Sobolev spaces

JUHA KINNUNEN
Sobolev spaces
Department of Mathematics and Systems Analysis, Aalto University
2017

Contents
1 SOBOLEV SPACES 1
1.1 Weak derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Sobolev spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.3 Properties of weak derivatives . . . . . . . . . . . . . . . . . . . . . . . 8
1.4 Completeness of Sobolev spaces . . . . . . . . . . . . . . . . . . . . 9
1.5 Hilbert space structure . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.6 Approximation by smooth functions . . . . . . . . . . . . . . . . . . . 12
1.7 Local approximation in Sobolev spaces . . . . . . . . . . . . . . . . . 16
1.8 Global approximation in Sobolev spaces . . . . . . . . . . . . . . . . 17
1.9 Sobolev spaces with zero boundary values . . . . . . . . . . . . . . . 18
1.10 Chain rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
1.11 Truncation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
1.12 Sequential weak compactness of Sobolev spaces . . . . . . . . . . 25
1.13 Difference quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
1.14 Absolute continuity on lines . . . . . . . . . . . . . . . . . . . . . . . . 31
2 SOBOLEV INEQUALITIES 37
2.1 Gagliardo-Nirenberg-Sobolev inequality . . . . . . . . . . . . . . . . 38
2.2 Sobolev-Poincaré inequalities . . . . . . . . . . . . . . . . . . . . . . . 43
2.3 Morrey’s inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
2.4 Lipschitz functions and W1,∞ . . . . . . . . . . . . . . . . . . . . . . . . 52
2.5 Summary of the Sobolev embeddings . . . . . . . . . . . . . . . . . . 54
2.6 Direct methods in the calculus of variations . . . . . . . . . . . . . . 55
3 MAXIMAL FUNCTION APPROACH TO SOBOLEV SPACES 62
3.1 Representation formulas and Riesz potentials . . . . . . . . . . . . . 63
3.2 Sobolev-Poincaré inequalities . . . . . . . . . . . . . . . . . . . . . . . 70
3.3 Sobolev inequalities on domains . . . . . . . . . . . . . . . . . . . . . 78
3.4 A maximal function characterization of Sobolev spaces . . . . . . . 81
3.5 Pointwise estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
3.6 Approximation by Lipschitz functions . . . . . . . . . . . . . . . . . . . 88
3.7 Maximal operator on Sobolev spaces . . . . . . . . . . . . . . . . . . 93

CONTENTS ii
4 POINTWISE BEHAVIOUR OF SOBOLEV FUNCTIONS 97
4.1 Sobolev capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
4.2 Capacity and measure . . . . . . . . . . . . . . . . . . . . . . . . . . 100
4.3 Quasicontinuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
4.4 Lebesgue points of Sobolev functions . . . . . . . . . . . . . . . . . . 110
4.5 Sobolev spaces with zero boundary values . . . . . . . . . . . . . . . 115

1Sobolev spaces
In this chapter we begin our study of Sobolev spaces. The Sobolev space is a
vector space of functions that have weak derivatives. Motivation for studying
these spaces is that solutions of partial differential equations, when they exist,
belong naturally to Sobolev spaces.
1.1 Weak derivatives
Notation. Let Ω ⊂ Rn
be open, f : Ω → R and k = 1,2,.... Then we use the following
notations:
C(Ω) = {f : f continuous in Ω}
supp f = {x ∈ Ω : f (x) = 0} = the support of f
C0(Ω) = {f ∈ C(Ω) : supp f is a compact subset of Ω}
Ck
(Ω) = {f ∈ C(Ω) : f is k times continuously diferentiable}
Ck
0(Ω) = Ck
(Ω)∩C0(Ω)
C∞
=
∞
k=1
Ck
(Ω) = smooth functions
C∞
0 (Ω) = C∞
(Ω)∩C0(Ω)
= compactly supported smooth functions
= test functions
W A R N I N G : In general, supp f Ω.
Examples 1.1:
(1) Let u : B(0,1) → R, u(x) = 1−|x|. Then suppu = B(0,1).
1

CHAPTER 1. SOBOLEV SPACES 2
(2) Let f : R → R be
f (x) =



x2
, x 0,
−x2
, x < 0.
Now f ∈ C1
(R)C2
(R) although the graph looks smooth.
(3) Let us define ϕ : Rn
→ R,
ϕ(x) =



e
1
|x|2−1 , x ∈ B(0,1),
0, x ∈ Rn
B(0,1).
Now ϕ ∈ C∞
0 (Rn
) and suppϕ = B(0,1) (exercise).
Let us start with a motivation for definition of weak derivatives. Let Ω ⊂ Rn
be open, u ∈ C1
(Ω) and ϕ ∈ C∞
0 (Ω). Integration by parts gives
ˆ
Ω
u
ϕ
xj
dx = −
ˆ
Ω
u
xj
ϕdx.
There is no boundary term, since ϕ has a compact support in Ω and thus vanishes
near Ω.
Let then u ∈ Ck
(Ω), k = 1,2,..., and let α = (α1,α2,...,αn) ∈ Nn
(we use the
convention that 0 ∈ N) be a multi-index such that the order of multi-index |α| =
α1 +...+αn is at most k. We denote
Dα
u =
|α|
u
x
α1
1 ... x
αn
n
=
α1
x
α1
1
...
αn
x
αn
n
u.
T H E M O R A L : A coordinate of a multi-index indicates how many times a
function is differentiated with respect to the corresponding variable. The order of
a multi-index tells the total number of differentiations.
Successive integration by parts gives
ˆ
Ω
uDα
ϕdx = (−1)|α|
ˆ
Ω
Dα
uϕdx.
Notice that the left-hand side makes sense even under the assumption u ∈ L1
loc
(Ω).
Definition 1.2. Assume that u ∈ L1
loc
(Ω) and let α ∈ Nn
be a multi-index. Then
v ∈ L1
loc
(Ω) is the αth weak partial derivative of u, written Dα
u = v, if
ˆ
Ω
uDα
ϕdx = (−1)|α|
ˆ
Ω
vϕdx
for every test function ϕ ∈ C∞
0 (Ω). We denote D0
u = D(0,...,0)
= u. If |α| = 1, then
Du = (D1u,D2u...,Dnu)
is the weak gradient of u. Here
Dju =
u
xj
= D(0,...,1,...,0)
u, j = 1,...,n,
(the jth component is 1).

T H E M O R A L : Classical derivatives are defined as pointwise limits of differ-
ence quotients, but the weak derivatives are defined as a functions satisfying the
integration by parts formula. Observe, that changing the function on a set of
measure zero does not affect its weak derivatives.
W A R N I N G We use the same notation for the weak and classical derivatives. It
should be clear from the context which interpretation is used.
Remarks 1.3:
(1) If u ∈ Ck
(Ω), then the classical weak derivatives up to order k are also
the corresponding weak derivatives of u. In this sense, weak derivatives
generalize classical derivatives.
(2) If u = 0 almost everywhere in an open set, then Dα
u = 0 almost everywhere
in the same set.
Lemma 1.4. A weak αth partial derivative of u, if it exists, is uniquely defined
up to a set of measure zero.
Proof. Assume that v,v ∈ L1
loc
(Ω) are both weak αth partial derivatives of u, that
is, ˆ
Ω
uDα
ϕdx = (−1)|α|
ˆ
Ω
vϕdx = (−1)|α|
ˆ
Ω
vϕdx
for every ϕ ∈ C∞
0 (Ω). This implies that
ˆ
Ω
(v− v)ϕdx = 0 for every ϕ ∈ C∞
0 (Ω). (1.1)
Claim: v = v almost everywhere in Ω.
Reason. Let Ω Ω (i.e. Ω is open and Ω is a compact subset of Ω). The
space C∞
0 (Ω ) is dense in Lp
(Ω ) (we shall return to this later). There exists a
sequence of functions ϕi ∈ C∞
0 (Ω ) such that |ϕi| 2 in Ω and ϕi → sgn(v − v)
almost everywhere in Ω as i → ∞. Here sgn is the signum function.
Identity (1.1) and the dominated convergence theorem, with the majorant
|(v− v)ϕi| 2(|v|+|v|) ∈ L1
(Ω ), give
0 = lim
i→∞
ˆ
Ω
(v− v)ϕi dx =
ˆ
Ω
lim
i→∞
(v− v)ϕi dx
=
ˆ
Ω
(v− v)sgn(v− v)dx =
ˆ
Ω
|v− v|dx
This implies that v = v almost everywhere in Ω for every Ω Ω. Thus v = v
almost everywhere in Ω.
From the proof we obtain a very useful corollary.

