate mathematics, especially differential equations, linear algebra, and general mathematical analysis. Students may, for the first time, perceive these branches of mathematics as being essential to the foundations of applied mathematics. The book could just as well have been titled Prolegomena to Applied Math- ematics, inasmuch as it is not about applied mathematics itself but rather about topics in analysis that impinge on applied mathematics. Of course, there is no end to the list of topics that could lay claim to inclusion in such a book. Who is bold enough to predict what branches of mathematics will be useful in applications over the next decade? A look at the past would certainly justify my favorite algorithm for creating an applied mathematician: Start with a pure mathematician, and turn him or her loose on real-world problems. As in some other books I have been involved with, lowe a great debt of gratitude to Ms. Margaret Combs, our departmental 'lEX-pert. She typeset and kept up-to-date the notes for the course over many years, and her resourcefulness made my burden much lighter. The staff of Springer-Verlag has been most helpful in seeing this book to completion. In particular, I worked closely with Dr. Ina Lindemann and Ms. Terry Kornak on editorial matters, and I thank them for their efforts on my behalf. I am indebted to David Kramer for his meticulous copy-editing of the manuscript; it proved to be very helpful in the final editorial process. I thank my wife, Victoria, for her patience and assistance during the period of work on the book, especially the editorial phase. I dedicate the book to her in appreciation. I will be pleased to hear from readers having questions or suggestions for improvements in the book. For this purpose, electronic mail is efficient: cheney(Qmath. utexas .edu. I will also maintain a web site for material related to the book at http://www .math. utexas .edu/users/cheney/ AAMbook Ward Cheney Department of Mathematics University of Texas at Austin
- 12. Contents Preface .................................................................... v Chapter 1. Normed Linear Spaces ..................................... 1 1.1 Definitions and Examples ............................................ 1 1.2 Convexity, Convergence, Compactness, Completeness ................. 6 1.3 Continuity, Open Sets, Closed Sets .................................. 15 1.4 More About Compactness .......................................... 19 1.5 Linear Transformations ............................................. 24 1.6 Zorn's Lemma, Hamel Bases, and the Hahn-Banach Theorem ....... 30 1.7 The Baire Theorem and Uniform Boundedness ...................... 40 1.8 The Interior Mapping and Closed Mapping Theorems ............... 47 1.9 Weak Convergence ................................................. 53 1.10 Reflexive Spaces .................................................... 58 Chapter 2. Hilbert Spaces . ............................................ 61 2.1 Geometry .......................................................... 61 2.2 Orthogonality and Bases ............................................ 70 2.3 Linear Functionals and Operators ................................... 81 2.4 Spectral Theory .................................................... 91 2.5 Sturm-Liouville Theory ........................................... 105 Chapter 3. Calculus in Banach Spaces .............................. 115 3.1 The Frechet Derivative ............................................ 115 3.2 The Chain Rule and Mean Value Theorems ........................ 121 3.3 Newton's Method ................................................. 125 3.4 Implicit Function Theorems ....................................... 135 3.5 Extremum Problems and Lagrange Multipliers ..................... 145 3.6 The Calculus of Variations ........................................ 152 Chapter 4. Basic Approximate Methods of Analysis . ............. .170 4.1 Discretization ..................................................... 170 4.2 The Method of Iteration ........................................... 176 4.3 Methods Based on the Neumann Series ........................... 186 4.4 Projections and Projection Methods ............................... 191 4.5 The Galerkin Method ............................................. 198 4.6 The Rayleigh-Ritz Method ........................................ 205 4.7 Collocation Methods .............................................. 213 4.8 Descent Methods .................................................. 226 4.9 Conjugate Direction Methods ...................................... 232 4.10 Methods Based on Homotopy and Continuation .................... 237 vii
- 13. viii Contents Chapter 5. Distributions.............................................. 246 5.1 Definitions and Examples .......................................... 246 5.2 Derivatives of Distributions ........................................ 253 5.3 Convergence of Distributions ...................................... 257 5.4 Multiplication of Distributions by Functions ....................... 260 5.5 Convolutions ...................................................... 268 5.6 Differential Operators ............................................. 273 5.7 Distributions with Compact Support .............................. 280 Chapter 6. The Fourier Transform . ................................. 287 6.1 Definitions and Basic Properties ................................... 287 6.2 The Schwartz Space .............................................. 294 6.3 The Inversion Theorems ........................................... 301 6.4 The Plancherel Theorem .......................................... 305 6.5 Applications of the Fourier Transform ............................. 310 6.6 Applications to Partial Differential Equations ...................... 318 6.7 Tempered Distributions ........................................... 321 6.8 Sobolev Spaces .................................................... 325 Chapter 7. Additional Topics .. ...................................... 333 7.1 Fixed-Point Theorems ............................................ 333 7.2 Selection Theorems ................................................ 339 7.3 Separation Theorems .............................................. 342 7.4 The Arzela-Ascoli Theorems ...................................... 347 7.5 Compact Operators and the Fredholm Theory ..................... 351 7.6 Topological Spaces ................................................ 361 7.7 Linear Topological Spaces ......................................... 367 7.8 Analytic Pitfalls ................................................... 373 Chapter 8. Measure and Integration . ............................... 381 8.1 Extended Reals, Outer Measures, Measurable Spaces ............... 381 8.2 Measures and Measure Spaces ..................................... 386 8.3 Lebesgue Measure ................................................. 391 8.4 Measurable Functions ............................................. 394 8.5 The Integral for Nonnegative Functions ............................ 399 8.6 The Integral, Continued ........................................... 404 8.7 The LP-Spaces .................................................... 409 8.8 The Radon-Nikodym Theorem .................................... 413 8.9 Signed Measures .................................................. 417 8.10 Product Measures and Fubini's Theorem .......................... .420 References ................................................. _. __ ....... .429 Index ......................................................... _......... 437 Symbols . ___ ............................... _... _....................... 443
- 14. Chapter 1 N ormed Linear Spaces 1.1 Definitions and Examples 1 1.2 Convexity, Convergence, Compactness, Completeness 6 1.3 Continuity, Open Sets, Closed Sets 15 1.4 More about Compactness 19 1.5 Linear Transformations 24 1.6 Zorn's Lemma, Hamel Bases, and the Hahn-Banach Theorem 30 1.7 The Baire Theorem and Uniform Boundedness 40 1.8 The Interior Mapping and Closed Mapping Theorems 47 1.9 Weak Convergence 53 1.10 Reflexive Spaces 58 1.1 Definitions and Examples This chapter gives an introduction to the theory of normed linear spaces. A skeptical reader may wonder why this topic in pure mathematics is useful in applied mathematics. The reason is quite simple: Many problems of applied mathematics can be formulated as a search for a certain function, such as the function that solves a given differential equation. Usually the function sought must belong to a definite family of acceptable functions that share some useful properties. For example, perhaps it must possess two continuous derivatives. The families that arise naturally in formulating problems are often linear spaces. This means that any linear combination of functions in the family will be another member of the family. It is common, in addition, that there is an appropriate means of measuring the "distance" between two functions in the family. This concept comes into play when the exact solution to a problem is inaccessible, while approximate solutions can be computed. We often measure how far apart the exact and approximate solutions are by using a norm. In this process we are led to a normed linear space, presumably one appropriate to the problem at hand. Some normed linear spaces occur over and over again in applied mathematics, and these, at least, should be familiar to the practitioner. Examples are the space of continuous functions on a given domain and the space of functions whose squares have a finite integral on a given domain. A knowledge of function spaces enables an applied mathematician to consider a problem from a more 1
- 15. 2 Chapter 1 Normed Linear Spaces lofty viewpoint, from which he or she may have the advantage of being more aware of significant features as distinguished from less significant details. We begin by reviewing the concept of a vector space, or linear space. (These terms are interchangeable.) The reader is probably already familiar with these spaces, or at least with the example of vectors in JRn. However, many function spaces are also linear spaces, and much can be learned about these function spaces by exploiting their similarity to the more elementary examples. Here, as a reminder, we include the axioms for a vector space or linear space. A real vector space is a triple (X, +, .), in which X is a set, and + and· are binary operations satisfying certain axioms. Here are the axioms: (i) If x and y belong to X then so does x + y (closure axiom). (ii) x + y = y + x (commutativity). (iii) x + (y + z) = (x +y) + z (associativity). (iv) X contains a unique element, 0, such that x +0 = x for all x in X. (v) With each element x there is associated a unique element, -x, such that x +(-x) = O. (vi) If x E X and AE JR, then A.x E X (JR denotes the set of real numbers.) (closure axiom) (vii) A· (x +y) = A· x +A· y (A E JR), (distributivity). (viii) (A+J,t)·X=A·X+J,t·X (A,J,tEJR), (distributivity). (ix) A· (J,t. x) = (AJ,t) . x (associativity). (x) 1· x = x. These axioms need not be intimidating. The essential feature of a linear space is that there is an addition defined among the elements of X, and when we add two elements, the result is again in the space X. One says that the space is closed (algebraically) under the operation of addition. A similar remark holds true for multiplication of an element by a real number. The remaining axioms simply tell us that the usual rules of arithmetic are valid for the two operations. Most rules that you expect to be true are indeed true, but if they do not appear among the axioms it is because they follow from the axioms. The effort to keep the axioms minimal has its rewards: When one must verify that a given system is a real vector space there will be a minimum of work involved! In this set of axioms, the first five define an (additive) Abelian group. In axiom (iv), the uniqueness of 0 need not be mentioned, for it can be proved with the aid of axiom (ii). Usually, if A E JR and x E X, we write AX in place of A. x. The reader will note the ambiguity in the symbol + and the symbol o. For example, when we write Ox = 0 two different zeros are involved, and in axiom (viii) the plus signs are not the same. We usually write x - y in place of x + (-y). Furthermore, we are not going to belabor elementary consequences of the axioms such as AL:~ Xi = L:~ Axi. We usually refer to X as the linear space rather than (X, +, .). Observe that in a linear space, we have no way of assigning a meaning to expressions that involve a limiting process, such as L:;'" Xi. This drawback will disappear soon, upon the introduction of a norm. From time to time we will prefer to deal with a complex vector space. In such a space A·X is defined (and belongs to X) whenever AE C and x E X. (The
- 16. Section 1.1 Definitions and Examples 3 symbol C denotes the set of complex numbers.) Other fields can be employed in place of JR and C, but they are rarely useful in applied mathematics. The field elements are often termed scalars, and the elements of X are often called vectors. Let X be a vector space. A norm on X is a real-valued function, denoted by II II, that fulfills three axioms: (i) Ilxll > 0 for each nonzero element in X. (ii) IIAxl1 = IAlllxl1 for each Ain JR and each x in X. (iii) Ilx + YII ~ Ilxll + IIYII for all x, YE X. (Triangle Inequality) A vector space in which a norm has been introduced is called a normed linear space. Here are eleven examples. Example 1. function. Let X = JR, and define Ilxll = lxi, the familiar absolute value • Example 2. Let X = C, where the scalar field is also C. Use Ilxll = lxi, where Ixl has its usual meaning for a complex number x. Thus if x = a + ib (where a and b are real), then Ixl = v'a2 + b2 . • Example 3. Let X = C, and take the scalar field to be lR. The terminology we have adopted requires that this be called a real vector space, since the scalar field is lR. • Example 4. Let X = JRn . Here the elements of X are n-tuples of real numbers that we can display in the form x = [x(l), x(2), ... ,x(n)] or x = [Xl, X2, . .. ,xn ]. A useful norm is defined by the equation IIxlioo = max Ix(i)1 l,;;;.';;;n Note that an n-tuple is a function on the set {l, 2, ... ,n}, and so the notation x(i) is consistent with that interpretation. (This is the "sup" norm.) • Example 5. Let X = JRn , and define a norm by the equation Ilxll = L~l Ix(i)l· Observe that in Examples 4 and 5 we have two distinct normed linear spaces, although each involves the same linear space. This shows the ad- vantage of being more formal in the definition and saying that a normed linear space is a pair (X, II II) etc. etc., but we refrain from doing this unless it is necessary. • Example 6. Let X be the set of all real-valued continuous functions defined on a fixed compact interval [a, b]. The norm usually employed here is (The notation maxa~s';;;b Ix(s)1 denotes the maximum of the expression Ix(s)1 as s runs over the interval [a, b].) The space X described here is often denoted by C[a, b]. Sticklers would insist on C([a, b]), because C(S) will be used for the continuous functions on some general domain S. (This again is the "sup" norm.) •
- 17. 4 Chapter 1 Normed Linear Spaces Example 7. Let X be the set of all Lebesgue-integrable functions defined on a fixed interval [a, bJ. The usual norm for this space is IIxll = J:Jx(s)Jds. In this space, the vectors are actually equivalence classes of functions, two functions being regarded as equivalent if they differ only on a set of measure O. (The reader who is unfamiliar with the Lebesgue integral can substitute the Riemann integral in this example. The resulting spaces are different, one being complete and the other not. This is a rather complicated matter, best understood after the study of measure theory and Lebesgue integration. Chapter 8 is devoted to this branch of analysis. The notion of completeness of a space is taken up in the next section.) • Example 8. Let X = f, the space of all sequences in R x = [x(1),x(2), ... J in which only a finite number of terms are nonzero. (The number of nonzero terms is not fixed but can vary with different sequences.) Define IIxll = maXn Jx(n)J. • Example 9. Let X = foo, the space of all real sequences x for which sUPn Jx(n)J < 00. Define IIxll to be that supremum, as in Example 8. • Example 10. Let X = II, the space of all polynomials having real coefficients. A typical element of II is a function x having the form One possible norm on II is x H maxi lail. Others are x H maxO:s;t:S;l Ix(t)1 or x H J; Jx(t)J dt or x H (L:~ JXJ3)1/3. • Example 11. Let X = Rn , and use the familiar Euclidean norm, defined by IIxll2 = (I)x(iW) 1/2 • i=l In all of these examples (as well as in others to come) it is regarded as obvious how the algebraic structure is defined. A complete development would define x +y, AX, 0, and -x, and then verify the axioms for a linear space. After that, the alleged norm would be shown to satisfy the axioms for a norm. Thus, in Example 6, the zero element is the function denoted by 0 and defined by O(s) = 0 for all s E [a, bJ. The operation of addition is defined by the equation (x + y)(s) = x(s) + y(s) and so on. The concept of linear independence is of central importance. Recall that a subset S in a linear space is linearly independent if it is not possible to find a finite, nonempty, set of distinct vectors Xl, X2, ... ,Xm in S and nonzero scalars C1, C2,' .. ,Cm for which
