![4 1. Linear Spaces
Given vectors v1, . . . , vn ∈ V and scalars α1, . . . , αn ∈ K , we call
n
i=1
αivi = α1v1 + · · · + αnvn
a linear combination of v1, . . . , vn. It is meaningful to remove “redundant”
vectors from the linear combination. Thus we introduce the concepts of
linear dependence and independence.
Definition 1.1.5 We say v1, . . . , vn ∈ V are linearly dependent if there
are scalars αi ∈ K , 1 ≤ i ≤ n, with at least one αi non-zero such that
n
i=1
αivi = 0. (1.1.1)
We say v1, . . . , vn ∈ V are linearly independent if they are not linearly
dependent, meaning that the only choice of scalars {αi} for which (1.1.1)
is valid is αi = 0 for i = 1, 2, . . . , n.
We observe that v1, . . . , vn are linearly dependent if and only if at least
one of the vectors can be expressed as a linear combination of the rest of
the vectors. In particular, a set of vectors containing the zero element is
always linearly dependent. Similarly, v1, . . . , vn are linearly independent if
and only if none of the vectors can be expressed as a linear combination of
the rest of the vectors; in other words, none of the vectors is “redundant.”
Example 1.1.6 In Rd
, d vectors x(i)
= (x
(i)
1 , . . . , x
(i)
d )T
, 1 ≤ i ≤ d, are
linearly independent if and only if the determinant
x
(1)
1 · · · x
(d)
1
.
.
.
...
.
.
.
x
(1)
d · · · x
(d)
d
is non-zero. This follows from a standard result in linear algebra. The con-
dition (1.1.1) is equivalent to a homogeneous system of linear equations,
and a standard result of linear algebra says that this system has (0, . . . , 0)T
as its only solution if and only if the above determinant is non-zero.
Example 1.1.7 Within the space C[0, 1], the vectors 1, x, x2
, . . . , xn
are
linearly independent. This can be proven in several ways. Assuming
n
j=0
αjxj
= 0, 0 ≤ x ≤ 1,
we can form its first n derivatives. Setting x = 0 in this polynomial and its
derivatives will lead to αj = 0 for j = 0, 1, . . . , n.](https://image.slidesharecdn.com/444877-250518161545-79673a8a/85/Theoretical-Numerical-Analysis-A-Functional-Analysis-Framework-1st-Edition-Kendall-Atkinson-25-320.jpg)
![1.1. Linear spaces 5
Definition 1.1.8 The span of v1, . . . , vn ∈ V is defined to be the set of all
the linear combinations of these vectors:
span{v1, . . . , vn} =
n
i=1
αivi
αi ∈ K , 1 ≤ i ≤ n
.
Evidently, span {v1, . . . , vn} is a linear subspace of V . Most of the
time, we apply this definition for the case where v1, . . . , vn are linearly
independent.
Definition 1.1.9 A linear space V is said to be finite dimensional if there
exists a finite maximal set of independent vectors {v1, . . . , vn}; i.e., the
set {v1, . . . , vn} is linearly independent, but {v1, . . . , vn, vn+1} is linearly
dependent for any vn+1 ∈ V . The set {v1, . . . , vn} is called a basis of the
space. If such a finite basis for V does not exist, then V is said to be infinite
dimensional.
We see that a basis is a set of independent vectors such that any vector
in the space can be written as a linear combination of them. Obviously a
basis is not unique, yet we have the following important result.
Theorem 1.1.10 For a finite-dimensional linear space, every basis for V
contains exactly the same number of vectors. This number is called the
dimension of the space.
A proof of this result can be found in most introductory textbooks on
linear algebra; for example, see [3, Section 5.4].
Example 1.1.11 The space Rd
is d-dimensional. There are infinitely
many possible choices for a basis of the space. A canonical basis for
this space is {ei = (0, . . . , 0, 1i, 0, . . . , 0)T
}d
i=1 in which the single 1 is in
component i.
We introduce the concept of a linear function.
Definition 1.1.12 Let L be a function from the linear space V to the linear
space W. We say L is a linear function if
(a) for all u, v ∈ V ,
L(u + v) = L(u) + L(v);
(b) for all v ∈ V and all α ∈ K ,
L(αv) = αL(v).
For such a linear function, we often write
L(v) = Lv, v ∈ V.
This definition is extended and discussed extensively in Chapter 2.
Other common notations are linear mapping, linear operator, and linear
transformation.](https://image.slidesharecdn.com/444877-250518161545-79673a8a/85/Theoretical-Numerical-Analysis-A-Functional-Analysis-Framework-1st-Edition-Kendall-Atkinson-26-320.jpg)
![6 1. Linear Spaces
Definition 1.1.13 Two linear spaces U and V are said to be isomorphic
if there is a linear bijective (i.e., one-to-one and onto) function l : U → V .
Many properties of a linear space U hold for any other linear space V
that is isomorphic to U; and then the explicit contents of the space do
not matter in the analysis of these properties. This usually proves to be
convenient. One such example is that if U and V are isomorphic and are
finite dimensional, then their dimensions are equal, a basis of V can be
obtained from a basis of U by applying the mapping l, and a basis of U
can be obtained from a basis of V by applying the inverse mapping of l.
Example 1.1.14 The set Pk of all polynomials of degree less than or equal
to k is a subspace of continuous function space C[0, 1]. An element in the
space Pk has the form a0 + a1x + · · · + akxk
. The mapping l : a0 + a1x +
· · · + akxk
→ (a0, a1, . . . , ak)T
is bijective from Pk to Rk+1
. Thus, Pk is
isomorphic to Rk+1
.
Definition 1.1.15 Let U and V be two linear spaces. The Cartesian
product of the spaces, W = U × V , is defined by
W = {w = (u, v) | u ∈ U, v ∈ V }
endowed with componentwise addition and scalar multiplication
(u1, v1) + (u2, v2) = (u1 + u2, v1 + v2) ∀ (u1, v1), (u2, v2) ∈ W,
α (u, v) = (α u, α v) ∀ (u, v) ∈ W, ∀ α ∈ K .
It is easy to verify that W is a linear space. The definition can be extended
in a straightforward way for the Cartesian product of any finite number of
linear spaces.
Example 1.1.16 The real plane can be viewed as the Cartesian product
of two real lines: R2
= R × R. In general,
Rd
= R × · · · × R
d times
.
Exercise 1.1.1 Show that the set of all continuous solutions of the differ-
ential equation u
(x) + u(x) = 0 is a finite-dimensional linear space. Is the
set of all continuous solutions of u
(x) + u(x) = 1 a linear space?
Exercise 1.1.2 When is the set {u ∈ C[0, 1] | u(0) = a} a linear space?
Exercise 1.1.3 Show that in any linear space V , a set of vectors is always
linearly dependent if one of the vectors is zero.
Exercise 1.1.4 Assume U and V are finite-dimensional linear spaces, and
let {u1, . . . , un} and {v1, . . . , vm} be bases for them, respectively. Using
these bases, create a basis for W = U × V .](https://image.slidesharecdn.com/444877-250518161545-79673a8a/85/Theoretical-Numerical-Analysis-A-Functional-Analysis-Framework-1st-Edition-Kendall-Atkinson-27-320.jpg)
![8 1. Linear Spaces
x1
x2
S
2
x1
x2
S
1
x1
x2
S
∞
Figure 1.1. The unit ball Sp = x ∈ R2
: xp ≤ 1
for p = 1, 2, ∞
(b) More generally, for 1 ≤ p ≤ ∞, the formulas
xp =
d
i=1
|xi|p
1/p
for 1 ≤ p ∞, (1.2.2)
x∞ = max
1≤i≤d
|xi| (1.2.3)
define norms in the space Rd
(cf. Exercise 1.2.5). The norm · p is called
the p-norm, and · ∞ is called the maximum or infinity norm. It is
straightforward to show that
x∞ = lim
p→∞
xp
by using the inequality (1.2.5) given below. Again, when d = 1, all these
norms coincide with the absolute value: xp = |x|, x ∈ R. Over Rd
, the
most commonly used norms are · p, p = 1, 2, ∞. The unit ball in R2
for
each of these norms is shown in Figure 1.1.
Example 1.2.4 (a) The standard norm for C[a, b] is the maximum norm
f∞ = max
a≤x≤b
|f(x)| , f ∈ C[a, b].
This is also the norm for Cp(2π) (with a = 0 and b = 2π), the space of
continuous 2π-periodic functions introduced in Example 1.1.2(f).](https://image.slidesharecdn.com/444877-250518161545-79673a8a/85/Theoretical-Numerical-Analysis-A-Functional-Analysis-Framework-1st-Edition-Kendall-Atkinson-29-320.jpg)
![1.2. Normed spaces 9
(b) For an integer k 0, the standard norm for Ck
[a, b] is
fk,∞ = max
0≤j≤k
f(j)
∞, f ∈ Ck
[a, b].
This is also the standard norm for Ck
p (2π).
With the use of a norm for V we can introduce a topology for V , a set
of open and closed sets for V .
Definition 1.2.5 Let (V, ·) be a normed space. Given v0 ∈ V and r 0,
the sets
B(v0, r) = {v ∈ V | v − v0 r},
B(v0, r) = {v ∈ V | v − v0 ≤ r}
are called the open and closed balls centered at v0 with radius r. When r = 1
and v0 = 0, we have unit balls.
Definition 1.2.6 Let A ⊆ V, a normed linear space. The set A is open if
for every v ∈ A, there is a radius r 0 such that B(v, r) ⊆ A. The set A
is closed in V if its complement V − A is open in V .
1.2.1 Convergence
With the notion of a norm at our disposal, we can define the important
concept of convergence.
Definition 1.2.7 Let V be a normed space with the norm ·. A sequence
{un} ⊆ V is convergent to u ∈ V if
lim
n→∞
un − u = 0.
We say that u is the limit of the sequence {un}, and write un → u as
n → ∞, or limn→∞ un = u.
It can be verified that any sequence can have at most one limit.
Definition 1.2.8 A function f : V → R is said to be continuous at u ∈ V
if for any sequence {un} with un → u, we have f(un) → f(u) as n → ∞.
The function f is said to be continuous on V if it is continuous at every
u ∈ V .
Proposition 1.2.9 The norm function · is continuous.
Proof. We need to show that if un → u, then un → u. This follows
from the backward triangle inequality
| u − v | ≤ u − v ∀ u, v ∈ V (1.2.4)
derived from the triangle inequality.
We have seen that on a linear space various norms can be defined.
Different norms give different measures of size for a given vector in the](https://image.slidesharecdn.com/444877-250518161545-79673a8a/85/Theoretical-Numerical-Analysis-A-Functional-Analysis-Framework-1st-Edition-Kendall-Atkinson-30-320.jpg)
![10 1. Linear Spaces
space. Consequently, different norms may give rise to different forms of
convergence.
Definition 1.2.10 We say two norms · (1) and · (2) are equivalent if
there exist positive constants c1, c2 such that
c1u(1) ≤ u(2) ≤ c2u(1) ∀ u ∈ V.
With two such equivalent norms, a sequence {un} converges in one norm
if and only if it converges in the other norm:
lim
n→∞
un − u(1) = 0 ⇐⇒ lim
n→∞
un − u(2) = 0.
Example 1.2.11 For the norms (1.2.2)–(1.2.3) on Rd
, it is straightforward
to show
x∞ ≤ xp ≤ d1/p
x∞ ∀ x ∈ Rd
. (1.2.5)
Thus all the norms xp, 1 ≤ p ≤ ∞, on Rd
are equivalent.
More generally, we have the following well-known result. For a proof, see
[11, p. 483].
Theorem 1.2.12 Over a finite-dimensional space, any two norms are
equivalent.
Thus, on a finite-dimensional space, different norms lead to the same
convergence notion. Over an infinite-dimensional space, however, such a
statement is no longer valid.
Example 1.2.13 Let V be the space of all bounded functions on [0, 1]. For
u ∈ V , in analogy with Example 1.2.3, we may define the following norms:
up =
1
0
|u(x)|p
dx
1/p
, 1 ≤ p ∞, (1.2.6)
u∞ = sup
0≤x≤1
|u(x)|. (1.2.7)
Now consider a sequence of functions {un} ⊆ V , defined by
un(x) =
1 − nx, 0 ≤ x ≤
1
n
,
0,
1
n
x ≤ 1.
It is easy to show that
unp = [n(p + 1)]−1/p
, 1 ≤ p ∞.
Thus we see that the sequence {un} converges to u = 0 in the norm · p,
1 ≤ p ∞. On the other hand,
un∞ = 1, n ≥ 1,
so {un} does not converge to u = 0 in the norm · ∞.](https://image.slidesharecdn.com/444877-250518161545-79673a8a/85/Theoretical-Numerical-Analysis-A-Functional-Analysis-Framework-1st-Edition-Kendall-Atkinson-31-320.jpg)
![1.2. Normed spaces 11
As we have seen in the last example, in an infinite-dimensional space,
some norms are not equivalent. Convergence defined by one norm can be
stronger than that by another.
Example 1.2.14 Consider again the space of all bounded functions on
[0, 1], and the family of norms · p, 1 ≤ p ∞, and · ∞. We have, for
any p ∈ [1, ∞),
up ≤ u∞, u ∈ V.
Thus, convergence in · ∞ implies convergence in · p, 1 ≤ p ∞, but
not conversely. Convergence in ·∞ is usually called uniform convergence.
1.2.2 Banach spaces
The concept of a normed space is usually too general, and special attention
is given to a particular type of normed space called a Banach space.
Definition 1.2.15 Let V be a normed space. A sequence {un} ⊆ V is
called a Cauchy sequence if
lim
m,n→∞
um − un = 0.
Obviously, a convergent sequence is a Cauchy sequence. In the finite-
dimensional space Rd
, any Cauchy sequence is convergent. However, in a
general infinite-dimensional space, a Cauchy sequence may fail to converge;
see Example 1.2.18 below.
Definition 1.2.16 A normed space is said to be complete if every Cauchy
sequence from the space converges to an element in the space. A complete
normed space is called a Banach space.
Example 1.2.17 Let Ω ⊆ Rd
be a bounded open set. For v ∈ C(Ω) and
1 ≤ p ∞, define the p-norm
vp =
Ω
|v(x)|p
dx
1/p
. (1.2.8)
Here, x = (x1, . . . , xd)T
and dx = dx1dx2 · · · dxd. In addition, define the
∞-norm or maximum norm
v∞ = max
x∈Ω
|v(x)|.
The space C(Ω) with · ∞ is a Banach space; i.e., the uniform limit of
continuous functions is itself continuous.
Example 1.2.18 The space C(Ω) with the norm · p, 1 ≤ p ∞, is
not a Banach space. To illustrate this, we consider the space C[0, 1] and a](https://image.slidesharecdn.com/444877-250518161545-79673a8a/85/Theoretical-Numerical-Analysis-A-Functional-Analysis-Framework-1st-Edition-Kendall-Atkinson-32-320.jpg)
![12 1. Linear Spaces
sequence in C[0, 1] defined as follows:
un(x) =
0, 0 ≤ x ≤ 1
2 − 1
2n ,
n x − 1
2 (n − 1), 1
2 − 1
2n ≤ x ≤ 1
2 + 1
2n ,
1, 1
2 + 1
2n ≤ x ≤ 1.
Let
u(x) =
0, 0 ≤ x 1
2 ,
1, 1
2 x ≤ 1.
Then un − up → 0 as n → ∞; i.e., the sequence {un} converges to u in
the norm · p. But obviously no matter how we define u(1/2), the limit
function u is not continuous.
1.2.3 Completion of normed spaces
It is important to be able to deal with function spaces using a norm of our
choice, as such a norm is often important or convenient in the formulation of
a problem or in the analysis of a numerical method. The following theorem
allows us to do this. A proof is discussed in [88, p. 84].
Theorem 1.2.19 Let V be a normed space. Then there is a complete
normed space W with the following properties:
(a) There is a subspace
V ⊆ W and a bijective (one-to-one and onto) linear
function I : V →
V with
IvW = vV ∀ v ∈ V.
The function I is called an isometric isomorphism of the spaces V and
V .
(b) The subspace
V is dense in W; i.e., for any w ∈ W, there is a sequence
{
vn} ⊆
V such that
w −
vnW → 0 as n → ∞.
The space W is called the completion of V , and W is unique up to an
isometric isomorphism.
The spaces V and
V are generally identified, meaning no distinction is
made between them. However, we also consider cases where it is important
to note the distinction. An important example of the theorem is to let V
be the rational numbers and W be the real numbers R . One way in which
R can be defined is as a set of equivalence classes of Cauchy sequences of
rational numbers, and
V can be identified with those equivalence classes of
Cauchy sequences whose limit is a rational number. A proof of the above
theorem can be made by mimicking this commonly used construction of
the real numbers from the rational numbers.](https://image.slidesharecdn.com/444877-250518161545-79673a8a/85/Theoretical-Numerical-Analysis-A-Functional-Analysis-Framework-1st-Edition-Kendall-Atkinson-33-320.jpg)
![1.2. Normed spaces 13
Example 1.2.20 Theorem 1.2.19 guarantees the existence of a unique ab-
stract completion of an arbitrary normed vector space. However, it is often
possible, and indeed desirable, to give a more concrete definition of the
completion of a given normed space; much of the subject of real analysis
is concerned with this topic. In particular, the subject of Lebesgue mea-
sure and integration deals with the completion of C(Ω) under the norms of
(1.2.8), · p for 1 ≤ p ∞. A complete development of Lebesgue mea-
sure and integration is given in any standard textbook on real analysis; for
example, see Royden [141] or Rudin [142]. We do not introduce formally
and rigorously the concepts of measurable set and measurable function.
Rather we think of measure theory intuitively as described in the following
paragraphs. Our rationale for this is that the details of Lebesgue measure
and integration can often be bypassed in most of the material we present in
this text.
Measurable subsets of R include the standard open and closed intervals
with which we are familiar. Multivariable extensions of intervals to Rd
are
also measurable, together with countable unions and intersections of them.
Intuitively, the measure of a set D ⊆ Rd
is its “length,” “area,” “volume,”
or suitable generalization; and we denote the measure of D by meas(D).
For a formal discussion of measurable set, see Royden [141] or Rudin [142].
To introduce the concept of measurable function, we begin by defining a
step function. A function v on a measurable set D is a step function if
D can be decomposed into a finite number of pairwise disjoint measurable
subsets D1, . . . , Dk with v(x) constant over each Dj. We say a function v
on D is a measurable function if it is the pointwise limit of a sequence of
step functions. This includes, for example, all continuous functions on D.
For each such measurable set Dj, we define a characteristic function
χj(x) =
1, x ∈ Dj,
0, x /
∈ Dj.
A general step function over the decomposition D1, . . ., Dk of D can then
be written
v(x) =
k
j=1
αjχj(x), x ∈ D (1.2.9)
with α1, . . . , αk scalars. For a general measurable function v over D, we
write it as a limit of step functions vk over D:
v(x) = lim
k→∞
vk(x), x ∈ D. (1.2.10)
We say two measurable functions are equal almost everywhere if the set
of points on which they differ is a set of measure zero. For notation, we
write
v = w (a.e.)](https://image.slidesharecdn.com/444877-250518161545-79673a8a/85/Theoretical-Numerical-Analysis-A-Functional-Analysis-Framework-1st-Edition-Kendall-Atkinson-34-320.jpg)
![14 1. Linear Spaces
to indicate that v and w are equal almost everywhere. Given a measur-
able function v on D, we introduce the concept of an equivalence class of
equivalent functions:
[v] = {w | w measurable on D and v = w (a.e.)} .
For most purposes, we generally consider elements of an equivalence class
[v] as being a single function v.
We define the Lebesgue integral of a step function v over D, given in
(1.2.9), by
D
v(x) dx =
k
j=1
αj meas(Dj).
For a general measurable function, given in (1.2.10), define the Lebesgue
integral of v over D by
D
v(x) dx = lim
k→∞
D
vk(x) dx.
There are a great many properties of Lebesgue integration, and we refer
the reader to any text on real analysis for further details. Note that the
Lebesgue integrals of elements of an equivalence class [v] are identical.
Let Ω be an open measurable set in Rd
. Introduce
Lp
(Ω) = [v] | v measurable on Ω and vp ∞
.
The norm vp is defined as in (1.2.8), although now we use Lebesgue
integration rather than Riemann integration.
L∞
(Ω) = {[v] | v measurable on Ω and v∞ ∞} .
For v measurable on Ω, define
v∞ = ess sup
x∈Ω
|v(x)|
≡ inf
meas(Ω)=0
sup
x∈ΩΩ
|v(x)| ,
where “ meas(Ω
) = 0” means Ω
is a measurable set with measure zero.