Corollary 1.5 (Fundamental lemma of the calculus of variations). If f ∈ L1
loc
(Ω)
satisﬁes ˆ
Ω
f ϕdx = 0
0 (Ω), then f = 0 almost everywhere in Ω.
T H E M O R A L : This is an integral way to say that a function is zero almost
everywhere.
Example 1.6. Let n = 1 and Ω = (0,2). Consider
u(x) =



x, 0 < x 1,
1, 1 x < 2,
and
v(x) =



1, 0 < x 1,
0, 1 x < 2.
We claim that u = v in the weak sense. To see this, we show that
ˆ 2
0
uϕ dx = −
ˆ 2
0
vϕdx
0 ((0,2)).
Reason. An integration by parts and the fundamental theorem of calculus give
ˆ 2
0
u(x)ϕ (x)dx =
ˆ 1
0
xϕ (x)dx+
ˆ 2
1
ϕ (x)dx
= xϕ(x)
1
0
=ϕ(1)
−
ˆ 1
0
ϕ(x)dx+ϕ(2)
=0
−ϕ(1)
= −
ˆ 1
0
ϕ(x)dx = −
ˆ 2
0
vϕ(x)dx
0 ((0,2)).
1.2 Sobolev spaces
Deﬁnition 1.7. Assume that Ω is an open subset of Rn
. The Sobolev space
Wk,p
(Ω) consists of functions u ∈ Lp
(Ω) such that for every multi-index α with
|α| k, the weak derivative Dα
u exists and Dα
u ∈ Lp
(Ω). Thus
Wk,p
(Ω) = {u ∈ Lp
(Ω) : Dα
u ∈ Lp
(Ω), |α| k}.

If u ∈ Wk,p
(Ω), we deﬁne its norm
u Wk,p(Ω) =
|α| k
ˆ
Ω
|Dα
u|p
dx
1
p
, 1 p < ∞,
and
u Wk,∞(Ω) =
|α| k
esssup
Ω
|Dα
u|.
Notice that D0
u = D(0,...,0)
u = u. Assume that Ω is an open subsets of Ω. We say
that Ω is compactly contained in Ω, denoted Ω Ω, if Ω is a compact subset of
Ω. A function u ∈ W
k,p
loc
(Ω), if u ∈ Wk,p
(Ω ) for every Ω Ω.
T H E M O R A L : Thus Sobolev space Wk,p
(Ω) consists of functions in Lp
(Ω) that
have weak partial derivatives up to order k and they belong to Lp
(Ω).
Remarks 1.8:
(1) As in Lp
spaces we identify Wk,p
functions which are equal almost every-
where.
(2) There are several ways to deﬁne a norm on Wk,p
(Ω). The norm · Wk,p(Ω)
is equivalent, for example, with the norm
|α| k
Dα
u Lp(Ω), 1 p ∞.
and · Wk,∞(Ω) is also equivalent with
max
|α| k
Dα
u L∞(Ω).
(3) For k = 1 we use the norm
u W1,p(Ω) = u
p
Lp(Ω)
+ Du
p
Lp(Ω)
1
p
=
ˆ
Ω
|u|p
dx+
ˆ
Ω
|Du|p
dx
1
p
, 1 p < ∞,
and
u W1,∞(Ω) = esssup
Ω
|u|+esssup
Ω
|Du|.
We may also consider equivalent norms
u W1,p(Ω) = u
p
Lp(Ω)
+
n
j=1
Dju
p
Lp(Ω)
1
p
,
u W1,p(Ω) = u Lp(Ω) +
n
j=1
Dju Lp(Ω) ,
and
u W1,p(Ω) = u Lp(Ω) + Du Lp(Ω)
when 1 p < ∞ and
u W1,∞(Ω) = max u L∞(Ω), D1 u L∞(Ω) ,..., Dnu L∞(Ω) .

Example 1.9. Let u : B(0,1) → [0,∞], u(x) = |x|−α
, α > 0. Clearly u ∈ C∞
(B(0,1)
{0}), but u is unbounded in any neighbourhood of the origin.
We start by showing that u has a weak derivative in the entire unit ball. When
x = 0 , we have
u
xj
(x) = −α|x|−α−1 xj
|x|
= −α
xj
|x|α+2
, j = 1,...,n.
Thus
Du(x) = −α
x
|x|α+2
.
Gauss’ theorem gives
ˆ
B(0,1)B(0,ε)
Dj(uϕ)dx =
ˆ
(B(0,1)B(0,ε))
uϕνj dS,
where ν = (ν1,...,νn) is the outward pointing unit (|ν| = 1) normal of the boundary
and ϕ ∈ C∞
0 (B(0,1)). As ϕ = 0 on B(0,1), this can be written as
ˆ
B(0,1)B(0,ε)
Djuϕdx+
ˆ
B(0,1)B(0,ε)
uDjϕdx =
ˆ
B(0,ε)
uϕνj dS.
By rearranging terms, we obtain
ˆ
B(0,1)B(0,ε)
uDjϕdx = −
ˆ
B(0,1)B(0,ε)
Djuϕdx+
ˆ
B(0,ε)
uϕνj dS. (1.2)
Let us estimate the last term on the right-hand side. Since ν(x) = − x
|x| , we have
νj(x) = −
xj
|x| , when x ∈ B(0,ε). Thus
ˆ
B(0,ε)
uϕνj dS ϕ L∞(B(0,1))
ˆ
B(0,ε)
ε−α
dS
= ϕ L∞(B(0,1))ωn−1εn−1−α
→ 0 as ε → 0,
if n−1−α > 0. Here ωn−1 = H n−1
( B(0,1)) is the (n−1)-dimensional measure of
the sphere B(0,1).
Next we study integrability of Dju. We need this information in order to be
able to use the dominated convergence theorem. A straightforward computation
gives
ˆ
B(0,1)
Dju dx
ˆ
B(0,1)
|Du| dx = α
ˆ
B(0,1)
|x|−α−1
dx
= α
ˆ 1
0
ˆ
B(0,r)
|x|−α−1
dS dr = αωn−1
ˆ 1
0
r−α−1+n−1
dr
= αωn−1
ˆ 1
0
rn−α−2
dr =
αωn−1
n−α−1
rn−α−1
1
0
< ∞,
if n−1−α > 0.

The following argument shows that Dju is a weak derivative of u also in a
neighbourhood of the origin. By the dominated convergence theorem
ˆ
B(0,1)
uDjϕdx =
ˆ
B(0,1)
lim
ε→0
uDjϕχB(0,1)B(0,ε)
dx
= lim
ε→0
ˆ
B(0,1)B(0,ε)
uDjϕdx
= −lim
ε→0
ˆ
B(0,1)B(0,ε)
Djuϕdx+lim
ε→0
ˆ
B(0,ε)
uϕνj dS
= −
ˆ
B(0,1)
lim
ε→0
DjuϕχB(0,1)B(0,ε)
dx
= −
ˆ
B(0,1)
Djuϕdx.
Here we used the dominated convergence theorem twice: First to the function
uDjϕχB(0,1)B(0,ε)
,
which is dominated by |u| Dϕ ∞ ∈ L1
(B(0,1)), and then to the function
DjuϕχB(0,1)B(0,ε)
,
which is dominated by |Du| ϕ ∞ ∈ L1
(B(0,1)). We also used (1.2) and the fact that
the last term there converges to zero as ε → 0.
Now we have proved that u has a weak derivative in the unit ball. We note
that u ∈ Lp
(B(0,1)) if and only if −pα + n > 0, or equivalently, α < n
p . On the
other hand, |Du| ∈ Lp
(B(0,1), if −p(α+1)+ n > 0, or equivalently, α <
n−p
p . Thus
u ∈ W1,p
(B(0,1)) if and only if α <
n−p
p .
Let (ri) be a countable and dense subset of B(0,1) and deﬁne u : B(0,1) → [0,∞],
u(x) =
∞
i=1
1
2i
|x− ri|−α
.
Then u ∈ W1,p
(B(0,1)) if α <
n−p
p .
Reason.
|x− ri|−α
W1,p(B(0,1))
∞
i=1
1
2i
|x− ri|−α
W1,p(B(0,1))
=
∞
i=1
1
2i
|x|−α
W1,p(B(0,1))
= |x|−α
W1,p(B(0,1)) < ∞.
Note that if α > 0, then u is unbounded in every open subset of B(0,1) and not
differentiable in the classical sense in a dense subset.
T H E M O R A L : Functions in W1,p
, 1 p < n, n 2, may be unbounded in every
open subset.

Example 1.10. Observe, that u(x) = |x|−α
, α > 0, does not belong to W1,n
(B(0,1).
However, there are unbounded functions in W1,n
, n 2. Let u : B(0,1) → R,
u(x) =



log log 1+ 1
|x| , x = 0,
0, x = 0.
Then u ∈ W1,n
(B(0,1)) when n 2, but u ∉ L∞
(B(0,1)). This can be used to
construct a function in W1,n
(B(0,1) that is unbounded in every open subset of
B(0,1) (exercise).
T H E M O R A L : Functions in W1,p
, 1 p n, n 2, are not continuous. Later
we shall see, that every W1,p
function with p > n coincides with a continuous
function almost everywhere.
Example 1.11. The function u : B(0,1) → R,
u(x) = u(x1,...,xn) =



1, xn > 0,
0, xn < 0,
does not belong to W1,p
(B(0,1) for any 1 p ∞ (exercise).
1.3 Properties of weak derivatives
The following general properties of weak derivatives follow rather directly from
the deﬁnition.
Lemma 1.12. Assume that u,v ∈ Wk,p
(Ω) and |α| k. Then
(1) Dα
u ∈ Wk−|α|,p
(Ω),
(2) Dβ
(Dα
u) = Dα
(Dβ
u) for all multi-indices α,β with |α|+|β| k,
(3) for every λ,µ ∈ R, λu +µv ∈ Wk,p
(Ω) and
Dα
(λu +µv) = λDα
u +µDα
v,
(4) if Ω ⊂ Ω is open, then u ∈ Wk,p
(Ω ),
(5) (Leibniz’s formula) if η ∈ C∞
0 (Ω), then ηu ∈ Wk,p
(Ω) and
Dα
(ηu) =
β α
α
β
Dβ
ηDα−β
u,
where
α
β
=
α!
β!(α−β)!
, α! = α1!...αn!
and β α means that βj αj for every j = 1,...,n.