- 18. Section 1.1 Definitions and Examples 5 (Linear independence is not a property of a point; it is a property of a set of points. Because of this, the usage "the vectors... are independent" is mis- leading.) The reader probably recalls how this notion enters into the theory of nth-order ordinary differential equations: A general solution must involve a linearly independent set of n solutions. Some other basic concepts to recall from linear algebra are mentioned here. The span of a set S in a vector space X is denoted by span(S), and consists of all vectors in X that are expressible as linear combinations of vectors in S. Remember that linear combinations are always finite expressions of the form L~=l AiXi' We say that "S spans X" when X = span(S). A base or basis for a vector space X is any set that is linearly independent and spans X. Both properties are essential. Any set that is linearly independent is contained in a basis, and any set that spans the space contains a basis. A vector space is said to be finite dimensional if it has a finite basis. An important theorem states that if a space is finite dimensional, then every basis for that space has the same number of elements. This common number is then called the dimension of the space. (There is an infinite-dimensional version of this theorem as well.) The material of this chapter is accessible in many textbooks and treatises, such as: [Au], [Av], [BN], [Ban], [Bea], rep], [Day], [Dies], [Dieu], [DS], [Edw], [Frie2], [Fried], [GP], [Gre], [Gri], [HS], [HP], [Hoi], [Horv], [Jam], [KA], [Kee], [KF], [Kre], [LanI], [Lo], [Moo], [NaSn], rOD], [Ped], [Red], [RS], [RN], [Roy], [Rul], [Sim], [Tay2], [Yo], and [Ze]. Problems 1.1 Here is a Chinese proverb that is pertinent to the problems: I hear, I forget; I see, I remember; I do, I understand! 1. Let X be a linear space over the complex field. Let XT be the space obtained from X by restricting the scalars to the real field. Prove that XT is a real linear space. Show by an example that not every real linear space is of the form XT for some complex linear space X. Caution: When we say that a linear space is a real linear space, this has nothing to do with the elements of the space. It means only that the scalar field is IR and not IC. 2. Prove the norm axioms for Examples 4-7. 3. Prove that in any normed linear space, 11011 =0 and !llxll - Ilyll! ~ Ilx - yll 4. Denote the norms in Examples 4 and 5 by II IL", and II Ill' respectively. Find the best constants in the inequality Prove that your constants are the best. (The "constants" a and (3 will depend on n but not x.) 5. In Examples 4, 5, 6, and 7 find the precise conditions under which we have Ilx +yll = IIxll + Ilyll· 6. Prove that in any normed linear space, if x # 0, then x/llxli is a vector of norm 1. 7. The Euclidean norm on IRn is defined in Example 11. Find the best constants in the inequality ollxlloc ~ IIxl12 ~ (3llxllx'
- 19. 6 Chapter 1 Normed Linear Spaces 8. What theorems in elementary analysis are needed to prove the closure axioms for Example 6? 9. What is the connection between the normed linear spaces f and II defined in Examples 8 and 1O? 10. For any t in the open interval (0,1), let t be the sequence [t, t2 , t3 , .. .J. Notice that t E foe. Prove that the set {t: 0 < t < I} is linearly independent. 11. In the space II we define special elements called monomials. They are given by xn(t) = tn where n =0, 1,2, ... Prove that {Xn : n =0, 1,2,3 ...} is linearly independent. 12. Let T be a set of real numbers. We say that T is bounded above if there is an M in ]R such that t ~ M for all t in T. We say that M is an upper bound of T. The completeness axiom for ]R asserts that if a set T is bounded above, then the set of all its upper bounds is an interval of the form [b,oo). The number b is the least upper bound, or supremum of T, written b = l.u.b.(T) = sup(T). Prove that if x < b, then (x, oo)nT is nonempty. Give examples to show that [b, oo)nT can be empty or nonempty. There are corresponding concepts of bounded below, lower bound, greatest lower bound, and infimum. 13. Which of these expressions define norms on ]R2? Explain. (a) max{lx(l)l, Ix(l) + x(2)1} (b) Ix(2) - x(l)1 (c) Ix(l)1 + Ix(2) - x(l)1 + Ix(2)1 14. Prove that in any normed linear space the conditions Ilxll = 1 and IIx - yll < £ < 1 imply that Ilx - Y/llylill < 2£. 15. Prove that if NI and N2 are norms on a linear space, then so are olNI + 02N2 (when Or > 0 and 02 > 0) and (N'f + Ni)I/2. 16. Is the following set of axioms for a norm equivalent to the set given in the text? (a) Ilxll ¥ oif x ¥ 0, (b) IIAXII = -Allxll if A ~ 0, (c) IIx + yll ~ IIxll + lIyll· 17. Prove that in a normed linear space, if Ilx+yll = Ilxll+llyll, then Ilox+,Byll = lIoxll+ll,Byll for all nonnegative 0 and ,B. 18. Why is the word "distinct" essential in our definition of linear independence on page 4? 19. Is the set of functions J;(x) = Ix - ii, where i = 1,2 ..., linearly independent? 20. One example of an "exotic" vector space is described as follows. Let X be the set of positive real numbers. We define an "addition", Ell, by x Ell y = xy and a "scalar multiplication" by a 0 x = xa. Prove that (X, Ell, 0) is a vector space. 21. In Example 10, two norms (say NI and N2) were suggested. Do there exist constants such that NI ~ ON2 or N2 ~ ,BNI? 22. In Examples 4 and 5, let n = 2, and draw sketches of the sets {x E]R2 : Ilxll = I}. (Symmetries can be exploited.) 1.2 Convexity, Convergence, Compactness, Completeness A subset K in a linear space is said to be convex if it contains every line segment connecting two of its elements. Formally, convexity is expressed as follows: [XEK & yEK & O~"~l] ~ ..x+(l-..)YEK The notion of convexity arises frequently in optimization problems. For example, the theory of linear programming (optimization of linear functions) is based on
- 20. Section 1.2 Convexity, Convergence, Compactness, Completeness 7 the fact that a linear function on a convex polyhedral set must attain its extrema at the vertices of the set. Thus, to locate the maxima of a linear function over a convex polyhedral set, one need only test the vertices. The central idea of Dantzig's famous simplex method is to move from vertex to vertex, always improving the value of the objective function. Another application of convexity occurs in studying deformations of a physi- cal body. The "yield surface" of an object is generally convex. This is the surface in 6-dimensional space that gives the stresses at which an object will fail struc- turally. Six dimensions are needed to account for all the variables. See [Mar], pages 100-104. Among examples of convex sets in a linear space X we have: (i) the space X itself; (ii) any set consisting of a single point; (iii) the empty set; (iv) any linear subspace of X; (v) any line segment; i.e. a set of the following form in which a and bare fixed: {Aa+(I-A)b: O~A~I} In a normed linear space, another important convex set is the unit cell or unit ball: {x EX: Ilxll ~ I} In order to see that the unit ball is convex, let Ilxll ~ 1, IIYII ~ 1, and 0 ~ A~ 1. Then, with Jl = 1 - A, If we let n = 2 in Examples 4 and 5 of Section 1.1, then we can draw pictures of the unit balls. They are shown in Figures 1.1 and 1.2. 1 -1 1 -1 -1 -1 Figures 1.1 and 1.2. Unit balls There is a family of norms on ]Rn, known as the t'p-norms, of which the norms in Examples 4 and 5 are special cases. The general formula, for 1 ~ p < 00, is ( n ) lip Ilxllp = ~ Ix(i}IP
- 21. 8 Chapter 1 Normed Linear Spaces The case p = 00 is special; for it we use the formula IIXlloo = max Ix(i)1 l~t~n It can be shown (Problem 1) that limp --+oo Ilxllp = Ilxlloo. (This explains the notation.) The unit balls (in ]R2) for II lip are shown for p = 1, 2, and 7, in Figure 1.3. 0.5 ·0.5 0.5 -0.5 Figure 1.3. The unit balls in fp, for p = 1, 2, and 7. In any normed linear space there exists a metric (and its corresponding topology) that arises by defining the distance between two points as d(x, y) = Ilx - YII All the topological notions from the theory of metric spaces then become avail- able in a normed linear space. (See Problem 23.) In Chapter 7, Section 6, the theory of general topological spaces is broached. But we shall discuss here topological concepts restricted to metric spaces or to normed linear spaces. A sequence Xl, X2, ... in a normed linear space is said to converge to a point X (and we write Xn --+ x) if lim Ilxn - xii = 0 n--+oo For example, in the space of continuous functions on [0,1J furnished with the max-norm (as in Example 6 of Section 1, page 3), the sequence of functions xn(t) = sin(t/n) converges to 0, since IIXn - 011 = sup Isin(t/n)1 = sin(1/n) --+ 0 O~t~l The notion of convergence is often needed in applied mathematics. For example, the solution to a problem may be a function that is difficult to find but can be approached by a suitable sequence of functions that are easier to obtain. (Maybe they can be explicitly calculated.) One then would need to know exactly in what sense the sequence was approaching the actual solution to the problem. A subset K in a normed space is said to be compact if each sequence in K has a subsequence that converges to a point in K. (Caution: In general topology, this concept would be called sequential compactness. Refer to Section 7.6.) A subsequence of a sequence Xl, X2, ... is of the form x nl ' x n2 ' ... , where the integers ni satisfy nl < n2 < n3 < .... Our notation for a sequence is [xn J, or [xn : n E NJ, or [Xl, X2, .. .J. With this meagre equipment we can already prove some interesting results.
- 22. Section 1.2 Convexity, Convergence, Compactness, Completeness 9 Theorem 1. Let K be a compact set in a normed linear space X. To each x in X there corresponds at least one point in K of minimum distance from x. Proof. Let x be any member of X. The distance from x to K is defined to be the number dist (x, K) = inf Ilx - zll zEK By the definition of an infimum (Problem 12 in Section 1.1, page 6), there exists a sequence [Yn] in K such that Ilx - Ynll-+ dist (x, K). Since K is compact, there is a subsequence converging to a point in K, say Yni -+ Y E K. Since Ilx - YII ~ Ilx - Yni II + IIYni - YII we have in the limit Ilx-YII ~ dist (x, K) ~ Ilx-YII. (The final inequality follows from the definition of the distance function.) • The preceding theorem can be useful in problems involving noisy measure- ments. For example, suppose that a noisy measurement of a single entity x is available. If a set K of admissible noise-free values for x is prescribed, then the best noise-free estimate of x can be taken to be a point of K as close as possible to x. Theorem 1 is also important in approximation theory, a branch of analysis that provides the theoretical underpinning for many areas of applied mathematics. Example 1. On the real line, an open interval (a, b) is not compact, for we can take a sequence in the interval that converges to the endpoint b, say. Then every subsequence will also converge to b. Since b is not in the interval, the interval cannot be compact. On the other hand, a closed and bounded interval, say [a, b], is compact. This is a special case of the Heine-Borel theorem. See the discussion before Lemma 1 in Section 1.4, page 20. • Given a sequence [xn] in a normed linear space (or indeed in any metric space), is it possible to determine, from the sequence alone, whether it con- verges? This is certainly an important matter for practical purposes, since we often use algorithms to generate sequences that should converge to a solution of a given problem. The answer to the posed question is that we cannot infer convergence, in general, solely from the sequence itself. If we confine ourselves to the information contained in the sequence, we can construct the doubly indexed sequence Cnm = Ilxn- xmll. If [cnm] does not converge to zero, then the given sequence [xn] cannot converge, as is easily proved: For any x in the space, write This shows that if Cnm does not converge to 0, then [xn] cannot converge. On the other hand, if Cnm converges to zero, one intuitively thinks that the sequence ought to converge, and if it does not, there must be a flaw in the space itself: The limit of the sequence should exist, but the limiting point is somehow missing from the space. Think of the rational numbers as an example. The missing ingredient is completeness of the space, to which we now turn.
- 23. 10 Chapter 1 Normed Linear Spaces A sequence [xn] in a normed linear space X is said to have the Cauchy property or to be a Cauchy sequence if lim sup Ilxi - Xj II = ° n~oo i~n j~n If every Cauchy sequence in the space X is convergent (to a point of X, of course), then the space X is said to be complete. A complete normed linear space is termed a Banach space, in honor of Stefan Banach, who lived from 1892 to 1945. His book [Ban] stimulated the study of functional analysis for several decades. Examples 1-7, 9, and 11, given previously, are all Banach spaces. The real number field IR is complete, and so is the complex number field C. The rational field <Q is not complete. These facts are established in elementary analysis courses. Completeness is important in constructing solutions to a problem by taking the limit of successive approximations. One often wants information about the limit (i.e., the solution). Does it have the same properties as the approximations? For example, if all the approximating functions are continuous, must the limit also be continuous? If all the approximating functions are bounded, is the limit also bounded? The answers to such questions depend on the sense in which the limit is achieved; in other words, they depend on the norm that has been chosen and the function space that goes with it. Typically, one wants a norm that leads to a complete normed linear space, i.e., a Banach space. Here is an example of a normed linear space that is not a Banach space: Example 2. Let the space be the one described in Example 8 of Section 1.1, page 4. This is e, the space of "finitely-nonzero sequences," with the "sup norm" Ilxll = maxi Ix(i)l· Define a sequence [Xk] in eby the equation Xk = [1,~,~, ... ,~,0, 0, ...J If m > n, then Xm - Xn = [0, ... ,0, n : l' ... , ~,0, ...J Since IIXm -Xnll = 1/(n+ 1), we conclude that the sequence [Xk] has the Cauchy property. If the space were complete, we would have Xn -+ y, where y E e. The point y would be finitely nonzero, say y(n) = 0 for n > N. Then for m > N, Xm would have as its Nth term the value liN, while the Nth term of y is O. Thus IIXm - YII ~ liN, and convergence cannot take place. • Theorem 2. Banach space. The space C[a,b] with norm Ilxll = maxs Ix(s)1 is a Proof. Let [xn] be a Cauchy sequence in C[a, b]. (This space is described in Example 6, page 3.) Then for each s, [xn(s)] is a Cauchy sequence in R Since IR is complete, this latter sequence converges to a real number that we may denote
- 24. Section 1.2 Convexity, Convergence, Compactness, Completeness 11 by x(s). The function x thus defined must now be shown to be continuous, and we must also show that Ilxn - xii -+ o. Let t be fixed as the point at which continuity is to be proved. We write This inequality should suggest to the reader how the proof must proceed. Let e > o. Select N so that Ilxn - xmll ~ e/3 whenever m ~ n ~ N (Cauchy property). Then for m ~ n ~ N, Ixn(s) - xm(s)1 ~ e/3. By letting m -+ 00 we get Ixn(s) - x(s)1 ~ e/3 for all s. This shows that Ilxn- xii ~ e/3 and that the sequence Ilxn - xii converges to o. By the continuity of Xn there exists a 6 > 0 such that IXn(s) - xn(t)1 < e/3 whenever It - sl < 6. Inequality (1) now shows that Ix(s) - x(t)1 < e when It - sl < 6. (This proof illustrates what is sometimes called "an e/3 argument.") • Remarks. Theorem 2 is due to Weierstrass. It remains valid if the interval [a, b] is replaced by any compact Hausdorff space. (For topological notions, refer to Section 7.6, starting on page 361.) The traditional formulation of this theorem states that a uniformly convergent sequence of continuous functions on a closed and bounded interval must have a continuous limit. A sequence of functions [In] converges uniformly to I if (2) Ve 3n Vk Vs [k>n ====> IIk(s)-I(s)l<e] (In this succinct description, it is understood that e > 0, n E N, kEN, and s is in the domain of the functions.) By contrast, pointwise convergence is defined by Vs Ve 3n Vk [k>n ====> 1!k(s)-I(s)l<e] Our use of the austere and forbidding logical notation is to bring out clearly and to emphasize the importance of the order of the quantifiers. Thus, in the definition of uniform convergence, n does not (cannot) depend on s, while in the definition of pointwise convergence, n may depend on s. Notice that by the definition of the norm being used, (2) can be written or simply as limn .....oo IIIn - 11100 = O. The latter is conceptually rather simple, if one is already comfortable with this norm (called the "supremum norm" or the "maximum norm"). The (perhaps) simplest example of a sequence of continuous functions that converges pointwise but not uniformly to a continuous function is the sequence [In] described as follows. The value of In(x) is 1 everywhere except on the interval [0,2/n], where its value is given by Inx - 11. Problems 1.2 1. Prove that limp-tx Ilxlip = maxU:;i~n Ix(i)1 for every x in IRn.
- 25. 12 Chapter 1 Normed Linear Spaces 2. Is this property of a sequence equivalent to the Cauchy property? lim sup IIXk - xnll =0 n---too k~n Answer the same question for this property: For every positive E there is a natural number n such that Ilxm - xnll < E whenever m ~ n. 3. Prove that if a sequence [Xn] in a Banach space satisfies 2::'=1 Ilxnll < 00, then the series 2::'=1 Xn converges. 4. Prove that Theorem 2 is not true for the norm JIx(t)1 dt. 5. Prove that the union of a finite number of compact sets is compact. Give an example to show that the union of an infinite family of compact sets can fail to be compact. 6. Prove that II lip on IRn does not satisfy the triangle inequality if 0 < p < 1 and n ~ 2. 7. Prove that if Xn -+ x, then the set {X,X1, X2, ... } is compact. 8. A cluster point (or accumulation point) of a sequence is the limit of any convergent subsequence. Prove that if a sequence lies in a compact set and has only one cluster point, then it is convergent. 9. Prove that the convergence in Problem 1 above is monotone. 10. Give an example of a countable compact set in IR having infinitely many accumulation points. If your example has more than a countable number of accumulation points, give another example, having no more than a countable number. 11. Let Xo and Xl be any two points in a normed linear space. Define X2, X3, ... inductively by putting n =0,1,2, ... Prove that the resulting sequence is a Cauchy sequence. 12. A particular Banach space of great importance is the space '-=(S), consisting of all bounded real-valued functions on a given set S. For X E loo(8) we define Ilxli oo =sup Ix(s)1 sES Prove that this space is complete. Cultural note: The space loc (ll) is of special interest. Every separable metric space can be embedded isometrically in it! You might enjoy trying to prove this, but that is not part of problem 12. 13. Prove that in a normed linear space a sequence cannot converge to two different points. 14. How does a sequence [Xn : n E !I] differ from a countable set {Xn : n E !I}? 15. Is there a norm that makes the space of all real sequences a Banach space? 16. Let Co denote the space of all real sequences that converge to zero. Define Ilxll sUPn Ix(n)l. Prove that Co is a Banach space. 17. If K is a convex set in a linear space, then these two sets are also convex: u + K = {u + X : x E K} and AK = {AX: x E K} 18. Let A be a subset of a linear space. Put AC = {tAiai : n E!I , Ai ~ 0 , ai E A, t Ai =1} Prove that A C AC Prove that AC is convex. Prove that AC is the smallest convex set containing A. This latter assertion means that if A is contained in a convex set B, then AC is also contained in B. The set AC is the convex hull of A.