The spaces Lp
(Ω), 1 ≤ p ∞, are Banach spaces, and they are concrete
realizations of the abstract completion of C(Ω) under the norm of (1.2.8).
The space L∞
(Ω) is also a Banach space, but it is much larger than the
space C(Ω) with the ∞-norm · ∞. Additional discussion of the spaces
Lp
(Ω) is given in Section 1.5.
Example 1.2.21 More generally, let w be a positive measurable function
on Ω, called a weight function. We can define weighted spaces Lp
w(Ω) as](https://image.slidesharecdn.com/444877-250518161545-79673a8a/85/Theoretical-Numerical-Analysis-A-Functional-Analysis-Framework-1st-Edition-Kendall-Atkinson-35-320.jpg)
![1.2. Normed spaces 15
follows:
Lp
w(Ω) =
v measurable
Ω
w(x) |v(x)|p
dx ∞
, p ∈ [1, ∞),
L∞
w (Ω) = {v measurable | ess supΩ w(x) |v(x)| ∞} .
These are Banach spaces with the norms
vp,w =
Ω
w(x) |v(x)|p
dx
1/p
, p ∈ [1, ∞),
vp,∞ = ess sup
x∈Ω
w(x)|v(x)|.
The space C(Ω) of Example 1.1.2(c) with the norm
vC(Ω) = max
x∈Ω
|v(x)|
is also a Banach space, and it can be considered as a proper subset of
L∞
(Ω). See Example 2.5.3 for a situation where it is necessary to distin-
guish between C(Ω) and the subspace of L∞
(Ω) to which it is isometric and
isomorphic.
Example 1.2.22 (a) For any integer m ≥ 0, the normed spaces Cm
[a, b]
and Ck
p (2π) of Example 1.2.4(b) are Banach spaces.
(b) Let 1 ≤ p ∞. As an alternative norm on Cm
[a, b], introduce
f =
m
j=0
f(j)
p
p
1
p
.
The space Cm
[a, b] is not complete with this norm. Its completion is denoted
by Wm,p
(a, b), and it is an example of a Sobolev space. It can be shown
that if f ∈ Wm,p
(a, b), then f, f
, . . . , f(m−1)
are continuous, and f(m)
exists almost everywhere and belongs to Lp
(a, b). This and its multivariable
generalizations are discussed at length in Chapter 6.
A knowledge of the theory of Lebesgue measure and integration is very
helpful in dealing with problems defined on spaces of such functions.
Nonetheless, many results can be proven by referring to only the origi-
nal space and its associated norm, say, C(Ω) with · p, from which a
Banach space is obtained by a completion argument, say Lp
(Ω). We return
to this in Theorem 2.4.1 of Chapter 2.
Exercise 1.2.1 Prove the backward triangle inequality of (1.2.4).
Exercise 1.2.2 Show that · ∞ is a norm on C(Ω), with Ω a bounded
open set in Rd
.
Exercise 1.2.3 Show that · ∞ is a norm on L∞
(Ω), with Ω a bounded
open set in Rd
.](https://image.slidesharecdn.com/444877-250518161545-79673a8a/85/Theoretical-Numerical-Analysis-A-Functional-Analysis-Framework-1st-Edition-Kendall-Atkinson-36-320.jpg)
![16 1. Linear Spaces
Exercise 1.2.4 Show that · 1 is a norm on L1
(Ω), with Ω a bounded
open set in Rd
.
Exercise 1.2.5 Show that for 1 ≤ p ≤ ∞, xp defined by (1.2.2)–(1.2.3)
is a norm in the space Rd
. The main task is to verify the triangle inequality,
which can be done by first proving the Hölder inequality, |x·y| ≤ xpyp ,
x, y ∈ Rd
. Here p
is the conjugate of p defined through the relation 1/p
+
1/p = 1; by convention, p
= 1 if p = ∞, p
= ∞ if p = 1.
Exercise 1.2.6 Define Cα
[a, b], 0 α ≤ 1, as the set of all f ∈ C[a, b]
for which
Mα(f) ≡ sup
a≤x,y≤b
x=y
|f(x) − f(y)|
|x − y|
α ∞.
Define fα = f∞ + Mα(f). Show Cα
[a, b] with this norm is complete.
Exercise 1.2.7 Define Cb[0, ∞) as the set of all functions f that are
continuous on [0, ∞) and satisfy
f∞ ≡ sup
x≥0
|f(x)| ∞.
Show Cb[0, ∞) with this norm is complete.
Exercise 1.2.8 Does the formula (1.2.2) define a norm on Rd
for 0 p
1?
Exercise 1.2.9 Prove the equivalence on C1
[0, 1] of the following norms:
fa ≡ |f(0)| +
1
0
|f
(x)| dx,
fb ≡
1
0
|f(x)| dx +
1
0
|f
(x)| dx.
Hint: You may need to use the integral mean value theorem: Given g ∈
C[0, 1], there is ξ ∈ [0, 1] such that
1
0
g(x) dx = g(ξ).
Exercise 1.2.10 Let V1 and V2 be normed spaces with norms · 1 and
· 2. Recall that the product space V1 × V2 is defined by
V1 × V2 = {(v1, v2) | v1 ∈ V1, v2 ∈ V2}.
Show that the quantities max{v11, v22} and (v1p
1 + v2p
2)1/p
, 1 ≤
p ∞ all define norms on the space V1 × V2.
Exercise 1.2.11 Over the space C1
[0, 1], determine which of the following
is a norm, and which is only a semi-norm:
(a) max
0≤x≤1
|u(x)|;](https://image.slidesharecdn.com/444877-250518161545-79673a8a/85/Theoretical-Numerical-Analysis-A-Functional-Analysis-Framework-1st-Edition-Kendall-Atkinson-37-320.jpg)
![1.2. Normed spaces 17
(b) max
0≤x≤1
[|u(x)| + |u
(x)|];
(c) max
0≤x≤1
|u
(x)|;
(d) |u(0)| + max
0≤x≤1
|u
(x)|;
(e) max
0≤x≤1
|u
(x)| +
0.2
0.1
|u(x)| dx.
Exercise 1.2.12 Over a normed space (V, · ), we define a function of
two variables d(u, v) = u−v. Then d(·, ·) is a distance function; in other
words, d(·, ·) has the following properties of an ordinary distance between
two points:
(a) d(u, v) ≥ 0 for any u, v ∈ V , and d(u, v) = 0 if and only if u = v;
(b) d(u, v) = d(v, u) for any u, v ∈ V ;
(c) (the triangle inequality) d(u, w) ≤ d(u, v) + d(v, w) for any u, v, w ∈
V .
A linear space endowed with a distance function is called a metric space.
Certainly a normed space can be viewed as a metric space. There are ex-
amples of metrics (distance functions) that are not generated by any norm,
though.
Exercise 1.2.13 Let V be a normed space and {un} a Cauchy sequence.
Suppose there is a subsequence {un } ⊆ {un} and some element v ∈ V such
that un → u as n
→ ∞. Show that un → u as n → ∞.
Exercise 1.2.14 Let V be a normed space, V0 ⊆ V a subspace. The
quotient space V/V0 is defined to be the space of all the classes [v] =
{v + v0 | v0 ∈ V0}. Prove that the formula
[v]V/V0
= inf
v0∈V0
v + v0V
defines a norm on V/V0. Show that if V is a Banach space and V0 ⊆ V is
a closed subspace, then V/V0 is a Banach space.
Exercise 1.2.15 Assuming a knowledge of Lebesgue integration, show that
W1,2
(a, b) ⊆ C[a, b].
Generalize this result to the space Wm,p
(a, b) with other values of m and
p.
Hint: For v ∈ W1,2
(a, b), use
v(x) − v(y) =
y
x
v
(z) dz .](https://image.slidesharecdn.com/444877-250518161545-79673a8a/85/Theoretical-Numerical-Analysis-A-Functional-Analysis-Framework-1st-Edition-Kendall-Atkinson-38-320.jpg)
![18 1. Linear Spaces
Exercise 1.2.16 On C1
[0, 1], define
(u, v)∗ = u(0) v(0) +
1
0
u
(x) v
(x) dx
and
u∗ =
(u, u)∗ .
Show that
u∞ ≤ c u∗ ∀ u ∈ C1
[0, 1]
for a suitably chosen constant c.
1.3 Inner product spaces
In studying linear problems, inner product spaces are usually used. These
are the spaces where a norm can be defined through the inner product and
the notion of orthogonality of two elements can be introduced. The inner
product in a general space is a generalization of the usual scalar product
in the plane R2
or the space R3
.
Definition 1.3.1 Let V be a linear space over K = R or C . An inner
product (·, ·) is a function from V × V to K with the following properties.
1. For any u ∈ V , (u, u) ≥ 0 and (u, u) = 0 if and only if u = 0.
2. For any u, v ∈ V , (u, v) = (v, u).
3. For any u, v, w ∈ V , any α, β ∈ K, (α u+β v, w) = α (u, w)+β (v, w).
The space V together with the inner product (·, ·) is called an inner product
space. When the definition of the inner product (·, ·) is clear from the con-
text, we simply say V is an inner product space. When K = R, V is called
a real inner product space, while if K = C, V is a complex inner product
space.
In the case of a real inner product space, the second axiom reduces to
the symmetry of the inner product:
(u, v) = (v, u) ∀ u, v ∈ V.
For an inner product, there is an important property called the Schwarz
inequality.
Theorem 1.3.2 (Schwarz inequality) If V is an inner product space,
then
|(u, v)| ≤
(u, u) (v, v) ∀ u, v ∈ V,
and the equality holds if and only if u and v are linearly dependent.](https://image.slidesharecdn.com/444877-250518161545-79673a8a/85/Theoretical-Numerical-Analysis-A-Functional-Analysis-Framework-1st-Edition-Kendall-Atkinson-39-320.jpg)
![1.3. Inner product spaces 19
Proof. We give the proof only for the real case. The result is obviously
true if either u = 0 or v = 0. Now suppose u = 0, v = 0. Define
φ(t) = (u + t v, u + t v) = (u, u) + 2 (u, v) t + (v, v) t2
, t ∈ R .
The function φ is quadratic and non-negative, so its discriminant must be
non-positive,
[2 (u, v)]2
− 4 (u, u) (v, v) ≤ 0;
i.e., the Schwarz inequality is valid. For v = 0, the equality holds if and
only if u = −t v for some t ∈ R .
An inner product (·, ·) induces a norm through the formula
u =
(u, u), u ∈ V.
In verifying the triangle inequality for the quantity thus defined, we need
to use the above Schwarz inequality.
Proposition 1.3.3 An inner product is continuous with respect to its in-
duced norm. In other words, if · is the norm defined by u =
(u, u),
then un − u → 0 and vn − v → 0 as n → ∞ imply
(un, vn) → (u, v) as n → ∞.
In particular, if un → u, then for any v,
(un, v) → (u, v) as n → ∞.
Proof. Since {un} and {vn} are convergent, they are bounded; i.e., for
some M ∞, un ≤ M, vn ≤ M for any n. We write
(un, vn) − (u, v) = (un − u, vn) + (u, vn − v).
Using the Schwarz inequality, we have
|(un, vn) − (u, v)| ≤ un − u vn + u vn − v
≤ M un − u + u vn − v.
Hence the result holds.
Commonly seen inner product spaces are usually associated with their
canonical inner products. As an example, the canonical inner product for
the space Rd
is
(x, y) =
d
i=1
xiyi = yT
x, ∀ x = (x1, . . . , xd)T
, y = (y1, . . . , yd)T
∈ Rd
.
This inner product induces the Euclidean norm
x =
!
!
d
i=1
|xi|2 =
(x, x).](https://image.slidesharecdn.com/444877-250518161545-79673a8a/85/Theoretical-Numerical-Analysis-A-Functional-Analysis-Framework-1st-Edition-Kendall-Atkinson-40-320.jpg)
![20 1. Linear Spaces
When we talk about the space Rd
, implicitly we understand the inner
product and the norm are the ones defined above, unless stated otherwise.
For the complex space Cd
, the inner product and the corresponding norm
are
(x, y) =
d
i=1
xiyi = y∗
x, ∀ x = (x1, . . . , xd)T
, y = (y1, . . . , yd)T
∈ Cd
and
x =
!
!
d
i=1
|xi|2 =
(x, x).
The space L2
(Ω) is an inner product space with the canonical inner
product
(u, v) =
Ω
u(x) v(x) dx.
This inner product induces the standard L2
(Ω)-norm
u2 =
#
Ω
|u(x)|2dx =
(u, u).
We have seen that an inner product induces a norm, which is always the
norm we use on the inner product space unless stated otherwise. It is easy
to show that on a complex inner product space,
(u, v) =
1
4
[u + v2
− u − v2
+ iu + iv2
− iu − iv2
],
and on a real inner product space,
(u, v) =
1
4
[u + v2
− u − v2
]. (1.3.1)
These relations are called the polarization identities. Thus in any normed
linear space, there can exist at most one inner product that generates the
norm.
On the other hand, not every norm can be defined through an inner
product. We have the following characterization for any norm induced by
an inner product.
Theorem 1.3.4 A norm · on a linear space V is induced by an inner
product if and only if it satisfies the parallelogram law:
u + v2
+ u − v2
= 2u2
+ 2v2
∀ u, v ∈ V. (1.3.2)
Note that if u and v form two adjacent sides of a parallelogram, then
u + v and u − v represent the lengths of the diagonals of the parallelo-
gram. This theorem can be considered to be a generalization of the theorem
of Pythagoras for right triangles.](https://image.slidesharecdn.com/444877-250518161545-79673a8a/85/Theoretical-Numerical-Analysis-A-Functional-Analysis-Framework-1st-Edition-Kendall-Atkinson-41-320.jpg)
![1.3. Inner product spaces 21
Proof. We prove the result for the case of a real space only. Assume
· =
(·, ·) for some inner product (·, ·). Then for any u, v ∈ V ,
u + v2
+ u − v2
= (u + v, u + v) + (u − v, u − v)
=
$
u2
+ 2(u, v) + v2
%
+
$
u2
− 2(u, v) + v2
%
= 2u2
+ 2v2
.
Conversely, assume the norm · satisfies the parallelogram law. For
u, v ∈ V , let us define
(u, v) =
1
4
$
u + v2
− u − v2
%
and show that it is an inner product. First,
(u, u) =
1
4
2u2
= u2
≥ 0
and (u, u) = 0 if and only if u = 0. Second,
(u, v) =
1
4
$
v + u2
− v − u2
%
= (v, u).
Finally, we show the linearity, which is equivalent to the following two
relations:
(u + v, w) = (u, w) + (v, w) ∀ u, v, w ∈ V
and
(α u, v) = α (u, v) ∀ u ∈ V, α ∈ R .
We have
(u, w) + (v, w)
= 1
4
$
u + w2
− u − w2
+ v + w2
− v − w2
%
= 1
4
$
(u + w2
+ v + w2
) − (u − w2
+ v − w2
)
%
= 1
4
$1
2 (u + v + 2 w2
+ u − v2
) − 1
2 (u + v − 2 w2
+ u − v2
)
%
= 1
8
$
u + v + 2 w2
− u + v − 2 w2
%
= 1
8 [ 2 (u + v + w2
+ w2
) − u + v2
− 2 (u + v − w2
+ w2
) + u + v2
]
= 1
4
$
u + v + w2
− u + v − w2
%
= (u + v, w).
The proof of the second relation is more involved. For fixed u, v ∈ V , let
us define a function of a real variable
f(α) = α u + v2
− α u − v2
.](https://image.slidesharecdn.com/444877-250518161545-79673a8a/85/Theoretical-Numerical-Analysis-A-Functional-Analysis-Framework-1st-Edition-Kendall-Atkinson-42-320.jpg)
![1.3. Inner product spaces 23
defines an inner product on l2
. Furthermore, l2
becomes a Hilbert space
under this inner product.
(c) The space L2
(0, 1) is a Hilbert space with the inner product
(u, v) =
1
0
u(x) v(x) dx.
(d) The space L2
(Ω) is a Hilbert space with the inner product
(u, v) =
Ω
u(x) v(x) dx.
More generally, if w(x) is a positive function on Ω, then the space
L2
w(Ω) =
v measurable
Ω
|v(x)|2
w(x) dx ∞
is a Hilbert space with the inner product
(u, v)w =
Ω
u(x) v(x) w(x) dx.
This space is a weighted L2
space.
Example 1.3.7 Recall the Sobolev space Wm,p
(a, b) defined in Example
1.2.22. If we choose p = 2, then we obtain a Hilbert space. It is usually
denoted by Hm
(a, b) ≡ Wm,2
(a, b). The associated inner product is defined
by
(f, g)Hm =
m
j=0
(
f(j)
, g(j)
)
, f, g ∈ Hm
(a, b)
using the standard inner product (·, ·) of L2
(a, b). Recall from Exercise
1.2.15 that H1
(a, b) ⊆ C[a, b].
1.3.2 Orthogonality
With the notion of an inner product at our disposal, we can define the
angle between two vectors u and v as follows:
θ = arccos
*
(u, v)
u v
+
.
This definition makes sense because, by the Schwarz inequality (Theorem
1.3.2), the argument of arccos is between −1 and 1. The case of a right
angle is particularly important. We see that two vectors u and v form a
right angle if and only if (u, v) = 0.
Definition 1.3.8 Two vectors u and v are said to be orthogonal if (u, v) =
0. An element v ∈ V is said to be orthogonal to a subset U ⊆ V , if (u, v) = 0
for any u ∈ U.](https://image.slidesharecdn.com/444877-250518161545-79673a8a/85/Theoretical-Numerical-Analysis-A-Functional-Analysis-Framework-1st-Edition-Kendall-Atkinson-44-320.jpg)
![24 1. Linear Spaces
Definition 1.3.9 Let U be a subset of an inner product space V . We define
its orthogonal complement to be the set
U⊥
= {v ∈ V | (v, u) = 0 ∀ u ∈ U}.
The orthogonal complement of any set is a closed subspace (cf. Exercise
1.3.7).
Definition 1.3.10 Let V be an inner product space.
(a) Suppose V is finite dimensional. A basis {v1, . . . , vn} of V is said to be
an orthogonal basis if
(vi, vj) = 0, 1 ≤ i = j ≤ n.
If, additionally, vi = 1, 1 ≤ i ≤ n, then we say the basis is orthonormal,
and we combine these conditions as
(vi, vj) = δij ≡
1, i = j,
0, i = j.
(b) Suppose V is infinite dimensional normed space. We say V has a count-
ably infinite basis if there is a sequence {vi}i≥1 ⊆ V for which the following
is valid: For each v ∈ V , we can find scalars {αn,i}n
i=1, n = 1, 2, . . . , such
that
,
,
,
,
,
v −
n
i=1
αn,ivi
,
,
,
,
,
→ 0 as n → ∞.
The space V is also said to be separable. The sequence {vi}i≥1 is called a
basis if any finite subset of the sequence is linearly independent. If V is an
inner product space, and if the sequence {vi}i≥1 also satisfies
(vi, vj) = δij, i, j ≥ 1, (1.3.3)
then {vi}i≥1 is called an orthonormal basis for V .
(c) We say that an infinite dimensional normed space V has a Schauder
basis {vn}n≥1 if for each v ∈ V , it is possible to write
v =
∞
n=1
αnvn (1.3.4)
as a convergent series in V for a unique choice of scalars {αn}n≥1.
For a discussion of the distinction between V having a Schauder basis
and V being separable, see [103, p.68]. For V an inner product space, it
is straightforward to show that an orthonormal basis {vn}n≥1 is also a
Schauder basis, and therefore (1.3.4) is valid for an orthonormal basis. The
advantage of using an orthogonal or an orthonormal basis is that it is easy to
decompose a vector as a linear combination of the basis elements. Assuming
{vn}n≥1 is an orthonormal basis of V , let us determine the coefficients](https://image.slidesharecdn.com/444877-250518161545-79673a8a/85/Theoretical-Numerical-Analysis-A-Functional-Analysis-Framework-1st-Edition-Kendall-Atkinson-45-320.jpg)
![28 1. Linear Spaces
1
−1
x
y
v
1
(x)
v2
(x)
v3
(x)
v4
(x)
Figure 1.2. Graphs on [−1, 1] of the orthonormal Legendre polynomials of degrees
0,1,2,3
As we see from Example 1.3.13, it is cumbersome to construct orthonor-
mal (or orthogonal) polynomials directly. Fortunately, for many important
cases of the weighted function w(x) and integration interval (a, b), formulas
of orthogonal polynomials in the weighted space L2
w(a, b) are known (see
Section 3.4).