T H E M O R A L : Weak derivatives have the same properties as classical deriva-
tives of smooth functions.
Proof. (1) Follows directly from the deﬁnition of weak derivatives. See also (2).
(2) Let ϕ ∈ C∞
0 (Ω). Then Dβ
ϕ ∈ C∞
0 (Ω). Therefore
(−1)|β|
ˆ
Ω
Dβ
(Dα
u)ϕdx =
ˆ
Ω
Dα
uDβ
ϕdx
= (−1)|α|
ˆ
Ω
uDα+β
ϕdx
= (−1)|α|
(−1)|α+β|
ˆ
Ω
Dα+β
uϕdx
for all test functions ϕ ∈ C∞
0 (Ω). Notice that
|α|+|α+β| =α1 +...+αn +(α1 +β1)+...+(αn +βn)
=2(α1 +...+αn)+β1 +...+βn
=2|α|+|β|.
As 2|α| is an even number, the estimate above, together with the uniqueness
results Lemma 1.4 and Corollary 1.5, implies that Dβ
(Dα
u) = Dα+β
u.
(3) and (4) Clear.
(5) First we consider the case |α| = 1. Let ϕ ∈ C∞
0 (Ω). By Leibniz’s rule for
differentiable functions and the deﬁnition of weak derivative
ˆ
Ω
ηuDα
ϕdx =
ˆ
Ω
(uDα
(ηϕ)− u(Dα
η)ϕ)dx
= −
ˆ
Ω
(ηDα
u + uDα
η)ϕdx
for all ϕ ∈ C∞
0 (Ω). The case |α| > 1 follows by induction (exercise).
1.4 Completeness of Sobolev spaces
One of the most useful properties of Sobolev spaces is that they are complete.
Thus Sobolev spaces are closed under limits of Cauchy sequences.
A sequence (ui) of functions ui ∈ Wk,p
(Ω), i = 1,2,..., converges in Wk,p
(Ω) to
a function u ∈ Wk,p
(Ω), if for every ε > 0 there exists iε such that
ui − u Wk,p(Ω) < ε when i iε.
Equivalently,
lim
i→∞
ui − u Wk,p(Ω) = 0.
A sequence (ui) is a Cauchy sequence in Wk,p
(Ω), if for every ε > 0 there exists
iε such that
ui − uj Wk,p(Ω) < ε when i, j iε.

W A R N I N G : This is not the same condition as
ui+1 − ui Wk,p(Ω) < ε when i iε.
Indeed, the Cauchy sequence condition implies this, but the converse is not true
(exercise).
Theorem 1.13 (Completeness). The Sobolev space Wk,p
(Ω), 1 p ∞, k =
1,2,..., is a Banach space.
T H E M O R A L : The spaces Ck
(Ω), k = 1,2,..., are not complete with respect
to the Sobolev norm, but Sobolev spaces are. This is important in existence
arguments for PDEs.
Proof. Step 1: · Wk,p(Ω) is a norm.
Reason. (1) u Wk,p(Ω) = 0 ⇐⇒ u = 0 almost everywhere in Ω.
=⇒ u Wk,p(Ω) = 0 implies u Lp(Ω) = 0, which implies that u = 0 almost every-
where in Ω.
⇐= u = 0 almost everywhere in Ω implies
ˆ
Ω
Dα
uϕdx = (−1)|α|
ˆ
Ω
uDα
ϕdx = 0
for all ϕ ∈ C∞
0 (Ω). This together with Corollary 1.5 implies that Dα
u = 0 almost
everywhere in Ω for all α, |α| k.
(2) λu Wk,p(Ω) = |λ| u Wk,p(Ω), λ ∈ R. Clear.
(3) The triangle inequality for 1 p < ∞ follows from the elementary inequal-
ity (a+ b)α
aα
+ bα
, a, b 0, 0 < α 1, and Minkowski’s inequality, since
u + v Wk,p(Ω) =
|α| k
Dα
u + Dα
v
p
Lp(Ω)
1
p
|α| k
Dα
u Lp(Ω) + Dα
v Lp(Ω)
p
1
p
|α| k
Dα
u
p
Lp(Ω)
1
p
+
|α| k
Dα
v
p
Lp(Ω)
1
p
= u Wk,p(Ω) + v Wk,p(Ω).
Step 2: Let (ui) be a Cauchy sequence in Wk,p
(Ω). As
Dα
ui − Dα
uj Lp(Ω) ui − uj Wk,p(Ω), |α| k,
it follows that (Dα
ui) is a Cauchy sequence in Lp
(Ω), |α| k. The completeness
of Lp
(Ω) implies that there exists uα ∈ Lp
(Ω) such that Dα
ui → uα in Lp
(Ω). In
particular, ui → u(0,...,0) = u in Lp
(Ω).

Step 3: We show that Dα
u = uα, |α| k. We would like to argue
ˆ
Ω
uDα
ϕdx = lim
i→∞
ˆ
Ω
uiDα
ϕdx
= lim
i→∞
(−1)|α|
ˆ
Ω
Dα
uiϕdx
= (−1)|α|
ˆ
Ω
Dα
uϕdx
0 (Ω). On the second line we used the definition of the weak
derivative. Next we show how to conclude the fist and last inequalities above.
1 < p < ∞ Let ϕ ∈ C∞
0 (Ω). By Hölder’s inequality we have
ˆ
Ω
uiDα
ϕdx−
ˆ
Ω
uDα
ϕdx =
ˆ
Ω
(ui − u)Dα
ϕdx
ui − u Lp(Ω) Dα
ϕ Lp (Ω) → 0
and consequently we obtain the first inequality above. The last inequality follows
in the same way, since
ˆ
Ω
Dα
uiϕdx−
ˆ
Ω
Dα
uϕdx Dα
ui − uα Lp(Ω) ϕ Lp (Ω) → 0.
p = 1, p = ∞ A similar argument as above (exercise).
This means that the weak derivatives Dα
u exist and Dα
u = uα, |α| k. As
we also know that Dα
ui → uα = Dα
u, |α| k, we conclude that ui −u Wk,p(Ω) → 0.
Thus ui → u in Wk,p
(Ω).
Remark 1.14. Wk,p
(Ω), 1 p < ∞ is separable. In the case k = 1 consider the
mapping u → (u,Du) from W1,p
(Ω) to Lp
(Ω)× Lp
(Ω))n
and recall that a subset of
a separable space is separable (exercise). However, W1,∞
(Ω) is not separable.
1.5 Hilbert space structure
The space Wk,2
(Ω) is a Hilbert space with the inner product
〈u,v〉Wk,2(Ω) =
|α| k
〈Dα
u,Dα
v〉L2(Ω),
where
〈Dα
u,Dα
v〉L2(Ω) =
ˆ
Ω
Dα
uDα
vdx.
Observe that
u Wk,2(Ω) = 〈u,u〉
1
2
Wk,2(Ω)
.

1.6 Approximation by smooth functions
This section deals with the question whether every function in a Sobolev space
can be approximated by a smooth function.
Define φ ∈ C∞
0 (Rn
) by
φ(x) =



c e
1
|x|2−1 , |x| < 1,
0, |x| 1,
where c > 0 is chosen so that
ˆ
Rn
φ(x)dx = 1.
For ε > 0, set
φε(x) =
1
εn
φ
x
ε
.
The function φ is called the standard mollifier. Observe that φε 0, suppφε =
B(0,ε) and
ˆ
Rn
φε(x)dx =
1
εn
ˆ
Rn
φ
x
ε
dx =
1
εn
ˆ
Rn
φ(y)εn
dy =
ˆ
Rn
φ(x)dx = 1
for all ε > 0. Here we used the change of variable y = x
ε
, dx = εn
dy.
Notation. If Ω ⊂ Rn
is open with Ω = , we write
Ωε = {x ∈ Ω : dist(x, Ω) > ε}, ε > 0.
If f ∈ L1
loc
(Ω), we obtain its standard convolution mollification fε : Ωε → [−∞,∞],
fε(x) = (f ∗φε)(x) =
ˆ
Ω
f (y)φε(x− y)dy.
T H E M O R A L : Since the convolution is a weighted integral average of f over
the ball B(x,ε) for every x, instead of Ω it is well defined only in Ωε. If Ω = Rn
, we
do not have this problem.
Remarks 1.15:
(1) For every x ∈ Ωε,
fε(x) =
ˆ
Ω
f (y)φε(x− y)dy =
ˆ
B(x,ε)
f (y)φε(x− y)dy.
(2) By a change of variables z = x− y we have
ˆ
Ω
f (y)φε(x− y)dy =
ˆ
Ω
f (x− z)φε(z)dz