- 26. Section 1.2 Convexity, Convergence, Compactness, Completeness 13 19. If A and B are convex sets, is their vector sum convex? The vector sum of these two sets isA+B={a+b: aEA,bEB}. 20. Can a norm be recovered from its unit ball? Hint: If x E X, then X/A is in the unit ball whenever IAI ~ Ilxli. (Prove this.) On the other hand, X/A is not in the unit ball if IAI < Ilxll· (Prove this.) 21. What are necessary and sufficient conditions on a set 8 in a linear space X in order that 8 be the unit ball for some norm on X? 22. Prove that the intersection of a family of convex sets (all contained in one linear space) is convex. 23. A metric space is a pair (X, d) in which X is a set and d is a function (called a metric) from X x X to IR such that (i) d(x, y) ~ 0 (ii) d(x, y) =0 if and only if X = y (iii) d(x, y) =d(y, x) (iv) d(x, y) ~ d(x, z) + d(z, y) Prove that a normed linear space is a metric space if d(x, y) is defined as II x - y II. 24. For this problem only, we use the following notation for a line segment in a linear space: (a, b) = {Aa+ (1-A)b: 0 ~ A ~ 1} A polygonal path joining points a and b is any finite union of line segments U~=l (ai, ai+l), where al =a and an+l =b. If the linear space has a norm, the length of the polygonal path is L.:~=l Ila; - ai+lli. Give an example of a pair of points a, bin a normed linear space and a polygonal path joining them such that the polygonal path is not identical to (a, b) but has the same length. A path of length Iia - bll connecting a and b is called a geodesic path. Prove that any geodesic polygonal path connecting a and b is contained in the set {x: Ilx - all ~ lib - all}. 25. If Xn -+ x and if the Cesaro means are defined by an = (Xl +...+xn)/n, then an -+ x. (This is to be proved in an arbitrary normed linear space.) 26. Prove that a Cauchy sequence that contains a convergent subsequence must converge. 27. A compact set in a normed linear space must be bounded; i.e., contained in some multiple of the unit ball. 28. Prove that the equation f(x) = L.:;;"=o ak cos bkx defines a continuous function on IR, provided that 0 ~ a < l. The parameter b can be any real number. You will find useful Theorem 2 and Problem 3. Cultural Note: If 0 < a < 1 and if b is an odd integer greater than a-I, then f is differentiable nowhere. This is the famous Weierstrass nondifferentiable function. (See Section 7.8, page 374, for more information about this function.) 29. Prove that a sequence [xn] in a normed linear space converges to a point x if and only if every subsequence of [Xn] converges to x. 30. Prove that if ¢ is a strictly increasing function from N into N, then ¢(n) ~ n for all n. 3l. Let 8 be a subset of a linear space. Let 81 be the union of all line segments that join pairs of points in 8. Is 81 necessarily convex?
- 27. 14 Chapter 1 Normed Linear Spaces 32. (continuation) What happens if we repeat the process and construct 82 , 83 , ...? (Thus, for example, 82 is the union of line segments joining points in 81.) 33. Let I be a compact interval in JR, I = [a, b]. Let X be a Banach space. The notation C(I, X) denotes the linear space of all continuous maps I : I --t X. We norm C(I, X) by putting 11111 = SUPtEI III(t)lI. Prove that C(I, X) is a Banach space. 34. Define In(x) =e-nx . Show that this sequence of functions converges pointwise on [0,1] to the function 9 such that g(O) = 1 and g(t) = 0 for t -# O. Show that in the L2-norm on [0,1], In converges to O. The L2-norm is defined by 11111 = {fo1 II(t)i2dt}1/2. 35. Let [Xn] be a sequence in a Banach space. Suppose that for every c > 0 there is a convergent sequence [Yn] such that sUPn IIxn - Ynll < c. Prove that [xn] converges. 36. In any normed linear space, define K(x, r) = {y : IIx - yll ~ r}. Prove that if K(x, ~) c K(O, 1) then 0 E K(x, ~). 37. Show that the closed unit ball in a normed linear space cannot contain a disjoint pair of closed balls having radius ~. 38. (Converse of Problem 3) Prove that if every absolutely convergent series converges in a normed linear space, then the space is complete. (A series 2:Xn is absolutely convergent if 2: IIxnll < 00.) 39. Let X be a compact Hausdorff space, and let C(X) be the space of all real-valued continuous functions on X, with norm 11111 =supII(x)l. Let [In] be a Cauchy sequence in C(X). Prove that lim lim In(x) = lim lim In(x) X---+Xo n-+oo n-+<Xl x---t-XQ Give examples to show why compactness, continuity, and the Cauchy property are needed. 40. The space £1 consists of all sequences x = [x(1),x(2), ... ] in which x(n) E JR and 2: Ix(n)1 < 00. The space £2 consists of sequences for which 2: Ix(n)12 < 00. Prove that £1 C £2 by establishing the inequality 2: Ix(n)12 ~ (2: Ix(n)1)2. 41. Let X be a normed linear space, and 8 a dense subset of X. Prove that if each Cauchy sequence in 8 has a limit in X, then X is complete. A set 8 is dense in X if each point of X is the limit of some sequence in 8. 42. Give an example of a linearly independent sequence [xo, Xl, X2,"'] of vectors in loo such that 2::'=0 Xn =O. Don't forget to prove that 2: Xn =O. 43. Prove, in a normed space, that if Xn --t X and Ilxn - Ynll --t 0, then Yn --t X. If Xn --t X and IIxn - Yn II --t 1, what is lim Yn ? 44. Whenever we consider real-valued or complex-valued functions, there is a concept of absolute value of a function. For example, if x E C[O, 1], we define Ixl by writing Ixl(t) = /x(t)/. A norm on a space of functions is said to be monotone if IIxll ;;:: IIYII whenever Ixl ;;:: IYI· Prove that the norms II 1100 and II lip are monotone norms. 45. (Continuation) Prove that there is no monotone norm on the space of all real-valued sequences. 46. Why isn't the example of this section a counterexample to Theorem 2?
- 28. Section 1.3 Continuity, Open Sets, Closed Sets 15 47. Any normed linear space X can be embedded as a dense subspace in a complete normed linear space X. The latter is fully determined by the former, and is called the completion of X. A more general assertion of the same sort is true for metric spaces. Prove that the completion of the space f. in Example 8 of Section 1.1 (page 4) is the space Co described in Problem 16. Further remarks about the process of completion occur in Section 1.8, page 60. 48. Metric spaces were defined in Problem 23, page 13. In a metric space, a Cauchy sequence is one that has the property limn,m d(xn,xm ) = O. A metric space is complete if every Cauchy sequence converges to some point in the space. For the discrete metric space mentioned in Problem 11 (page 19), identify the Cauchy sequences and determine whether the space is complete. 1.3 Continuity, Open Sets, Closed Sets Consider a function f, defined on a subset D of a normed linear space X and taking values in another normed linear space Y. We say that f is continuous at a point x in D if for every sequence [xnl in D converging to x, we have also f(xn} -t f(x}. Expressed otherwise, A function that is continuous at each point of its domain is said simply to be continuous. Thus a continuous function is one that preserves the convergence of sequences. Example. The norm in a normed linear space is continuous. To see that this is so, just use Problem 3, page 5, to write IIXnll-llxll ~ IIXn - xii Thus, if Xn -t x, it follows that IlxnII -t Ilxll. • With these definitions at our disposal, we can prove a number of important (yet elementary) theorems. Theorem 1. Let f be a continuous mapping whose domain D is a compact set in a normed linear space and whose range is contained in another normed linear space. Then f(D} is compact. Proof. To show that f(D} is compact, we let [Ynl be any sequence in f(D}, and prove that this sequence has a convergent subsequence whose limit is in f(D}. There exist points Xn ED such that f(xn} = Yn' Since D is compact, the sequence [xnl has a subsequence [xnil that converges to a point xED. Since f is continuous,
- 29. 16 Chapter 1 Normed Linear Spaces Thus the subsequence [Yni 1converges to a point in f(D). • The following is a generalization to normed linear spaces of a theorem that should be familiar from elementary calculus. It provides a tool for optimization problems-even those for which the solution is a function. Theorem 2. A continuous real-valued function whose domain is a compact set in a normed linear space attains its supremum and infi- mum; both of these are therefore finite. Proof. Let f be a continuous real-valued function whose domain is a compact set D in a normed linear space. Let M = sup{J(x) : XED}. Then there is a sequence [xnl in D for which f(xn) -+ M. (At this stage, we admit the possibility that M may be +00.) By compactness, there is a subsequence [xnil converging to a point xED. By continuity, f(xni ) -+ f(x). Hence f(x) = M, and of course M < 00. The proof for the infimum is similar. • A function f whose domain and range are subsets of normed linear spaces is said to be uniformly continuous if there corresponds to each positive c a positive 8 such that IIf (x) - f (y) II < c for all pairs of points (in the domain of f) satisfying Ilx - YII < 8. The crucial feature of this definition is that 8 serves simultaneously for all pairs of points. The definition is global, as distinguished from local. Theorem 3. A continuous function whose domain is a compact subset ofa normed space and whose values lie in another normed space is uniformly continuous. Proof. Let f be a function (defined on a compact set) that is not uniformly continuous. We shall show that f is not continuous. There exists an c > 0 for which there is no corresponding 8 to fulfill the condition of uniform continuity. That implies that for each n there is a pair of points (xn,Yn) satisfying the condition Ilxn - Ynll < lin and II!(xn) - f(Yn)11 ? c. By compactness the sequence [xnl has a subsequence [xniJ that converges to a point x in the domain of f. Then Yni -+ x also because IIYni - xii ~ IIYni - xni II + Ilxni - xii· Now the continuity of f at x fails because c ~ Ilf(xni ) - f(Yni)11 ~ Ilf(xni ) - f(x)11 + Ilf(x) - f(Yni)11 • A subset F in a normed space is said to be closed if the limit of every convergent sequence in F is also in F. Thus, for all sequences this implication is valid: [xn E F & Xn -+ xl ==> x E F As is true of the notion of completeness, the concept of a closed set is useful when the solution of a problem is constructed as a limit of an approximating sequence. By Problem 4, the intersection of any family of closed sets is closed. There- fore, the intersection of all the closed sets containing a given set A is a closed set containing A, and it is the smallest such set. It is commonly written as II or cl(A), and is called the closure of A.
- 30. Section 1.3 Continuity, Open Sets, Closed Sets Theorem 4. The inverse image of a closed set by a continuous map is closed. 17 Preof. Recall that the inverse image of a set A by a map f is defined to be f-' (A) = {x : f(x) E A}. Let f : X -t Y, where X and Yare normed spaces and I is continuous. Let K be a closed set in Y. To show that f-1(K) is closed, we start by letting [xn ] be a convergent sequence in f-1(K). Thus Xn -t x and f(xn ) E K. By continuity, f(xn ) -t f(x). Since K is closed, f(x) E K. Hence x E f-1(K). • As an example, consider the unit ball in a normed space: {x: Ilxll ~ I} This is the inverse image of the closed interval [0,1] by the function x H Ilxll. This function is continuous, as shown above. Hence, the unit ball is closed. Likewise, each of the sets {x: IIx-all ~r} is closed. {x: IIx-all ~r} {x: Q ~ IIx - all ~ ,6} An open set is a set whose complement is closed. Thus, from the preceding remarks, the so-called "open unit ball," i.e., the set u = {x : Ilxll < I} is open, because its complement is closed. Likewise, all of these sets are open: {x: Ilxll > I} {x: IIx - all < r} {x: Q < Ilx - all <,6} An alternative way of describing the open sets, closer to the spirit of general topology, will now be discussed. The open c-cell or c-ball about a point Xo is the set B(xo,c) = {x : Ilx - xoll < c} Sometimes this is called the c-neighborhood of Xo. A useful characterization of open sets is the following: A subset U in X is open if and only if for each x E U there is an c > 0 such that B(x, c) C U. The collection of open sets is called the topology of X. One can verify easily that the topology T for a normed linear space has these characteristic properties: (1) the empty set, 0 , belongs to T; (2) the space itself, X, belongs to T; (3) the intersection of any two members of T belongs to T; (4) the union of any subfamily of T belongs to T. These are the axioms for any topology. One section of Chapter 7 provides an introduction to general topology. A series 2::%"=1 xk whose elements are in a normed linear space is conver- gent if the sequence of partial sums Sn = 2::Z=l Xk converges. The given series is said to be absolutely convergent if the series of real numbers 2::%"=1 Ilxkll is convergent. That means simply that 2::%"=1 IIXkl1 < 00. Problem 3, page 13, asks for a proof that absolute convergence implies convergence, provided that the space is complete. See also Problem 38, page 14. The following theorem gives another important property of absolutely convergent series.
- 31. 18 Chapter 1 Normed Linear Spaces Theorem 5. If a series in a Banach space is absolutely convergent, then all rearrangements of the series converge to a common value. Proof. Let L~l Xi be such a series and L~l Xki a rearrangement of it. Put X = L~l Xi, Sn = L7=1 Xi, Sn = L7=1 Xki , and M = L~llixili. Then L~l IlxkiII ~ M. This proves that L~l Xki is absolutely convergent and hence convergent. (Here we require the completeness of the space.) Put y = L~l Xki· Let c > o. Select n such that Li~n Ilxill < c and such that IISm - xii < c when m ~ n. Select T so that IISr - YII < c and so that {I, ... , n} C {kl,...,kr }. Select m such that {kl , ... , kr } C {I, ... ,m}. Then m ~ nand m IISm - Srll = II(xl + ... + xm) - (Xkl + ... + xkr)11 ~ L Ilxill < c i=n+l Hence • In using a series that is not absolutely convergent, some caution must be exercised. Even in the case of a series of real numbers, bizarre results can arise if the series is randomly re-ordered. A good example of a series of real numbers that converges yet is not absolutely convergent is the series Ln(_1)n In. The series of corresponding absolute values is the divergent harmonic series. There is a remarkable theorem that includes this example: Riemann's Theorem. If a series of real numbers is convergent but not absolutely so, then for every real number, some rearrangement of the series converges to that real number. Proof. Let the series L Xn satisfy the hypotheses. Then lim Xn = 0 and L Xn - L Xn = L IXnl = 00 Xn>O xn<O Since the series L Xn converges, the two series on the left of the preceding equation must diverge to +00 and -00, respectively. (See Problems 12 and 13.) Now let T be any real number. Select positive terms (in order) from the series until their sum exceeds T. Now add negative terms (chosen in order) until the new partial sum is less than T. Continue in this manner. Since limxn = 0, the partial sums thus created differ from T by quantities that tend to zero. • Problems 1.3 1. Prove that the sequential definition of continuity of f at x is equivalent to the "e,8" definition, which is "Ie> 0 38> 0 Vu [ IIx - ull < 8 ~ IIf(x) - f(u) II < eJ
- 32. Section 1.4 More about Compactness 19 2. Let U be an arbitrary subset of a normed space. Prove that the function x ...... dist(x, U) is continuous. This function was defined in the proof of Theorem 1 in Section 1.2, page 9. Prove, in fact, that it is "nonexpansive": Idist(x, U) - dist(y, U)I ::::; Ilx - yll 3. Let X be a normed space. We make X x X into a normed linear space by defining lI(x,y)1I = IIxll + Ilyll· Show that the map (x,y) ...... x + y is continuous. Show that the norm is continuous. Show that the map (A, x) ...... AX is continuous when lR x X is normed by II(A,x)11 = IAI + IIxll· 4. Prove that the intersection of a family of closed sets is closed. 5. If x # 0, put x= x/llxll. This defines the mdial projection of x onto the surface of the unit ball. Prove that if x and y are not zero, then Ilx - ilil ::::; 211x - yll/llxli 6. Use Theorem 2 and Problem 2 in this section to give a brief proof of Theorem 1 in Section 2, page 9. 7. Using the definition of an open set as given in this section, prove that a set U is open if and only if for each x in U there is a positive c such that B(x, c) C U. 8. Prove that the inverse image of an open set by a continuous map is open. 9. The (algebraic) sum of two sets in a linear space is defined by A + B = {a + b: a E A, b E B}. Is the sum of two closed sets (in a normed linear space) closed? (Cf. Problem 19, page 13.) 10. Prove that if the series L:':l Xi converges (in some normed linear space), then Xi --t 0. 11. A common misconception about metric spaces is that the closure of an open ball S = {x : d(a, x) < r} is the closed ball S' = {x : d(a, x) ::::; r}. Investigate whether this is correct in a discrete metric space (X,d), where d(x,y) = 1 if x # y. What is the situation in a normed linear space? (Refer to Problem 23, page 13.) 12. Let L Xn and L Yn be two series of nonnegative terms. Prove that if one of these series converges but the other does not, then the series L(Xn - Yn) diverges. Can you improve this result by weakening the hypotheses? 13. Let L Xn be a convergent series of real numbers such that L IXn I = 00. Prove that the series of positive terms extracted from the series L Xn diverges to 00. It may be helpful to introduce Un = max(xn, 0) and Vn = min(xn,O). By using the partial sums of series, one reduces the question to matters concerning the convergence of sequences. 14. Refer to Problem 12, page 12, for the space too(S). We write::::; to signify a pointwise inequality between two members of this space. Let 9n and In be elements of this space, for n = 1,2, ... Let 9n ~ 0, In-l - 9n-l ::::; In ::::; M, and L~ 9i ::::; M for all n. Prove that the sequence [In] converges pointwise. Give an example to show that convergence in norm may fail. 1.4 More About Compactness We continue our study of compactness in normed linear spaces. The starting point for the next group of theorems is the Heine-Borel theorem, which states that every closed and bounded subset of the real line is compact, and conversely. We assume that the reader is familiar with that theorem.