Exercise 1.3.1 Given an inner product, show that the formula u =
(u, u) defines a norm.
Exercise 1.3.2 Assume f : R → R is a continuous function, satisfying
f(α) = f(β) + f(α − β) for any α, β ∈ R , and f(0) = 0. Then f(α) =
α f(1).
Solution: From f(α) = f(β) + f(α − β) and f(0) = 0, by an induction
argument, we have f(n α) = n f(α) for any integer n. Then from f(α) =
2 f(α/2), we have f(1/2n
) = (1/2n
) f(1) for any integer n ≥ 0. Finally,
for any integer m, any non-negative integer n, f(m 2−n
) = m f(2−n
) =
(m 2−n
) f(1). Now any rational can be represented as a finite sum q =
i mi 2−i
. Hence, f(q) =
i f(mi 2−i
) =
i mi 2−i
f(1) = q f(1). Since
the set of the rational numbers is dense in R and f is a continuous function,
we see that for any real ξ, f(ξ) = ξ f(1).](https://image.slidesharecdn.com/444877-250518161545-79673a8a/85/Theoretical-Numerical-Analysis-A-Functional-Analysis-Framework-1st-Edition-Kendall-Atkinson-49-320.jpg)
![1.3. Inner product spaces 29
Exercise 1.3.3 The norms · p, 1 ≤ p ≤ ∞, over the space Rd
are
defined in Example 1.2.3. Find all the values of p for which the norm · p
is induced by an inner product.
Hint: Apply Theorem 1.3.4.
Exercise 1.3.4 Let w1, . . . , wd be positive constants. Show that the for-
mula
(x, y) =
d
i=1
wixiyi
defines an inner product on Rd
. This is an example of a weighted inner
product. What happens if we only assume wi ≥ 0, 1 ≤ i ≤ d?
Exercise 1.3.5 Let A ∈ Rd×d
be a symmetric, positive definite matrix
and let (·, ·) be the Euclidean inner product on Rd
. Show that the quantity
(Ax, y) defines an inner product on Rd
.
Exercise 1.3.6 Show that in an inner product space, u + v = u +
v for some u, v ∈ V if and only if u and v are non-negatively linearly
dependent (i.e., for some c0 ≥ 0, either u = c0v or v = c0u).
Exercise 1.3.7 Prove that the orthogonal complement of a subset is a
closed subspace.
Exercise 1.3.8 Let V0 be a subset of a Hilbert space V . Show that the
following statements are equivalent:
(a) V0 is dense in V ; i.e., for any v ∈ V , there exists {vn}n≥1 ⊆ V0 such
that v − vnV → 0 as n → ∞.
(b) V ⊥
0 = {0}.
(c) If u ∈ V satisfies (u, v) = 0 ∀ v ∈ V0, then u = 0.
(d) For every 0 = u ∈ V , there is a v ∈ V0 such that (u, v) = 0.
Exercise 1.3.9 On C1
[a, b], define
(f, g)∗ = f(a)g(a) +
1
0
f
(x)g
(x) dx, f, g ∈ C1
[a, b]
and f∗ =
(f, f)∗. Show that
f∞ ≤ c f∗ ∀ f ∈ C1
[a, b]
for a suitable constant c.
Exercise 1.3.10 Consider the Fourier series (1.3.9) for a function v ∈
Cm
p (2π) with m ≥ 2. Show that
,
,
,
,
,
v −
N
n=−N
αnvn
,
,
,
,
,
∞
≤
cm(v)
Nm−1
, N ≥ 1.](https://image.slidesharecdn.com/444877-250518161545-79673a8a/85/Theoretical-Numerical-Analysis-A-Functional-Analysis-Framework-1st-Edition-Kendall-Atkinson-50-320.jpg)
![32 1. Linear Spaces
Lipschitz constant of v, and is denoted by Lip(v). The Lipschitz constant
is characterized by the relation
Lip(v) = sup
|v(x) − v(y)|
x − y
x, y ∈ Ω, x = y
.
More generally, a function v is said to be Hölder continuous with exponent
β ∈ (0, 1] if for some constant c,
|v(x) − v(y)| ≤ c x − y
β
for x, y ∈ Ω.
The Hölder space C0,β
(Ω) is defined to be the subspace of C(Ω) that con-
sists of functions that are Hölder continuous with the exponent β. With
the norm
vC0,β (Ω) = vC(Ω) + sup
|v(x) − v(y)|
x − y
β
x, y ∈ Ω, x = y
the space C0,β
(Ω) becomes a Banach space. When β = 1, the Hölder space
C0,1
(Ω) consists of all the Lipschitz continuous functions.
For m ∈ Z+ and β ∈ (0, 1], we similarly define the Hölder space
Cm,β
(Ω) =
1
v ∈ Cm
(Ω) | Dα
v ∈ C0,β
(Ω) for all α with |α| = m
2
;
this is a Banach space with the norm
vCm,β (Ω) = vCm(Ω)
+
|α|=m
sup
|Dα
v(x) − Dα
v(y)|
x − y
β
x, y ∈ Ω, x = y
.
Exercise 1.4.1 Show that C(Ω) with the norm vC(Ω) is a Banach space.
Exercise 1.4.2 Show that the space C1
(Ω) with the norm vC(Ω) is not
a Banach space.
Exercise 1.4.3 Let vn(x) = 1
n sin nx. Show that vn → 0 in C0,β
[0, 1] for
any β ∈ (0, 1), but vn 0 in C0,1
[0, 1].
Exercise 1.4.4 Discuss whether it is meaningful to use the Hölder space
C0,β
(Ω) with β 1.
Exercise 1.4.5 Consider v(s) = sα
for some 0 α 1. For which β ∈
(0, 1] is it true that v ∈ C0,β
[0, 1]?
1.5 Lp
spaces
In the study of Lp
(Ω) spaces, we identify functions (i.e., such functions
are considered identical) that are equal almost everywhere (a.e.) on Ω. For](https://image.slidesharecdn.com/444877-250518161545-79673a8a/85/Theoretical-Numerical-Analysis-A-Functional-Analysis-Framework-1st-Edition-Kendall-Atkinson-53-320.jpg)
![1.5. Lp
spaces 33
p ∈ [1, ∞), Lp
(Ω) is the linear space of measurable functions v : Ω → R
such that
vLp(Ω) =
Ω
|v(x)|p
dx
1/p
∞. (1.5.1)
The space L∞
(Ω) consists of all essentially bounded measurable functions
v : Ω → R ,
vL∞(Ω) = inf
meas (Ω)=0
sup
x∈ΩΩ
|v(x)| ∞. (1.5.2)
Some basic properties of the Lp
spaces are summarized in the following
theorem.
Theorem 1.5.1 Let Ω be an open bounded set in Rd
.
(a) For p ∈ [1, ∞], Lp
(Ω) is a Banach space.
(b) For p ∈ [1, ∞], every Cauchy sequence in Lp
(Ω) has a subsequence
that converges pointwise a.e. on Ω.
(c) If 1 ≤ p ≤ q ≤ ∞, then Lq
(Ω) ⊆ Lp
(Ω),
vLp(Ω) ≤ meas (Ω)
1
p − 1
q vLq(Ω) ∀ v ∈ Lq
(Ω),
and
vL∞(Ω) = lim
p→∞
vLp(Ω) ∀ v ∈ L∞
(Ω).
(d) If 1 ≤ p ≤ r ≤ q ≤ ∞ and we choose θ ∈ [0, 1] such that
1
r
=
θ
p
+
(1 − θ)
q
,
then
vLr(Ω) ≤ vθ
Lp(Ω)v1−θ
Lq(Ω) ∀ v ∈ Lq
(Ω).
In (c), when q = ∞, 1/q is understood to be 0. The result (d) is called
an interpolation property of the Lp
spaces. To prove (c) and (d), we need
to use the Hölder inequality. We first prove Young’s inequality.
Lemma 1.5.2 (Young’s inequality) Let a, b ≥ 0, p, q 1, 1/p+1/q =
1. Then
ab ≤
ap
p
+
bq
q
.
Proof. For any fixed b ≥ 0, define a function
f(a) =
ap
p
+
bq
q
− ab
on [0, ∞). From f
(a) = 0 we obtain a = b1/(p−1)
. We have f(b1/(p−1)
) = 0.
Since f(0) ≥ 0, lima→∞ f(a) = ∞ and f is continuous on [0, ∞), we see](https://image.slidesharecdn.com/444877-250518161545-79673a8a/85/Theoretical-Numerical-Analysis-A-Functional-Analysis-Framework-1st-Edition-Kendall-Atkinson-54-320.jpg)
Theoretical Numerical Analysis A Functional Analysis Framework 1st Edition Kendall Atkinson
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- 6. Texts in Applied Mathematics 9 Editors J.E. Marsden L. Sirovich M. Golubitsky Advisors G. Iooss P. Holmes D. Barkley M. Dellnitz P. Newton Springer New York Berlin Heidelburg Barcelona Hong Kong London Milan Paris Singapore Tokyo
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- 8. Kendall Atkinson Weimin Han Theoretical Numerical Analysis A Functional Analysis Framework With 25 Illustrations 1 3
- 9. Kendall Atkinson Weimin Han Department of Mathematics Department of Mathematics Department of Computer Science University of Iowa University of Iowa Iowa City, IA 52242 Iowa City, IA 52242 USA USA whan@math.uiowa.edu Kendall-Atkinson@uiowa.edu and Department of Mathematics Zhejiang University Hangzhou People’s Republic of China Series Editors J.E. Marsden L. Sirovich Control and Dynamical Systems, 107-81 Division of Applied Mathematics California Institute of Technology Brown University Pasadena, CA 91125 Providence, RI 02912 USA USA M. Golubitsky Department of Mathematics University of Houston Houston, TX 77204-3476 USA Mathematics Subject Classification (2000): 65-01, 73V05, 45L05, 4601 Library of Congress Cataloging-in-Publication Data Atkinson, Kendall E. Theoretical numerical analysis: a functional analysis framework / Kendall Atkinson, Weimin Han. p. cm.—(Texts in applied mathematics; 39) Includes bibliographical references and index. ISBN 0-387-95142-3 (alk. paper) 1. Functional analysis. I. Han, Weimin. II. Title. III Series. QA320.A85 2001 515—dc21 00–061920 Printed on acid-free paper. c 2001 Springer-Verlag New York, Inc. All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer-Verlag New York, Inc., 175 Fifth Avenue, New York, NY 10010, USA), 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 Michael Koy; manufacturing supervised by Joe Quatela. Typeset by The Bartlett Press, Inc., Marietta, GA. Printed and bound by Maple-Vail Book Manufacturing Group, York, PA. Printed in the United States of America. 9 8 7 6 5 4 3 2 1 ISBN 0-387-95142-3 SPIN 10780644 Springer-Verlag New York Berlin Heidelberg A member of BertelsmannSpringer Science + Business Media GmbH
- 10. Dedicated to Daisy and Clyde Atkinson Hazel and Wray Fleming and Daqing Han, Suzhen Qin Huidi Tang, Elizabeth Jing Han
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- 12. Series Preface Mathematics is playing an ever more important role in the physical and biological sciences, provoking a blurring of boundaries between scientific disciplines and a resurgence of interest in the modern as well as the clas- sical techniques of applied mathematics. This renewal of interest, both in research and teaching, has led to the establishment of the series: Texts in Applied Mathematics (TAM). The development of new courses is a natural consequence of a high level of excitement on the research frontier as newer techniques, such as numerical and symbolic computer systems, dynamical systems, and chaos, mix with and reinforce the traditional methods of applied mathematics. Thus, the purpose of this textbook series is to meet the current and future needs of these advances and encourage the teaching of new courses. TAM will publish textbooks suitable for use in advanced undergraduate and beginning graduate courses, and will complement the Applied Math- ematical Sciences (AMS) series, which will focus on advanced textbooks and research level monographs. California Institute of Technology J.E. Marsden Brown University L. Sirovich University of Houston M. Golubitsky
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- 14. Preface This textbook has grown out of a course that we teach periodically at the University of Iowa. We have beginning graduate students in mathematics who wish to work in numerical analysis from a theoretical perspective, and they need a background in those “tools of the trade” that we cover in this text. Ordinarily, such students would begin with a one-year course in real and complex analysis, followed by a one- or two-semester course in func- tional analysis and possibly a graduate level course in ordinary differential equations, partial differential equations, or integral equations. We still ex- pect our students to take most of these standard courses, but we also want to move them more rapidly into a research program. The course based on this book is designed to facilitate this goal. The textbook covers basic results of functional analysis and also some additional topics that are needed in theoretical numerical analysis. Ap- plications of this functional analysis are given by considering, at length, numerical methods for solving partial differential equations and integral equations. The material in the text is presented in a mixed manner. Some topics are treated with complete rigor, whereas others are simply presented without proof and perhaps illustrated (e.g., the principle of uniform boundedness). We have chosen to avoid introducing a formalized framework for Lebesgue measure and integration and also for distribution theory. Instead we use standard results on the completion of normed spaces and the unique ex- tension of densely defined bounded linear operators. This permits us to introduce the Lebesgue spaces formally and without their concrete realiza- tion using measure theory. The weak derivative can be introduced similarly
- 15. x Preface using the unique extension of densely defined linear operators, avoiding the need for a formal development of distribution theory. We describe some of the standard material on measure theory and distribution theory in an intuitive manner, believing this is sufficient for much of subsequent mathematical development. In addition, we give a number of deeper re- sults without proof, citing the existing literature. Examples of this are the open mapping theorem, the Hahn-Banach theorem, the principle of uniform boundedness, and a number of the results on Sobolev spaces. The choice of topics has been shaped by our research program and inter- ests at the University of Iowa. These topics are important elsewhere, and we believe this text will be useful to students at other universities as well. The book is divided into chapters, sections, and subsections where appro- priate. Mathematical relations (equalities and inequalities) are numbered by chapter, section, and their order of occurrence. For example, (1.2.3) is the third-numbered mathematical relation in Section 1.2 of Chapter 1. Defi- nitions, examples, theorems, lemmas, propositions, corollaries, and remarks are numbered consecutively within each section, by chapter and section. For example, in Section 1.1, Definition 1.1.1 is followed by Example 1.1.2. The first three chapters cover basic results of functional analysis and approximation theory that are needed in theoretical numerical analysis. Early on, in Chapter 4, we introduce methods of nonlinear analysis, as stu- dents should begin early to think about both linear and nonlinear problems. Chapter 5 is a short introduction to finite difference methods for solving time-dependent problems. Chapter 6 is an introduction to Sobolev spaces, giving different perspectives of them. Chapters 7 through 10 cover material related to elliptic boundary value problems and variational inequalities. Chapter 11 is a general introduction to numerical methods for solving in- tegral equations of the second kind, and Chapter 12 gives an introduction to boundary integral equations for planar regions with a smooth boundary curve. We give exercises at the end of most sections. The exercises are numbered consecutively by chapter and section. At the end of each chapter, we provide some short discussions of the literature, including recommendations for additional reading. During the preparation of the book, we received helpful suggestions from numerous colleagues and friends. We particularly thank P.G. Ciar- let, William A. Kirk, Wenbin Liu, and David Stewart. We also thank the anonymous referees whose suggestions led to an improvement of the book. Kendall Atkinson Weimin Han Iowa City, IA
- 16. Contents Series Preface vii Preface ix 1 Linear Spaces 1 1.1 Linear spaces . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Normed spaces . . . . . . . . . . . . . . . . . . . . . . . . 7 1.2.1 Convergence . . . . . . . . . . . . . . . . . . . . . 9 1.2.2 Banach spaces . . . . . . . . . . . . . . . . . . . . 11 1.2.3 Completion of normed spaces . . . . . . . . . . . 12 1.3 Inner product spaces . . . . . . . . . . . . . . . . . . . . 18 1.3.1 Hilbert spaces . . . . . . . . . . . . . . . . . . . . 22 1.3.2 Orthogonality . . . . . . . . . . . . . . . . . . . . 23 1.4 Spaces of continuously differentiable functions . . . . . . 30 1.4.1 Hölder spaces . . . . . . . . . . . . . . . . . . . . 31 1.5 Lp spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 1.6 Compact sets . . . . . . . . . . . . . . . . . . . . . . . . 35 2 Linear Operators on Normed Spaces 38 2.1 Operators . . . . . . . . . . . . . . . . . . . . . . . . . . 39 2.2 Continuous linear operators . . . . . . . . . . . . . . . . 41 2.2.1 L(V, W) as a Banach space . . . . . . . . . . . . . 45 2.3 The geometric series theorem and its variants . . . . . . 46 2.3.1 A generalization . . . . . . . . . . . . . . . . . . . 49 2.3.2 A perturbation result . . . . . . . . . . . . . . . . 50
- 17. xii Contents 2.4 Some more results on linear operators . . . . . . . . . . . 55 2.4.1 An extension theorem . . . . . . . . . . . . . . . 55 2.4.2 Open mapping theorem . . . . . . . . . . . . . . . 57 2.4.3 Principle of uniform boundedness . . . . . . . . . 58 2.4.4 Convergence of numerical quadratures . . . . . . 59 2.5 Linear functionals . . . . . . . . . . . . . . . . . . . . . . 62 2.5.1 An extension theorem for linear functionals . . . 63 2.5.2 The Riesz representation theorem . . . . . . . . . 64 2.6 Adjoint operators . . . . . . . . . . . . . . . . . . . . . . 67 2.7 Types of convergence . . . . . . . . . . . . . . . . . . . . 72 2.8 Compact linear operators . . . . . . . . . . . . . . . . . 73 2.8.1 Compact integral operators on C(D) . . . . . . . 74 2.8.2 Properties of compact operators . . . . . . . . . . 76 2.8.3 Integral operators on L2 (a, b) . . . . . . . . . . . 78 2.8.4 The Fredholm alternative theorem . . . . . . . . 79 2.8.5 Additional results on Fredholm integral equations . . . . . . . . . . . . . . . . . . 83 2.9 The resolvent operator . . . . . . . . . . . . . . . . . . . 87 2.9.1 R(λ) as a holomorphic function . . . . . . . . . . 89 3 Approximation Theory 92 3.1 Interpolation theory . . . . . . . . . . . . . . . . . . . . . 93 3.1.1 Lagrange polynomial interpolation . . . . . . . . 94 3.1.2 Hermite polynomial interpolation . . . . . . . . . 98 3.1.3 Piecewise polynomial interpolation . . . . . . . . 98 3.1.4 Trigonometric interpolation . . . . . . . . . . . . 101 3.2 Best approximation . . . . . . . . . . . . . . . . . . . . . 105 3.2.1 Convexity, lower semicontinuity . . . . . . . . . . 105 3.2.2 Some abstract existence results . . . . . . . . . . 107 3.2.3 Existence of best approximation . . . . . . . . . . 110 3.2.4 Uniqueness of best approximation . . . . . . . . . 111 3.3 Best approximations in inner product spaces . . . . . . . 113 3.4 Orthogonal polynomials . . . . . . . . . . . . . . . . . . 117 3.5 Projection operators . . . . . . . . . . . . . . . . . . . . 121 3.6 Uniform error bounds . . . . . . . . . . . . . . . . . . . 124 3.6.1 Uniform error bounds for L2 -approximations . . . 126 3.6.2 Interpolatory projections and their convergence . . . . . . . . . . . . . . . . . . 128 4 Nonlinear Equations and Their Solution by Iteration 131 4.1 The Banach fixed-point theorem . . . . . . . . . . . . . . 131 4.2 Applications to iterative methods . . . . . . . . . . . . . 135 4.2.1 Nonlinear equations . . . . . . . . . . . . . . . . . 135 4.2.2 Linear systems . . . . . . . . . . . . . . . . . . . 136 4.2.3 Linear and nonlinear integral equations . . . . . . 139