(3) For every x ∈ Ωε,
|fε(x)|
ˆ
B(x,ε)
f (y)φε(x− y)dy φε ∞
ˆ
B(x,ε)
|f (y)|dy < ∞.
(4) If f ∈ C0(Ω), then fε ∈ C0(Ωε), whenever
0 < ε < ε0 =
1
2
dist(supp f , Ω).
Reason. If x ∈ Ωε s.t. dist(x,supp f ) > ε0 (in particular, for every x ∈ Ωε
Ωε0 ) then B(x,ε)∩supp f = , which implies that fε(x) = 0.
Lemma 1.16 (Properties of molliﬁers).
(1) fε ∈ C∞
(Ωε).
(2) fε → f almost everywhere as ε → 0.
(3) If f ∈ C(Ω), then fε → f uniformly in every Ω Ω.
(4) If f ∈ L
p
loc
(Ω), 1 p < ∞, then fε → f in Lp
W A R N I N G : (4) does not hold for p = ∞, since there are functions in L∞
(Ω)
that are not continuous.
Proof. (1) Let x ∈ Ωε, j = 1,...,n, e j = (0,...,1,...,0) (the jth component is 1).
Choose h0 > 0 such that B(x,h0) ⊂ Ωε and let h ∈ R, |h| < h0. Then
fε(x+ he j)− fε(x)
h
=
1
εn
ˆ
B(x+he j,ε)∪B(x,ε)
1
h
φ
x+ he j − y
ε
−φ
x− y
ε
f (y)dy
Let us set Ω = B(x,h0 +ε). Now Ω Ω and B(x+ he j,ε)∪B(x,ε) ⊂ Ω .
Claim:
1
h
φ
x+ he j − y
ε
−φ
x− y
ε
→
1
ε
φ
xj
x− y
ε
for all y ∈ Ω as h → 0.
Reason. Let ψ(x) = φ
x−y
ε
. Then
ψ
xj
(x) =
1
ε
φ
xj
x− y
ε
, j = 1,...,n
and
ψ(x+ he j)−ψ(x) =
ˆ h
0 t
(ψ(x+ te j))dt =
ˆ h
0
Dψ(x+ te j)· e j dt.
Thus
|ψ(x+ he j)−ψ(x)|
ˆ |h|
0
|Dψ(x+ te j)· e j|dt
1
ε
ˆ |h|
0
Dφ
x+ te j − y
ε
dt
|h|
ε
Dφ L∞(Rn).

This estimate shows that we can use the Lebesgue dominated convergence theorem
(on the third row) to obtain
fε
xj
(x) = lim
h→0
fε(x+ he j)− fε(x)
h
= lim
h→0
1
εn
ˆ
Ω
1
h
φ
x+ he j − y
ε
−φ
x− y
ε
f (y)dy
=
1
εn
ˆ
Ω
1
ε
φ
xj
x− y
ε
f (y)dy
=
ˆ
Ω
φε
xj
(x− y) f (y)dy
=
φε
xj
∗ f (x).
A similar argument shows that Dα
fε exists and
Dα
fε = Dα
φε ∗ f in Ωε
for every multi-index α.
(2) Recall that
´
B(x,ε) φε(x− y)dy = 1. Therefore we have
|fε(x)− f (x)| =
ˆ
B(x,ε)
φε(x− y)f (y)dy− f (x)
ˆ
B(x,ε)
φε(x− y)dy
=
ˆ
B(x,ε)
φε(x− y)(f (y)− f (x)dy
1
εn
ˆ
B(x,ε)
φ
x− y
ε
|f (y)− f (x)|dy
Ωn φ L∞(Rn)
1
|B(x,ε)|
ˆ
B(x,ε)
|f (y)− f (x)|dy → 0
for almost every x ∈ Ω as ε → 0. Here Ωn = |B(0,1)| and the last convergence
follows from the Lebesgue’s differentiation theorem.
(3) Let Ω Ω Ω, 0 < ε < dist(Ω , Ω ), and x ∈ Ω . Because Ω is compact
and f ∈ C(Ω), f is uniformly continuous in Ω , that is, for every ε > 0 there exists
δ > 0 such that
|f (x)− f (y)| < ε for all x, y ∈ Ω with |x− y| < δ.
By combining this with an estimate from the proof of (ii), we conclude that
|fε(x)− f (x)| Ωn φ L∞(Rn)
1
|B(x,ε)|
ˆ
B(x,ε)
|f (y)− f (x)|dy < Ωn φ L∞(Rn) ε
for all x ∈ Ω if ε < δ.
(4) Let Ω Ω Ω.
Claim: ˆ
Ω
|fε|p
dx
ˆ
Ω
|f |p
dx
whenever 0 < ε < dist(Ω , Ω ) and 0 < ε < dist(Ω , Ω).

Reason. Take x ∈ Ω . By Hölder’s inequality implies
|fε(x)| =
ˆ
B(x,ε)
φε(x− y)f (y)dy
ˆ
B(x,ε)
φε(x− y)
1− 1
p φε(x− y)
1
p |f (y)|dy
ˆ
B(x,ε)
φε(x− y)dy
1
p
ˆ
B(x,ε)
φε(x− y)|f (y)|p
dy
1
p
By raising the previous estimate to power p and by integrating over Ω , we obtain
ˆ
Ω
|fε(x)|p
dx
ˆ
Ω
ˆ
B(x,ε)
dydx
=
ˆ
Ω
ˆ
Ω
dxdy
=
ˆ
Ω
|f (y)|p
ˆ
Ω
φε(x− y)dxdy
=
ˆ
Ω
|f (y)|p
dy.
Here we used Fubini’s theorem and once more the fact that the integral of φε is
one.
Since C(Ω ) is dense in Lp
(Ω ). Therefore for every ε > 0 there exists g ∈ C(Ω )
such that
ˆ
Ω
|f − g|p
dx
1
p ε
3
.
By (2), we have gε → g uniformly in Ω as ε → 0. Thus
ˆ
Ω
|gε − g|p
dx
1
p
sup
Ω
|gε − g| Ω
1
p <
ε
3
,
when ε > 0 is small enough. Now we use the Minkowski’s inequality and the
previous claim to conclude that
ˆ
Ω
|fε − f |p
dx
1
p
ˆ
Ω
|fε − gε|p
dx
1
p
+
ˆ
Ω
|gε − g|p
dx
1
p
+
ˆ
Ω
|g − f |p
dx
1
p
2
ˆ
Ω
|g − f |p
dx
1
p
+
ˆ
Ω
|gε − g|p
dx
1
p
2
ε
3
+
ε
3
= ε .
Thus fε → f in Lp
(Ω ) as ε → 0.

1.7 Local approximation in Sobolev spaces
Next we show that the convolution approximation converges locally in Sobolev
spaces.
Theorem 1.17. Let u ∈ Wk,p
(Ω), 1 p < ∞. then
(1) Dα
uε = Dα
u ∗φε in Ωε and
(2) uε → u in Wk,p
T H E M O R A L : Smooth functions are dense in local Sobolev spaces. Thus
every Sobolev function can be locally approximated with a smooth function in the
Sobolev norm.
Proof. (1) Fix x ∈ Ωε. Then
Dα
uε(x) = Dα
(u ∗φε)(x) = (u ∗ Dα
φε)(x)
=
ˆ
Ω
Dα
x φε(x− y)u(y)dy
= (−1)|α|
ˆ
Ω
Dα
y (φε(x− y))u(y)dy.
Here we ﬁrst used the proof of Lemma 1.16 (1) and then the fact that
xj
φ
x− y
ε
= −
xj
φ
y− x
ε
= −
yj
φ
x− y
ε
.
For every x ∈ Ωε, the function ϕ(y) = φε(x− y) belongs to C∞
0 (Ω). Therefore
ˆ
Ω
Dα
y (φε(x− y))u(y)dy = (−1)|α|
ˆ
Ω
Dα
u(y)φε(x− y)dy.
By combining the above facts, we see that
Dα
uε(x) = (−1)|α|+|α|
ˆ
Ω
Dα
u(y)φε(x− y)dy = (Dα
u ∗φε)(x).
Notice that (−1)|α|+|α|
= 1.
(2) Let Ω Ω, and choose ε > 0 s.t. Ω ⊂ Ωε. By (i) we know that Dα
uε =
Dα
u ∗φε in Ω , |α| k. By Lemma 1.16, we have Dα
uε → Dα
u in Lp
(Ω ) as ε → 0,
|α| k. Consequently
uε − u Wk,p(Ω ) =
|α| k
Dα
uε − Dα
u
p
Lp(Ω )
1
p
→ 0.

1.8 Global approximation in Sobolev spaces
The next result shows that the convolution approximation converges also globally
in Sobolev spaces.
Theorem 1.18 (Meyers-Serrin). If u ∈ Wk,p
(Ω), 1 p < ∞, then there exist
functions ui ∈ C∞
(Ω)∩Wk,p
(Ω) such that ui → u in Wk,p
(Ω).
T H E M O R A L : Smooth functions are dense in Sobolev spaces. Thus every
Sobolev function can be approximated with a smooth function in the Sobolev norm.
In particular, this holds true for the function with a dense inﬁnity set in Example
1.9.
Proof. Let Ω0 = and
Ωi = x ∈ Ω : dist(x, Ω) >
1
i
∩B(0, i), i = 1,2,....
Then
Ω =
∞
i=1
Ωi and Ω1 Ω2 ... Ω.
Claim: There exist ηi ∈ C∞
0 (Ωi+2 Ωi−1), i = 1,2,..., such that 0 ηi 1 and
∞
i=1
ηi(x) = 1 for every x ∈ Ω.
This is a partition of unity subordinate to the covering {Ωi}.
Reason. By using the distance function and convolution approximation we can
construct ηi ∈ C∞
0 (Ωi+2Ωi−1) such that 0 ηi 1 and ηi = 1 in Ωi+1Ωi (exercise).
Then we deﬁne
ηi(x) =
ηi(x)
∞
j=1 ηj(x)
, i = 1,2,....
Observe that the sum is only over four indices in a neighbourhood of a given
point.
Now by Lemma 1.12 (5), ηiu ∈ Wk,p
(Ω) and
supp(ηiu) ⊂ Ωi+2 Ωi−1.
Let ε > 0. Choose εi > 0 so small that
supp(φεi
∗(ηiu)) ⊂ Ωi+2 Ωi−1
(see Remark 1.15 (4)) and
φεi
∗(ηiu)−ηiu Wk,p(Ω) <
ε
2i
, i = 1,2,....