- 33. 20 Chapter 1 Normed Linear Spaces Our first goal in this section is to show that the Heine-Borel theorem is true for a normed linear space if and only if the space is finite-dimensional. Since most interesting function spaces are infinite-dimensional, verifying the compactness of a set in these spaces requires information beyond the simple properties of being bounded and closed. Many important theorems in functional analysis address the question of identifying the compact sets in various normed linear spaces. Examples of such theorems will appear in Chapter 7. Lemma 1. each ball {x In the space IRn with norm Ilxlloo = maxl~i~n Ix(i)1 Ilxlloo ~ c} is compact. Proof. Let [xkl be a sequence of points in IRn satisfying Ilxklloo ~ c. Then the components obey the inequality -c ~ xk(i) ~ c. By the compactness of the interval [-c, c], there exists an increasing sequence heN having the property that lim [Xk(1) : k E Id exists. Next, there exists another increasing sequence 12 C h such that lim [xk(2) : k E 12l exists. Then lim [Xk(1) : k E 12l exists also, because h C h. Continuing in this way, we obtain at the nth step an increasing sequence In such that lim[xdi) : k E Inl exists for each i = 1, ... , n. Denoting that limit by x'(i), we have defined a vector x' such that Ilxk - x'iloo ---+ 0 as k runs through the sequence of integers In. • Lemma 2. A closed subset of a compact set is compact. Proof. If F is a closed subset of a compact set K, and if [xnl is a sequence in F, then by the compactness of K a subsequence converges to a point of K. The limit point must be in F, since F is closed. • A subset S in a normed linear space is said to be bounded if there is a constant c such that Ilxll ~ c for all XES. Expressed otherwise, sUPrES Ilxll < 00. Theorem 1. In a finite-dimensional normed linear space, each closed and bounded set is compact. Proof. Let X be a finite-dimensional normed linear space. Select a basis for X, say {Xl, ... ,xn}. Define a mapping T : IRn ---+ X by the equation n Ta = La(i)Xi a= (a(l), ... ,a(n)) EIRn i=l If we assign the norm II 1100 to IRn , then T is continuous because IITa - Tbll = II t(a(i) - b(i))Xill ~ tla(i) - b(i)1 Ilxill n n ~ mfx la(i) - b(i)I' L IIXjl1 = Iia - blloo L IIXj II j=l j=l
- 34. Section 1.4 More about Compactness 21 Now let F be a closed and bounded set in X. Put M = T-1(F). Then M is closed by Theorem 4 in Section 1.3, page 17. Since F = T(M), we can use Theorem 1 in Section 1.3, page 15, to conclude that F is compact, provided that M is compact. To show that M is compact, we can use Lemmas 1 and 2 above if we can show that for some c, In other words, we have only to prove that M is bounded. To this end, define {3 = inf{ IITal1 :Iialloo = I} This is the infimum of a continuous map on a compact set (prove that). Hence the infimum is attained at some point b. Thus Ilblloo = 1 and {3 = IITbl1 = II ~b(i)Xill Since the points Xi constitute a linearly independent set, and since b i- 0, we conclude that Tb i- 0 and that {3 > O. Since F is bounded, there is a constant csuch that Ilxll ~ cfor all x E F. Now, if a E ]Rn and a i- 0, then ajllalloo is a vector of norm 1; consequently, IIT(ajllalloo)11 ~ {3, or This is obviously true for a = 0 also. For a E M we have Ta E F, and {3llalloo ~ IITal1 ~ c, whence Iialloo ~ cj{3. Thus, M is indeed bounded. • Corollary 1. Every finite-dimensional normed linear space is com- plete. Proof. Let [xnl be a Cauchy sequence in such a space. Let us prove that the sequence is bounded. Select an index m such that Ilxi - Xjll < 1 whenever i, j ~ m. Then we have (i ~ m) Hence for all i, Since the ball of radius c is compact, our sequence must have a convergent subsequence, say xni ~ x'. Given E: > 0, select N so that Ilxi - Xj II < E: when i,j ~ N. Then IIXj - XnJ < E: when i,j ~ N, because ni > i. By taking the limit as i ~ 00, we conclude that IIXj - x'il ~ E: when j ~ N. This shows that Xj ~ x. •
- 35. 22 Chapter 1 Normed Linear Spaces Corollary 2. Every finite-dimensional subspace in a normed linear space is closed. Proof. Recall that a subset Y in a linear space is a subspace if it is a linear space in its own right. (The only axioms that require verification are the ones concerned with algebraic closure of Y under addition and scalar multiplication.) Let Y be a finite-dimensional subspace in a normed space. To show that Y is closed, let Yn E Y and Yn -+ y. We want to know that Y E Y. The preceding corollary establishes this: The convergent sequence has the Cauchy property and hence converges to a point in Y, because Y is complete. • Riesz's Lemma. If U is a closed and proper subspace (U is neither onor the entire space) in a normed linear space, and if0 < A < 1, then there exists a point x such that 1 = Ilxll and dist(x, U) > A. Proof. Since U is proper, there exists a point z EX" U. Since U is closed, dist(z, U) > o. (See Problem 11.) By the definition of dist(z, U) there is an element u in U satisfying the inequality liz - ull < A-I dist(z, U). Put x = (z - u)/llz - ull. Obviously, Ilxll = 1. Also, with the help of Problem 7, we have dist(x, U) = dist(z - u, U)/llz - ull = dist(z, U)/llz - ull > A • Theorem 2. If the unit ball in a normed linear space is compact, then the space has finite dimension. Proof. If the space is not finite dimensional, then a sequence [xn ] can be defined inductively as follows. Let Xl be any point such that IlxI11 = 1. If Xl, ... ,xn-l have been defined, let Un - l be the subspace that they span. By Corollary 2, above, Un - l is closed. Use Riesz's Lemma to select Xn so that IIXnl1 = 1 and dist(xn,Un-I) > 1· Then IIXn - xiii> 1whenever i < n. This sequence cannot have any convergent subsequence. • Putting Theorems 1 and 2 together, we have the following result. Theorem 3. A normed linear space is finite dimensional ifand only if its unit ball is compact. In any normed linear space, a compact set is necessarily closed and bounded. In a finite-dimensional space, these two conditions are also sufficient for compact- ness. In any infinite-dimensional space, some additional hypothesis is required to imply compactness. For many spaces, necessary and sufficient conditions for compactness are known. These invariably involve some uniformity hypothesis. See Section 7.4, page 347, for some examples, and [DS] (Section IV.14) for many others. Problems 1.4 1. A real-valued function f defined on a normed space is said to be lower semicontinuous if each set {x : f{x) ~ >.} is closed (>. E JR). Prove that every continuous function is lower
- 36. Section 1.4 More about Compactness 23 semicontinuous. Prove that if f and -fare lower semicontinuous, then f is continuous. Prove that a lower semicontinuous function attains its infimum on a compact set. 2. Prove that the collection of open sets (as we have defined them) in a normed linear space fulfills the axioms for a topology. 3. Two norms, Nl and N2, on a vector space X are said to be equivalent if there exist positive constants 0 and (3 such that ONl ~ N2 ~ (3Nl . Show that this is an equivalence relation. Show that the topologies engendered by a pair of equivalent norms are identical. 4. Prove that a Cauchy sequence converges if and only if it has a convergent subsequence. 5. Let X be the linear subspace of all real sequences x = [x(l), x(2), ... J such that only a finite number of terms are nonzero. Is there a norm for X such that (X, II II) is a Banach space? 6. Using the notation in the proof of Theorem 1, prove in detail that F = T(M). 7. Prove these properties of the distance function dist(x, U) (defined in Section 1.2, page 9) when U is a linear subspace in a normed linear space: (a) dist(Ax, U) = IAI dist(x, U) (b) dist(x - u, U) = dist(x, U) (u E U) (c) dist(x + y, U) ~ dist(x, U) +dist(y, U) 8. Prove this version of Riesz's Lemma: If U is a finite-dimensional proper subspace in a normed linear space X, then there exists a point x for which IIxll = dist(x, U) = 1. 9. Prove that if the unit ball in a normed linear space is complete, then the space is complete. 10. Let U be a finite-dimensional subspace in a normed linear space X. Show that for each x E X there exists a u E U satisfying IIx - ull =dist(x, U). 11. Let U be a closed subspace in a normed space X. Prove that the distance functional has the property that for x EX" U, dist(x,U) > O. 12. In any infinite-dimensional normed linear space, the open unit ball contains an infinite disjoint family of open balls all having radius ~ (!!) (Prove it, of course. While you're at it, try to improve the number ~.) 13. In the proof of Theorem 1, show that M is bounded as follows. If it is not bounded, let ak E M and lIaklloc --+ 00. Put a~ = ak/llakiloc. Prove that the sequence [a~] has a convergent subsequence whose limit is nonzero. By considering Ta~, obtain a contradiction of the injective nature of T. 14. Prove that the sequence [xnJ constructed in the proof of Theorem 2 is linearly indepen- dent. 15. Prove that in any infinite-dimensional normed linear space there is a sequence [xn] in the unit ball such that IIxn - Xm II > 1 when n # m. If you don't succeed, prove the same result with the weaker inequality IIxn - Xm II ~ 1. (Use the proof of Theorem 2 and Problem 8 above.) Also prove that the unit ball in eoe contains a sequence satisfying Ilxn - xmll =2 when n # m. Reference: [DiesJ. 16. Let S be a subset of a normed linear space such that IIx - yll ~ 1 when x and y are different points in S. Prove that S is closed. Prove that if S is an infinite set then it cannot be compact. Give an example of such a set that is bounded and infinite in the space e[O, 1J. 17. Let A and B be nonempty closed sets in a normed linear space. Prove that if A + B is compact, then so are A and B. Why do we assume that the sets are nonempty? Prove that if A is compact, then A +B is closed.
- 37. 24 Chapter 1 Normed Linear Spaces 1.5 Linear Transformations Consider two vector spaces X and Y over the same scalar field. A mapping f : X -t Y is said to be linear if f(au +(3v) = af(u) +(3f(v) for all scalars a and (3 and for all vectors u, v in X. A linear map is often called a linear transformation or a linear operator. If Y happens to be the scalar field, the linear map is called a linear functional. By taking a = (3 = awe see at once that a linear map f must have the property f(O) = O. This meaning of the word "linear" differs from the one used in elementary mathematics, where a linear function of a real variable x means a function of the form x >--+ ax +b. Example 1. If X = IRn and Y = IRm , then each linear map of X into Y is of the form f(x) = y, n y(i) = LaijX(j) (1 ~ i ~ m) j=1 where the aij are certain real numbers that form an m x n matrix. • Example 2. Let X = C[O, 1J and Y = IR. One linear functional is defined by f(x) = f; x(s) ds. • Example 3. Let X be the space of all functions on [O,lJ that possess n continuous derivatives, x', x", ... ,x(n). Let ao, a1, ... ,an be fixed elements of X. Then a linear operator D is defined by i=O Such an operator is called a differential operator. • Example 4. Let X = C[O, 1J = Y. Let k be a continuous function on [O,lJ x [O,lJ. Define K by (Kx)(s) = 11k(s,t)x(t)dt This is a linear operator, in fact a linear integral operator. • Example 5. Let X be the set of all bounded continuous functions on IR+ = {t E IR: t ~ a}. Put (Lx)(s) = 100 e-stx(t)dt This linear operator is called the Laplace Transform. • Example 6. Let X be the set of all continuous functions on IR for which f~oo Ix(t)1 dt < 00. Define (Fx)(s) = 1:e-27ristx(t) dt This linear operator is called the Fourier Transform. • If a linear transformation T acts between two normed linear spaces, then the concept of continuity becomes meaningful.
- 38. Section 1.5 Linear Transformations Theorem 1. A linear transformation acting between normed linear spaces is continuous if and only if it is continuous at zero. 25 Proof. Let T : X -t Y be such a linear transformation. If it is continuous, then of course it is continuous at O. For the converse, suppose that T is continuous atO. For each € > 0 there is a 8 > 0 such that for all x, jjxjj < 8 => jjTxjj < € Hence jjx - yjj < 8 => jjTx - Tyjj = jjT(x - y)jj < € • A linear transformation T acting between two normed linear spaces is said to be bounded if it is bounded in the usual sense on the unit ball: sup{ jjTxjj : jjxjj ~ I} < 00 Example 7. Let X = C 1[0,1]' the space of all continuously differentiable functions on [0,1]. Give X the norm jjxjjoo = sup Ix(s)l. Let f be the linear functional defined by f (x) == x'(1). This functional is not bounded, as is seen by considering the vectors Xn (s) = sn. On the other hand, the functional in Example 2 is bounded since If(x)1 ~ J; Ix(s)1 ds ~ jjxjjoo· • Theorem 2. A linear transformation acting between normed linear spaces is continuous if and only if it is bounded. Proof. Let T : X -t Y be such a map. If it is continuous, then there is a 8 > 0 such that jjxjj ~ 8 => jjTxjj ~ 1 If jjxjj ~ 1, then 8x is a vector of norm at most 8. Consequently, jjT(8x)jj ~ 1, whence jjTxjj ~ 1/8. Conversely, if jjTxjj ~ M whenever jjxjj ~ 1, then jjxll ~ :r => lI~xll ~ 1 => liT (~x) II ~ M => IITxl1 ~ € This proves continuity at 0, which suffices, by the preceding theorem. • If T : X -t Y is a bounded linear transformation, we define IITII = sup{ IITxl1 : Ilxll ~ I} It can be shown that this defines a norm on the family of all bounded linear transformations from X into Y; this family is a vector space, and it now becomes a normed linear space, denoted by £(X, Y). The definition of jjTjjleads at once to the important inequality jjTxjj ~ jjTjj jjxjj To prove this, notice first that it is correct for x = 0, since TO = O. On the other hand, if x#- 0, then x/jjxjj is a vector of norm 1. By the definition of jjTjj, we have jjT(x/jjxjj)jj ~ jjTjj, which is equivalent to the inequality displayed above. That inequality contains three distinct norms: the ones defined on X, Y, and £(X, Y).
- 39. 26 Chapter 1 Normed Linear Spaces Theorem 3. A linear functional on a normed space is continuous if and only if its kernel ("null space") is closed. Proof. Let f : X -t lR be a linear functional. Its kernel is ker(f) = {x : f(x) = O} This is the same as f-1({O}). Thus if f is continuous, its kernel is closed, by Theorem 4 in Section 1.3, page 17. Conversely, if f is discontinuous, then it is not bounded. Let Ilxnll ::::;; 1 and f(xn ) -t 00. Take any x not in the kernel and consider the points x - EnXn, where En = f(x)/ f(xn ). These points belong to the kernel of f and converge to x, which is not in the kernel, so the latter is not closed. • Corollary 1. Every linear functional on a finite-dimensional normed linear space is continuous. Proof. If f is such a functional, its null space is a subspace, which, by Corollary 2 in Section 1.4, page 22, must be closed. Then Theorem 3 above implies that f is continuous. • Corollary 2. Every linear transformation from a finite-dimensional normed space to another normed space is continuous. Proof. Let T : X -t Y be such a transformation. Let {b1 , . .. , bn } be a basis for X. Then each x E X has a unique expression as a linear combination of basis elements. The coefficients depend on x, and so we write x = L~=1 Ai(X)bi. These functionals Ai are in fact linear. Indeed, from the previous equation and the equation u = L Ai(u)bi we conclude that n ax + j3u = Z)aAi(X) + j3Ai(U)] bi i=1 Since we have also n ax + j3u = L Ai(ax + j3u)bi i=1 we may conclude (by the uniqueness of the representations) that Now use the preceding corollary to infer that the functionals Ai are continuous. Getting back to T, we have and this is obviously continuous. •
- 40. Section 1.5 Linear Transformations 27 Corollary 3. All norms on a finite-dimensional vector space are equivalent, as defined in Problem 3, page 23. Proof. Let X be a finite-dimensional vector space having two norms 11111 and 11112· The identity map J from (X, II Ill) to (X, II IU is continuous by the preceding result. Hence it is bounded. This implies that By the symmetry in the hypotheses, there is a (3 such that IIxl11 ~ (3llxI12. • Recall that if X and Yare two normed linear spaces, then the notation £(X, Y) denotes the set of all bounded linear maps of X into Y. We have seen that boundedness is equivalent to continuity for linear maps in this context. The space £(X, Y) has, in a natural way, all the structure of a normed linear space. Specifically, we define (aA +(3B)(x) = a(Ax) +(3(Bx) IIAII = sup{ IIAxlly :x EX, Ilxllx ~ l} In these equations, A and B are elements of £(X, Y), and x is any member of X. Theorem 4. If X is a normed linear space and Y is a Banach space, then £(X, Y) is a Banach space. Proof. The only issue is the completeness of £(X, Y). Let [An] be a Cauchy sequence in £(X, Y). For each x E X, we have This shows that [Anx] is a Cauchy sequence in Y. By the completeness of Y we can define Ax = lim Anx. The linearity of A follows by letting n -+ 00 in the equation The boundedness of A follows from the boundedness of the Cauchy sequence [An]. If IIAnl1 ~ M then IIAnXl1 ~ Mllxll for all x, and in the limit we have IIAxl1 ~ Mllxll· Finally, we have IIAn- AII-+ 0 because if IIAn- Am II ~ c when m,n;;;: N, then for all x of norm 1 we have IIAnx - AmXl1 ~ c when m,n;;;: N. Then we can let m -+ 00 to get IIAnx - Axil ~ c and IIAn - All ~ c. • The composition of two linear mappings A and B is conventionally written as AB rather than AoB. Thus, (AB)x = A(Bx). If AA is well-defined (Le., the range of A is contained in its domain), then we write it as A2. All nonnegative powers are then defined recursively by writing AD = J, An+1 = AAn.