- 18. Contents xiii 4.2.4 Ordinary differential equations in Banach spaces . . . . . . . . . . . . . . . . . . . . 143 4.3 Differential calculus for nonlinear operators . . . . . . . . 146 4.3.1 Fréchet and Gâteaux derivatives . . . . . . . . . . 146 4.3.2 Mean value theorems . . . . . . . . . . . . . . . . 149 4.3.3 Partial derivatives . . . . . . . . . . . . . . . . . . 151 4.3.4 The Gâteaux derivative and convex minimization . . . . . . . . . . . . . . . . 152 4.4 Newton’s method . . . . . . . . . . . . . . . . . . . . . . 154 4.4.1 Newton’s method in a Banach space . . . . . . . 155 4.4.2 Applications . . . . . . . . . . . . . . . . . . . . . 157 4.5 Completely continuous vector fields . . . . . . . . . . . . 159 4.5.1 The rotation of a completely continuous vector field . . . . . . . . . . . . . . . . . . . . . . 161 4.6 Conjugate gradient iteration . . . . . . . . . . . . . . . . 162 5 Finite Difference Method 171 5.1 Finite difference approximations . . . . . . . . . . . . . . 171 5.2 Lax equivalence theorem . . . . . . . . . . . . . . . . . . 177 5.3 More on convergence . . . . . . . . . . . . . . . . . . . . 186 6 Sobolev Spaces 193 6.1 Weak derivatives . . . . . . . . . . . . . . . . . . . . . . 193 6.2 Sobolev spaces . . . . . . . . . . . . . . . . . . . . . . . . 198 6.2.1 Sobolev spaces of integer order . . . . . . . . . . 199 6.2.2 Sobolev spaces of real order . . . . . . . . . . . . 204 6.2.3 Sobolev spaces over boundaries . . . . . . . . . . 206 6.3 Properties . . . . . . . . . . . . . . . . . . . . . . . . . . 207 6.3.1 Approximation by smooth functions . . . . . . . . 207 6.3.2 Extensions . . . . . . . . . . . . . . . . . . . . . . 208 6.3.3 Sobolev embedding theorems . . . . . . . . . . . 208 6.3.4 Traces . . . . . . . . . . . . . . . . . . . . . . . . 210 6.3.5 Equivalent norms . . . . . . . . . . . . . . . . . . 211 6.3.6 A Sobolev quotient space . . . . . . . . . . . . . . 215 6.4 Characterization of Sobolev spaces via the Fourier transform . . . . . . . . . . . . . . . . . . . . . . 219 6.5 Periodic Sobolev spaces . . . . . . . . . . . . . . . . . . . 222 6.5.1 The dual space . . . . . . . . . . . . . . . . . . . 225 6.5.2 Embedding results . . . . . . . . . . . . . . . . . 226 6.5.3 Approximation results . . . . . . . . . . . . . . . 227 6.5.4 An illustrative example of an operator . . . . . . 228 6.5.5 Spherical polynomials and spherical harmonics . . . . . . . . . . . . . . . . . 229 6.6 Integration by parts formulas . . . . . . . . . . . . . . . 234
- 19. xiv Contents 7 Variational Formulations of Elliptic Boundary Value Problems 238 7.1 A model boundary value problem . . . . . . . . . . . . . 239 7.2 Some general results on existence and uniqueness . . . . 241 7.3 The Lax-Milgram lemma . . . . . . . . . . . . . . . . . . 244 7.4 Weak formulations of linear elliptic boundary value problems . . . . . . . . . . . . . . . . . . . . . . . . 248 7.4.1 Problems with homogeneous Dirichlet boundary conditions . . . . . . . . . . . . . . . . 249 7.4.2 Problems with non-homogeneous Dirichlet boundary conditions . . . . . . . . . . . . . . . . 249 7.4.3 Problems with Neumann boundary conditions . . . . . . . . . . . . . . . . 251 7.4.4 Problems with mixed boundary conditions . . . . 253 7.4.5 A general linear second-order elliptic boundary value problem . . . . . . . . . . . . . . 254 7.5 A boundary value problem of linearized elasticity . . . . 257 7.6 Mixed and dual formulations . . . . . . . . . . . . . . . . 260 7.7 Generalized Lax-Milgram lemma . . . . . . . . . . . . . . 264 7.8 A nonlinear problem . . . . . . . . . . . . . . . . . . . . 265 8 The Galerkin Method and Its Variants 270 8.1 The Galerkin method . . . . . . . . . . . . . . . . . . . . 270 8.2 The Petrov-Galerkin method . . . . . . . . . . . . . . . . 276 8.3 Generalized Galerkin method . . . . . . . . . . . . . . . 278 9 Finite Element Analysis 281 9.1 One-dimensional examples . . . . . . . . . . . . . . . . . 283 9.1.1 Linear elements for a second-order problem . . . 283 9.1.2 High-order elements and the condensation technique . . . . . . . . . . . . . . . 286 9.1.3 Reference element technique, non-conforming method . . . . . . . . . . . . . . 288 9.2 Basics of the finite element method . . . . . . . . . . . . 291 9.2.1 Triangulation . . . . . . . . . . . . . . . . . . . . 291 9.2.2 Polynomial spaces on the reference elements . . . 293 9.2.3 Affine-equivalent finite elements . . . . . . . . . . 295 9.2.4 Finite element spaces . . . . . . . . . . . . . . . . 296 9.2.5 Interpolation . . . . . . . . . . . . . . . . . . . . . 298 9.3 Error estimates of finite element interpolations . . . . . . 300 9.3.1 Interpolation error estimates on the reference element . . . . . . . . . . . . . . . . . . 300 9.3.2 Local interpolation error estimates . . . . . . . . 301 9.3.3 Global interpolation error estimates . . . . . . . . 304 9.4 Convergence and error estimates . . . . . . . . . . . . . . 308
- 20. Contents xv 10 Elliptic Variational Inequalities and Their Numerical Approximations 313 10.1 Introductory examples . . . . . . . . . . . . . . . . . . . 313 10.2 Elliptic variational inequalities of the first kind . . . . . . 319 10.3 Approximation of EVIs of the first kind . . . . . . . . . . 323 10.4 Elliptic variational inequalities of the second kind . . . . 326 10.5 Approximation of EVIs of the second kind . . . . . . . . 331 10.5.1 Regularization technique . . . . . . . . . . . . . . 333 10.5.2 Method of Lagrangian multipliers . . . . . . . . . 335 10.5.3 Method of numerical integration . . . . . . . . . . 337 11 Numerical Solution of Fredholm Integral Equations of the Second Kind 342 11.1 Projection methods: General theory . . . . . . . . . . . . 343 11.1.1 Collocation methods . . . . . . . . . . . . . . . . 343 11.1.2 Galerkin methods . . . . . . . . . . . . . . . . . . 345 11.1.3 A general theoretical framework . . . . . . . . . . 346 11.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 351 11.2.1 Piecewise linear collocation . . . . . . . . . . . . . 351 11.2.2 Trigonometric polynomial collocation . . . . . . . 354 11.2.3 A piecewise linear Galerkin method . . . . . . . . 356 11.2.4 A Galerkin method with trigonometric polynomials . . . . . . . . . . . . . 358 11.3 Iterated projection methods . . . . . . . . . . . . . . . . 362 11.3.1 The iterated Galerkin method . . . . . . . . . . . 364 11.3.2 The iterated collocation solution . . . . . . . . . 366 11.4 The Nyström method . . . . . . . . . . . . . . . . . . . . 372 11.4.1 The Nyström method for continuous kernel functions . . . . . . . . . . . . . . . . . . . 373 11.4.2 Properties and error analysis of the Nyström method . . . . . . . . . . . . . . . . . . 376 11.4.3 Collectively compact operator approximations . . . . . . . . . . . . . . 383 11.5 Product integration . . . . . . . . . . . . . . . . . . . . . 385 11.5.1 Error analysis . . . . . . . . . . . . . . . . . . . . 388 11.5.2 Generalizations to other kernel functions . . . . . 390 11.5.3 Improved error results for special kernels . . . . . 392 11.5.4 Product integration with graded meshes . . . . . 392 11.5.5 The relationship of product integration and collocation methods . . . . . . . . . . . . . . . . . 396 11.6 Projection methods for nonlinear equations . . . . . . . . 398 11.6.1 Linearization . . . . . . . . . . . . . . . . . . . . 398 11.6.2 A homotopy argument . . . . . . . . . . . . . . . 401 11.6.3 The approximating finite-dimensional problem . . . . . . . . . . . . . 402
- 21. xvi Contents 12 Boundary Integral Equations 405 12.1 Boundary integral equations . . . . . . . . . . . . . . . . 406 12.1.1 Green’s identities and representation formula . . 407 12.1.2 The Kelvin transformation and exterior problems . . . . . . . . . . . . . . . . . . 409 12.1.3 Boundary integral equations of direct type . . . 413 12.2 Boundary integral equations of the second kind . . . . . 419 12.2.1 Evaluation of the double layer potential . . . . . 421 12.2.2 The exterior Neumann problem . . . . . . . . . . 425 12.3 A boundary integral equation of the first kind . . . . . . 431 12.3.1 A numerical method . . . . . . . . . . . . . . . . 433 References 436 Index 445
- 22. 1 Linear Spaces Linear (or vector) spaces are the standard setting for studying and solving a large proportion of the problems in differential and integral equations, approximation theory, optimization theory, and other topics in applied mathematics. In this chapter, we gather together some concepts and re- sults concerning various aspects of linear spaces, especially some of the more important linear spaces such as Banach spaces, Hilbert spaces, and certain other function spaces that are used frequently in this work and in applied mathematics generally. 1.1 Linear spaces A linear space is a set of elements equipped with two binary operations, called vector addition and scalar multiplication, in such a way that the operations behave linearly. Definition 1.1.1 Let V be a set of objects, to be called vectors; and let K be a set of scalars, either R the set of real numbers, or C the set of complex numbers. Assume there are two operations: (u, v) → u+v ∈ V and (α, v) → αv ∈ V , called addition and scalar multiplication, respectively, defined for any u, v ∈ V and any α ∈ K . These operations are to satisfy the following rules. 1. u + v = v + u for any u, v ∈ V (commutative law); 2. (u + v) + w = u + (v + w) for any u, v, w ∈ V (associative law);
- 23. 2 1. Linear Spaces 3. there is an element 0 ∈ V such that 0+u = u for any u ∈ V (existence of the zero element); 4. for any u ∈ V , there is an element −u ∈ V such that u + (−u) = 0 (existence of negative elements); 5. 1u = u for any u ∈ V ; 6. α(βu) = (αβ)u for any u ∈ V , any α, β ∈ K (associative law for scalar multiplication); 7. α(u + v) = αu + αv and (α + β)u = αu + βu for any u, v ∈ V , and any α, β ∈ K (distributive laws). Then V is called a linear space, or a vector space. When K is the set of the real numbers, V is a real linear space; and when K is the set of the complex numbers, V becomes a complex linear space. In this work, most of the time we only deal with real linear spaces. So when we say V is a linear space, the reader should usually assume V is a real linear space, unless explicitly stated otherwise. Some remarks are in order concerning the definition of a linear space. From the commutative law and the associative law, we observe that to add several elements, the order of summation does not matter, and it does not cause any ambiguity to write expressions such as u + v + w or n i=1 ui. By using the commutative law and the associative law, it is not difficult to verify that the zero element and the negative element (−u) of a given element u ∈ V are unique, and they can be equivalently defined through the relations v + 0 = v for any v ∈ V , and (−u) + u = 0. Below, we write u − v for u + (−v). Example 1.1.2 (a) The set of the real numbers R is a real linear space when the addition and scalar multiplication are the usual addition and mul- tiplication. Similarly, the set of complex numbers C is a complex linear space. (b) Let d be a positive integer. The letter d is used generally in this work for the spatial dimension. The set of all vectors with d real components, with the usual vector addition and scalar multiplication, forms a linear space Rd . A typical element in Rd can be expressed as x = (x1, . . . , xd)T , where x1, . . . , xd ∈ R. Similarly, Cd is a complex linear space. (c) Let Ω ⊆ Rd be an open subset of Rd . In this work, the symbol Ω always stands for an open subset of Rd . The set of all the continuous func- tions on Ω forms a linear space C(Ω), under the usual addition and scalar multiplication of functions: For f, g ∈ C(Ω), the function f + g defined by (f + g)(x) = f(x) + g(x) x ∈ Ω,
- 24. 1.1. Linear spaces 3 belongs to C(Ω); so does the scalar multiplication function α f defined through (α f)(x) = α f(x) x ∈ Ω. Similarly, C(Ω) denotes the space of continuous functions on the closed set Ω. Clearly, C(Ω) ⊆ C(Ω). (d) A related function space is C(D), containing all functions f : D → K that are continuous on a general set D ⊆ Rd . The arbitrary set D can be an open or closed set in Rd , or perhaps neither; and it can be a lower dimensional set such as a portion of the boundary of an open set in Rd . When D is a closed and bounded subset of Rd , a function from the space C(D) is necessarily bounded. (e) For any non-negative integer m, we may define the space Cm (Ω) as the space of all the functions that together with their derivatives of orders up to m are continuous on Ω. We may also define the space Cm (Ω) to be the space of all the functions that together with their derivatives of orders up to m are continuous on Ω. These function spaces are discussed at length in Section 1.4. (f) The space of continuous 2π-periodic functions is denoted by Cp(2π). It is the set of all f ∈ C(−∞, ∞) for which f(x + 2π) = f(x) − ∞ x ∞. For an integer k ≥ 0, the space Ck p (2π) denotes the set of all functions in Cp(2π) that have k continuous derivatives on (−∞, ∞). We usually write C0 p (2π) as simply Cp(2π). These spaces are used in connection with problems in which periodicity plays a major role. Definition 1.1.3 A subspace W of the linear space V is a subset of V that is closed under the addition and scalar multiplication operations of V , i.e., for any u, v ∈ W and any α ∈ K , we have u + v ∈ W and αv ∈ W. It can be verified that W itself, equipped with the addition and scalar multiplication operations of V , is a linear space. Example 1.1.4 In the linear space R3 , W = {x = (x1, x2, 0)T | x1, x2 ∈ R} is a subspace, consisting of all the vectors on the x1x2-plane. In contrast, W = {x = (x1, x2, 1)T | x1, x2 ∈ R} is not a subspace. Nevertheless, we observe that W is a translation of the subspace W, W = x0 + W, where x0 = (0, 0, 1)T . The set W is an example of an affine set.
- 25. 4 1. Linear Spaces Given vectors v1, . . . , vn ∈ V and scalars α1, . . . , αn ∈ K , we call n i=1 αivi = α1v1 + · · · + αnvn a linear combination of v1, . . . , vn. It is meaningful to remove “redundant” vectors from the linear combination. Thus we introduce the concepts of linear dependence and independence. Definition 1.1.5 We say v1, . . . , vn ∈ V are linearly dependent if there are scalars αi ∈ K , 1 ≤ i ≤ n, with at least one αi non-zero such that n i=1 αivi = 0. (1.1.1) We say v1, . . . , vn ∈ V are linearly independent if they are not linearly dependent, meaning that the only choice of scalars {αi} for which (1.1.1) is valid is αi = 0 for i = 1, 2, . . . , n. We observe that v1, . . . , vn are linearly dependent if and only if at least one of the vectors can be expressed as a linear combination of the rest of the vectors. In particular, a set of vectors containing the zero element is always linearly dependent. Similarly, v1, . . . , vn are linearly independent if and only if none of the vectors can be expressed as a linear combination of the rest of the vectors; in other words, none of the vectors is “redundant.” Example 1.1.6 In Rd , d vectors x(i) = (x (i) 1 , . . . , x (i) d )T , 1 ≤ i ≤ d, are linearly independent if and only if the determinant x (1) 1 · · · x (d) 1 . . . ... . . . x (1) d · · · x (d) d is non-zero. This follows from a standard result in linear algebra. The con- dition (1.1.1) is equivalent to a homogeneous system of linear equations, and a standard result of linear algebra says that this system has (0, . . . , 0)T as its only solution if and only if the above determinant is non-zero. Example 1.1.7 Within the space C[0, 1], the vectors 1, x, x2 , . . . , xn are linearly independent. This can be proven in several ways. Assuming n j=0 αjxj = 0, 0 ≤ x ≤ 1, we can form its first n derivatives. Setting x = 0 in this polynomial and its derivatives will lead to αj = 0 for j = 0, 1, . . . , n.
- 26. 1.1. Linear spaces 5 Definition 1.1.8 The span of v1, . . . , vn ∈ V is defined to be the set of all the linear combinations of these vectors: span{v1, . . . , vn} = n i=1 αivi αi ∈ K , 1 ≤ i ≤ n . Evidently, span {v1, . . . , vn} is a linear subspace of V . Most of the time, we apply this definition for the case where v1, . . . , vn are linearly independent. Definition 1.1.9 A linear space V is said to be finite dimensional if there exists a finite maximal set of independent vectors {v1, . . . , vn}; i.e., the set {v1, . . . , vn} is linearly independent, but {v1, . . . , vn, vn+1} is linearly dependent for any vn+1 ∈ V . The set {v1, . . . , vn} is called a basis of the space. If such a finite basis for V does not exist, then V is said to be infinite dimensional. We see that a basis is a set of independent vectors such that any vector in the space can be written as a linear combination of them. Obviously a basis is not unique, yet we have the following important result. Theorem 1.1.10 For a finite-dimensional linear space, every basis for V contains exactly the same number of vectors. This number is called the dimension of the space. A proof of this result can be found in most introductory textbooks on linear algebra; for example, see [3, Section 5.4]. Example 1.1.11 The space Rd is d-dimensional. There are infinitely many possible choices for a basis of the space. A canonical basis for this space is {ei = (0, . . . , 0, 1i, 0, . . . , 0)T }d i=1 in which the single 1 is in component i. We introduce the concept of a linear function. Definition 1.1.12 Let L be a function from the linear space V to the linear space W. We say L is a linear function if (a) for all u, v ∈ V , L(u + v) = L(u) + L(v); (b) for all v ∈ V and all α ∈ K , L(αv) = αL(v). For such a linear function, we often write L(v) = Lv, v ∈ V. This definition is extended and discussed extensively in Chapter 2. Other common notations are linear mapping, linear operator, and linear transformation.
- 27. 6 1. Linear Spaces Definition 1.1.13 Two linear spaces U and V are said to be isomorphic if there is a linear bijective (i.e., one-to-one and onto) function l : U → V . Many properties of a linear space U hold for any other linear space V that is isomorphic to U; and then the explicit contents of the space do not matter in the analysis of these properties. This usually proves to be convenient. One such example is that if U and V are isomorphic and are finite dimensional, then their dimensions are equal, a basis of V can be obtained from a basis of U by applying the mapping l, and a basis of U can be obtained from a basis of V by applying the inverse mapping of l. Example 1.1.14 The set Pk of all polynomials of degree less than or equal to k is a subspace of continuous function space C[0, 1]. An element in the space Pk has the form a0 + a1x + · · · + akxk . The mapping l : a0 + a1x + · · · + akxk → (a0, a1, . . . , ak)T is bijective from Pk to Rk+1 . Thus, Pk is isomorphic to Rk+1 . Definition 1.1.15 Let U and V be two linear spaces. The Cartesian product of the spaces, W = U × V , is defined by W = {w = (u, v) | u ∈ U, v ∈ V } endowed with componentwise addition and scalar multiplication (u1, v1) + (u2, v2) = (u1 + u2, v1 + v2) ∀ (u1, v1), (u2, v2) ∈ W, α (u, v) = (α u, α v) ∀ (u, v) ∈ W, ∀ α ∈ K . It is easy to verify that W is a linear space. The definition can be extended in a straightforward way for the Cartesian product of any finite number of linear spaces. Example 1.1.16 The real plane can be viewed as the Cartesian product of two real lines: R2 = R × R. In general, Rd = R × · · · × R d times . Exercise 1.1.1 Show that the set of all continuous solutions of the differ- ential equation u (x) + u(x) = 0 is a finite-dimensional linear space. Is the set of all continuous solutions of u (x) + u(x) = 1 a linear space? Exercise 1.1.2 When is the set {u ∈ C[0, 1] | u(0) = a} a linear space? Exercise 1.1.3 Show that in any linear space V , a set of vectors is always linearly dependent if one of the vectors is zero. Exercise 1.1.4 Assume U and V are finite-dimensional linear spaces, and let {u1, . . . , un} and {v1, . . . , vm} be bases for them, respectively. Using these bases, create a basis for W = U × V .