By Theorem 1.17 (2), this is possible. Define
v =
∞
i=1
φεi
∗(ηiu).
This function belongs to C∞
(Ω), since in a neighbourhood of any point x ∈ Ω, there
are at most finitely many nonzero terms in the sum. Moreover,
v− u Wk,p(Ω) =
∞
i=1
φεi
∗(ηiu)−
∞
i=1
ηiu
Wk,p(Ω)
∞
i=1
φεi
∗(ηiu)−ηiu Wk,p(Ω)
∞
i=1
ε
2i
= ε.
Remarks 1.19:
(1) The Meyers-Serrin theorem 1.18 gives the following characterization for
the Sobolev spaces W1,p
(Ω), 1 p < ∞: u ∈ W1,p
(Ω) if and only if there exist
functions ui ∈ C∞
(Ω)∩Wk,p
(Ω), i = 1,2,..., such that ui → u in Wk,p
(Ω) as
i → ∞. In other words, W1,p
(Ω) is the completion of C∞
(Ω) in the Sobolev
norm.
Reason. =⇒ Theorem 1.18.
⇐= Theorem 1.13.
(2) The Meyers-Serrin theorem 1.18 is false for p = ∞. Indeed, if ui ∈ C∞
(Ω)∩
W1,∞
(Ω) such that ui → u in W1,∞
(Ω), then u ∈ C1
(Ω) (exercise). Thus
special care is required when we consider approximations in W1,∞
(Ω).
(3) Let Ω Ω. The proof of Theorem 1.17 and Theorem 1.18 shows that for
every ε > 0 there exists v ∈ C∞
0 (Ω) such that v− u W1,p(Ω ) < ε.
(4) The proof of Theorem 1.18 shows that not only C∞
(Ω) but also C∞
0 (Ω) is
dense in Lp
(Ω), 1 p < ∞.
1.9 Sobolev spaces with zero boundary
values
In this section we study definitions and properties of first order Sobolev spaces
with zero boundary values in an open subset of Rn
. A similar theory can be
developed for higher order Sobolev spaces as well. Recall that, by Theorem 1.18,
the Sobolev space W1,p
(Ω) can be characterized as the completion of C∞
(Ω) with
respect to the Sobolev norm when 1 p < ∞.

Deﬁnition 1.20. Let 1 p < ∞. The Sobolev space with zero boundary values
W
1,p
0 (Ω) is the completion of C∞
0 (Ω) with respect to the Sobolev norm. Thus
u ∈ W
1,p
0 (Ω) if and only if there exist functions ui ∈ C∞
0 (Ω), i = 1,2,..., such that
ui → u in W1,p
(Ω) as i → ∞. The space W
1,p
0 (Ω) is endowed with the norm of
W1,p
(Ω).
T H E M O R A L : The only difference compared to W1,p
(Ω) is that functions in
W
1,p
0 (Ω) can be approximated by C∞
0 (Ω) functions instead of C∞
(Ω) functions,
that is,
W1,p
(Ω) = C∞(Ω) and W
1,p
0 (Ω) = C∞
0 (Ω),
where the completions are taken with respect to the Sobolev norm. A function
in W
1,p
0 (Ω) has zero boundary values in Sobolev’s sense. We may say that u, v ∈
W1,p
(Ω) have the same boundary values in Sobolev’s sense, if u−v ∈ W
1,p
0 (Ω). This
is useful, for example, in Dirichlet problems for PDEs.
W A R N I N G : Roughly speaking a function in W1,p
(Ω) belongs to W
1,p
0 (Ω), if it
vanishes on the boundary. This is a delicate issue, since the function does not
have to be zero pointwise on the boundary. We shall return to this question later.
Remark 1.21. W
1,p
0 (Ω) is a closed subspace of W1,p
(Ω) and thus complete (exer-
cise).
Remarks 1.22:
(1) Clearly C∞
0 (Ω) ⊂ W
1,p
0 (Ω) ⊂ W1,p
(Ω) ⊂ Lp
(Ω).
(2) If u ∈ W
1,p
0 (Ω), then the zero extension u : Rn
→ [−∞,∞],
u(x) =



u(x), x ∈ Ω,
0, x ∈ Rn
Ω,
belongs to W1,p
(Rn
) (exercise).
Lemma 1.23. If u ∈ W1,p
(Ω) and suppu is a compact subset of Ω, then u ∈
W
1,p
0 (Ω).
Proof. Let η ∈ C∞
0 (Ω) be a cutoff function such that η = 1 on the support of u.
Claim: If ui ∈ C∞
(Ω), i = 1,2,..., such that ui → u in W1,p
(Ω), then ηui ∈ C∞
0 (Ω)
converges to ηu = u in W1,p
(Ω).
Reason. We observe that
ηui −ηu W1,p(Ω) = ηui −ηu
p
Lp(Ω)
+ D(ηui −ηu)
p
Lp(Ω)
1
p
ηui −ηu Lp(Ω) + D(ηui −ηu) Lp(Ω),

where
ηui −ηu Lp(Ω) =
ˆ
Ω
|ηui −ηu|p
dx
1
p
=
ˆ
Ω
|η|p
|ui − u|p
dx
1
p
η L∞(Ω)
ˆ
Ω
|ui − u|p
dx
1
p
→ 0
and by Lemma 1.12 (5)
D(ηui −ηu) Lp(Ω) =
ˆ
Ω
|D(ηui −ηu)|p
dx
1
p
=
ˆ
Ω
|(ui − u)Dη+(Dui − Du)η|p
dx
1
p
ˆ
Ω
|(ui − u)Dη|p
dx
1
p
+
ˆ
Ω
|(Dui − Du)η|p
dx
1
p
Dη L∞(Ω)
ˆ
Ω
|ui − u|p
dx
1
p
+ η L∞(Ω)
ˆ
Ω
|Dui − Du|p
dx
1
p
→ 0
as i → ∞.
Since ηui ∈ C∞
0 (Ω), i = 1,2,..., and ηui → u in W1,p
(Ω), we conclude that
u ∈ W
1,p
0 (Ω).
Since W
1,p
0 (Ω) ⊂ W1,p
(Ω), functions in these spaces have similar general prop-
erties and they will not be repeated here. Thus we shall focus on properties that
are typical for Sobolev spaces with zero boundary values.
Lemma 1.24. W1,p
(Rn
) = W
1,p
0 (Rn
), 1 p < ∞.
T H E M O R A L : The standard Sobolev space and the Sobolev space with zero
boundary value coincide in the whole space.
W A R N I N G : W1,p
(B(0,1)) = W
1,p
0 (B(0,1)), 1 p < ∞. Thus the spaces are not
same in general.
Proof. Assume that u ∈ W1,p
(Rn
). Let ηk ∈ C∞
0 (B(0,k + 1)) such that η = 1 on
B(0,k), 0 ηk 1 and |Dηk| c. Lemma 1.23 implies uηk ∈ W
1,p
0 (Rn
).
Claim: uηk → u in W1,p
(Rn
) as k → ∞.

Proof. By Theorem 1.18, there exist a sequence of functions ui ∈ C∞
(Ω)∩W1,p
(Ω),
i = 1,2,..., such that ui → u in W1,p
(Ω) as i → ∞. Let ϕ ∈ C∞
0 (Ω).
Claim:
ˆ
Ω
(f ◦ u)Djϕdx = lim
i→∞
ˆ
Ω
f (ui)Djϕdx.
Reason. 1 < p < ∞ By Hölder’s inequality
ˆ
Ω
f (u)Djϕdx−
ˆ
Ω
f (ui)Djϕdx
ˆ
Ω
|f (u)− f (ui)||Dϕ|dx
ˆ
Ω
|f (u)− f (ui)|p
dx
1
p
ˆ
Ω
|Dϕ|p
dx
1
p
f ∞
ˆ
Ω
|u − ui|p
dx
1
p
ˆ
Ω
|Dϕ|p
dx
1
p
→ 0.
On the last row, we used the fact that
|f (u)− f (ui)| =
ˆ u
uui
f (t)dt f ∞|u − ui|.
Finally, the convergence to zero follows, because the ﬁrst and the last term are
bounded and ui → u in Lp
(Ω).
p = 1, p = ∞ A similar argument as above (exercise).
Next, we use the claim above, integration by parts for smooth functions and
the chain rule for smooth functions to obtain
ˆ
Ω
(f ◦ u)Djϕdx = lim
i→∞
ˆ
Ω
f (ui)Dj ϕdx
= − lim
i→∞
ˆ
Ω
Dj(f (ui))ϕdx
= − lim
i→∞
ˆ
Ω
f (ui)Djuiϕdx
= −
ˆ
Ω
f (u)Djuϕdx
= −
ˆ
Ω
(f ◦ u)Djuϕdx, j = 1,...,n,
0 (Ω). We leave it as an exercise to show the fourth inequality in
the display above.
Finally, we need to show that f (u) and f (u) u
xj
are in Lp
(Ω). Since
|f (u)| = |f (u)− f (0)| =
ˆ u
0
|f (t)dt f ∞|u|,
we have
ˆ
Ω
|f (u)|p
dx
1/p
f ∞
ˆ
Ω
|u|p
dx
1
p
< ∞,

and similarly,
ˆ
Ω
f (u)Dju
p
dx
1
p
f ∞
ˆ
Ω
|Du|p
dx
1
p
< ∞.
1.11 Truncation
The truncation property is an important property of ﬁrst order Sobolev spaces,
which means that we can cut the functions at certain level and the truncated
function is still in the same Sobolev space. Higher order Sobolev spaces do not
enjoy this property, see Example 1.6.
Theorem 1.26. If u ∈ W1,p
(Ω), then u+
= max{u,0} ∈ W1,p
(Ω), u−
= −min{u,0} ∈
W1,p
(Ω), |u| ∈ W1,p
(Ω) and
Du+
=