- 41. 28 Chapter 1 Normed Linear Spaces Theorem 5. The Neumann Theorem. Let A be a bounded linear operator on a Banach space X (and taking values in X). If IIAll < 1, then I - A is invertible, and 00 (I - A)-I = L Ak k=O Proof. Put Bn = 2:~=o Ak. The sequence [Bn] has the Cauchy property, for if n > m, then IIBn - Bmll = 1 kf+1Akll ~ kf+IIIAkll ~ ~IIAllk 00 = IIAllmL IIAllk = IIAllm/(1-IIAII) k=O (In this calculation we used Problem 20.) Since the space of all bounded linear operators on X into X is complete (Theorem 4), the sequence [Bn] converges to a bounded linear operator B. We have n n+1 (I - A)Bn = Bn - ABn = L Ak - L Ak = J - An+1 k=O k=1 Taking a limit, we obtain (J - A)B = I. Similarly, B(I - A) = I. Hence B = (I - A)-I. • The Neumann Theorem is a powerful tool, having applications to many applied problems, such as integral equations and the solving of large systems of linear equations. For examples, see Section 4.3, which is devoted to this theorem, and Section 3.3, which has an example of a nonlinear integral equation. Problems 1.5 1. Prove that the closure of a linear subspace in a normed linear space is also a subspace. (The closure operation is defined on page 16.) 2. Prove that the operator norm defined here has the three properties required of a norm. 3. Prove that the kernel of a linear functional is either closed or dense. (A subset in a topological space X is dense if its closure is X.) 4. Let {Xl, ... , Xk} be a linearly independent finite set in a normed linear space. Show that there exists a 8 > 0 sum that the condition max IIXi-Yill<8 l~i~k implies that {Yl, ... ,Yk} is also linearly independent. 5. Prove directly that if T is an unbounded linear operator, then it is discontinuous at O. (Start with a sequence [xnJ such that Ilxnll ~ 1 and IITxnll -+ 00.)
- 42. Section 1.5 Linear Transformations 29 6. Let A be an m x n matrix. Let X = IRn , with norm IIxllco =maxl ';:; i';:;n Ix(i)l. Let Y = IRm , with norm lIyllco = maxl ';:;i';:;m ly(i)l· Define a linear transformation T from X to Y by putting (Tx)(i) =2::7= 1 a;j x(j), 1 ~ i ~ m. Prove that IITII = max; 2::7= 1Ia;jl. 7. Prove that a linear map is injective (i.e., one-to-one) if and only if its kernel is the 0 subspace. (The kernel of a map T is {x : Tx = O}.) 8. Prove that the norm of a linear transformation is the infimum of all the numbers l'v! that satisfy the inequality IITxl1 ~ Mllxll for all x. 9. Prove the (surprising) result that a linear transformation is continuous if and only if it transforms every sequence converging to zero into a bounded sequence. 10. If f is a linear functional on X and N is its kernel, then there exists a one-dimensional subspace Y such that X = Y EEl N. (For two sets in a linear space, we define U + V as the set of all sums u + v when u ranges over U and v ranges over V. If U and V are subspaces with only 0 in common we write this sum as U EEl V.) 11. The space eco(S) was defined in Problem 12 of Section 1.2, page 12. Let S = N, and define T: eoc(N) -+ c[-!,!l by the equation (Tx)(s) = 2::;;"=1 x(k)sk. Prove that T is linear and continuous. 12. Prove or disprove: A linear map from a normed linear space into a finite-dimensional normed linear space must be continuous. 13. Addition of sets in a vector space is defined by A + B = {a + b : a E A , b E B} . Better: A + B = {x : 3 a E A & 3 b E B such that x = a + b}. Scalar multiplication is AA = {Aa : a E A}. Does the family of all subsets of a vector space X form a vector space with these definitions? 14. Let Y be a closed subspace in a Banach space X. A "coset" is a set of the form x +Y = {x +y : y E V}. Show that the family of all cosets is a normed linear space if we use the norm III x + Y III = dist(x, V). 15. Refer to Problem 12 in the preceding section, page 23. Show that the assertion there is not true if ~ is replaced by !. 16. Prove that for a bounded linear transformation T : X -+ Y IITII = sup IITxll = sup IITxll/llxll IIrll=1 #0 17. Prove that a bounded linear transformation maps Cauchy sequences into Cauchy se- quences. 18. Prove that if a linear transformation maps some nonvoid open set of the domain space to a bounded set in the range space, then it is continuous. 19. On the space C[O, 1] we define "point-evaluation functionals" by t*(x) = x(t). Here t E (0, 1] and x E C(O, 1]. Prove that IIt'li = 1. Prove that if I/> = 2::7=1 A;t: , where tl , t2, · .·,tn are distinct points in [0, 1], then 111/>11 = 2::7=1 IA; I· 20. In the proof of the Neumann Theorem we used the inequality IIAk II ~ IIAlik . Prove this. 21. Prove that if {<PI, ... ,qln} is a linearly independent set of linear functionals, then for suitable Xj we have qI;(Xj) =8ij for 1 ~ i,j ~ n.
- 43. 30 Chapter 1 Normed Linear Spaces 22. Prove that if a linear transformation is discontinuous at one point, then it is discontinuous everywhere. 23. Linear transformations on infinite-dimensional spaces do not always behave like their counterparts on finite-dimensional spaces. The space Co was defined in Problem 1.2.16 (page 12). On the space Co define Ax =A[x(I), x(2), ...j = [x(2), x(3), ...j Bx = B[x(I), x(2), ...j = [0, x(I), x(2), ...j Prove that A is surjective but not invertible. Prove that B is injective but not invertible. Determine whether right or left inverses exist for A and B. 24. What is meant by the assertion that the behavior of a linear map at any point of its domain is exactly like its behavior at o? 25. Prove that every linear functional f on IRn has the form f(x) = 2:~=1 oix(i), where x(I), x(2), ... ,x(n) are the coordinates of x. Let 0 = [01,02, ... , On] and show that the relationship f ...... 0 is linear, injective, and surjective (hence, an isomorphism). 26. Is it true for linear operators in general that continuity follows from the null space being closed? 27. Let <Po,(Pl, ... ,<Pn be linear functionals on a linear space. Prove that if the kernel of <Po contains the kernels of all <Pi for 1 :(; i :(; n, then <Po is a linear combination of <PI,... ,<Pn. 28. If L is a bounded linear map from a normed space X to a Banach space Y, then L has a unique continuous linear extension defined on the completion of X and taking values in Y. (Refer to Problem 1.2.47, page 15.) Prove this assertion as well as the fact that the norm of the extension equals the norm of the original L. 29. Let A be a continuous linear operator on a Banach space X. Prove that the series 2::'=0 An In! converges in £(X, X). The resulting sum can be denoted by eA. Is eA invertible? 30. Investigate the continuity of the Laplace transform (in Example 5, page 24). 1.6 Zorn's Lemma, Hamel Bases, and the Hahn-Banach Theorem This section is devoted to two results that require the Axiom of Choice for their proofs. These are a theorem on existence of Hamel bases, and the Hahn-Banach Theorem. The first of these extends to all vector spaces the notion of a base, which is familiar in the finite-dimensional setting. The Hahn-Banach Theorem is needed at first to guarantee that on a given normed linear space there can be defined continuous maps into the scalar field. There are many situations in applied mathematics where the Hahn-Banach Theorem plays a crucial role; convex optimization theory is a prime example. The Axiom of Choice is an axiom that most mathematicians use unre- servedly, but is nonetheless controversial. Its status was clarified in 1940 by a famous theorem of Godel [Go]. His theorem can be stated as follows.
- 44. Section 1.6 Zorn's Lemma, Hamel Bases, Hahn-Banach Theorem 31 Theorem 1. If a contradiction can be derived from the Zermelo- Fraenkel axioms of set theory (which include the Axiom of Choice), then a contradiction can be derived within the restricted set theory based on the Zermelo-Fraenkel axioms without the Axiom of Choice. In other words, the Axiom of Choice by itself cannot be responsible for intro- ducing an inconsistency in set theory. That is why most mathematicians are willing to accept it. In 1963, Paul Cohen [Coh] proved that the Axiom of Choice is independent of the remaining axioms in the Zermelo-F'raenkel system. Thus it cannot be proved from them. The statement of this axiom is as follows: Axiom of Choice. If A is a set and / a function on A such that /(a) is a nonvoid set for each a E A, then / has a "choice function." That means a function c on A such that c(a) E /(a) for all a E A. For example, suppose that A is a finite set: A = {a1, ... ,an}. For each i in {1,2, ... ,n} a nonempty set /(ai) is given. In n steps, we can select "repre- sentatives" Xl E /(a1), X2 E /(a2), etc. Having done so, define c(ad = Xi for i = 1,2, ... , n. Attempting the same construction for an infinite set such as A = JR, with accompanying infinite sets /(a), leads to an immediate difficulty. To get around the difficulty, one might try to order the elements of each set /(a) in such a way that there is always a "first" element in /(a). Then c(a) can be defined to be the first element in /(a). But the proposed ordering will require another axiom at least as strong as the Axiom of Choice! For a second example, see Problem 45, page 40. A number of other set-theoretic axioms are equivalent to the Axiom of Choice. See [Kel] and [RR]. Among these equivalent axioms, we single out Zorn's Lemma as being especially useful. First, we require some definitions. Definition 1. A partially ordered set is a pair (X, -<) in which X is a set and -< is a relation on X such that (i) X -< X for all X (ii) If x -< y and y -< Z, then x -< Z Definition 2. A chain, or totally ordered set, is a partially ordered set in which for any two elements x and y, either x -< y or y -< x. Definition 3. In a partially ordered set X, an upper bound for a subset A in X is any point x in X such that a -< x for all a E A. Example 1. Let S be any set, and denote by 28 the family of all subsets of S, including the empty set 0 and S itself. This is often called the "power set" of S. Order 28 by the inclusion relation c. Then (28 , C) is a partially ordered set. It is not totally ordered. An upper bound for any subset of 28 is S. • Example 2. In JR2, define x -< y to mean that x(i) ~ y(i) for i = 1 and 2. This is a partial ordering but not a total ordering. Which quadrants in JR2 have upper bounds? • Example 3. Let:F be a family of functions (whose ranges and domains need not be specified). For / and g in :F we write / -< g if two conditions are fulfilled:
- 45. 32 Chapter 1 Normed Linear Spaces (i) dom(f) C dom(g) (ii) J(x) = g(x) for all x in dom(f) When this occurs, we say that "g is an extension of J." Notice that this is equivalent to the assertion J C g, provided that we interpret (as ultimately we must) J and 9 as sets of pairs of elements. • Definition 4. An element m in a partially ordered set X is said to be a maximal element if every x in X that satisfies the condition m -< x also satisfies x -< m. Zorn's Lemma. A partially ordered set contains a maximal element if each totally ordered subset has an upper bound. Definition 5. Let X be a linear space. A subset H of X is called a Hamel base, or Hamel basis, if each point in X has a unique expression as a finite linear combination of elements of H. Example 4. Let X be the space of all polynomials defined on lR. A Hamel base for X is given by the sequence [hnJ where hn(s) = sn, n =0, 1,2,.... • Theorem 2. Every nontrivial vector space has a Hamel base. Proof. Let X be a nontrivial vector space. To show that X has a Hamel base we first prove that X has a maximal linearly independent set, and then we show that any such set is necessarily a Hamel base. Consider the collection of all linearly independent subsets of X, and partially order this collection by inclusion, C. In order to use Zorn's Lemma, we verify that every chain in this partially ordered set has an upper bound. Let C be a chain. Consider S' = U{S : SEC}. This certainly satisfies S C S· for all SEC. But is S· linearly independent? Suppose that '2::7=1 niSi = 0 for some scalars ni and for some distinct points Si in S·. Each Si belongs to some Si E C. Since C is a chain (and since there are only finitely many Si), one of these sets (say Sj) contains all the others. Since Sj is linearly independent, we conclude that '2:: Inil = O. This establishes the linear independence of S· and the fact that every chain in our partially ordered set has an upper bound. Now by Zorn's Lemma, the collection of all linearly independent sets in X has a maximal element, H. To see that H is a Hamel base, let x be any element of X. By the maximality of H, either H U {x} is linearly dependent or H U {x} C H (and then x E H). In either case, x is a linear combination of elements of H. If x can be represented in two different ways as a linear combination of members of H, then by subtraction, we obtain 0 as a nontrivial linear combination of elements of H, contradicting the linear independence of H. • In the next theorem, when we say that one real-valued function, J, is dom- inated by another, p, we mean simply that J(x) ~ p(x) for all x. Hahn-Banach Theorem. Let X be a real linear space, and let p be a function from X to IR such that p(x + y) ~ p(x) + p(y) and p(>..x) = >..p(x) if>.. ;?; O. Any linear functional defined on a subspace of
- 46. Section 1.6 Zorn's Lemma, Hamel Bases, Hahn-Banach Theorem 33 X and dominated by p has an extension that is linear, defined on X, and dominated by p. Proof. Let f be such a functional, and let Xo be its domain. Thus Xo is a linear subspace of X. In approaching the theorem for the first time and wonder- ing how to discover a proof, one naturally asks how to extend the functional f to a domain containing Xo that is only one dimension larger than Xo. If that is impossible, then the theorem itself cannot be true. Accordingly, let y be a point not in the original domain. To extend f to Xo +span(y) it suffices to specify a value for f(y) because of the necessary equation f(x + AY) = f(x) + Af(y) (x E Xo , AE JR) The value of f(y) must be assigned in such a way that f(x) + Af(y) ~ p(x + AY) (x E Xo , AE JR) If A= 0, this inequality is certainly valid. If A> 0, we must have f (~) +f (y) ~ p (~ + y) (x E Xo) or If A < 0, we must have 1 f(X2) + f(y) ~ -;xp(x + AY) = -p(-X2 - y) These two conditions on f(y) can be written together as In order to see that there is a number satisfying this inequality, we compute f(Xl) - f(X2) = f(Xl - X2) ~ P(XI - X2) = P(XI + y - X2 - y) ~ P(XI +y) +P(-X2 - y) This completes the extension by one dimension. Next, we partially order by the inclusion relation (C) all the linear exten- sions of f that are dominated by p. Thus h c 9 if and only if the domain of 9 contains the domain of h, and g(x) = h(x) on the domain of h. In order to use Zorn's Lemma, we must verify that each chain in this partially ordered set has an upper bound. But this is true, since the union of all the elements in such a chain is an upper bound !9r the chain. (Problem 2.) By Zor~s Lemma, there exists a maximal element f in our partially ordered set. Then f ~ a linear functional that is an extension of f and is dominated by p. Finally, f must be defined on all of X, for if it were not, a further extension would be possible, as shown in the first part of the proof. •
- 47. 34 Chapter 1 Normed Linear Spaces Corollary 1. Let <I> be a linear Functional defined on a subspace Y in a normed linear space X and satisFying I<I>(Y) I ~ MIIYII (y E y) Then <I> has a linear extension defined on all of X and satisFying the above inequality on X. Proof. Use the Hahn-Banach Theorem with p(x) = Mllxll. Corollary 2. Let Y be a subspace in a normed linear space X. IF w E X and dist(w, Y) > 0, then there exists a continuous linear Functional <I> defined on X such that <I>(Y) = 0 For all Y E Y, <I>(w) = 1, and 11<1>11 = 1/dist(w, Y). • Proof. Let Z be the subspace generated by Y and w. Each element of Z has a unique representation as Y+AW, where Y E Y and AE R It is clear that <I> must be defined on Z by writing <I>(y +AW) = A. The norm of <I> on Z is computed as follows, in which the supremum is over all nonzero vectors in Z: 11<1>11 = sup I<I>(Y + Aw)/lly + Awll = sup IAI/IIY + Awll = sup lillY/A + wll =11inf IIY + wll = 11dist(w, Y) By Corollary 1, we can extend the functional <I> to all of X without increase of its norm. • Corollary 3. To each point w in a normed linear space there corresponds a continuous linear functional <I> such that 11<1>11 = 1 and <I>(w) = 11wI1· Proof. In Corollary 2, take Y to be the O-subspace. • At this juncture, it makes sense to associate with any normed linear space X a normed space X' consisting of all continuous linear functionals defined on X. Corollary 3 shows that X' is not trivial. The space X' is called the conjugate space of X, or the dual space or the adjoint of X. Example 4. Let X = Rn , endowed with the max-norm. Then X' is (or can be identified with) Rn with the norm II 111' To see that this is so, recall (Problem 1.5.25, page 30) that if <I> E X*, then <I>(x) = 2:~=1 u(i)x(i) for a suitable u E Rn. Then n n 11<1>11 = sup IL U(i)X(i)1 = L l'u(i)1 = Ilulll Ilxlloo';;;l i=l i=l • Example 5. Let Co denote the Banach space of all real sequences that converge to zero, normed by putting Ilxlloo = sup Ix(n)l. Let f1 denote the Banach space of all real sequences u for which 2:::"=1 lu(n)1 < 00, normed by putting Ilulll = 2:::"=1 lu(n)l· With each u E f1 we associate a functional <l>u E Co by means of the equation <l>u(x) = 2:::"=1 u(n)x(n). (The connection between these two spaces is the subject of the next result.) •