- 28. 1.2. Normed spaces 7 1.2 Normed spaces In numerical analysis, we frequently need to examine the closeness of a numerical solution to the exact solution. To answer the question quanti- tatively, we need to have a measure on the magnitude of the difference between the numerical solution and the exact solution. A norm of a vector in a linear space provides such a measure. Definition 1.2.1 Given a linear space V , a norm · is a function from V to R with the following properties. 1. v ≥ 0 for any v ∈ V , and v = 0 if and only if v = 0; 2. αv = |α| v for any v ∈ V and α ∈ K ; 3. u + v ≤ u + v for any u, v ∈ V . The space V equipped with the norm · is called a normed linear space or a normed space. We usually say V is a normed space when the definition of the norm is clear from the context. Some remarks are in order on the definition of a norm. The three axioms in the definition mimic the principal properties of the notion of the ordinary length of a vector in R2 or R3 . The first axiom says the norm of any vector must be non-negative, and the only vector with zero norm is zero. The second axiom is usually called positive homogeneity. The third axiom is also called the triangle inequality, which is a direct extension of the triangle inequality on the plane: The length of any side of a triangle is not greater than the sum of the lengths of the other two sides. With the definition of a norm, we can use the quantity u − v as a measure for the distance between u and v. Definition 1.2.2 Given a linear space V , a semi-norm | · | is a function from V to R with the properties of a norm except that |v| = 0 does not necessarily imply v = 0. One place in this work where the notion of a semi-norm plays an important role is in estimating the error of polynomial interpolation. Example 1.2.3 (a) For x = (x1, . . . , xd)T , the formula x2 = d i=1 x2 i 1/2 (1.2.1) defines a norm in the space Rd (cf. Exercise 1.2.5), called the Euclidean norm, which is the usual norm for the space Rd . When d = 1, the norm coincides with the absolute value: x2 = |x| for x ∈ R .
- 29. 8 1. Linear Spaces x1 x2 S 2 x1 x2 S 1 x1 x2 S ∞ Figure 1.1. The unit ball Sp = x ∈ R2 : xp ≤ 1 for p = 1, 2, ∞ (b) More generally, for 1 ≤ p ≤ ∞, the formulas xp = d i=1 |xi|p 1/p for 1 ≤ p ∞, (1.2.2) x∞ = max 1≤i≤d |xi| (1.2.3) define norms in the space Rd (cf. Exercise 1.2.5). The norm · p is called the p-norm, and · ∞ is called the maximum or infinity norm. It is straightforward to show that x∞ = lim p→∞ xp by using the inequality (1.2.5) given below. Again, when d = 1, all these norms coincide with the absolute value: xp = |x|, x ∈ R. Over Rd , the most commonly used norms are · p, p = 1, 2, ∞. The unit ball in R2 for each of these norms is shown in Figure 1.1. Example 1.2.4 (a) The standard norm for C[a, b] is the maximum norm f∞ = max a≤x≤b |f(x)| , f ∈ C[a, b]. This is also the norm for Cp(2π) (with a = 0 and b = 2π), the space of continuous 2π-periodic functions introduced in Example 1.1.2(f).
- 30. 1.2. Normed spaces 9 (b) For an integer k 0, the standard norm for Ck [a, b] is fk,∞ = max 0≤j≤k f(j) ∞, f ∈ Ck [a, b]. This is also the standard norm for Ck p (2π). With the use of a norm for V we can introduce a topology for V , a set of open and closed sets for V . Definition 1.2.5 Let (V, ·) be a normed space. Given v0 ∈ V and r 0, the sets B(v0, r) = {v ∈ V | v − v0 r}, B(v0, r) = {v ∈ V | v − v0 ≤ r} are called the open and closed balls centered at v0 with radius r. When r = 1 and v0 = 0, we have unit balls. Definition 1.2.6 Let A ⊆ V, a normed linear space. The set A is open if for every v ∈ A, there is a radius r 0 such that B(v, r) ⊆ A. The set A is closed in V if its complement V − A is open in V . 1.2.1 Convergence With the notion of a norm at our disposal, we can define the important concept of convergence. Definition 1.2.7 Let V be a normed space with the norm ·. A sequence {un} ⊆ V is convergent to u ∈ V if lim n→∞ un − u = 0. We say that u is the limit of the sequence {un}, and write un → u as n → ∞, or limn→∞ un = u. It can be verified that any sequence can have at most one limit. Definition 1.2.8 A function f : V → R is said to be continuous at u ∈ V if for any sequence {un} with un → u, we have f(un) → f(u) as n → ∞. The function f is said to be continuous on V if it is continuous at every u ∈ V . Proposition 1.2.9 The norm function · is continuous. Proof. We need to show that if un → u, then un → u. This follows from the backward triangle inequality | u − v | ≤ u − v ∀ u, v ∈ V (1.2.4) derived from the triangle inequality. We have seen that on a linear space various norms can be defined. Different norms give different measures of size for a given vector in the
- 31. 10 1. Linear Spaces space. Consequently, different norms may give rise to different forms of convergence. Definition 1.2.10 We say two norms · (1) and · (2) are equivalent if there exist positive constants c1, c2 such that c1u(1) ≤ u(2) ≤ c2u(1) ∀ u ∈ V. With two such equivalent norms, a sequence {un} converges in one norm if and only if it converges in the other norm: lim n→∞ un − u(1) = 0 ⇐⇒ lim n→∞ un − u(2) = 0. Example 1.2.11 For the norms (1.2.2)–(1.2.3) on Rd , it is straightforward to show x∞ ≤ xp ≤ d1/p x∞ ∀ x ∈ Rd . (1.2.5) Thus all the norms xp, 1 ≤ p ≤ ∞, on Rd are equivalent. More generally, we have the following well-known result. For a proof, see [11, p. 483]. Theorem 1.2.12 Over a finite-dimensional space, any two norms are equivalent. Thus, on a finite-dimensional space, different norms lead to the same convergence notion. Over an infinite-dimensional space, however, such a statement is no longer valid. Example 1.2.13 Let V be the space of all bounded functions on [0, 1]. For u ∈ V , in analogy with Example 1.2.3, we may define the following norms: up = 1 0 |u(x)|p dx 1/p , 1 ≤ p ∞, (1.2.6) u∞ = sup 0≤x≤1 |u(x)|. (1.2.7) Now consider a sequence of functions {un} ⊆ V , defined by un(x) = 1 − nx, 0 ≤ x ≤ 1 n , 0, 1 n x ≤ 1. It is easy to show that unp = [n(p + 1)]−1/p , 1 ≤ p ∞. Thus we see that the sequence {un} converges to u = 0 in the norm · p, 1 ≤ p ∞. On the other hand, un∞ = 1, n ≥ 1, so {un} does not converge to u = 0 in the norm · ∞.
- 32. 1.2. Normed spaces 11 As we have seen in the last example, in an infinite-dimensional space, some norms are not equivalent. Convergence defined by one norm can be stronger than that by another. Example 1.2.14 Consider again the space of all bounded functions on [0, 1], and the family of norms · p, 1 ≤ p ∞, and · ∞. We have, for any p ∈ [1, ∞), up ≤ u∞, u ∈ V. Thus, convergence in · ∞ implies convergence in · p, 1 ≤ p ∞, but not conversely. Convergence in ·∞ is usually called uniform convergence. 1.2.2 Banach spaces The concept of a normed space is usually too general, and special attention is given to a particular type of normed space called a Banach space. Definition 1.2.15 Let V be a normed space. A sequence {un} ⊆ V is called a Cauchy sequence if lim m,n→∞ um − un = 0. Obviously, a convergent sequence is a Cauchy sequence. In the finite- dimensional space Rd , any Cauchy sequence is convergent. However, in a general infinite-dimensional space, a Cauchy sequence may fail to converge; see Example 1.2.18 below. Definition 1.2.16 A normed space is said to be complete if every Cauchy sequence from the space converges to an element in the space. A complete normed space is called a Banach space. Example 1.2.17 Let Ω ⊆ Rd be a bounded open set. For v ∈ C(Ω) and 1 ≤ p ∞, define the p-norm vp = Ω |v(x)|p dx 1/p . (1.2.8) Here, x = (x1, . . . , xd)T and dx = dx1dx2 · · · dxd. In addition, define the ∞-norm or maximum norm v∞ = max x∈Ω |v(x)|. The space C(Ω) with · ∞ is a Banach space; i.e., the uniform limit of continuous functions is itself continuous. Example 1.2.18 The space C(Ω) with the norm · p, 1 ≤ p ∞, is not a Banach space. To illustrate this, we consider the space C[0, 1] and a
- 33. 12 1. Linear Spaces sequence in C[0, 1] defined as follows: un(x) = 0, 0 ≤ x ≤ 1 2 − 1 2n , n x − 1 2 (n − 1), 1 2 − 1 2n ≤ x ≤ 1 2 + 1 2n , 1, 1 2 + 1 2n ≤ x ≤ 1. Let u(x) = 0, 0 ≤ x 1 2 , 1, 1 2 x ≤ 1. Then un − up → 0 as n → ∞; i.e., the sequence {un} converges to u in the norm · p. But obviously no matter how we define u(1/2), the limit function u is not continuous. 1.2.3 Completion of normed spaces It is important to be able to deal with function spaces using a norm of our choice, as such a norm is often important or convenient in the formulation of a problem or in the analysis of a numerical method. The following theorem allows us to do this. A proof is discussed in [88, p. 84]. Theorem 1.2.19 Let V be a normed space. Then there is a complete normed space W with the following properties: (a) There is a subspace V ⊆ W and a bijective (one-to-one and onto) linear function I : V → V with IvW = vV ∀ v ∈ V. The function I is called an isometric isomorphism of the spaces V and V . (b) The subspace V is dense in W; i.e., for any w ∈ W, there is a sequence { vn} ⊆ V such that w − vnW → 0 as n → ∞. The space W is called the completion of V , and W is unique up to an isometric isomorphism. The spaces V and V are generally identified, meaning no distinction is made between them. However, we also consider cases where it is important to note the distinction. An important example of the theorem is to let V be the rational numbers and W be the real numbers R . One way in which R can be defined is as a set of equivalence classes of Cauchy sequences of rational numbers, and V can be identified with those equivalence classes of Cauchy sequences whose limit is a rational number. A proof of the above theorem can be made by mimicking this commonly used construction of the real numbers from the rational numbers.
- 34. 1.2. Normed spaces 13 Example 1.2.20 Theorem 1.2.19 guarantees the existence of a unique ab- stract completion of an arbitrary normed vector space. However, it is often possible, and indeed desirable, to give a more concrete definition of the completion of a given normed space; much of the subject of real analysis is concerned with this topic. In particular, the subject of Lebesgue mea- sure and integration deals with the completion of C(Ω) under the norms of (1.2.8), · p for 1 ≤ p ∞. A complete development of Lebesgue mea- sure and integration is given in any standard textbook on real analysis; for example, see Royden [141] or Rudin [142]. We do not introduce formally and rigorously the concepts of measurable set and measurable function. Rather we think of measure theory intuitively as described in the following paragraphs. Our rationale for this is that the details of Lebesgue measure and integration can often be bypassed in most of the material we present in this text. Measurable subsets of R include the standard open and closed intervals with which we are familiar. Multivariable extensions of intervals to Rd are also measurable, together with countable unions and intersections of them. Intuitively, the measure of a set D ⊆ Rd is its “length,” “area,” “volume,” or suitable generalization; and we denote the measure of D by meas(D). For a formal discussion of measurable set, see Royden [141] or Rudin [142]. To introduce the concept of measurable function, we begin by defining a step function. A function v on a measurable set D is a step function if D can be decomposed into a finite number of pairwise disjoint measurable subsets D1, . . . , Dk with v(x) constant over each Dj. We say a function v on D is a measurable function if it is the pointwise limit of a sequence of step functions. This includes, for example, all continuous functions on D. For each such measurable set Dj, we define a characteristic function χj(x) = 1, x ∈ Dj, 0, x / ∈ Dj. A general step function over the decomposition D1, . . ., Dk of D can then be written v(x) = k j=1 αjχj(x), x ∈ D (1.2.9) with α1, . . . , αk scalars. For a general measurable function v over D, we write it as a limit of step functions vk over D: v(x) = lim k→∞ vk(x), x ∈ D. (1.2.10) We say two measurable functions are equal almost everywhere if the set of points on which they differ is a set of measure zero. For notation, we write v = w (a.e.)
- 35. 14 1. Linear Spaces to indicate that v and w are equal almost everywhere. Given a measur- able function v on D, we introduce the concept of an equivalence class of equivalent functions: [v] = {w | w measurable on D and v = w (a.e.)} . For most purposes, we generally consider elements of an equivalence class [v] as being a single function v. We define the Lebesgue integral of a step function v over D, given in (1.2.9), by D v(x) dx = k j=1 αj meas(Dj). For a general measurable function, given in (1.2.10), define the Lebesgue integral of v over D by D v(x) dx = lim k→∞ D vk(x) dx. There are a great many properties of Lebesgue integration, and we refer the reader to any text on real analysis for further details. Note that the Lebesgue integrals of elements of an equivalence class [v] are identical. Let Ω be an open measurable set in Rd . Introduce Lp (Ω) = [v] | v measurable on Ω and vp ∞ . The norm vp is defined as in (1.2.8), although now we use Lebesgue integration rather than Riemann integration. L∞ (Ω) = {[v] | v measurable on Ω and v∞ ∞} . For v measurable on Ω, define v∞ = ess sup x∈Ω |v(x)| ≡ inf meas(Ω)=0 sup x∈ΩΩ |v(x)| , where “ meas(Ω ) = 0” means Ω is a measurable set with measure zero. The spaces Lp (Ω), 1 ≤ p ∞, are Banach spaces, and they are concrete realizations of the abstract completion of C(Ω) under the norm of (1.2.8). The space L∞ (Ω) is also a Banach space, but it is much larger than the space C(Ω) with the ∞-norm · ∞. Additional discussion of the spaces Lp (Ω) is given in Section 1.5. Example 1.2.21 More generally, let w be a positive measurable function on Ω, called a weight function. We can define weighted spaces Lp w(Ω) as
- 36. 1.2. Normed spaces 15 follows: Lp w(Ω) = v measurable Ω w(x) |v(x)|p dx ∞ , p ∈ [1, ∞), L∞ w (Ω) = {v measurable | ess supΩ w(x) |v(x)| ∞} . These are Banach spaces with the norms vp,w = Ω w(x) |v(x)|p dx 1/p , p ∈ [1, ∞), vp,∞ = ess sup x∈Ω w(x)|v(x)|. The space C(Ω) of Example 1.1.2(c) with the norm vC(Ω) = max x∈Ω |v(x)| is also a Banach space, and it can be considered as a proper subset of L∞ (Ω). See Example 2.5.3 for a situation where it is necessary to distin- guish between C(Ω) and the subspace of L∞ (Ω) to which it is isometric and isomorphic. Example 1.2.22 (a) For any integer m ≥ 0, the normed spaces Cm [a, b] and Ck p (2π) of Example 1.2.4(b) are Banach spaces. (b) Let 1 ≤ p ∞. As an alternative norm on Cm [a, b], introduce f = m j=0 f(j) p p 1 p . The space Cm [a, b] is not complete with this norm. Its completion is denoted by Wm,p (a, b), and it is an example of a Sobolev space. It can be shown that if f ∈ Wm,p (a, b), then f, f , . . . , f(m−1) are continuous, and f(m) exists almost everywhere and belongs to Lp (a, b). This and its multivariable generalizations are discussed at length in Chapter 6. A knowledge of the theory of Lebesgue measure and integration is very helpful in dealing with problems defined on spaces of such functions. Nonetheless, many results can be proven by referring to only the origi- nal space and its associated norm, say, C(Ω) with · p, from which a Banach space is obtained by a completion argument, say Lp (Ω). We return to this in Theorem 2.4.1 of Chapter 2. Exercise 1.2.1 Prove the backward triangle inequality of (1.2.4). Exercise 1.2.2 Show that · ∞ is a norm on C(Ω), with Ω a bounded open set in Rd . Exercise 1.2.3 Show that · ∞ is a norm on L∞ (Ω), with Ω a bounded open set in Rd .
- 37. 16 1. Linear Spaces Exercise 1.2.4 Show that · 1 is a norm on L1 (Ω), with Ω a bounded open set in Rd . Exercise 1.2.5 Show that for 1 ≤ p ≤ ∞, xp defined by (1.2.2)–(1.2.3) is a norm in the space Rd . The main task is to verify the triangle inequality, which can be done by first proving the Hölder inequality, |x·y| ≤ xpyp , x, y ∈ Rd . Here p is the conjugate of p defined through the relation 1/p + 1/p = 1; by convention, p = 1 if p = ∞, p = ∞ if p = 1. Exercise 1.2.6 Define Cα [a, b], 0 α ≤ 1, as the set of all f ∈ C[a, b] for which Mα(f) ≡ sup a≤x,y≤b x=y |f(x) − f(y)| |x − y| α ∞. Define fα = f∞ + Mα(f). Show Cα [a, b] with this norm is complete. Exercise 1.2.7 Define Cb[0, ∞) as the set of all functions f that are continuous on [0, ∞) and satisfy f∞ ≡ sup x≥0 |f(x)| ∞. Show Cb[0, ∞) with this norm is complete. Exercise 1.2.8 Does the formula (1.2.2) define a norm on Rd for 0 p 1? Exercise 1.2.9 Prove the equivalence on C1 [0, 1] of the following norms: fa ≡ |f(0)| + 1 0 |f (x)| dx, fb ≡ 1 0 |f(x)| dx + 1 0 |f (x)| dx. Hint: You may need to use the integral mean value theorem: Given g ∈ C[0, 1], there is ξ ∈ [0, 1] such that 1 0 g(x) dx = g(ξ). Exercise 1.2.10 Let V1 and V2 be normed spaces with norms · 1 and · 2. Recall that the product space V1 × V2 is defined by V1 × V2 = {(v1, v2) | v1 ∈ V1, v2 ∈ V2}. Show that the quantities max{v11, v22} and (v1p 1 + v2p 2)1/p , 1 ≤ p ∞ all define norms on the space V1 × V2. Exercise 1.2.11 Over the space C1 [0, 1], determine which of the following is a norm, and which is only a semi-norm: (a) max 0≤x≤1 |u(x)|;
- 38. 1.2. Normed spaces 17 (b) max 0≤x≤1 [|u(x)| + |u (x)|]; (c) max 0≤x≤1 |u (x)|; (d) |u(0)| + max 0≤x≤1 |u (x)|; (e) max 0≤x≤1 |u (x)| + 0.2 0.1 |u(x)| dx. Exercise 1.2.12 Over a normed space (V, · ), we define a function of two variables d(u, v) = u−v. Then d(·, ·) is a distance function; in other words, d(·, ·) has the following properties of an ordinary distance between two points: (a) d(u, v) ≥ 0 for any u, v ∈ V , and d(u, v) = 0 if and only if u = v; (b) d(u, v) = d(v, u) for any u, v ∈ V ; (c) (the triangle inequality) d(u, w) ≤ d(u, v) + d(v, w) for any u, v, w ∈ V . A linear space endowed with a distance function is called a metric space. Certainly a normed space can be viewed as a metric space. There are ex- amples of metrics (distance functions) that are not generated by any norm, though. Exercise 1.2.13 Let V be a normed space and {un} a Cauchy sequence. Suppose there is a subsequence {un } ⊆ {un} and some element v ∈ V such that un → u as n → ∞. Show that un → u as n → ∞. Exercise 1.2.14 Let V be a normed space, V0 ⊆ V a subspace. The quotient space V/V0 is defined to be the space of all the classes [v] = {v + v0 | v0 ∈ V0}. Prove that the formula [v]V/V0 = inf v0∈V0 v + v0V defines a norm on V/V0. Show that if V is a Banach space and V0 ⊆ V is a closed subspace, then V/V0 is a Banach space. Exercise 1.2.15 Assuming a knowledge of Lebesgue integration, show that W1,2 (a, b) ⊆ C[a, b]. Generalize this result to the space Wm,p (a, b) with other values of m and p. Hint: For v ∈ W1,2 (a, b), use v(x) − v(y) = y x v (z) dz .
- 39. 18 1. Linear Spaces Exercise 1.2.16 On C1 [0, 1], define (u, v)∗ = u(0) v(0) + 1 0 u (x) v (x) dx and u∗ = (u, u)∗ . Show that u∞ ≤ c u∗ ∀ u ∈ C1 [0, 1] for a suitably chosen constant c. 1.3 Inner product spaces In studying linear problems, inner product spaces are usually used. These are the spaces where a norm can be defined through the inner product and the notion of orthogonality of two elements can be introduced. The inner product in a general space is a generalization of the usual scalar product in the plane R2 or the space R3 . Definition 1.3.1 Let V be a linear space over K = R or C . An inner product (·, ·) is a function from V × V to K with the following properties. 1. For any u ∈ V , (u, u) ≥ 0 and (u, u) = 0 if and only if u = 0. 2. For any u, v ∈ V , (u, v) = (v, u). 3. For any u, v, w ∈ V , any α, β ∈ K, (α u+β v, w) = α (u, w)+β (v, w). The space V together with the inner product (·, ·) is called an inner product space. When the definition of the inner product (·, ·) is clear from the con- text, we simply say V is an inner product space. When K = R, V is called a real inner product space, while if K = C, V is a complex inner product space. In the case of a real inner product space, the second axiom reduces to the symmetry of the inner product: (u, v) = (v, u) ∀ u, v ∈ V. For an inner product, there is an important property called the Schwarz inequality. Theorem 1.3.2 (Schwarz inequality) If V is an inner product space, then |(u, v)| ≤ (u, u) (v, v) ∀ u, v ∈ V, and the equality holds if and only if u and v are linearly dependent.