Du almost everywhere in {x ∈ Ω : u(x) > 0},
0 almost everywhere in {x ∈ Ω : u(x) 0},
Du−
=



0 almost everywhere in {x ∈ Ω : u(x) 0},
−Du almost everywhere in {x ∈ Ω : u(x) < 0},
and
Du =



Du almost everywhere in {x ∈ Ω : u(x) > 0},
0 almost everywhere in {x ∈ Ω : u(x) = 0},
−Du almost everywhere in {x ∈ Ω : u(x) < 0}.
T H E M O R A L : In contrast with C1
, the Sobolev space W1,p
are closed under
taking absolute values.
Proof. Let ε > 0 and let fε : R → R, fε(t) = t2 +ε2 − ε. The function fε has the
following properties: fε ∈ C1
(R), fε(0) = 0
lim
ε→0
fε(t) = |t| for every t ∈ R,
(fε) (t) =
1
2
(t2
+ε2
)−1/2
2t =
t
t2 +ε2
for every t ∈ R,
and (fε) ∞ 1 for every ε > 0. From Lemma 1.25, we conclude that fε ◦ u ∈
W1,p
(Ω) and
ˆ
Ω
(fε ◦ u)Djϕdx = −
ˆ
Ω
(fε) (u)Djuϕdx, j = 1,...,n,

0 (Ω). We note that
lim
ε→0
(fε) (t) =



1, t > 0,
0, t = 0,
−1, t < 0,
and consequently
ˆ
Ω
|u|Djϕdx = lim
ε→0
ˆ
Ω
(fε ◦ u)Djϕdx
= −lim
ε→0
ˆ
Ω
(fε) (u)Djuϕdx
= −
ˆ
Ω
Dj|u|ϕdx j = 1,...,n,
0 (Ω), where Dj|u| is as in the statement of the theorem. We leave
it as an exercise to prove that the ﬁrst equality in the display above holds.
The other claims follow from formulas
u+
=
1
2
(u +|u|) and u−
=
1
2
(|u|− u).
Remarks 1.27:
(1) If u,v ∈ W1,p
(Ω), then max{u,v} ∈ W1,p
(Ω) and min{u,v} ∈ W1,p
(Ω). More-
over,
D max{u,v} =



Du almost everywhere in {x ∈ Ω : u(x) v(x)},
Dv almost everywhere in {x ∈ Ω : u(x) v(x)},
and
D min{u,v} =



Du almost everywhere in {x ∈ Ω : u(x) v(x)},
Dv almost everywhere in {x ∈ Ω : u(x) v(x)}.
If u, v ∈ W
1,p
0 (Ω), then max{u,v} ∈ W
1,p
0 (Ω) and min{u,v} ∈ W
1,p
0 (Ω) (exer-
cise).
Reason.
max{u,v} =
1
2
(u + v+|u − v|) and min{u,v} =
1
2
(u + v−|u − v|).
(2) If u ∈ W1,p
(Ω) and λ ∈ R, then Du = 0 almost everywhere in {x ∈ Ω : u(x) =
λ} (exercise).
(3) If u ∈ W1,p
(Ω) and λ ∈ R, then min{u,λ} ∈ W
1,p
loc
(Ω) and
D min{u,λ} =



Du almost everywhere in {x ∈ Ω : u(x) < λ},
0 almost everywhere in {x ∈ Ω : u(x) λ}.

A similar claim also holds for max{u,λ}. This implies that a function
u ∈ W1,p
(Ω) can be approximated by the truncated functions
uλ = max{−λ,min{u,λ}}
=



λ almost everywhere in {x ∈ Ω : u(x) λ},
u almost everywhere in {x ∈ Ω : −λ < u(x) < λ},
−λ almost everywhere in {x ∈ Ω : u(x) −λ},
in W1,p
(Ω). (Here λ > 0.)
Reason. By applying the dominated convergence theorem to
|u − uλ|p
2p
(|u|p
+|uλ|p
) 2p+1
|u|p
∈ L1
(Ω),
we have
lim
λ→∞
ˆ
Ω
|u − uλ|p
dx =
ˆ
Ω
lim
λ→∞
|u − uλ|p
dx = 0,
and by applying the dominated convergence theorem to
|Du − Duλ|p
|Du|p
∈ L1
(Ω),
we have
lim
λ→∞
ˆ
Ω
|Du − Duλ|p
dx =
ˆ
Ω
lim
λ→∞
|Du − Duλ|p
dx = 0.
T H E M O R A L : Bounded W1,p
functions are dense in W1,p
.
1.12 Sequential weak compactness of Sobolev
spaces
In this section we assume that 1 < p < ∞ and that Ω ⊂ Rn
. We recall the deﬁnition
of weak convergence in Lp
.
Deﬁnition 1.28. A sequence (fi) of functions in Lp
(Ω) converges weakly in Lp
(Ω)
to a function f ∈ Lp
(Ω), if
lim
i→∞
ˆ
Ω
fi g dx =
ˆ
Ω
f g dx for every g ∈ Lp
(Ω),
where p =
p
p−1 is the conjugate exponent of p.
If fi → f weakly in Lp
(Ω), then (fi) is bounded in Lp
(Ω), that is, fi ∈ Lp
(Ω),
i = 1,2,..., and
sup
i
fi Lp(Ω) < ∞.

Moreover,
f Lp(Ω) liminf
i→∞
fi Lp(Ω). (1.3)
T H E M O R A L : The Lp
-norm is lower semicontinuous with respect to the weak
convergence.
Thus a weakly converging sequence is bounded. The next result shows that
the converse is true up to a subsequence.
Theorem 1.29. Let 1 < p < ∞. Assume that the sequence (fi) is bounded in
Lp
(Ω). Then there exists a subsequence (fik
) and f ∈ Lp
(Ω) such that fik
→ f
weakly in Lp
(Ω) as k → ∞.
T H E M O R A L : This shows that Lp
with 1 < p < ∞ is weakly sequentially
compact, that is, every bounded sequence in Lp
has a weakly converging subse-
quence. One of the most useful applications of weak convergence is in compactness
arguments. A bounded sequence in Lp
does not need to have any convergent sub-
sequence with convergence interpreted in the standard Lp
sense. However, there
exists a weakly converging subsequence.
Remark 1.30. Theorem 1.29 is equivalent to the fact that Lp
spaces are reﬂexive
for 1 < p < ∞.
Weak convergence is a very weak mode of convergence and sometimes we need
a tool to upgrade it to strong convergence.
Theorem 1.31 (Mazur’s lemma). ] Assume that X is a normed space and that
a sequence (xi) converges weakly to x as i → ∞ in X . Then for every ε > 0, there
exists k ∈ N and a convex combination k
i=1 ai xi such that
x−
k
i=1
ai xi < ε.
Recall, that in a convex combination k
i=1 ai xi we have ai 0 and k
i=1 ai = 1.
Observe that some of the coefﬁcients ai may be zero so that the convex combination
is essentially for a subsequence.
T H E M O R A L : For every weakly converging sequence, there is a subsequence of
convex combinations that converges strongly. Thus weak convergence is upgraded
to strong convergence for a subsequence of convex combinations.
Remark 1.32. Mazur’s lemma can be used to give a proof for (1.3) (exercise).
Theorem 1.33. Let 1 < p < ∞. Assume that (ui) is a bounded sequence in
W1,p
(Ω). Then there exists a subsequence (uik
) and u ∈ W1,p
(Ω) such that
uik
→ u weakly in Lp
(Ω) and Duik
→ Du weakly in Lp
(Ω) as k → ∞. Moreover, if
ui ∈ W
1,p
0 (Ω), i = 1,2..., then u ∈ W
1,p
0 (Ω).