- 48. Section 1.6 Zorn's Lemma, Hamel Bases, Hahn-Banach Theorem 35 Proposition. The mapping u f--t 1>u is an isometric isomorphism between £1 and co' Thus we can say that Co "is" £1. Proof. Perhaps we had better give a name to this mapping. Let A : £1 -+ Co be defined by Au =: 1>u. It is to be shown that for each u, Au is linear and continuous on co. Then it is to be shown that A is linear, surjective, and isometric. Isometric means IIAul1 =: Ilulll. That 1>u is well-defined follows from the absolute convergence of the series defining 1>u(x): The linearity of 1>u is obvious: 1>u(ax +(3y) =: L u(n) [ax(n) +(3y(n)] =: a L u(n)x(n) +(3 L u(n)y(n) =: a1>u(x) +(31)u(Y) The continuity or boundedness of 1>u is easy: By taking a supremum in this last inequality, considering only x for which Ilxlloo ::;; 1, we get On the other hand, if c > 0 is given, we can select N so that I:~N+l lu(n)1 < c. Then we define x by putting x(n) =: sgn u(n) for n ::;; N, and by setting x(n) =: 0 for n > N. Clearly, x E Co and Ilxlloo =: 1. Hence N N II1>ull ~ 1>u(x) =: Lx(n)u(n) =: L lu(n)1 > Ilulll - c n=l n=l Since c was arbitrary, II1>ull ~ Ilulll. Hence we have proved Next we show that A is surjective. Let '¢ E co. Let <5n be the element of Co that has a 1 in the nth coordinate and zeros elsewhere. Then for any x, 00 x=: L x(n)<5n n=l Since '¢ is continuous and linear,
- 49. 36 Chapter 1 Normed Linear Spaces Consequently, if we put u(n) ='I/J(8n ), then 'I/J(x) =¢u(x) and 'I/J =¢u' To verify that u E £1, we define (as above) x(n) = sgnu(n) for n ~ Nand x(n) = 0 for n> N. Then N 2:lu(n)1 =2: x(n)u(n) ='I/J(x) ~ 11'l/Jllllxll = 1I'l/J1l n=l Thus Ilull l ~ 1I'l/J1l. Finally, the linearity of A follows from writing ¢O:U+{JV(x) = 2:(au +,Bv)(n)x(n) =a 2:u(n)x(n) +,B 2:v(n)x(n) = (a¢u +,B¢v)(x) • Corollary 4. For each x in a normed linear space X, we have IIxll = max{I¢(x)1 : ¢ E X' , II¢II = 1} Proof. If ¢ E X' and II¢II = 1, then 1¢(x)1 ~ 1I¢lIlIxll = IIxll Therefore, sup{I¢(x)1 : ¢ E X' , II¢II = 1} ~ IIxll For the reverse inequality, note first that it is trivial if x = o. Otherwise, use Corollary 3. Then there is a functional 'I/J E X' such that 'I/J(x) = IIxll and 1I'l/J1l = 1. Note that the supremum is attained. • A subset Z in a normed space X is said to be fundamental if the set of all linear combinations of elements in Z is dense in X. Expressed otherwise, for each x E X and for each e > 0 there is a vector L~=l .Zi such that Zi E Z, Ai E JR, and IIx - 2:AiZili < e We could also state that dist(x,span Z) = 0 for all x E X. As an example, the vectors 81 = (1,0,0, ... ) 82 = (0,1,0, ... ) etc. form a fundamental set in the space Co. Example 6. In the space C[a, b], with the usual supremum norm, an important fundamental set is the sequence of monomials uo(t) = 1 , U1(t) = t , U2(t) = t2 , ... The Weierstrass Approximation Theorem asserts the fundamentality of this se- quence. Thus, for any x E C[a, bJ and any e > 0 there is a polynomial u for which IIx - ulloo < e. Of course, uis of the form L~o AiUi' • Definition 5. If A is a subset of a normed linear space X, then the annihilator of A is the set AJ.. = {¢ E X' : ¢(a) = 0 for all a E A}
- 50. Section 1.6 Zorn's Lemma, Hamel Bases, Hahn-Banach Theorem 37 Theorem 3. A subset in a normed space is fundamental if and only if its annihilator is {a}. Proof. Let X be the space and Z the subset in question. Let Y be the closure of the linear span of Z. If Y i= X, let x EX" Y. Then by Corollary 2, there exists </J EX' such that </J(x) = 1 and </J E y.l.. Hence </J E Z.l. and Z.l. i= O. If Y = X, then any element of Z.l. annihilates the span of Z as well as Y and X. Thus it must be the zero functional; i.e., Z.l. = O. • Theorem 4. IfX is a normed linear space (not necessarily complete) then its conjugate space X' is complete. Proof. This follows from Theorem 4 in Section 1.5, page 27, by letting Y = lR in that theorem. • Problems 1.6 1. Let X and Y be sets. A function from a subset of X to Y is a subset f of X x Y such that for each x E X there is at most one y E Y satisfying (x, y) E f. We write then f(x) = y. The set of all such functions is denoted by S. Prove or disprove the following: (a) S is partially ordered by inclusion. (b) The union of two elements of S is a member of S. (c) The intersection of two elements of S is a member of S. (d) The union of any chain in S is a member of S. 2. In the proof of the Hahn-Banach theorem, show that the union of the elements in a chain is an upper bound for the chain. (There are five distinct things to prove.) 3. Denote by CO the normed linear space of all functions x : N --t IR having the property limn-too x(n) = 0, with norm given by IIxll = sUPn Jx(n)J. Do the vectors em defined by em(n) =8nm form a Hamel base for CO? 4. If {h" : a E I} is a Hamel base for a vector space X, then each element x in X has a representation x = L" A(a)h" in which A : I --t IR and {a: A(a) # O} is finite. (Prove this.) 5. Prove that every real vector space is isomorphic to a vector space whose elements are real-valued functions. ("Function spaces are all there are.") 6. Prove that any linearly independent set in a vector space can be extended to produce a Hamel base. 7. If U is a linear subspace in a vector space X, then U has an "algebraic complement," which is a subspace V such that X = U +V, Un V = O. ("0" denotes the zero subspace.) (Prove this.) FIVE EXERCISES (8-12) ON BANACH LIMITS 8. The space loo consists of all bounded sequences, with norm IIxlioo = sUPn Jx(n)J. Define T : lOO --t loo by putting Tx = [x(l), x(2) - x(l), x(3) - x(2), x(4) - x(3) ...J Let M denote the range of T, and put u =[1,1,1, ...J. Prove that dist(u,M) =1. 9. Prove that there exists a continuous linear functional </> E M.l such that 11</>11 = </>(u) = 1. The functional </> is called a Banach limit, and is sometimes written LIM. 10. Prove that if x E loo and x ;;;: 0, then </>(x) ;;;: O. 11. Prove that </>(x) =limn x(n) when the limit exists. 12. Prove that if y = [x(2), x(3), ...J then </>(x) =</>(y).
- 51. 38 Chapter 1 Normed Linear Spaces 13. Let loo denote the normed linear space of all bounded real sequences, with norm given by IIxIL", =sUPn Ix{n)l. Prove that loe is complete, and therefore a Banach space. Prove that lj = loo, where the equality here really means isometrically isomorphic. 14. A hyperplane in a normed space is any translate of the null space of a continuous, linear, nontrivial functional. Prove that a set is a hyperplane if and only if it is of the form {x : ¢(x) = ,X}, where ¢ E X' "0 and ,X E lIt A translate of a set S in a vector space is a set of the form v +S = {v +s : S E S}. 15. A half-space in a normed linear space X is any set of the form {x : ¢(x) ~ ,X}, where ¢ E X' "0 and ,X E JR. Prove that for every x satisfying IIxll = 1 there exists a half-space such that x is on the boundary of the half-space and the unit ball is contained in the half-space. 16. Prove that a linear functional ¢ is a linear combination of linear functionals ¢1, ... , ¢n if and only if N(¢) :l n::"1 N(¢i). Here N(¢) denotes the null space of ¢. (Use induction and trickery.) 17. Prove that a linear map transforms convex sets into convex sets. 18. Prove that in a normed linear space, the closure of a convex set is convex. 19. Let Y be a linear subspace in a normed linear space X. Prove that dist(x, Y) =sup{¢(x) : ¢ E X' , ¢ 1. Y , II¢II = I} Here the notation ¢ 1. Y means that ¢(y) =0 for all y E Y. 20. Let Y be a subset of a normed linear space X. Prove that Y -L is a closed linear subspace in X'. 21. If Z is a linear subspace in X', where X is a normed linear space, we define Z-L ={x EX: ¢(x) =0 for all ¢ E Z} Prove that for any closed subspace Y in X, (Y -L h = Y. Generalize. 22. Let J(z) = L.::;;"=o anzn, where [an] is a sequence of complex numbers for which nan --t O. Prove the famous theorem of Tauber that L.:: an converges if and only if limz--+1 J(z) exists. (See [DS], page 78.) 23. Do the vectors tln defined just after Corollary 4 form a fundamental set in the space lX) consisting of bounded sequences with norm Ilxlloo =maxn Ix(n)l? THREE EXERCISES (24-26) ON SCHAUDER BASES (See [Sem] and [Sing].) 24. A Schauder base (or basis) for a Banach space X is a sequence [un] in X such that eacil x in X has a unique representation 00 x = L,XnUn n=l This equation means, of course, that limN--+ocllx - L.::~=1 'xnunll = O. Show that one Schauder base for Co is given by un(m) =tlnm (n, m = 1,2,3, ...).
- 52. Section 1.6 Zorn's Lemma, Hamel Bases, Hahn-Banach Theorem 39 25. Prove that the An in the preceding problem are functions of x and must be, in fact, linear and continuous. 26. Prove that if the Banach space X possesses a Schauder base, then X must be separable. That is, X must contain a countable dense set. 27. Prove that for any set A in a normed linear space all these sets are the same: A-L, (closure A)-L, (span A)-L, [closure (span A)]-L,. 28. Prove that for x E co, 29. Use the Axiom of Choice to prove that for any set S having at least 2 points there is a function f : S --t S that does not have a fixed point. 30. An interesting Banach space is the space C consisting of all convergent sequences. The norm is IIxlloc = sUPn Jx(n)J. Obviously, we have these set inclusions among the examples encountered so far: Prove that Co is a hyperplane in c. Identify in concrete terms the conjugate space c'. 31. Prove that if H is a Hamel base for a normed linear space, then so is {h/llhil :h E H}. 32. Let X and Y be linear spaces. Let H be a Hamel base for X. Prove that a linear map from X to Y is completely determined by its values on H, and that these values can be arbitrarily-assigned elements of Y. 33. Prove that on every infinite-dimensional normed linear space there exist discontinuous linear functionals. (The preceding two problems can be useful here.) 34. Using Problem 33 and Problem 1.5.3, page 28, prove that every infinite-dimensional normed linear space is the union of a disjoint pair of dense convex sets. 35. Let two equivalent norms be defined on a single linear space. (See Problem 1.4.3, page 23.) Prove that if the space is complete with respect to one of the norms, then it is complete with respect to the other. Prove that this result fails (in general) if we assume only that one norm is less than or equal to a constant multiple of the other. 36. Let Y be a subspace of a normed space X. Prove that there is a norm-preserving injective map J : Y' --t X' such that for each <P E Y', J<p is an extension of <p. 37. Let Y be a subspace of a normed space X. Prove that if Y -L =0, then Y is dense in X. 38. Let T be a bounued linear map of Co into Co. Show that T must have the form (Tx)(n) = 2.::':1 ani X (i) for a suitable infinite matrix [ani]. Prove that sUPn 2.::':1 Jani J = IITII· 39. Prove that if #S =n, then #2s = 2n. 40. What implications exist among these four properties of a set S in a normed linear space X? (a) S is fundamental in X; (b) S is linearly independent; (c) S is a Schauder base for X; (d) S is a Hamel base for X.
- 53. 40 Chapter 1 Normed Linear Spaces 41. A "spanning set" in a linear space is a set S such that each point in the space is a linear combination of elements from S. Prove that every linear space has a minimal spanning set. 42. Let f : IR --t IR. Define x -< y to mean f(x) :::; f(y). Under what conditions is this a partial order or a total order? 43. Criticize the following "proof' that if X and Y are any two normed linear spaces, then X' == Y·. We can assume that X and Yare subspaces of a third normed space z. (For example, we could use Z == X EI1 Y, a direct sum.) Clearly, X' is a subspace of Z', since the Hahn-Banach Theorem asserts that an element of X' can be extended, without increasing its norm, to Z. Clearly, Z· is a subspace of Y', since each element of Z· can be restricted to become an element of Y·. So, we have X' C Z· C Y·. By symmetry, Y' C X'. So X· == Y' . 44. Let K be a subset of a linear space X, and let f : K --t IR. Establish necessary and sufficient conditions in order that f be the restriction to K of a linear functional on X. 45. For each a in a set A, let f(a) be a subset of N. Without using the Axiom of Choice, prove that f has a choice function. 1.7 The Baire Theorem and Uniform Boundedness This section is devoted to the first consequences of completeness in a normed linear space. These are stunning and dramatic results that distinguish Banach spaces from other normed linear spaces. Once we have these theorems (in this section and the next), it will be clear why it is always an advantage to be working with a complete space. The reader has undoubtedly seen this phenomenon when studying the real number system (which is complete). When we compare the real and the rational number systems, we notice that the latter has certain deficiencies, which indeed had already been encountered by the ancient Greeks. For example, they knew that no square could have rational sides and rational diagonal! Put another way, certain problems posed within the realm of rational numbers do not have solutions among the rational numbers; rather, we must expect solutions sometimes to be irrational. The simplest example, of course, is x2 = 2. Our story begins with a purely metric-space result. Theorem 1. Baire's Theorem. In a complete metric space, the intersection of a countable family of open dense sets is dense. Proof. (A set is "dense" if its closure is the entire space.) Let 0 1 , O2, • •• be open dense sets in a complete metric space X. In order to show that n:=lOn is dense, it is sufficient to prove that this set intersects an arbitrary nonvoid open ball 81 in X. For each n we will define an open ball and a closed ball: Select any Xl E X and let T1 > O. We want to prove that 81intersects n:=lOn. Since 0 1 is open and dense, 0 1 n 81 is open and nonvoid. Take 8~ C 81 n 0 1•
- 54. Section 1.7 The Baire Theorem and Uniform Boundedness 41 Then take S~ C S2 n O2 , S~ C S3 n 0 3 , and so on. At the same time we can insist that Tn .J.. o. Then for all n, The points Xn form a Cauchy sequence because Xi, Xj E Sn if i,j > n, and so Since X is complete, the sequence [xnl converges to some point X'. Since for i > n, Xi E S~+1 C S1 nOn we can let i -+ 00 to conclude that x' E S~+1 C S1 n On. Since this is true for all n, the set n:=1On does indeed intersect S1. • Corollary. If a complete metric space is expressed as a countable union of closed sets, then one of the closed sets must have a nonempty interior. Proof. Let Xbe a complete metric space, and suppose that X= U:=1 Fn , where each Fn is a closed set having empty interior. The sets On = X "Fn are open and dense. Hence by Baire's Theorem, n:=1On is dense. In particular, it is nonempty. If X E n:=1On, then X EX" U:=1 Fn, a contradiction. • A subset in a metric space X (or indeed in any topological space) is said to be nowhere dense in X if its closure has an empty interior. Thus the set of irrational points on the horizontal axis in ]R2 is nowhere dense in 1R2 . A set that is a countable union of nowhere dense sets is said to be of category I in X. A set that is not of category I is said to be of category II in X. Observe that all three of these notions are dependent on the space. Thus one can have E C X C Z, where E is of category II in X and of category I in Z. For a concrete example, the one in the preceding paragraph will serve. The Corollary implies that if X is a complete metric space, then X is of the second category in X. Intuitively, we think of sets of the first category as being "thin," and those of the second category as "fat." (See Problems 5, 6, 7, for example.) Theorem 2. The Banach-Steinhaus Theorem. Let {Ao} be a family of continuous linear transformations defined on a Banach space X and taking values in a normed linear space. In order that sUPo IIAol1 < 00, it is necessary and sufficient that the set {x EX: sUPo IIAoxl1 < oo} be of the second category in X. Proof. Assume first that c = sUPo IIAo II < 00. Then every x satisfies IIAoxl1 :::;; cllxll, and every x belongs to the set F = {x : sUPo IIAoxl1 < oo}. Since F = X, the preceding corollary implies that F is of the second category in X. For the sufficiency, define Fn = {x EX: sup IIAoxl1 :::;; n} o
- 55. Another Random Document on Scribd Without Any Related Topics
- 56. to Georgia, westward to Western New York, Eastern Ohio, Kentucky, and Eastern Tennessee. It has little value as timber, because it does not grow large enough. Scotch and Austrian Pines. In the same manner other pines may be studied. Fig. 238 shows a cone and a bit of foliage of the Scotch pine, and Fig. 239 the Austrian pine. These cones grew the past season and are not yet mature. After they ripen and shed the seeds which they contain, they will look somewhat like the cone in Fig 235. The Scotch pine has short and blue-green needles. The Austrian pine is coarser, and has long dark- green needles. There are but two leaves in a cluster on these kinds of pines and we shall find that the sheath which incloses the base of the leaf-cluster is more conspicuous than in either the white or the pitch pine. Do the leaves persist in the Scotch and Austrian pines longer than they do in the others we have examined? Study the cones of these and other pines.