- 40. 1.3. Inner product spaces 19 Proof. We give the proof only for the real case. The result is obviously true if either u = 0 or v = 0. Now suppose u = 0, v = 0. Define φ(t) = (u + t v, u + t v) = (u, u) + 2 (u, v) t + (v, v) t2 , t ∈ R . The function φ is quadratic and non-negative, so its discriminant must be non-positive, [2 (u, v)]2 − 4 (u, u) (v, v) ≤ 0; i.e., the Schwarz inequality is valid. For v = 0, the equality holds if and only if u = −t v for some t ∈ R . An inner product (·, ·) induces a norm through the formula u = (u, u), u ∈ V. In verifying the triangle inequality for the quantity thus defined, we need to use the above Schwarz inequality. Proposition 1.3.3 An inner product is continuous with respect to its in- duced norm. In other words, if · is the norm defined by u = (u, u), then un − u → 0 and vn − v → 0 as n → ∞ imply (un, vn) → (u, v) as n → ∞. In particular, if un → u, then for any v, (un, v) → (u, v) as n → ∞. Proof. Since {un} and {vn} are convergent, they are bounded; i.e., for some M ∞, un ≤ M, vn ≤ M for any n. We write (un, vn) − (u, v) = (un − u, vn) + (u, vn − v). Using the Schwarz inequality, we have |(un, vn) − (u, v)| ≤ un − u vn + u vn − v ≤ M un − u + u vn − v. Hence the result holds. Commonly seen inner product spaces are usually associated with their canonical inner products. As an example, the canonical inner product for the space Rd is (x, y) = d i=1 xiyi = yT x, ∀ x = (x1, . . . , xd)T , y = (y1, . . . , yd)T ∈ Rd . This inner product induces the Euclidean norm x = ! ! d i=1 |xi|2 = (x, x).
- 41. 20 1. Linear Spaces When we talk about the space Rd , implicitly we understand the inner product and the norm are the ones defined above, unless stated otherwise. For the complex space Cd , the inner product and the corresponding norm are (x, y) = d i=1 xiyi = y∗ x, ∀ x = (x1, . . . , xd)T , y = (y1, . . . , yd)T ∈ Cd and x = ! ! d i=1 |xi|2 = (x, x). The space L2 (Ω) is an inner product space with the canonical inner product (u, v) = Ω u(x) v(x) dx. This inner product induces the standard L2 (Ω)-norm u2 = # Ω |u(x)|2dx = (u, u). We have seen that an inner product induces a norm, which is always the norm we use on the inner product space unless stated otherwise. It is easy to show that on a complex inner product space, (u, v) = 1 4 [u + v2 − u − v2 + iu + iv2 − iu − iv2 ], and on a real inner product space, (u, v) = 1 4 [u + v2 − u − v2 ]. (1.3.1) These relations are called the polarization identities. Thus in any normed linear space, there can exist at most one inner product that generates the norm. On the other hand, not every norm can be defined through an inner product. We have the following characterization for any norm induced by an inner product. Theorem 1.3.4 A norm · on a linear space V is induced by an inner product if and only if it satisfies the parallelogram law: u + v2 + u − v2 = 2u2 + 2v2 ∀ u, v ∈ V. (1.3.2) Note that if u and v form two adjacent sides of a parallelogram, then u + v and u − v represent the lengths of the diagonals of the parallelo- gram. This theorem can be considered to be a generalization of the theorem of Pythagoras for right triangles.
- 42. 1.3. Inner product spaces 21 Proof. We prove the result for the case of a real space only. Assume · = (·, ·) for some inner product (·, ·). Then for any u, v ∈ V , u + v2 + u − v2 = (u + v, u + v) + (u − v, u − v) = $ u2 + 2(u, v) + v2 % + $ u2 − 2(u, v) + v2 % = 2u2 + 2v2 . Conversely, assume the norm · satisfies the parallelogram law. For u, v ∈ V , let us define (u, v) = 1 4 $ u + v2 − u − v2 % and show that it is an inner product. First, (u, u) = 1 4 2u2 = u2 ≥ 0 and (u, u) = 0 if and only if u = 0. Second, (u, v) = 1 4 $ v + u2 − v − u2 % = (v, u). Finally, we show the linearity, which is equivalent to the following two relations: (u + v, w) = (u, w) + (v, w) ∀ u, v, w ∈ V and (α u, v) = α (u, v) ∀ u ∈ V, α ∈ R . We have (u, w) + (v, w) = 1 4 $ u + w2 − u − w2 + v + w2 − v − w2 % = 1 4 $ (u + w2 + v + w2 ) − (u − w2 + v − w2 ) % = 1 4 $1 2 (u + v + 2 w2 + u − v2 ) − 1 2 (u + v − 2 w2 + u − v2 ) % = 1 8 $ u + v + 2 w2 − u + v − 2 w2 % = 1 8 [ 2 (u + v + w2 + w2 ) − u + v2 − 2 (u + v − w2 + w2 ) + u + v2 ] = 1 4 $ u + v + w2 − u + v − w2 % = (u + v, w). The proof of the second relation is more involved. For fixed u, v ∈ V , let us define a function of a real variable f(α) = α u + v2 − α u − v2 .
- 43. 22 1. Linear Spaces We show that f(α) is a linear function of α. We have f(α) − f(β) = α u + v2 + β u − v2 − α u − v2 − β u + v2 = 1 2 $ (α + β) u2 + (α − β) u + 2 v2 % −1 2 $ (α + β) u2 + (α − β) u − 2 v2 % = 1 2 $ (α − β) u + 2 v2 − (α − β) u − 2 v2 % = 2 α−β 2 u + v2 − α−β 2 u − v2 ' = 2 f ( α−β 2 ) . Taking β = 0 and noticing f(0) = 0, we find that f(α) = 2 f (α 2 ) . Thus we also have the relation f(α) − f(β) = f(α − β). From the above relations, the continuity of f, and the value f(0) = 0, one concludes that (see Exercise 1.3.2) f(α) = c0α = α f(1) = α $ u + v2 − u − v2 % from which, we get the second required relation. 1.3.1 Hilbert spaces Among the inner product spaces, of particular importance are the Hilbert spaces. Definition 1.3.5 A complete inner product space is called a Hilbert space. From the definition, we see that an inner product space V is a Hilbert space if V is a Banach space under the norm induced by the inner product. Example 1.3.6 (Some examples of Hilbert spaces) (a) The Cartesian space Cd is a Hilbert space with the inner product (x, y) = d i=1 xiyi. (b) The space l2 = {x = {xi}i≥1 | ∞ i=1 |xi|2 ∞} is a linear space with α x + β y = {α xi + β yi}i≥1. It can be shown that (x, y) = ∞ i=1 xiyi
- 44. 1.3. Inner product spaces 23 defines an inner product on l2 . Furthermore, l2 becomes a Hilbert space under this inner product. (c) The space L2 (0, 1) is a Hilbert space with the inner product (u, v) = 1 0 u(x) v(x) dx. (d) The space L2 (Ω) is a Hilbert space with the inner product (u, v) = Ω u(x) v(x) dx. More generally, if w(x) is a positive function on Ω, then the space L2 w(Ω) = v measurable Ω |v(x)|2 w(x) dx ∞ is a Hilbert space with the inner product (u, v)w = Ω u(x) v(x) w(x) dx. This space is a weighted L2 space. Example 1.3.7 Recall the Sobolev space Wm,p (a, b) defined in Example 1.2.22. If we choose p = 2, then we obtain a Hilbert space. It is usually denoted by Hm (a, b) ≡ Wm,2 (a, b). The associated inner product is defined by (f, g)Hm = m j=0 ( f(j) , g(j) ) , f, g ∈ Hm (a, b) using the standard inner product (·, ·) of L2 (a, b). Recall from Exercise 1.2.15 that H1 (a, b) ⊆ C[a, b]. 1.3.2 Orthogonality With the notion of an inner product at our disposal, we can define the angle between two vectors u and v as follows: θ = arccos * (u, v) u v + . This definition makes sense because, by the Schwarz inequality (Theorem 1.3.2), the argument of arccos is between −1 and 1. The case of a right angle is particularly important. We see that two vectors u and v form a right angle if and only if (u, v) = 0. Definition 1.3.8 Two vectors u and v are said to be orthogonal if (u, v) = 0. An element v ∈ V is said to be orthogonal to a subset U ⊆ V , if (u, v) = 0 for any u ∈ U.
- 45. 24 1. Linear Spaces Definition 1.3.9 Let U be a subset of an inner product space V . We define its orthogonal complement to be the set U⊥ = {v ∈ V | (v, u) = 0 ∀ u ∈ U}. The orthogonal complement of any set is a closed subspace (cf. Exercise 1.3.7). Definition 1.3.10 Let V be an inner product space. (a) Suppose V is finite dimensional. A basis {v1, . . . , vn} of V is said to be an orthogonal basis if (vi, vj) = 0, 1 ≤ i = j ≤ n. If, additionally, vi = 1, 1 ≤ i ≤ n, then we say the basis is orthonormal, and we combine these conditions as (vi, vj) = δij ≡ 1, i = j, 0, i = j. (b) Suppose V is infinite dimensional normed space. We say V has a count- ably infinite basis if there is a sequence {vi}i≥1 ⊆ V for which the following is valid: For each v ∈ V , we can find scalars {αn,i}n i=1, n = 1, 2, . . . , such that , , , , , v − n i=1 αn,ivi , , , , , → 0 as n → ∞. The space V is also said to be separable. The sequence {vi}i≥1 is called a basis if any finite subset of the sequence is linearly independent. If V is an inner product space, and if the sequence {vi}i≥1 also satisfies (vi, vj) = δij, i, j ≥ 1, (1.3.3) then {vi}i≥1 is called an orthonormal basis for V . (c) We say that an infinite dimensional normed space V has a Schauder basis {vn}n≥1 if for each v ∈ V , it is possible to write v = ∞ n=1 αnvn (1.3.4) as a convergent series in V for a unique choice of scalars {αn}n≥1. For a discussion of the distinction between V having a Schauder basis and V being separable, see [103, p.68]. For V an inner product space, it is straightforward to show that an orthonormal basis {vn}n≥1 is also a Schauder basis, and therefore (1.3.4) is valid for an orthonormal basis. The advantage of using an orthogonal or an orthonormal basis is that it is easy to decompose a vector as a linear combination of the basis elements. Assuming {vn}n≥1 is an orthonormal basis of V , let us determine the coefficients
- 46. 1.3. Inner product spaces 25 {αn}n≥1 in the decomposition (1.3.4) for any v ∈ V . By the continuity of the inner product and the orthonormality condition (1.3.3), we have (v, vk) = ∞ n=1 αn(vn, vk) = αk. Thus v = ∞ n=1 (v, vn) vn. (1.3.5) In addition, by direct computation using (1.3.3), , , , , , N n=1 (v, vn) vn , , , , , 2 = N n=1 |(v, vn)|2 . Using the convergence of (1.3.5) in V , we can let N → ∞ to obtain v = ! ! ∞ n=1 |(v, vn)|2. (1.3.6) A simple consequence of the identity (1.3.6) is the inequality N n=1 |(v, vn)|2 ≤ v2 , N ≥ 1, v ∈ V. (1.3.7) The decomposition (1.3.5) can be termed as the generalized Fourier series; then the identity (1.3.6) can be called the generalized Parseval identity, whereas (1.3.7) can be called the generalized Bessel inequality. Example 1.3.11 Let V = L2 (0, 2 π) with complex scalars. The complex exponentials vn(x) = 1 √ 2 π einx , n = 0, ±1, ±2, . . . (1.3.8) form an orthonormal basis. For any v ∈ L2 (0, 2 π), we have the Fourier series expansion v(x) = ∞ n=−∞ αnvn(x) (1.3.9) where αn = (v, vn) = 1 √ 2 π 2 π 0 v(x) e−inx dx. (1.3.10) Also (1.3.6) and (1.3.7) reduce to the ordinary Parseval identity and Bessel inequality.
- 47. 26 1. Linear Spaces When a non-orthogonal basis for an inner product space is given, there is a standard procedure to construct an orthonormal basis. Theorem 1.3.12 (Gram-Schmidt method) Let {wn}n≥1 be a basis of the inner product space V . Then there is an orthonormal basis {vn}n≥1 with the property that span {wn}N n=1 = span {vn}N n=1 ∀ N ≥ 1. Proof. The proof is done inductively. For N = 1, define v1 = w1 w1 , which satisfies v1 = 1. For N ≥ 2, assume {vn}N−1 n=1 have been constructed with (vn, vm) = δnm, 1 ≤ n, m ≤ N − 1, and span {wn}N−1 n=1 = span {vn}N−1 n=1 . Write ṽN = wN + N−1 n=1 αN,nvn. Now choose {αN,n}N−1 n=1 by setting (ṽN , vn) = 0, 1 ≤ n ≤ N − 1. This implies αN,n = −(wN , vn), 1 ≤ n ≤ N − 1. This procedure “removes” from wN the components in the directions of v1, . . . , vN−1. Finally, define vN = ṽN ṽN , which is meaningful since ṽN = 0. (Why?) Then the sequence {vn}N n=1 satisfies (vn, vm) = δnm, 1 ≤ n, m ≤ N and span {wn}N n=1 = span {vn}N n=1. The Gram-Schmidt method can be used, e.g., to construct an orthonor- mal basis in L2 (−1, 1) for a polynomial space of certain degrees. As a result we obtain the well-known Legendre polynomials (after a proper scaling), which play an important role in some numerical analysis problems.
- 48. 1.3. Inner product spaces 27 Example 1.3.13 Let us construct the first three orthonormal polynomials in L2 (−1, 1). For this purpose, we take w1(x) = 1, w2(x) = x, w3(x) = x2 . Then easily, v1(x) = w1(x) w1 = 1 √ 2 . To find v2(x), we write ṽ2(x) = w2(x) + α2,1v1(x) = x + 1 √ 2 α2,1 and choose α2,1 = −(x, 1 √ 2 ) = − 1 −1 1 √ 2 x dx = 0. So ṽ2(x) = x, and v2(x) = ṽ2(x) ṽ2 = - 3 2 x. Finally, we write ṽ3(x) = w3(x) + α3,1v1(x) + α3,2v2(x) = x2 + 1 √ 2 α3,1 + - 3 2 α3,2x. Then α3,1 = −(w3, v1) = − 1 −1 x2 1 √ 2 dx = − √ 2 3 , α3,2 = −(w3, v2) = − 1 −1 x2 - 3 2 x dx = 0. Hence ṽ3(x) = x2 − 1 3 . Since ṽ32 = 8 45 , we have v3(x) = 3 2 - 5 2 x2 − 1 3 . The fourth orthonormal polynomial is v4(x) = - 7 8 . 5x3 − 3x / . The graphs of these first four Legendre polynomials are given in Figure 1.2.
- 49. 28 1. Linear Spaces 1 −1 x y v 1 (x) v2 (x) v3 (x) v4 (x) Figure 1.2. Graphs on [−1, 1] of the orthonormal Legendre polynomials of degrees 0,1,2,3 As we see from Example 1.3.13, it is cumbersome to construct orthonor- mal (or orthogonal) polynomials directly. Fortunately, for many important cases of the weighted function w(x) and integration interval (a, b), formulas of orthogonal polynomials in the weighted space L2 w(a, b) are known (see Section 3.4). Exercise 1.3.1 Given an inner product, show that the formula u = (u, u) defines a norm. Exercise 1.3.2 Assume f : R → R is a continuous function, satisfying f(α) = f(β) + f(α − β) for any α, β ∈ R , and f(0) = 0. Then f(α) = α f(1). Solution: From f(α) = f(β) + f(α − β) and f(0) = 0, by an induction argument, we have f(n α) = n f(α) for any integer n. Then from f(α) = 2 f(α/2), we have f(1/2n ) = (1/2n ) f(1) for any integer n ≥ 0. Finally, for any integer m, any non-negative integer n, f(m 2−n ) = m f(2−n ) = (m 2−n ) f(1). Now any rational can be represented as a finite sum q = i mi 2−i . Hence, f(q) = i f(mi 2−i ) = i mi 2−i f(1) = q f(1). Since the set of the rational numbers is dense in R and f is a continuous function, we see that for any real ξ, f(ξ) = ξ f(1).
- 50. 1.3. Inner product spaces 29 Exercise 1.3.3 The norms · p, 1 ≤ p ≤ ∞, over the space Rd are defined in Example 1.2.3. Find all the values of p for which the norm · p is induced by an inner product. Hint: Apply Theorem 1.3.4. Exercise 1.3.4 Let w1, . . . , wd be positive constants. Show that the for- mula (x, y) = d i=1 wixiyi defines an inner product on Rd . This is an example of a weighted inner product. What happens if we only assume wi ≥ 0, 1 ≤ i ≤ d? Exercise 1.3.5 Let A ∈ Rd×d be a symmetric, positive definite matrix and let (·, ·) be the Euclidean inner product on Rd . Show that the quantity (Ax, y) defines an inner product on Rd . Exercise 1.3.6 Show that in an inner product space, u + v = u + v for some u, v ∈ V if and only if u and v are non-negatively linearly dependent (i.e., for some c0 ≥ 0, either u = c0v or v = c0u). Exercise 1.3.7 Prove that the orthogonal complement of a subset is a closed subspace. Exercise 1.3.8 Let V0 be a subset of a Hilbert space V . Show that the following statements are equivalent: (a) V0 is dense in V ; i.e., for any v ∈ V , there exists {vn}n≥1 ⊆ V0 such that v − vnV → 0 as n → ∞. (b) V ⊥ 0 = {0}. (c) If u ∈ V satisfies (u, v) = 0 ∀ v ∈ V0, then u = 0. (d) For every 0 = u ∈ V , there is a v ∈ V0 such that (u, v) = 0. Exercise 1.3.9 On C1 [a, b], define (f, g)∗ = f(a)g(a) + 1 0 f (x)g (x) dx, f, g ∈ C1 [a, b] and f∗ = (f, f)∗. Show that f∞ ≤ c f∗ ∀ f ∈ C1 [a, b] for a suitable constant c. Exercise 1.3.10 Consider the Fourier series (1.3.9) for a function v ∈ Cm p (2π) with m ≥ 2. Show that , , , , , v − N n=−N αnvn , , , , , ∞ ≤ cm(v) Nm−1 , N ≥ 1.
- 51. 30 1. Linear Spaces Hint: Use integration by parts in (1.3.10). 1.4 Spaces of continuously differentiable functions Spaces of continuous functions and continuously differentiable functions were introduced in Example 1.1.2. In this section, we provide a more detailed review of these spaces. Let Ω be an open bounded subset of Rd . A typical point in Rd is denoted by x = (x1, . . . , xd)T . For multivariable functions, it is convenient to use the multi-index notation for partial derivatives. A multi-index is an ordered collection of d non-negative integers, α = (α1, . . . , αd). The quantity |α| = d i=1 αi is said to be the length of α. If v is an m-times differentiable function, then for any α with |α| ≤ m, Dα v(x) = ∂|α| v(x) ∂xα1 1 · · · ∂xαd d is the αth order partial derivative. This is a handy notation for partial derivatives. Some examples are ∂v ∂x1 = Dα v for α = (1, 0, . . . , 0), ∂d v ∂x1 · · · ∂xd = Dα v for α = (1, 1, . . . , 1). The set of all the derivatives of order m of a function v can be written as {Dα v | |α| = m}. For low-order partial derivatives, there are other commonly used notations; e.g., the partial derivative ∂v/∂xi is also written as ∂xi v, or ∂iv, or v,xi , or v,i. The space C(Ω) consists of all real-valued functions that are continuous on Ω. Since Ω is open, a function from the space C(Ω) is not necessarily bounded. For example, with d = 1 and Ω = (0, 1), the function v(x) = 1/x is continuous but unbounded on (0, 1). Indeed, a function from the space C(Ω) can behave “nastily” as the variable approaches the boundary of Ω. Usually, it is more convenient to deal with continuous functions that are continuous up to the boundary. Let C(Ω) be the space of functions that are uniformly continuous on Ω. Any function in C(Ω) is bounded. The notation C(Ω) is consistent with the fact that a uniformly continuous function on Ω has a unique continuous extension to Ω. The space C(Ω) is a Banach space with its canonical norm vC(Ω) = sup{|v(x)| | x ∈ Ω} ≡ max{|v(x)| | x ∈ Ω}. We have C(Ω) ⊆ C(Ω), and the inclusion is proper; i.e., there are functions v ∈ C(Ω) that cannot be extended to a continuous function on Ω. A simple example is v(x) = 1/x on (0, 1).