Proof. (1) Assume that u ∈ W1,p
(Ω) Since (ui) is a bounded sequence in Lp
(Ω), by
Theorem 1.29 there exists a subsequence (uik
) and u ∈ Lp
(Ω) such that uik
→ u
weakly in Lp
(Ω) as k → ∞. Since (Djui), j = 1,...,n, is a bounded sequence
in Lp
(Ω), by passing to subsequences successively, we obtain a subsequence
(Djuik
) and gj ∈ Lp
(Ω) such that Djuik
→ gj weakly in Lp
(Ω) as k → ∞ for every
j = 1,...,n.
Claim: gj = Dju, j = 1,...,n.
Reason. For every ϕ ∈ C∞
0 (Ω), we have
ˆ
Ω
uDjϕdx = lim
k→∞
ˆ
Ω
uik
Djϕdx
= − lim
k→∞
ˆ
Ω
Djuik
ϕdx
= −
ˆ
Ω
gjϕdx.
Here the first equality follows from the definition of weak convergence, the second
equality follows form the definition of weak derivative and the last equality follows
from weak convergence. Thus gj = Dju, j = 1,...,n. This shows that the weak
partial derivatives Dju, j = 1,...,n, exist and belong to Lp
(Ω). It follows that
u ∈ W1,p
(Ω).
(2) Then assume that u ∈ W
1,p
0 (Ω). We use the notation above. By Mazur’s
lemma, see Theorem 1.31, there there exists a subsequence of (uik
), denoted by
(ui), such that for convex combinations
k
i=1
aiui → u and
k
i=1
aiDui → Du in Lp
(Ω) as k → ∞.
This shows that
k
i=1
aiui → u in W1,p
(Ω) as k → ∞.
Since
k
j=1
aiui ∈ W
1,p
0 (Ω) and
k
j=1
aiDui ∈ W
1,p
0 (Ω),
completeness of W
1,p
0 (Ω) implies u ∈ W
1,p
0 (Ω)
Remarks 1.34:
(1) Theorem 1.33 is equivalent to the fact that W1,p
spaces are reflexive for
1 < p < ∞.
(2) Another way to see that W1,p
spaces are reflexive for 1 < p < ∞ is to
recall that a closed subspace of a reflexive space is reflexive. Thus it is
enough to find an isomorphism between W1,p
(Ω) and a closed subspace of
Lp
(Ω,Rn+1
) = Lp
(Ω,Rn
)×···× Lp
(Ω,Rn
). The mapping u → (u,Du) will do

for this purpose. This holds true for W
1,p
0 (Ω) as well. This approach can be
used to characterize elements in the dual space by the Riesz representation
theorem.
Theorem 1.35. Let 1 < p < ∞. Assume that (ui) is a bounded sequence in
W1,p
(Ω) and ui → u almost everywhere in Ω. Then u ∈ W1,p
(Ω), ui → u weakly in
Lp
(Ω) and Dui → Du weakly in Lp
(Ω). Moreover, if ui ∈ W
1,p
0 (Ω), i = 1,2..., then
u ∈ W
1,p
0 (Ω).
T H E M O R A L : In order to show that u ∈ W1,p
(Ω) it is enough to construct
functions ui ∈ W1,p
(Ω), i = 1,2,..., such that ui → u almost everywhere in Ω and
supi ui W1,p(Ω)) < ∞.
Proof. We pass to subsequences several times in the argument and we denote all
subsequences again by (ui). By Theorem 1.33 there exists a subsequence (ui) and
a function u ∈ W1,p
(Ω) such that ui → u weakly in Lp
(Ω) and Dui → Du weakly in
Lp
(Ω) as i → ∞. By Mazur’s lemma, see Theorem 1.31, there exists a subsequence
(ui) such that the convex combinations
k
i=1
aiui → u and
k
i=1
aiDui → Du in Lp
(Ω) as k → ∞.
Since convergence in Lp
(Ω) implies that there is a subsequence that converges
almost everywhere in Ω and ui → u almost everywhere in Ω implies that
k
j=1
aiui → u almost everywhere in Ω as k → ∞,
we conclude that u = u and Du = Du almost everywhere in Ω. This show that the
weak limit is independent of the choice of the subsequences, which implies that
ui → u weakly in Lp
(Ω) and Dui → Du weakly in Lp
(Ω).
Remark 1.36. Theorem 1.33 and Theorem 1.35 do not hold when p = 1 (exercise).
1.13 Difference quotients
In this section we give a characterization of W1,p
, 1 < p < ∞, in terms of difference
quotients. This approach is useful in regularilty theory for PDEs. Moreover, this
characterization does not involve derivatives.
Deﬁnition 1.37. Let u ∈ L1
loc
(Ω) and Ω Ω. The jth
difference quotient is
Dh
j u(x) =
u(x+ he j)− u(x)
h
, j = 1,...,n,
for x ∈ Ω and h ∈ R such that 0 < |h| < dist(Ω , Ω). We denote
Dh
u = (Dh
1 u,...,Dh
nu).

T H E M O R A L : Note that the definition of the difference quotient makes sense
at every x ∈ Ω whenever 0 < |h| < dist(x, Ω). If Ω = Rn
, then the definition makes
sense for every h = 0.
Theorem 1.38.
(1) Assume u ∈ W1,p
(Ω), 1 p < ∞. Then for every Ω Ω, we have
Dh
u Lp(Ω ) c Du Lp(Ω)
for some constant c = c(n, p) and all 0 < |h| < dist(Ω , Ω).
(2) If u ∈ Lp
(Ω ), 1 < p < ∞, and there is a constant c such that
Dh
u Lp(Ω ) c
for all 0 < |h| < dist(Ω , Ω), then u ∈ W1,p
(Ω ) and
Du Lp(Ω ) c.
T H E M O R A L : Pointwise derivatives are defined as limit of difference quotients
and Sobolev spaces can be characterized by integrated difference quotients.
W A R N I N G : Claim (2) does not hold for p = 1 (exercise).
Proof. (1) First assume that u ∈ C∞
(Ω)∩W1,p
(Ω). Then
u(x+ he j)− u(x) =
ˆ h
0 t
(u(x+ te j))dt
=
ˆ h
0
Du(x+ te j)· e j dt
=
ˆ h
0
u
xj
(x+ te j)dt, j = 1,...,n,
for all x ∈ Ω , 0 < |h| < dist(Ω , Ω). By Hölder’s inequality
|Dh
j u(x)| =
u(x+ he j)− u(x)
h
1
|h|
ˆ |h|
0
u
xj
(x+ te j) dt
1
|h|
ˆ |h|
0
u
xj
(x+ te j)
p
dt
1/p
|h|
1− 1
p ,
which implies
|Dh
j u(x)|p 1
|h|
ˆ |h|
0
u
xj
(x+ te j)
p
dt

Next we integrate over Ω and switch the order of integration by Fubini’s theorem
to conclude
ˆ
Ω
|Dh
j u(x)|p
dx
1
|h|
ˆ
Ω
ˆ |h|
0
u
xj
(x+ te j)
p
dtdx
=
1
|h|
ˆ |h|
0
ˆ
Ω
u
xj
(x+ te j)
p
dxdt
ˆ
Ω
u
xj
(x)
p
dx.
The last inequality follows from the fact that, for 0 < |h| < dist(Ω , Ω), we have
ˆ
Ω
u
xj
(x+ te j)
p
dx
ˆ
Ω
u
xj
(x)
p
dx.
Using the elementary inequality (a1 +···+ an)α
nα
(aα
1 +···+ aα
n), ai 0, α > 0,
we obtain
ˆ
Ω
|Dh
u(x)|p
dx =
ˆ
Ω
n
j=1
|Dh
j u(x)|2
p
2
dx n
p
2
ˆ
Ω
n
j=1
|Dh
j u(x)|p
dx
= c
n
j=1
ˆ
Ω
|Dh
j u(x)|p
dx c
n
j=1
ˆ
Ω
u
xj
(x)
p
dx
c
ˆ
Ω
|Du(x)|p
dx
The general case u ∈ W1,p
(Ω) follows by an approximation, see Theorem 1.18
(exercise).
(2) Let ϕ ∈ C∞
0 (Ω ). Then by a change of variables we see that, for 0 < |h| <
dist(suppϕ, Ω ), we have
ˆ
Ω
u(x)
ϕ(x+ he j)−ϕ(x)
h
dx = −
ˆ
Ω
−u(x)+ u(x− he j)
−h
ϕ(x)dx, j = 1,...,n.
This shows that
ˆ
Ω
uDh
j ϕdx = −
ˆ
Ω
(D−h
j u)ϕdx, j = 1,...,n. (1.4)
By assumption
sup
0<|h|<dist(Ω , Ω)
D−h
j u Lp(Ω ) < ∞.
From reﬂexivity of Lp
(Ω ), 1 < p < ∞, it follows that there exists gj ∈ Lp
(Ω ),
j = 1,...,n and a subsequence hi → 0 such that D
−hi
j
u → gj weakly in Lp
(Ω ).