- 57. Fig. 238. Scotch pine. Half natural size. The Scotch and Austrian pines are not native to this country, but are much grown for ornament. They can be found in almost any park and in many other places where ornamental trees are grown. The Norway Spruce. The leaves of spruce trees are borne very differently from those of the pines. Instead of being in clusters of two or more, they are single and without a sheath at the base; neither are there scale-like bodies on the branches where the leaves are borne. Notice, too, that the leaves have a very short stem or petiole. The leaves of the Norway spruce are about one inch long, although the length varies more or less in different parts of the tree and in different trees. They are rather stiff and rigid and sharp-pointed. In a general way, the leaves are four-sided, though indistinctly so.
- 58. Fig. 239. Austrian pine. One-third natural size. It will be interesting to study the position which the leaves take on the branches. A hasty glance might give us the impression that the leaves are not produced on the under side of the branches; but a more careful examination will convince us that there are nearly as many on the under side as on the upper. The leaves are all pointing outward from the branch and as nearly upward as is possible. In other words, the leaves grow toward the light. We must not forget to see how long the leaves of the Norway spruce persist and to find out when the leaf-scars disappear. We can find leaves that must surely be six or seven years old and sometimes we can find them even older than this. The leaf scars, too, remain a long time. The falling of the leaves is illustrated in Fig. 240. It shows the extremities of a limb which is eight years old. The part between the tip and A is last season's growth; between A and B it is two years old; and beyond B is a part that grew three seasons ago. The section beyond C is six years old; from C to D is seven years of age. The four years' growth of this limb not shown in the drawing was as densely covered with foliage as is the part shown in the upper figure; but there are not many leaves between C and D (seven years old) and none on the eight-year-old wood (except those on the branchlets, and these are younger). The cone of the Norway spruce is nearly as long as that of the white pine, but it is not so rough and coarse as the white pine cone is. The cones are usually borne on the tips of small branchlets, although
- 59. occasionally one is borne in the manner shown in Fig. 241. The cones usually fall the first winter. Fig. 240. Twig of the common Norway spruce. Half natural size. The Norway spruce is not a native of this country, but like the Scotch and Austrian pines, it was introduced from Europe and is grown very widely as an ornamental tree. It is the commonest evergreen in yards and parks. The Black Spruce and Its Kin. There are several different kinds of spruces which we find growing in our forests and swamps, and sometimes these are planted for ornament. A sprig of foliage and a cone of one of these,—the black spruce,—is shown in Fig. 242. The foliage is not very unlike that of the Norway spruce, but the cones are very small in comparison. They are about one inch long, though they vary considerably in size. Before they open they are oval or plum-shaped; but when mature and the scales of the cone have expanded, they are nearly globular. They are
- 60. Fig. 241. Cone of Norway spruce. Half size. often borne in clusters, as well as singly, and persist for many years after the seeds have fallen. The position of the cones will depend upon their age. When young they point upward, but they gradually turn downward. In general appearance the white spruce resembles the black very closely. The leaves of the white spruce have a whitish or dusty looking tinge of color and when crushed or bruised give forth a peculiar, disagreeable odor. The cones vary in length from an inch to two inches, and in shape are more cylindrical or finger-shaped than the cone of the black spruce. The foliage of the red spruce lacks the whitish tinge of color of the white spruce and the cones, which are from one inch to two inches in length, are obovate in shape—that is, the widest place is through the upper part of the cone, and from this point it gradually tapers to the tip. They seldom persist longer than the second summer. The leaves of all these different kinds of spruces vary greatly in length, thickness, and sharpness of point, according to the part of the tree on which they grow, and their surroundings. The shedding of the leaves on these or other spruces can be determined as easily as in the Norway spruce. These three spruces like a cold climate and grow in many sections of the northern United States and Canada and farther south in the mountains. They are sometimes all found growing together, but the black spruce likes best the damp, cold swamps, while the others grow best on the drier and better drained lands. The black spruce is commonest. The red spruce is least known. The Balsam Fir.
- 61. This is another evergreen tree which grows naturally in the cold, damp grounds of the northern United States and Canada, and to some extent in the eastern states as far south as West Virginia. The foliage is borne in much the same manner as that of the spruces; yet there are interesting differences in the characters of these two kinds of leaves. Perhaps the most noticeable difference is in the shape; and the color of the fir leaves will attract our attention because the under side is a silvery color, while the upper side is green. What is the nature of the tip of the leaf and how does it compare with the pines and spruces in this respect? Does the leaf have a stem or petiole or is it attached directly to the branch without any stem? How are the leaves shed? Fig. 242.—Black spruce. Half natural size. The cones are about three inches long and present a rather delicate appearance. It will be interesting to determine the position of the cones, that is, the direction in which they point, and to learn whether it is the same when they are young as it is after they have matured. The grayish colored bark of the trunk and limbs bears many "blisters" from which Canada balsam is obtained. The Hemlock. A hemlock twig is an interesting object. It may have many characters in common with the spruce and fir; yet the impression which we get
- 62. from it, or from a large hemlock tree, is entirely distinct. The arrangement of the leaves and the gracefulness of the drooping branchlets are most pleasing. We are led to examine it more closely. We notice that the leaves appear to be borne in two more or less regular rows,—one on each side of the branch or twig; but in reality they come from all sides of the branch, and it is the position which the leaves assume that gives this two-rowed appearance. The leaves have a short stalk or petiole, and this stalk rests along the side of the branchlet in such a direction that the leaves are placed in single rows on either side of the branch. The petioles of the leaves are nearly parallel with the branch while the leaves often make a decided angle with the petiole. This fact can best be brought out by carefully examining a small twig. While we are noting the arrangement of the leaves on the branchlets, we should also notice the points of similarity and difference between these leaves and those of the spruces and firs. We shall find that there is more in common, at least so far as shape and color are concerned, between the hemlock and the fir than between the hemlock and the spruce. Fig. 243. Spray of the hemlock. Two-thirds natural size.
- 63. The small, delicate cones, borne on the tips of the branchlets, will also attract our attention (Fig. 243.) We may wonder at their small size, for they are only about three-quarters of an inch long, and very delicate; yet a second glance at the tree will impress us with the number of cones which the tree bears, and we conclude that, although the cones may be small, yet there are so many of them that there will be no lack of seeds. It is more difficult to trace the age of a hemlock limb than of many other kinds of trees, yet we can easily determine that many of the leaves are several years old when they fall. The bark of the hemlock is used in tanning hides for leather. The tree is much used for lumber. Where does it grow? The Arbor-vitæ. One might almost wonder, at first sight, if the arbor-vitæ (often, but wrongly, called the white cedar) has any leaves at all. It does possess them, however, but they are very different in size and shape from any of the others that we have examined. They are small scale-like bodies, closely pressed together along the sides of the branchlets, in four rows. Leaves pressed to the branches in this manner are said to be "appressed." The leaves of the arbor-vitæ are so close together that they overlap one another. The leaves are of two distinct shapes, sometimes known as the surface leaves and the flank leaves. The former are located on what appears to be the flattened surface of the branchlets, while the latter are on the sides or edges. See Fig. 244.
- 64. Fig. 244. The Arbor-vitæ. Nearly full size. If we carefully look at the leaves, we shall notice a raised spot near the point or tip. This is said to be a resin gland. This gland can be seen more plainly on the surface leaves that are two years old. Most of the leaves persist for at least two and sometimes three years; but even older ones can be found. These older leaves, however, exist not as green, active leaves, but merely as dried and lifeless scales. These lifeless leaves are probably detached from the branches by the forces of nature. The cones are even smaller than the hemlock cones. They are borne in the axils of the leaves in the same manner as the branchlets and are not conspicuous unless one is close to the tree. The arbor-vitæ is much planted for hedges and screens, as well as for other ornamental purposes. There are many horticultural varieties. The tree is abundant in a wild state in New York. Summary of the Kinds of Common Evergreens. The white pine (Pinus Strobus).—Leaves in clusters of five, soft and slender; cones five or six inches long, slightly curved; bark smooth except on the trunks and larger limbs of old trees, where it is fissured. The pitch pine (Pinus rigida).—Leaves in clusters of three, from three to four inches long, rather rigid; cones two to three inches long, often in clusters of two or more but frequently borne singly, persisting long after the seeds have been shed; bark more or less rough on the young growth and deeply fissured on the trunks of old trees. The Scotch pine (Pinus sylvestris).—Leaves usually in clusters of two, from two to four inches long, rigid, of a bluish-green hue when
- 65. seen in a large mass on the tree; cones two to three inches long and the scales tipped with a beak or prickle. The Austrian pine (Pinus Austriaca).—Leaves in clusters of two, five or six inches long and somewhat rigid, dark green in color, and persisting for four or five years; cones about three inches long, conical in shape; and scales not beaked or pointed as in the Scotch pine. The Norway spruce (Picea excelsa).—Leaves borne singly, about one inch long, dark green, four-sided; cones about six inches long, and composed of thin scales, and usually borne on the tips of branchlets. The small branches mostly drooping. The black spruce (Picea nigra).—In general appearance, this is not very unlike the Norway spruce, but the small branches stand out more horizontally and the cones are only one or one and one-half inches long, recurving on short branches. The cones persist for several years after shedding the seed. The white spruce (Picea alba).—Leaves about one inch long, having a glaucous or whitish tinge; twigs stout and rigid, of a pale greenish-white color; cones from one to two and one-half inches long, more or less cylindrical or "finger-shaped," and easily crushed when dry. The red spruce (Picea rubra).—The foliage lacks the whitish tinge of the white spruce and is of a dark or dark yellowish color; twigs stouter than those of the black spruce and not so much inclined to droop; cones about one inch long, obovate, and usually falling by the second summer. The hemlock (Tsuga Canadensis).—Leaves about one-half inch long, flat with rounded point, green on the upper side, whitish beneath, and borne on short appressed petioles; cones about three-quarters of an inch long, oval or egg-shaped, and borne
- 66. on the ends of small branchlets and often persisting for some time. The balsam fir (Abies balsamea).—Leaves narrow, less than one inch long, borne singly, very numerous and standing out from the branchlets in much the way of the spruce; cones about three inches long, cylindrical, composed of thin scales, and standing upright on the branches, or recurved; bark smooth, light green with whitish tinge. The arbor-vitæ (Thuya occidentalis).—Leaves very small, scale-like, and over-lapping one another in four rows, adhering closely to the branchlets; the cones oblong and small,—a half-inch or less in length,—and composed of but few scales. LEAFLET XXXIV. THE CLOVERS AND THEIR KIN.[46] By ANNA BOTSFORD COMSTOCK. The pedigree of honey does not concern the bee, A clover any time to him is aristocracy. —Emily Dickinson. here is a deep-seated prejudice that usefulness and beauty do not belong together;—a prejudice based obviously on human selfishness, for if a thing is useful to us we emphasize that quality so much that we forget to look for its beauty. Thus it is that the clover
- 67. suffers great injustice; it has for centuries been a most valuable forage crop, and, therefore, we forget to note its beauty, or to regard it as an object worthy of æsthetic attention. This is a pitiful fact; but it cheats us more than it does the clover, for the clover blossoms not for us, but for the bees and butterflies as well as for itself. As I remember the scenes which have impressed me most, I find among them three in which clover was the special attraction. One was a well-cultivated thrifty orchard carpeted with the brilliant red of the crimson clover in bloom. One was a great field of alfalfa spread near the shore of the Great Salt Lake, which met our eyes as we came through the pass in the Wasatch Mountains after days of travel in dust-colored lands; the brilliant green of that alfalfa field in the evening sunlight refreshed our eyes as the draught of cold water refreshes the parched throat of the traveller in a desert. And another was a gently undulating field in our own State stretching away like a sea to the west, covered with the purple foam of the red clover in blossom; and the fragrance of that field settled like a benediction over the acres that margined it. But we do not need landscapes to teach us the beauty of clover. Just one clover blossom studied carefully and looked at with clear-seeing eyes, reveals each floweret beautiful in color, interesting in form, and perfect in its mechanism for securing pollination. The clover is especially renowned for its partnerships with members of the animal kingdom. It readily forms a partnership with man, thriftily growing in his pastures and meadows, while he distributes its seed. For ages it has been a special partner of the bees, giving them honey for carrying its pollen. Below the ground it has formed a mysterious partnership with microbes, and the clover seems to be getting the best of the bargain. For many years clover was regarded as a crop helpful to the soil, and one reason given was the great length of the roots. Thus the roots of red clover often reach the depth of several feet, even in heavy soil, which they thus aerate and drain, especially when they decay and leave channels. But this is only half the story; for a long time people had noted that on clover roots were little swollen places or nodules, which were supposed to have come from some disease or insect
- 68. injury. The scientists became interested in the supposed disease, and they finally ascertained that these nodules are filled with bacteria, which are the underground partners of the clovers and other legumes. These bacteria are able to fix the free nitrogen of the air, and make it available for plant-food. As nitrogen is the most expensive of the fertilizers, any agency which can extract it from the free air for the use of plants is indeed a valuable aid to the farmer. Thus it is that in the modern agriculture, clover or some other legume is put on the land once in three or four years in the regular rotation of crops, and it brings back to the soil the nitrogen which other crops have exhausted. An interesting fact about the partnership between the root bacteria and the clover-like plants is that the plants do not flourish without this partnership, and investigators have devised a method by which these bacteria may be scattered in the soil on which some kinds of clover are to be planted, and thus aid in growing a crop. This method is to-day being used for the introduction of alfalfa here in New York State. But the use of clover as a fertilizer is not limited to its root factory for capturing nitrogen; its leaves break down quickly and readily yield the rich food material of which they are composed, so that the farmer who plows under his second-crop clover instead of harvesting it, adds greatly to the fertility of his farm. The members of three distinct genera are popularly called clovers: The True Clovers (Trifolium), of which six or seven species are found in New York State, and more than sixty species are found in the United States. The Medics (Medicago), of which four species are found here. The Melilots (Melilotus), or sweet clovers, of which we have two species. The True Clovers. (Trifolium.) The Red Clover (Fig. 245). (Trifolium pratense.[47])—This beautiful dweller in our fields came to us from Europe, and it is also a native of Asia. It is the clover most widely cultivated in New York State for fodder, and is one of our most important crops. Clover hay often being a standard of excellence by which other hay is measured. The
- 69. Fig. 245. The common red clover. export of clover seed from the United States has sometimes reached the worth of two million dollars per year, and this great industry is supposed to be carried on with the aid of that other partner of the red clover, the bumblebee. Bumblebees had to be imported into Australia before clover seed could be produced there. The whole question of the relation of the bumblebee to the pollination of clover no doubt needs to be re-studied, for recent observations have led to the contesting of prevailing opinions. It has been supposed that the failure of the clover seed crop in some places is due to the destruction of bumblebees; whether this is true or not, we are certain that bumblebees visit clover blooms, and the teacher can observe for himself. There is a more perennial form of red clover, known as variety perenne. It is distinguished from the common form of red clover by its taller growth and mostly less hairy herbage, and by the fact that the flower-head is usually somewhat stalked. Some persons regard it as a hybrid of red and zig-zag clover. Zig-Zag Clover. (T. medium.)—This is another species of red clover, resembling the one just discussed, except that its flower-head rises on a long stalk above the upper leaves, while the red clover has the flower-head set close to these leaves. The color of the blossom is darker than in red clover, and the flower-head is looser. The stems of the zig-zag clover are likely to be bent at angles and thus it gets its name. It is a question whether this species is really grown on farms. It is probable that some or all of the clover that passes under this name is Trifolium pratense var. perenne.