- 52. 1.4. Spaces of continuously differentiable functions 31 Denote by Z+ the set of non-negative integers. For any m ∈ Z+, Cm (Ω) is the space of functions that, together with their derivatives of order less than or equal to m, are continuous on Ω; that is, Cm (Ω) = {v ∈ C(Ω) | Dα v ∈ C(Ω) for |α| ≤ m}. This is a linear space. The notation Cm (Ω) denotes the space of functions which, together with their derivatives of order less than or equal to m, are continuous up to the boundary, Cm (Ω) = {v ∈ C(Ω) | Dα v ∈ C(Ω) for |α| ≤ m}. The space Cm (Ω) is a Banach space with the norm vCm(Ω) = max |α|≤m Dα vC(Ω). Algebraically, Cm (Ω) ⊆ Cm (Ω). When m = 0, we usually write C(Ω) and C(Ω) instead of C0 (Ω) and C0 (Ω). We set C∞ (Ω) = ∞ 0 m=0 Cm (Ω) ≡ {v ∈ C(Ω) | v ∈ Cm (Ω) ∀ m ∈ Z+}, C∞ (Ω) = ∞ 0 m=0 Cm (Ω) ≡ {v ∈ C(Ω) | v ∈ Cm (Ω) ∀ m ∈ Z+}. These are spaces of infinitely differentiable functions. Given a function v on Ω, its support is defined to be support v = {x ∈ Ω | v(x) = 0}. We say that v has a compact support if support v is a proper subset of Ω: support v ⊂ Ω. Thus, if v has a compact support, then there is a neighbor- ing open strip about the boundary ∂Ω such that v is zero on the part of the strip that lies inside Ω. Later on, we need the space C∞ 0 (Ω) = {v ∈ C∞ (Ω) | support v ⊂ Ω}. Obviously, C∞ 0 (Ω) ⊆ C∞ (Ω). In the case Ω is an interval such that Ω ⊃ (−1, 1), a standard example of a non-analytic C∞ 0 (Ω) function is v(x) = e1/(x2 −1) , |x| 1, 0, otherwise. 1.4.1 Hölder spaces A function v defined on Ω is said to be Lipschitz continuous if for some constant c, there holds the inequality |v(x) − v(y)| ≤ c x − y ∀ x, y ∈ Ω. In this formula, x − y denotes the standard Euclidean distance between x and y. The smallest possible constant in the above inequality is called the
- 53. 32 1. Linear Spaces Lipschitz constant of v, and is denoted by Lip(v). The Lipschitz constant is characterized by the relation Lip(v) = sup |v(x) − v(y)| x − y x, y ∈ Ω, x = y . More generally, a function v is said to be Hölder continuous with exponent β ∈ (0, 1] if for some constant c, |v(x) − v(y)| ≤ c x − y β for x, y ∈ Ω. The Hölder space C0,β (Ω) is defined to be the subspace of C(Ω) that con- sists of functions that are Hölder continuous with the exponent β. With the norm vC0,β (Ω) = vC(Ω) + sup |v(x) − v(y)| x − y β x, y ∈ Ω, x = y the space C0,β (Ω) becomes a Banach space. When β = 1, the Hölder space C0,1 (Ω) consists of all the Lipschitz continuous functions. For m ∈ Z+ and β ∈ (0, 1], we similarly define the Hölder space Cm,β (Ω) = 1 v ∈ Cm (Ω) | Dα v ∈ C0,β (Ω) for all α with |α| = m 2 ; this is a Banach space with the norm vCm,β (Ω) = vCm(Ω) + |α|=m sup |Dα v(x) − Dα v(y)| x − y β x, y ∈ Ω, x = y . Exercise 1.4.1 Show that C(Ω) with the norm vC(Ω) is a Banach space. Exercise 1.4.2 Show that the space C1 (Ω) with the norm vC(Ω) is not a Banach space. Exercise 1.4.3 Let vn(x) = 1 n sin nx. Show that vn → 0 in C0,β [0, 1] for any β ∈ (0, 1), but vn 0 in C0,1 [0, 1]. Exercise 1.4.4 Discuss whether it is meaningful to use the Hölder space C0,β (Ω) with β 1. Exercise 1.4.5 Consider v(s) = sα for some 0 α 1. For which β ∈ (0, 1] is it true that v ∈ C0,β [0, 1]? 1.5 Lp spaces In the study of Lp (Ω) spaces, we identify functions (i.e., such functions are considered identical) that are equal almost everywhere (a.e.) on Ω. For
- 54. 1.5. Lp spaces 33 p ∈ [1, ∞), Lp (Ω) is the linear space of measurable functions v : Ω → R such that vLp(Ω) = Ω |v(x)|p dx 1/p ∞. (1.5.1) The space L∞ (Ω) consists of all essentially bounded measurable functions v : Ω → R , vL∞(Ω) = inf meas (Ω)=0 sup x∈ΩΩ |v(x)| ∞. (1.5.2) Some basic properties of the Lp spaces are summarized in the following theorem. Theorem 1.5.1 Let Ω be an open bounded set in Rd . (a) For p ∈ [1, ∞], Lp (Ω) is a Banach space. (b) For p ∈ [1, ∞], every Cauchy sequence in Lp (Ω) has a subsequence that converges pointwise a.e. on Ω. (c) If 1 ≤ p ≤ q ≤ ∞, then Lq (Ω) ⊆ Lp (Ω), vLp(Ω) ≤ meas (Ω) 1 p − 1 q vLq(Ω) ∀ v ∈ Lq (Ω), and vL∞(Ω) = lim p→∞ vLp(Ω) ∀ v ∈ L∞ (Ω). (d) If 1 ≤ p ≤ r ≤ q ≤ ∞ and we choose θ ∈ [0, 1] such that 1 r = θ p + (1 − θ) q , then vLr(Ω) ≤ vθ Lp(Ω)v1−θ Lq(Ω) ∀ v ∈ Lq (Ω). In (c), when q = ∞, 1/q is understood to be 0. The result (d) is called an interpolation property of the Lp spaces. To prove (c) and (d), we need to use the Hölder inequality. We first prove Young’s inequality. Lemma 1.5.2 (Young’s inequality) Let a, b ≥ 0, p, q 1, 1/p+1/q = 1. Then ab ≤ ap p + bq q . Proof. For any fixed b ≥ 0, define a function f(a) = ap p + bq q − ab on [0, ∞). From f (a) = 0 we obtain a = b1/(p−1) . We have f(b1/(p−1) ) = 0. Since f(0) ≥ 0, lima→∞ f(a) = ∞ and f is continuous on [0, ∞), we see
- 55. Exploring the Variety of Random Documents with Different Content
- 56. sorts of combustible material ready to be kindled at a moment’s notice. The moment he reached it he fell, deprived of consciousness. The Bohemian, who had been watching his every movement, then appeared just where the foot-path entered the esplanade, and advanced with the greatest circumspection. Hiding himself behind the cabin, he listened, and heard only the laboured breathing of the watchman. Certain of the effect of his soporific, he approached Peyrou, stooped down, and touched his hands and his forehead and found that they were cold. “The dose is strong,” said he, “perhaps too strong. So much the worse, I did not wish to kill him.” Then advancing to the edge of the precipice, he saw distinctly the three pirate vessels in the distance. Moving slowly and cautiously, for fear of being discovered, they made use of oars to reach the entrance of the port, where the Bohemian was to join them. The practised eye of Hadji recognised in front of the two galleys certain luminous points or flames, which were nothing else than torches designed to burn the city and the fishing-boats. “By Eblis! they are going to smoke these citizens like foxes in their burrows. It is time, perhaps, for this old man to go to sleep for ever; but we must visit his cabin. I will have time to descend. I will be on the beach soon enough to seize a boat and join Captain Pog, who expects me before he begins the attack. Let us enter; they say the old man hides a treasure here.” Hadji took a brand from the fireplace and lit a lamp. The first object which met his eye was a trunk or box of sculptured ebony placed near the watchman’s bed. “That is a costly piece of furniture for such a recluse.” Not finding a key, he took a hatchet, broke open the lock, and opened the two leaves of the door; the shelves were empty. “It is not natural to lock up nothing with so much precaution; time presses, but this key will open everything.” He took up the hatchet again, and in a moment the ebony case was in pieces.
- 57. A double bottom fell apart. The Bohemian uttered a cry of joy as he perceived the little embossed silver casket of which we have spoken, and on which was marked a Maltese cross. This casket, which was quite heavy, was fastened no doubt by a secret spring, as neither key nor lock could be discovered. “I have my fine part of the booty, now let us run to help Captain Pog in taking his. Ah, ah!” added he, with a diabolical laugh, as he beheld the bay and the city wrapped in profound stillness, “soon Eblis will shake his wings of fire over that scene. The sky will be in flames, and the waters will run with blood!” Then, as a last precaution, he emptied a tunnel of water on the signal pile, and descended in hot haste to join the pirate vessels.
- 58. CHAPTER XXIX. CHRISTMAS While so many misfortunes were threatening the city, the inhabitants were quietly keeping Christmas. Notwithstanding the uneasiness the opinion of the watchman had given, notwithstanding the alarm caused by terror of the pirates, in every house, poor or rich, preparations were being made for the patriarchal feast. We have spoken of the magnificent cradle which had long been in course of preparation through the untiring industry of Dame Dulceline. It was at last finished and placed in the hall of the dais, or hall of honour in Maison-Forte. Midnight had just sounded. The woman in charge was impatiently awaiting the return of Raimond V., his daughter, Honorât de Berrol, and other relations and guests whom the baron had invited to the ceremony. All the family and guests had gone to La Ciotat, to be present at the midnight mass. Abbé Mascarolus had said mass in the chapel of the castle for those who had remained at home. We will conduct the reader to the hall of the dais, which occupied two-thirds of the long gallery which communicated with the two wings of the castle. It was never opened except on solemn occasions. A splendid red damask silk covered its walls. To supply the place of flowers, quite rare in that season, masses of green branches, cut from trees and arranged in boxes, hid almost entirely the ten large arched windows of this immense hall.
- 59. At one end of the hall rose a granite chimneypiece, ten feet high and heavily sculptured. Notwithstanding the season was cold, no fire burned in this vast fireplace, but an immense pile, composed of branches of vine, beech, olive, and fir-apples, only waited the formality of custom to throw waves of light and heat into the grand and stately apartment. Two pine-trees with long green branches ornamented with ribbons, oranges, and bunches of grapes, were set up in boxes on each side of the chimney, and formed above the mantelpiece a veritable thicket of verdure. Six copper chandeliers with lighted yellow wax candles only partially dissipated the darkness of the immense room. At the other end, opposite the chimney, rose the dais, resembling somewhat the canopy of a bed, with curtains, hangings, and cushions of red damask, as were, too, the mantle and gloves, a part of the equipment of office. The red draperies covered, with their long folds, five wooden steps, which were hidden under a rich Turkey carpet. Ordinarily the armorial chair of Raimond V. was placed on this elevation, and here enthroned, the old gentleman, as lord of the manor, administered on rare occasions justice to high and low. On Christmas Day, however, the cradle of the infant Jesus occupied this place of honour. A table of massive oak, covered over with a rich oriental drapery, furnished the middle of the gallery. On this table could be seen an ebony box handsomely carved, with a coat of arms on its lid. This box contained the book of accounts, a sort of record in which were written the births and all other important family events. Armchairs and benches of carved oak, with twisted feet, completed the furniture of this hall, to which its size and severe bareness gave an imposing character.
- 60. Dame Dulceline and Abbé Mascarolus had just finished placing the cradle under the dais. This marvel was a picture in relief about three feet square at the base and three feet high. The faithful representation of the stable where the Saviour was born would have been too severe a limitation to the poetical conceptions of the good abbé. So, instead of a stable, the holy scene was pictured under a sort of arcade sustained by two half ruined supports. In the spaces between the stones, real little stones artistically cut, were hung long garlands of natural vines and leaves, most beautifully intertwined. A cloud of white wax seemed to envelope the upper part of the arcade. Five or six cherubs about a thumb high, modelled in wax painted a natural colour, and wearing azure wings made of the feathers of humming-birds, were here and there set in the cloud, and held a streamer of white silk, in the middle of which glittered the words, embroidered in letters of gold: Gloria in Jezcelriir. The supports of the arcade rested on a sort of carpet of fine moss, packed so closely as to resemble green velvet, and in front of this erection was placed the cradle of the Saviour of the world; a real, miniature cradle, covered over with the richest laces. In it reposed the infant Jesus. Kneeling by the cradle, the Virgin Mary bent over the Babe her maternal brow, the white veil of the Queen of Angels falling over her feet and hiding half of her azure coloured silk robe. The paschal lamb, his four feet bound with a rose coloured ribbon, was laid at the foot of the cradle; behind it the kneeling ox thrust his large head, and his eyes of enamel seemed to contemplate the divine Infant. The ass, on a more distant plane, and half hidden by the posts of the arcade, behind which it stood, also showed his meek and gentle head. The dog seemed to cringe near the cradle, while the shepherds, clothed in coarse cassocks, and the magi kings, dressed in rich robes of brocatelle, were offering their adoration.
- 61. A fourth row of little candles, made of rose-scented wax, burned around the cradle. An immense amount of work, and really great resources of imagination, had been necessary to perfect such an exquisite picture. For instance, the ass, which was about six thumbs in height, was covered in mouse-skin which imitated his own to perfection. The black and white ox owed his hair to an India pig of the same colour, and his short and polished black horns to the rounded nippers of an enormous beetle. The robes of the magi kings revealed a fairy-like skill and patience, and their long white hair was really veritable hair, which Dame Dulceline had cut from her own venerable head. As to the figures of the cherubs, the infant Jesus, and other actors in this holy scene, they had been purchased in Marseilles from one of those master wax-chandlers, who always kept assorted materials necessary in the construction of these cradles. Doubtless it was not high art, but there was, in this little monument of a laborious and innocent piety, something as simple and as pathetic as the divine scene which they tried to reproduce with such religious conscientiousness. The good old priest and Dame Dulceline, after having lit the last candles which surrounded the cradle, stood a moment, lost in admiration of their work. “Never, M. Abbé,” said Dame Dulceline, “have we had such a beautiful cradle at Maison-Forte.” “That is true, Dame Dulceline; the representation of the animals approaches nature as closely as is permitted man to approach the marvels of creation.” “Ah, M. Abbé, why did it have to be that the accursed Bohemian, who they say is an emissary of the pirates, should give us the secret of making glass eyes for these animals?” “What does it matter, Dame Dulceline? Perhaps some day the miscreant will learn the eternal truth. The Lord employs every arm to build his temple.”
- 62. “Pray tell me, M. Abbé, why we must put the cradle under the dais in the hall of honour. Soon it will be forty years since I began making cradles for Maison-Forte des Anbiez. My mother made them for Raimond IV., father of Raimond V., for as many years. Ah, well! I have never asked before, nor have I even asked myself why this hall was always selected for the blessed exposition.” “Ah, you see, Dame Dulceline, there is always, at the base of our ancient religious customs, something consoling for the humble, the weak, and the suffering, and also something imposing as a lesson for the happy and the rich and the powerful of this world. This cradle, for instance, is the symbol of the birth of the divine Saviour. He was the poor child of a poor artisan, and yet some day he was to be as far above the most powerful of men as the heavens are above the earth. So you see, Dame Dulceline, upon the anniversary day of the redemption, the poor and rustic cradle of the infant Saviour takes the place of honour in the ceremonial hall of the noble baron.” “Ah, I understand, M. Abbé, they put the infant Jesus in the place of the noble baron, to show that the lords of this world should be first to bow before the Saviour!” “Without doubt, Dame Dulceline, in thus doing homage to the Lord through the symbol of his power, the baron preaches by example the communion and equality of men before God.” Dame Dulceline remained silent a moment, thinking of the abbé's words, then, satisfied with his explanation, she proposed another question to him, which in her mind was more difficult of solution. “M. Abbé,” asked she, with an embarrassed air, “you say that at the base of all ancient customs there is always a lesson; can there be one, then, in the custom of Palm Sunday, when foundling children run about the streets of Marseilles with branches of laurel adorned with fruit? For instance, last year, on Palm Sunday,—I blush to think of it even now, M. Abbé,—I was walking on the fashionable promenade of Marseilles with Master Tale-bard-Talebardon, who was not then the declared enemy of monseigneur, and, lo! one of the unfortunate little foundlings stopped right before me and the consul,
- 63. and said, with a sweet voice, as he kissed our hands, ‘Good morning, mother! good morning, father!’ By St Dulceline, my patron saint, M. Abbé, I turned purple with shame, and Master Talebard- Talebardon did, too. I beg your pardon, respectfully, for alluding to the coarse jokes of Master Laramée, who accompanied us, on the subject of this poor foundling’s insult! But this Master Laramée has neither modesty nor shame. I could not help repulsing with horror this nursling of public charity, and I pinched his arm sharply, and said to him: ‘Will you be silent, you ugly little bastard?’ He felt his fault, for he began to weep, and when I complained of his indecent impudence to a grave citizen, he replied to me: ‘My good lady, such is the custom here; on Palm Sunday foundlings have the privilege of running through the streets, and saying, ‘father and mother,’ to all whom they may meet.” “That is really the custom, Dame Dulceline,” said the abbé. “Well, it may be the custom, M. Abbé, but is that not a very impertinent and improper custom, to permit unfortunate little children without father or mother to walk up and say ‘mother’ to honest, discreet persons like myself, for example, who prefer the peace of celibacy to the disquietudes of family? As to the morality of this custom, I pray you explain it, M. Abbé. I look for it in vain with all my eyes. I can see nothing in it but what is outrageously indecent!” “And you are mistaken, Dame Dulceline,” said Abbé Mascarolus; “this custom is worthy of respect, and you were wrong to treat that poor child so cruelly.” “I was wrong? That little rascal comes and calls me mother, and I permit it? Why, then, thanks to this custom, there would—” “Thanks to this custom,” interrupted the abbé, “thanks to the privilege that these little unfortunates have, of being able to say, one day in the year, ‘father and mother’ to those they meet,—those dear names that they never pronounce, which, perhaps, may have never passed their lips—alas! how many there are, and I have seen them, who say these words with tears in their eyes, as they remember
- 64. that, when that day is past, they cannot repeat the blessed words! And sometimes it happens, Dame Dulceline, that strangers, moved to pity by such innocence and sorrow, or being touched by the caressing words, have adopted some of these unfortunates; others have given abundant alms, because this innocent appeal for charity is almost always heard. You see, Dame Dulceline, that this custom, too, has a useful end,—a pious signification.” The old woman bowed her head in silence, and finally replied to the good chaplain: “You are a clever man, M. Abbé; you are right. See what it is to have knowledge! Now I repent of having repulsed the child so cruelly. Next Palm Sunday I will not fail to carry several yards of good, warm cloth, and nice linen, and this time, I promise you, I will not act the cruel stepmother with the poor children who call me mother! But if that old sot, Laramée, makes any indecent joke about me, as sure as he has eyes I will prove to him that I have claws!” “That would prove too much, Dame Dulceline. But, since monseigneur does not yet return, and since we are discussing the customs of our good old Provence, and their usefulness to poor people, come, now, what have you observed on the day of St Lazarus, concerning the dance of St Elmo?” “What do you want me to tell you, M. Abbé? Now I distrust myself; before your explanation I railed against the custom of foundlings on Palm Sunday, now I respect it.” “Say always, Dame Dulceline, that the sin of ignorance is excusable. But what is your opinion concerning the dance of St Elmo?” “Bless me, M. Abbé, I understand nothing about it! I sometimes ask myself what is the good, the day of the feast of St. Elmo, of dressing up, at the expense of the city or community, all the poor young boys and girls as handsomely as possible. That is not all. Not content with that, these young people go from house to house, among the rich citizens and the lords, asking to borrow something. This one wants a gold necklace, that one a pair of diamond earrings,
- 65. another a silver belt, another a hatband set with precious stones, or a sword-belt braided in gold. Ah, well! in my opinion,—but I may change it in an hour,—M. Abbé, it is wrong to lend all these costly articles to poor people and artisans who have not a cent.” “Why so? Since the feast of St. Lazarus has been celebrated here, have you ever heard, Dame Dulceline, that any of those precious jewels have been lost or stolen?” “Good God in Heaven! Never, M. Abbé, neither here, nor in Marseilles, nor in all Provence, I believe. Thank God, our youth is honest, after all! For instance, last year Mlle. Reine loaned her Venetian girdle, which Stephanette says cost more than two thousand crowns. Ah, well! Thereson, the daughter of the miller at Pointe-aux-Cailles, who wore this costly ornament during all the feast, came and brought it back before sunset, although she had permission to keep it till night. And for this same feast of St. Lazarus, monseigneur loaned to Pierron, the fisherman of Maison- Forte, his beautiful gold chain, and his medallion set with rubies, that Master Laramée cleans, as you told him to do, with teardrops of the vine.” “That is true; and if one can mix with these teardrops of the vine a tear of a stag killed in venison season, Dame Dulceline, the rubies will shine like sparks of fire.” “Ah, well, M. Abbé, Pierron, the fisherman, brought back faithfully that precious chain even before the appointed hour. I repeat, M. Abbé, our youth is an honest youth, but I do not see the use of risking the loss, not by theft, but by accident, of beautiful jewels, for the pleasure of seeing these young people dance the old Provençal dances in the streets and roads, to the sound of tambourines and cymbalettes and flutes, that play the national airs, ooubados and bedocheos, until you are deaf.” “Ah, well, Dame Dulceline,” said Mascarolus, smiling sweetly, “you are going to learn that you were wrong not to see in this custom, too, a lesson and a use. When mademoiselle loaned to Thereson, the poor daughter of a miller, a costly ornament, she showed a blind
- 66. confidence in the girl; now, Dame Dulceline, confidence begets honesty and prevents dishonesty. That is not all; in giving Thereson the pleasure of wearing this ornament for one day, our young mistress showed her at the same time the charm and the nothingness of it, and then, as this pleasure is not forbidden to the poor people, they do not look on it with jealousy. This custom, in fact, establishes delightful relations between rich and poor, which are based on probity, confidence, and community of interest What do you think now of the dance of St. Elmo, Dame Dulceline?” “I think, M. Chaplain, that, although I have no jewels but a cross and a gold chain, I will lend them with a good heart to young Madelon, the best worker in my laundry, on the next feast of St. Lazarus, because every time I take this gold cross out of its box the poor girl devours it with her eyes, and I am sure that she will be wild with joy. But I am getting bewildered, M. Abbé; I brought some pure oil to fill the two Christmas lamps, which mademoiselle is to light, and I was about to forget them.” “Speaking of oil, Dame Dulceline, do not forget to fill well with oil that jug in which I have steeped those two beautiful bunches of grapes. I wish to attempt the experiment cited by M. de Maucaunys.” “What experiment, M. Abbé?” “This erudite and veracious traveller pretends that by leaving bunches of grapes, gathered on the day which marks the middle of September, in a jug of pure oil for seven months, the oil will acquire such a peculiar property that whenever it burns in a lamp whose light is thrown on the wall or the floor, thousands of bunches of grapes will appear on this wall or floor, perfect in colour, but as deceptive as objects painted on glass.” Dame Dulceline was just about to testify her admiration for the good and credulous chaplain, when she heard in the court the sound of carriage and horses, which announced the return of Raimond V. She disappeared precipitately. The door opened, and Raimond V. entered the gallery with several ladies and gentlemen, friends and
- 67. their wives, who had also been present at the midnight mass in the parochial church of La Ciotat. The baron and the other men were in holiday attire, and the women in that dress which going and coming on horseback rendered necessary, inasmuch as carriages were very rare. Although the countenance of Raimond V. was always joyous and cordial when he welcomed his guests at Maison-Forte, an expression of sadness from time to time now came over his features, for he had relinquished all hope of seeing his brothers at this family festival. The guests of the baron all admired the cradle Dame Dulceline had prepared with so much skill, and the chaplain received the praises of the company with as much modesty as gratitude. Honorât de Berrol appeared more melancholy than ever. Reine, on the contrary, realising the necessity for making him forget the refusal of her hand, which she had at last decided upon, by means of various evidences of kindness and friendship, treated the young man with cousinly esteem and affection. Nevertheless, she was conscious of a painful embarrassment; she had not yet informed the baron of her determination not to marry Honorât de Berrol. She had only obtained her father’s consent to have the nuptials delayed until the return of the commander and Father Elzear, who, from what was implied in their last letters, might arrive at any moment. Eulogies on the cradle seemed inexhaustible, when the baron, approaching the company of admiring guests, said: “My opinion is, ladies, that we had better begin the cachofué, for this hall is very damp and cold, and the fire is only waiting to blaze!” The cachofué, or feu caché, was an old Provençal ceremony, which consisted of bringing in a Christmas log and lighting it every evening until the New Year. This log was lighted and extinguished, so that it would last the given time. “Yes, yes, the cachofué, baron!” exclaimed the ladies, gaily. “You are to be the actor in the ceremony, so the time to begin depends on you.”