- 70. Fig. 246. Crimson clover. At all events, the zig-zag clover seems to be imperfectly understood by botanists and others. Crimson Clover—Scarlet Clover (Fig. 246). (T. incarnatum.)—While this beautiful clover grows as a weed in the southern parts of our State, it has only recently begun to play an important part in our horticulture. It is an annual, and its home is the Mediterranean region of Europe. It thrives best in loose, sandy soils, and in our State is chiefly used as a cover-crop for orchards, and to plow under as a fertilizer. It usually has bright, crimson flowers, arranged in a long, pointed head, and its brilliant green fan-shaped leaves make it the most artistically decorative of all our clovers. Buffalo Clover (Fig. 247). (T. reflexum.)—This is sometimes taken for a variety of the red clover, but only a glance is needed to distinguish it. While the head is perhaps an inch in diameter the flowerets are not directed upward and set close as in the red clover, but each floweret is on a little stalk, and is bent abruptly backward. The flowers are not pink. The standard is red, while the wings and keel are nearly white. The leaves are blunt at the tip. It grows in meadows in western New York and westward. This species is native to this country. Alsike Clover. (T. hybridum.)—This is a perennial and grows in low meadows and waste places from Nova Scotia to Idaho. It was introduced from Europe. It is especially valuable in wet meadows, where the red clover would be drowned. The blossoms of the alsike look like those of the white clover except that they are a little larger and are pink; but the long branching mostly upright stems are very different in habit from the creeping stems of the white clover; the blossoms are very fragrant.
- 71. Fig. 247. Three clovers, respectively, Buffalo, Yellow, and Rabbit- foot clover. The White Clover. (T. repens.)—This beautiful little clover, whose leaves make a rug for our feet in every possible place, is well known to us all. It is the clover best beloved by honey-bees, and the person who does not know the distinct flavor of white clover honey has lost something out of life. While in hard soil the white clover lasts only two or three years, on rich, moist lands it is a true perennial. While it was probably a native in the northern part of America, yet it is truly cosmopolitan and may be found in almost all regions of the temperate zones. Very likely the common stock of it is an introduction from Europe. By many this is considered to be the original shamrock. The Yellow, or Hop Clover (Fig. 247). (T. agrarium.)—This friendly little plant, filling waste places with brilliant green leaves and small yellow flower-heads, is not considered a clover by those who are not observant. But if the flowerets in the small, dense heads are examined, they will be seen to resemble very closely those of the other clovers. The stems are many-branched and often grow a foot or more in height. The flowers are numerous, and on fading turn brown, and resemble the fruit of a pigmy hop vine, whence the name. Its leaves are much more pointed than those of the medics, with which it might be confused because of its yellow flowers.
- 72. Low Hop Clover, or Hop Trefoil. (T. procumbens.)—This resembles the above species, except that it is smaller and also more spreading, and the stems and leaves are more downy. The Least Hop Clover. (T. dubium.)—This may be readily distinguished from the above species by the fact that its yellow flowerets occur from three to ten in a head. This is said by some to be the true shamrock, although the white clover is also called the shamrock. The Rabbit-Foot, or Stone Clover (Fig. 247). (T. arvense.)—This is another clover not easily recognized as such. It grows a foot or more in height and has erect branches. The leaflets are narrow and all arise from the same point. The flowerets occur in long, dense heads. The calyx is very silky, and the lobes are longer than the white corollas, thus giving the flower-head a soft, hairy look, something like the early stages of the blossom of the pussy willow. Because of its appearance it is often called "pussy clover." The Medics. (Medicago.) Alfalfa (Fig. 248). (Medicago sativa.)—This is the veteran of all the clovers, for it has been under cultivation for twenty centuries. It is a native of the valleys of western Asia. In America it was first introduced into Mexico with the Spanish invasion. It was brought from Chile to California in 1854, where it has since been the most important hay crop. In fact, there is no better hay than that made from alfalfa. It was probably introduced into the Atlantic States from southern Europe, and has grown as a weed for many years in certain localities in New England and the Middle States; only recently has it been considered a practicable crop for this climate, although it was grown in Jefferson Co., N. Y., in 1791. Its special value is that it is a true perennial, and may be cut three times or more during a season, and when once established it withstands hot, dry weather. It is of marvelous value to the semi-arid regions. The alfalfa flower is blue or violet, and grows in a loose raceme. The plant grows tall and its
- 73. Fig. 248. Alfalfa, foliage and flowers. stems are many branched. This and all these medics are introduced from Europe. Black or Hop Medic. (M. lupulina.)—This would hardly be called a clover by the novice. The long stems lie along the ground, and the tiny yellow flower-heads do not much resemble the clover blossom. It is a common weed in waste places in our State. It is perennial. The Toothed Medic. (M. denticulata.)—Instead of having the yellow flowerets in a dense head, this species has them in pairs or perhaps fours, or sometimes more. It is widely distributed as a weed, and is also introduced as a pasture plant for early grazing. It is of little value as hay. The Spotted Medic. (M. Arabica.)—This very much resembles the preceding species except that the leaves are likely to have on them conspicuous dark spots near the center. Like the preceding species it is an annual and a weed, and has also been introduced as a plant for early grazing. This and the toothed medic are known to farmers under the name of bur-clover. The reason for this name is found in the seed-pod, which is twisted in a spiral and has an outer margin of curved prickles. The Melilots, Or Sweet Clovers. (Melilotus.) In driving or walking along the country roads, we may find ourselves suddenly immersed in a wave of delightful fragrance, and if we look for the source we may find this friendly plant flourishing in the most forbidding of soils. Growing as a weed, it brings sweet perfume to us, and at the same time nitrogen, aeration and drainage to the hopeless soil, making rich those places where other weeds have not the temerity to attempt to grow. When the soil is generous, the sweet clover often grows very tall, sometimes as high as ten feet. It is a
- 74. Fig. 249. White sweet clover. cheerful, adaptable and beneficial plant, and I never see it without giving it a welcome, which, I am sorry to say, I cannot always grant to other roadside wayfarers. The sweet clovers are European. The White Sweet Clover (M. alba) is sometimes called Bokhara clover and has white flowers (Fig. 249). The Yellow Sweet Clover (M. officinalis) has yellow blossoms. It has interesting old English names, such as Balsam Flowers, King's Clover and Heartwort. Questions on the Clovers. Two general kinds of types of studies are to be made of the clovers: identification studies, whereby you will come to know the kinds of clover; life history studies, whereby you will come to know under what conditions the plants live and thrive. The latter is the more important, but the former usually precedes it, for one is better able to discover and discuss the biological questions when he is acquainted with the species. The following questions will bring out some of the important biological aspects: 1. How many of the true clovers, the medics, and the sweet clovers do you know? 2. Send me properly labelled pressed specimens of the leaves and blossoms of the clovers that you have been able to find.
- 75. 3. Dig a root of red clover and find the nodules on it. Please describe them. 4. What methods does the U. S. Department of Agriculture employ to inoculate the soil with bacteria so that alfalfa may grow? 5. How do clover roots protect the land from the effects of heavy rains? 6. How do the clover plants conserve the moisture in the soil? 7. How does this conservation of moisture aid the farmer and orchardist? 8. What is a cover-crop, and what are its uses? 9. Why do farmers sow red clover with grass seed? 10. How do the habits of the stems of white clover differ from those of other clovers? 11. Why is white clover so desirable for lawns? 12. Compare the floweret of the red clover with the sweet pea blossom and describe the resemblance. 13. Study a head of white clover from the time it opens until it is brown, and tell what changes take place in it day by day. 14. What has happened to the flowerets that are bent downward around the stalk? 15. Watch one of these flowerets deflect, and describe the process. 16. How many flowerets do you find in a head of red clover? Of white clover? Of alsike? 17. Which flowerets open first in a head of red clover?
- 76. 18. Describe a clover seed. Describe a seed of alfalfa. 19. What insects do you find visiting the red clover blossoms? The white clover blossoms? Alfalfa, or Lucerne.[48] The alfalfa plant is just now coming into great prominence in New York State. Every teacher, particularly in the rural schools, will need to know the plant and to have some information about it. What alfalfa is.—It is a clover-like plant. It is perennial. It has violet- purple flowers. The leaves have three narrow leaflets. It sends up many stiff stems, 2 to 3 feet high. The roots go straight down to great depths. Why it is important.—It is an excellent cattle food, and cattle-raising for dairy purposes is the leading special agricultural industry in New York State. In fact, New York leads all the States in the value of its dairy products. Any plant that is more nutritious and more productive of pasture and hay than the familiar clovers and grasses will add immensely to the dairy industry, and therefore to the wealth of the State. Alfalfa is such a plant. It gives three cuttings of hay year after year in New York State, thereby yielding twice as much as clover does. In the production of digestible nutrients per acre ranks above clover as 24 ranks above 10. When once established it withstands droughts, for the roots grow deep. Alfalfa is South European. It was early introduced into North America. It first came into prominence in the semi-arid West because of its drought-resisting qualities, and now it has added millions of dollars to the wealth of the nation. Gradually it is working its way into the East. It is discussed in the agricultural press and before farmers' institutes. Last year the College of Agriculture offered to send a small packet of seeds to such school children in New York State as wanted to grow a little garden plat of it. About 5,000 children were supplied. The teacher must now learn what alfalfa is.
- 77. In nearly every rural community, sufficient alfalfa can be found for school purposes. In many places it has run wild along roadsides. On these plants make the following observations: 1. Under what conditions have you found alfalfa growing? How did the plant come to grow there,—sown, or run wild? 2. Describe the form of the root. How does the root branch? 3. Do you find the little tubercles or nodules on the roots? On what part of the roots? How large? How numerous? 4. The crown of the plant (at the surface of the ground),—describe it, and how the tops and the roots start from it. 5. The stems,—how many from each crown, whether erect or prostrate, how they branch. 6. The leaves,—simple or compound? Form? Edges entire or fine toothed? Do the leaves "sleep" at night, as those of clover do? 7. Do you find any distinct spots on the leaves? What do you think is the cause of them? 8. Flowers,—how borne (whether singly or in clusters), color, form, resemblance to any other flowers you may know. Do they vary in color? 9. If possible, find the seed-pods and seeds, and describe. 10. Make inquiries as to whether alfalfa is becoming well known in your vicinity. Agricultural Account of Alfalfa. You may be asked some practical questions about alfalfa; therefore we give you a brief agricultural account of it. If you desire further
- 78. information, write to the College of Agriculture, Ithaca, N. Y., for Bulletin 221, "Alfalfa in New York." Alfalfa is grown mostly for hay. It is not adapted to pasture, because the new growth springs from the crown at the surface of the ground, and if this is destroyed the growth will not be renewed vigorously. New York is a hay-producing State. Grain feeds can be grown more cheaply in the West. It is of great importance to the State, therefore, if a better hay-producing plant can be found. We have seen that New York leads the States in dairy cattle. Other livestock also is abundant. Last year more than half a million horses and mules were fed in the State. Success has not attended efforts to grow alfalfa in all parts of New York. This is due to two principal reasons: (1) farmers have not known the plant and its habits well enough to give it the care and treatment it demands; (2) the soils of many localities, because of their physical condition or composition, are not suitable for the plant. The alfalfa seedling is not a strong plant. It cannot compete with weeds nor overcome adverse conditions of moisture; it cannot adapt itself to conditions resulting from poor preparation of land, and it is not vigorous in its ability to get food from any source. Care must be given to the preparation of the land in order that sufficient moisture may be supplied during the early stages of growth and that there may be an abundance of quickly available plant-food. After growth has started, alfalfa has the power to get some of its nitrogen from the air through the nodules which grow upon its roots; yet during the early stages of growth it is essential that the soil be supplied with all elements of plant-food in available form. While alfalfa requires an abundance of moisture for its best growth and development, yet it will not grow in soils that hold water for any considerable length of time. Such soils are usually those with an impervious subsoil or hard-pan, or those of clay or silt structure which retain free water to the exclusion of air. Therefore, it is important that alfalfa soils be well and uniformly drained, either by natural conditions
- 79. or by underground drains. One other essential of prime importance is that the soil be neutral or alkaline in its reaction; in other words, that it contain no free acid. Limestone or blue-grass soils are ideal in this regard for alfalfa. If acid is present, the difficulty may be corrected either wholly or in part by the application of 500 to 2,000 pounds of lime per acre. As in most other legumes (members of the family Leguminosæ, including peas, beans, clovers), there is a peculiar relationship existing between the plant and excrescences or nodules upon its roots. These nodules are essential to the normal growth and development of the plant. They contain bacteria, and these bacteria have the power of "fixing" or appropriating the free atmospheric nitrogen in the soil. Legumes are "nitrogen-gatherers," whereas most other plants secure their nitrogen only from decomposing organic matter. Failure to have the soil inoculated with the proper bacteria for alfalfa is the cause for many failures with the crop. In most instances when the plants do not make satisfactory growth, or have a yellow, dwarfed appearance, the trouble can be traced to the absence of these bacteria from the soil, and hence to a lack of nodules on the roots. The relationship existing between the plant and the organism is one of mutual benefit. Each kind of leguminous plant seems to have its characteristic bacterium, which grows on no other plant, although this question is not thoroughly settled. Farmers are becoming aware of this requisite in alfalfa culture and usually supply it in two different ways. The older method is to take the surface soil from an old alfalfa field, where the plants have grown well and where nodules are to be found on the roots, and to sow it on the land to be seeded at the rate of one hundred or more pounds per acre. In this way the soil becomes inoculated with the bacteria, and as the young plants spring into growth the bacteria develop on the roots. Another method is to inoculate the seed before sowing with artificial cultures of the bacteria. Both of these methods are usually successful, and if soil conditions are right the chances for failure are few.
- 80. Alfalfa should be cut when it opens into flower. At this time the stems and leaves contain their highest percentage of nutrients, the leaves do not so easily fall off in curing, and the stems are not so woody. Besides these reasons, if cutting be delayed until after flowering, the plant may not spring quickly into subsequent growth. Disease does not spare the alfalfa plant. Both leaves and roots are attacked, the leaf spot being serious. The parasitic dodder is a serious enemy in some parts of New York State. LEAFLET XXXV HOW PLANTS LIVE TOGETHER.[49] By L. H. BAILEY. o the general observer, plants seem to be distributed in a promiscuous and haphazard way, without law or order. This is because he does not see and consider. The world is now full of plants. Every plant puts forth its supreme effort to multiply its kind. The result is an intense struggle for an opportunity to live. Seeds are scattered in profusion, but only the few can grow. The many do not find the proper conditions. They fall on stony ground. In Fig. 250 this loss is shown. The trunk of an elm tree stands in the background. The covering of the ground, except about the very base of the tree, is a mat of elm seedlings. There are thousands of them in the space shown in the picture, so many that they make a sod-like covering which shows little detail in the photograph. Not one of these thousands will ever make a tree.
- 81. Fig. 250. A carpet of young elms, all of which must perish. Fig. 251. A plant society waiting for the spring. Since there is intense competition for every foot of the earth's surface that is capable of raising plants, it follows that every spot will probably have many kinds of plant inhabitants. Plants must live together. They associate; they become adapted or accustomed to each other. Some can live in shade; they thrive in the forest, where sun-loving plants
- 82. Fig. 252. Weak, narrow-leaved grasses grow in the cat-tail forest. perish. Others prefer the sun, and thereby live together. There are plant societies. Every distinct or separate area has its own plant society. There is one association for the hard- tramped dooryard,—knot-weed and broad-leaved plantain with interspersed grass and dandelions; one for the fence- row,—briars and choke-cherries and hiding weeds; one for the dry open field,—wire-grass and mullein and scattered docks; one for the slattern roadside,—sweet clover, ragweed, burdock; one for the meadow swale,—smartweed and pitchforks; one for the barnyard,—rank pigweeds and sprawling barn-grass; one for the dripping rock-cliff,—delicate bluebells and hanging ferns and grasses. Indefinitely might these categories be extended. We all know the plant societies, but we have not considered them. In every plant society there is one dominant note. It is the individuality of one kind of plant which grows most abundantly or overtops the others. Certain plant-forms come to mind when one thinks of willows, others when he thinks of an apple orchard, still others when he thinks of a beech forest. The farmer may associate "pussly" with cabbages and beets, but not with wheat and oats. He associates cockle with wheat, but not with oats or corn. We all associate dandelions with grassy areas, but not with burdock or forests. It is impossible to open one's eyes out-of-doors, outside the paved streets of cities, without seeing a plant society. A lawn is a plant society. It may contain only grass, or it may contain weeds hidden
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