- 68. “Alas! my friends, I hoped indeed that this honoured ceremony of our fathers would have been more complete, and that my brother the commander would have brought with him my good brother Elzear. But that is not to be thought of for this night at least.” “The Lord grant that the commander may arrive soon with his black galley,” said one of the ladies to the baron. “These wicked pirates, whom we all dread, would not dare make a descent if they knew he was in port.” “The pirates to the devil, good cousin!” cried the baron, gaily. “The watchman is spying them from the height of Cape l’Aigle; at his first signal all the coast will be in arms. The port of La Ciotat is armed; the citizens and fishermen are keeping Christmas with only one hand, they have the other on their muskets; my cannon and small guns are loaded, and ready to fire on the entrance to the port, if these sea-robbers dare show themselves. Manjour! my guests and cousins, if I had obeyed the Marshal of Vitry, at this hour my house would be disarmed and out of condition to defend the city.” “And you did very bravely, baron,” said the lord of Signerol, “to act as you did. Now the example has been given and the marshal will meddle no longer with our affairs.” “Manjour! I hope so indeed. If he does, we will meddle with his,” said the baron. “But where is my young comrade of the cachofué?” added he. “I am the eldest, but I must have the youngest to go for the Christmas log.” “Here is the dear child, father,” said Reine, leading a beautiful boy of six years, with large blue eyes, rosy cheeks, and lovely curls, up to the baron. His mother, a cousin of the baron, looked at the boy with pride, not unmixed with fear, for she suspected that he might not be equal to the complicated rôle necessary to be played in this patriarchal ceremony. “Are you sure you understand what is to be done, my little Cæsar?” asked the baron, bending over the little boy. “Yes, yes, monseigneur. Last year, at grandfather’s house, I carried the Christmas log,” replied the child, with a capable and resolute air.
- 69. “The linnet will become a hawk, I promise you, my cousin,” said the baron to the mother, delighted with the child’s self-confidence. Raimond V. then took the little fellow by the hand, and, followed by his guests, he descended to the door of Maison-Forte, which opened into the inner court, before beginning the ceremony of the cachofué. All the inmates and dependents of the castle, labourers, farmers, fishermen, vine-dressers, servants, women, children, and old men, were assembled in the court. Although the light of the moon was quite bright, a large number of torches, made of resinous wood fastened to poles, illuminated the court and the interior buildings of Maison-Forte. In the middle of the court were collected the combustibles necessary to kindle an immense pile of wood, which was to be set on fire the same moment that the cachofué in the hall of the dais was lighted. Raimond V. appeared before the assembly attended by four lackeys in livery, who walked before him, bearing candlesticks with white wax candles. He was followed by his family and his guests. At the sight of the baron, cries of “Long live monseigneur!” resounded on all sides. In front of the door on the ground lay a large olive-tree, the trunk and branches. It was the Christmas log. Abbé Mascarolus, in cassock and surplice, commenced the ceremony by blessing the Christmas log, or the calignaou, as it was called in the Provençal language; then the child approached, followed by Laramée, who, in his costume of majordomo, bore on a silver tray a gold cup filled with wine. The child took the cup in his little hands and poured, three times, a few drops of wine on the calignaou, or Christmas log, and recited, in a sweet and silvery voice, the old Provençal verse, always said upon this solemn occasion: “‘Allègre, Diou nous allègre,
- 70. Cachofué ven, tou ben ven, Diou nous fague la grace de veire l’an que ven, Se si an pas mai, que signen pas men.’” “Oh, let us be joyful, God gives us all joy; Cachofué comes, and it comes all to bless; God grant we may live to see the New Year; But if we are no more, may we never be less!” These innocent words, recited by the child with charming grace, were listened to with religious solemnity. Then the child wet his lips with the wine in the cup, and presented it to Raimond V., who did likewise, and the cup passed from hand to hand, among all the members of the baron’s family, until each one had wet his lips with the consecrated beverage. Then twelve foresters in holiday dress lifted the calignaou, and carried it into the hall of the dais, while, in conformity to the law of the ceremony, Raimond V. held in his hand one of the roots of the tree, and the child held one of the branches; the old man saying, “Black roots are old age,” and the child answering, “Green branches are youth,” and the assistants adding in chorus, “God bless us all, who love him and serve him!” The log, borne into the hall on the robust shoulders of the foresters, was placed in the immense fireplace, whereupon the child took a pine torch, and held it to a pile of fir-apples and boughs; a tall white flame sparkled in the vast, black hearth, and threw a joyous radiance to the farther end of the gallery. “Christmas, Christmas!” cried the guests of the baron, clapping their hands. “Christmas! Christmas!” repeated the vassals assembled in the interior court.
- 71. At the same moment, the pile of wood outside was kindled, and the tall yellow flames mounted in the midst of enthusiastic shouts, and whirls of a Provençal dance. One other last ceremony was to take place, and then the guests would gather around the supper-table. Reine advanced to the cradle, and Stephanette brought to her a wooden bowl filled with the corn of St. Barbara, which was already green. For it was the custom in Provence, every fourth of December, St Barbara’s day, to sow grains of corn in a porringer filled with earth frequently watered. This wet earth was exposed to a very high temperature, and the com grew rapidly. If it was green, it predicted a good harvest, if it was yellow, the harvest would be bad. Mlle, des Anbiez placed the wooden bowl at the foot of the cradle, and on each side of this offering lit two little square silver lamps, called in the Provençal tongue the lamps of Calenos, or Christmas lamps. “St Barbara’s corn, green; fine harvests all the year!” cried the baron: “so may my harvests and your harvests be, my guests and cousins! Now to the table, yes, to the table, friends, and then come the Christmas presents for friends and relations!” Master Laramée opened the folding doors which led to the dining- room, and announced supper. It is needless to speak of the abundance of this meal, worthy in every respect of the hospitality of Raimond V. What, however, we must not fail to remark, is that there were three table-cloths, in conformity to another ancient custom. On the smallest, which was in the middle of the table, in the style of a centre-piece, were the presents of fruits and cakes that the members of the family made to their head. On the second, a little larger and lapping over the first, were arranged the national dishes of the simplest character, such as bouillabaisse, a fish-soup, famous in Provence, and broiled salt tunny.
- 72. Lastly, on the third cloth, which covered the rest of the table, were the choicest dishes in abundance, and artistically arranged. We will leave the guests of Raimond V. to the enjoyment of a patriarchal hospitality as they discussed old customs, and grew excited over arguments relating to freedom and ancient privileges, always so respected and so valiantly defended by those who remain faithful to the pathetic and religious traditions of the olden time. That happy, peaceful evening was but too soon interrupted by the events to which we will now introduce the reader.
- 73. CHAPTER XXX. THE ARREST While Raimond V. and his guests were supping gaily, the company of soldiers seen by the watchman, about fifty men belonging to the regiment of Guitry, had arrived almost at the door of Maison-Forte. The recorder Isnard, followed by his clerk, as usual, said to Captain Georges, who commanded the detachment: “It would be prudent, captain, to try a summons before attacking by force, in order to take possession of the person of Raimond V. There are about fifty well-armed men in his lair behind good walls.” “Eh! what matters the walls to me?” “But, besides the walls, there is a bridge, and you see, captain, it is up.” “Eh! what do I care for the bridge? If Raimond V. refuses to lower it—ah, well, zounds! my carabineers will assault the place; that happened more than once in the last war! If necessary, we will attach a petard to the door, but let it be understood, recorder, that, whatever happens, you are to follow us to make an official report.” “Hum! hum!” grunted the man of law. “Without doubt, I and my clerk must assist you; I shall be able, even under that circumstance, to note the good conduct and zeal of the aforesaid clerk in charging him with this honourable mission.” “But, Master Isnard, that is your office, and not mine!” said the unhappy clerk. “Silence, my clerk, we are here before Maison-Forte. The moments are precious. Do you prepare to follow the captain and obey me!” The company had, in fact, reached the end of the sycamore walk, which bordered the half-circle. The bridge was up, and the windows opening on the interior court were brilliant with light, as the baron’s guests had departed but a
- 74. little while. “You see, captain, the bridge is up, and more, the moat is wide and deep, and full of water,” said the recorder. Captain Georges carefully examined the entrances of the place; after a few moments of silence, he pulled his moustache on the left side violently,—a sure sign of his disappointment. A sentinel, standing inside the court, seeing the glitter of arms in the moonlight, cried, in a loud voice: “Who goes there? Answer, or I will fire!” The recorder jumped back three steps, hid himself behind the captain, and replied, in a high voice: “In the name of the king and the cardinal, I, Master Isnard, recorder of the admiralty of Toulon, command you to lower this bridge!” “You will not depart?” said the voice. At the same time a light shone from one of the loopholes for guns which defended the entrance. It was easy to judge that the sentinel was blowing the match of his musket. “Take care!” cried Isnard. “Your master will be held responsible for what you are going to do!” This warning made the soldier reflect; he fired his musket in the air, at the same time crying the word of alarm in a stentorian voice. “He has fired on the king’s soldiers!” cried the recorder, pale with anger and fright “It is an act of armed rebellion. I saw it. Clerk, make a note of that act!” “No, recorder,” said the captain, “he has barked, but he has not desired to murder. I saw the light, too, and he fired in the air to give the alarm.” In answer to the sentinel’s cries, several lights appeared above the walls; numerous precipitate steps, and a great clang of arms were heard in the court At last, Master Laramée, a helmet on his head and his breast armed with a cuirass, appeared at one of the embrasures of the gate.
- 75. “In the name of God, what do you want?” cried he. “Is this the time, pray, to come here and trouble good people who are keeping Christmas?” “We have an order from the king which we come to put into execution,” said the recorder, “and I—” “I have some wine left yet in my glass, recorder; good evening, I am going to empty it,” said Laramée, “only, remember the bulls, and know that a musket-ball reaches farther than their horns. So, now, good-night, recorder!” “Think well on what you are going to do, insolent scoundrel,” said Captain Georges; “you are not dealing this time with a wet hen of a recorder, but with a fight-ing-cock, who has a hard beak and sharp spurs, I warn you.” “The fact is, Master Isnard,” said the clerk, humbly, to the recorder, “we are to this soldier what a pumpkin is to an artillery ball.” The recorder, already very much offended by the captain’s comparison, rudely repulsed the clerk, and, addressing Laramée with great importance, said: “You have this time, at your door, the right and the power, the hand and the sword of justice. So, majordomo, I order you to open and to lower the bridge.” A well-known voice interrupted the recorder; it was that of Raimond V., who had been informed of the arrival of the captain. Escorted by Laramée, who carried a torch, the old gentleman appeared erect upon the little platform that formed the entablature of the gate masked by the drawbridge. The fluctuating light of the torch threw red reflections on the group of soldiers, and shone upon their steel collars and iron head- pieces; half of the scene being in the shade or lighted by the rays of the moon. Raimond V. wore his holiday attire, richly braided with gold, and his white hair fell over his lace collar. Nothing was more dignified, more imposing or manly than his attitude.
- 76. “What do you want?” said he, in a sonorous voice. Master Isnard repeated the formula of his speech, and concluded by declaring that Raimond V., Baron des Anbiez, was arrested, and would be conducted under a safe escort to the prison of the provost of Marseilles, for the crime of rebellion against the orders of the king. The baron listened to the recorder in profound silence. When the man of law had finished, cries of indignation, howls, and threats, uttered by the dependents of the baron, resounded through the interior court. Raimond V. turned around, commanded silence, and replied to the recorder: “You wished to visit my castle illegally, and to exercise in it an authority contrary to the rights of the Provençal nobility. I drove you away with my whip. I did what I ought to have done. Now, Manjour! I cannot allow myself to be arrested for having done what I ought to have done in chastising a villain of your species. Now, execute the orders with which you are charged,—I will not prevent you, any more than I prevented your visit to my magazine of artillery. I regret the departure of my guests, for they also, in their name, would have protested against the oppression of the tyranny of Marseilles.” This speech from the baron was welcomed with cries of joy by the garrison of Maison-Forte. Raimond V. was about to descend from his pedestal when Captain Georges, who had the rough language and abrupt manners of an old soldier, advanced on the other side of the moat; he took his hat in his hand, and said to Raimond V., in a respectful tone: “Monseigneur, I must inform you of one thing, which is, that I have with me fifty determined soldiers, and that I am resolved, though to my regret, to execute my orders.” “Execute them, my brave friend,” said the baron, smiling, with a jocose manner, “execute them. Your marshal wishes to know if my powder is good; he instructs you to be the gunpowder prover. We will begin the trial whenever you wish.”
- 77. “Captain, this is too much parley,” cried the recorder. “I order you this instant to employ force of arms to take possession of this rebel against the commands of the king, our master, and to—” “Recorder, I have no orders to receive from you; only take care not to put yourself between the lance and the cuirass,—you might come to grief,” said the captain, imperiously, to Master Isnard. Then, turning to the baron, he said, with as much firmness as deference: “For the last time, monseigneur, I beseech you to consider well: the blood of your vassals will flow; you are going to kill old soldiers who have no animosity against you or yours, and all that, monseigneur,—permit an old graybeard to speak to you frankly,—all that because you wish to rebel against the orders of the king. May God forgive you, monseigneur, for causing the death of so many brave men, and me, for drawing the sword against one of the most worthy gentlemen of the province; but I am a soldier, and I must obey the orders I have received.” This simple and noble language made a profound impression on Raimond V. He bowed his head in silence, remained thoughtful for some minutes, then he descended from the platform. Murmurs inside were distinctly heard, dominated by the ringing voice of the baron. At the same instant the bridge was lowered and the gate opened; Raimond V. appeared, and said to the captain, as he offered his hand with a dignified and cordial air: “Enter, sir, enter; you are a brave and honest soldier. Although my head is white, it is sometimes as foolish as a boy’s. I was wrong. It is true, you must obey the orders which have been given to you. It is not to you, it is to the Marshal of Vitry that I should express my opinion of his conduct toward the Provençal nobility. These brave men ought not to be the victims of my resistance. To-morrow at the break of day, if you will, we will depart for Marseilles.” “Ah, monseigneur,” said the captain, pressing the hand of Raimond V. with emotion, and bowing with respect, “it is now that I really despair of the mission that I am to fulfil.”
- 78. The baron was about to reply to the captain when a distant, but dreadful noise rose on the air, attracting the attention of all those who filled the court of Maison-Forte. It was like the hollow roar of the sea in its fury. Suddenly a tremendous light illuminated the horizon in the direction of La Ciotat, and the bells of the convent and the church began to sound the alarm. The first idea that entered the baron’s mind was that the city was on fire. “Fire!” cried he, “La Ciotat is on fire! Captain, you have my word, I am your prisoner, but let us run to the city. You with your soldiers, I with my people, we can be useful there.” “I am at your orders, monseigneur.” At that moment the prolonged, reverberating sound of artillery made the shore tremble with its echoes, and shook the windows of Maison-Forte. “Cannon! Those are the pirates! The watchman to the devil for allowing us to be surprised! The pirates! To arms, captain! to arms! These demons are attacking the city. Laramée, my sword! Captain, to horse! to horse! You can take me prisoner to-morrow, but to-night let us run to defend this unfortunate city.” “But, monseigneur, your house—” “The devil take them if they venture here! Laramée and twenty men could defend it against an entire army. But this unfortunate city is surprised. Quick! to horse! to horse!” The roar of the artillery became more and more frequent. All the bells were ringing,—a deep rumbling sound reached as far as Maison-Forte,—and the flames increased in number and intensity. Laramée, in all haste, brought the baron’s helmet and cuirass. Raimond V. took the helmet, but would not hear of the cuirass. “Manjour! what time have I to fasten that paraphernalia? Quick, bring Mistraon to me,” cried he, running to the stable.
- 79. He found Mistraon bridled, but, seeing that it required some time to saddle him, he mounted the horse barebacked, told Laramée to keep twenty men for the defence of Maison-Forte, commended his daughter to his care, and took, in hot haste, the road to La Ciotat.
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