Therefore
ˆ
Ω
u
ϕ
xj
dx =
ˆ
Ω
u lim
hi→0
D
hi
j
ϕ dx
= lim
hi→0
ˆ
Ω
uD
hi
j
ϕdx
= − lim
hi→0
ˆ
Ω
(D
−hi
j
u)ϕdx
= −
ˆ
Ω
gjϕdx
0 (Ω ). Here the second equality follows from the dominated
convergence theorem and the last equality is the weak convergence. Thus u
xj
= gj,
j = 1,...,n, in the weak sense and u ∈ W1,p
(Ω ). This is essentially the same
argument as in the proof of Theorem 1.33.
1.14 Absolute continuity on lines
In this section we relate weak derivatives to classical derivatives and give a
characterization W1,p
in terms of absolute continuity on lines.
Recall that a function u : [a,b] → R is absolutely continuous, if for every ε > 0,
there exists δ > 0 such that if a = x1 < y1 x2 < y2 ... xm < ym = b is a partition
of [a,b] with
m
i=1
(yi − xi) < δ,
then
m
i=1
|u(yi)− u(xi)| < ε.
Absolute continuity can be characterized in terms of the fundamental theorem of
calculus.
Theorem 1.39. A function u : [a,b] → R is absolutely continuous if and only if
there exists a function g ∈ L1
((a,b)) such that
u(x) = u(a)+
ˆ x
a
g(t)dt.
By the Lebesgue differentiation theorem g = u almost everywhere in (a,b).
T H E M O R A L : Absolutely continuous functions are precisely those functions
for which the fundamental theorem of calculus holds true.
Examples 1.40:
(1) Every Lipchitz continuous function u : [a,b] → R is absolutely continuous.
(2) The Cantor function u is continuous in [0,1] and differentiable almost
everywhere in (0,1), but not absolutely continuous in [0,1].

Reason.
u(1) = 1 = 0 = u(0)+
ˆ 1
0
u (t)
=0
dt.
The next result relates weak partial derivatives with the classical partial
derivatives.
Theorem 1.41 (Nikodym, ACL characterization). Assume that u ∈ W
1,p
loc
(Ω),
1 p ∞ and let Ω Ω. Then there exists u∗ : Ω → [−∞,∞] such that u∗
= u
almost everywhere in Ω and u∗
is absolutely continuous on (n−1)-dimensional
Lebesgue measure almost every line segments in Ω that are parallel to the
coordinate axes and the classical partial derivatives of u∗
coincide with the weak
partial derivatives of u almost everywhere in Ω. Conversely, if u ∈ L
p
loc
(Ω) and
there exists u∗
as above such that Diu∗
∈ L
p
loc
(Ω), i = 1,...,n, then u ∈ W
1,p
loc
(Ω).
T H E M O R A L : This is a very useful characterization of W1,p
, since many
claims for weak derivatives can be reduced to the one-dimensional claims for
absolute continuous functions. In addition, this gives a practical tool to show that
a function belongs to a Sobolev space.
Remarks 1.42:
(1) The ACL characterization can be used to give a simple proof of Example
1.9 (exercise).
(2) In the one-dimensional case we obtain the following characterization:
u ∈ W1,p
((a,b)), 1 p ∞, if u can be redeﬁned on a set of measure zero
in such a way that u ∈ Lp
((a,b)) and u is absolutely continuous on every
compact subinterval of (a,b) and the classical derivative exists and belongs
to u ∈ Lp
((a,b)). Moreover, the classical derivative equals to the weak
derivative almost everywhere.
(3) A function u ∈ W1,p
(Ω) has a representative that has classical partial
derivatives almost everywhere. However, this does not give any informa-
tion concerning the total differentiability of the function. See Theorem
2.16.
(4) The ACL characterization can be used to give a simple proof of the Leibniz
rule. If u ∈ W1,p
(Ω)∩ L∞
(Ω) and v ∈ W1,p
(Ω)∩ L∞
(Ω), then uv ∈ W1,p
(Ω)
and
Dj(uv) = vDju + uDjv, j = 1,...,n,
almost everywhere in Ω (exercise), compare to Lemma 1.12 (5).
(5) The ACL characterization can be used to give a simple proof for Lemma
1.25 and Theorem 1.26. The claim that if u,v ∈ W1,p
(Ω), then max{u,v} ∈
W1,p
(Ω) and min{u,v} ∈ W1,p
(Ω) follows also in a similar way (exercise).

(6) The ACL characterization can be used to show that if Ω is connected,
u ∈ W
1,p
loc
(Ω) and Du = 0 almost everywhere in Ω, then u is a constant
almost everywhere in Ω (exercise).
Proof. Since the claims are local, we may assume that Ω = Rn
and that u has a
compact support.
=⇒ Let ui = uεi
, i = 1,2,..., be a sequence of standard convolution approxima-
tions of u such that suppui ⊂ B(0,R) for every i = 1,2,... and
ui − u W1,1(Rn) <
1
2i
, i = 1,2,...
By Lemma 1.16 (2), the sequence of convolution approximations converges point-
wise almost everywhere and thus the limit limi→∞ ui(x) exists for every x ∈ Rn
E
for some E ⊂ Rn
with |E| = 0. We deﬁne
u∗
(x) =



lim
i→∞
ui(x), x ∈ Rn
E,
0, x ∈ E.
We ﬁx a standard base vector in Rn
and, without loss of generality, we may assume
that it is (0,...,0,1). Let
fi(x1,...,xn−1) =
ˆ
R
|ui+1 − ui|+
n
j=1
ui+1
xj
−
ui
xj
(x1,...,xn)dxn
and
f (x1,...,xn−1) =
∞
i=1
fi(x1,...,xn−1).
By the monotone convergence theorem and Fubini’s theorem
ˆ
Rn−1
f dx1 ... dxn−1 =
ˆ
Rn−1
∞
i=1
fi dx1 ... dxn−1
=
∞
i=1
ˆ
Rn−1
fi dx1 ... dxn−1
=
∞
i=1
ˆ
Rn
|ui+1 − ui|+
ui+1
xj
−
ui
xj
dx
<
∞
i=1
1
2i
< ∞.
This shows that f ∈ L1
(Rn−1
) and thus f < ∞ (n−1)-almost everywhere in Rn−1
.
Let x = (x1,...,xn−1) ∈ Rn−1
such that f (x) < ∞. Denote
gi(t) = ui(x,t) and g(t) = u∗
(x,t).
Claim: (gi) is a Cauchy sequence in C(R).

Reason. Note that
gi = g1 +
i−1
k=1
(gk+1 − gk), i = 1,2,...,
where
|gk+1(t)− gk(t)| =
ˆ t
−∞
(gk+1 − gk)(s)ds
ˆ
R
|gk+1(s)− gk(s)|ds
ˆ
R
uk+1
xn
(x,s)−
uk
xn
(x,s) ds fk(x).
Thus
i−1
k=1
(gk+1(t)− gk(t))
i−1
k=1
|gk+1(t)− gk(t)|
∞
k=1
fk(x) = f (x) < ∞
for every t ∈ R. This implies that (gi) is a Cauchy sequence in C(R). Since C(R) is
complete, there exists g ∈ C(R) such that gi → g uniformly in R. It follows that
{x}×R ⊂ Rn
E.
Claim: (gi
) is a Cauchy sequence in L1
(R).
Reason.
ˆ
R
i−1
k=1
(gk+1(t)− gk(t)) dt
i−1
k=1
ˆ
R
|gk+1(t)− gk(t)|dt
i−1
k=1
fk(x) f (x) < ∞.
This implies that (gi) is a Cauchy sequence in L1
(R). Since L1
(R) is complete,
there exists g ∈ L1
(R) such that gi
→ g in L1
(R) as i → ∞ .
Claim: g is absolutely continuous in R.
Reason.
g(t) = lim
i→∞
gi(t) = lim
i→∞
ˆ t
−∞
gi(s)ds =
ˆ t
−∞
g (s)ds
This implies that g is absolutely continuous in R and g = g almost everywhere in
R.
Claim: g is the weak derivative of g.
Reason. Let ϕ ∈ C∞
0 (R). Then
ˆ
R
gϕ dt = lim
i→∞
ˆ
R
giϕ dt = − lim
i→∞
ˆ
R
giϕdt = −
ˆ
R
gϕdt.

Thus for every ϕ ∈ C∞
0 (Rn
) we have
ˆ
R
u∗
(x,xn)
ϕ
xn
(x,xn)dxn = −
ˆ
R
u∗
xn
(x,xn)ϕ(x,xn)dxn
and by Fubini’s theorem
ˆ
Rn
u
ϕ
xn
dx = −
ˆ
Rn
u∗
xn
ϕdx.
This shows that u∗
has the classical partial derivatives almost everywhere in Rn
and that they coincide with the weak partial derivatives of u almost everywhere
in Rn
.
⇐= Assume that u has a representative u∗
as in the statement of the theo-
rem. For every ϕ ∈ C∞
0 (Rn
), the function u∗
ϕ has the same absolute continuity
properties as u∗
. By the fundamental theorem of calculus
ˆ
R
(u∗
ϕ)
xn
(x,t)dt = 0
for (n−1)-almost every x ∈ Rn−1
. Thus
ˆ
R
u∗
(x,t)
ϕ
xn
(x,t)dt = −
ˆ
R
u∗
xn
(x,t)ϕ(x,t)dt
and by Fubini’s theorem
ˆ
Rn
u∗ ϕ
xn
dx = −
ˆ
Rn
u∗
xn
ϕdx.
Since u∗
= u almost everywhere in Rn
, we see that u∗
xn
is the nth weak partial
derivative of u. The same argument applies to all other partial derivatives u∗
xj
,
j = 1,...,n as well.
Example 1.43. The radial projection u : B(0,1) → B(0,1), u(x) = x
|x| is discontinu-
ous at the origin. However, the coordinate functions
xj
|x| , j = 1,...,n, are absolutely
continuous on almost every lines. Moreover,
Di
xj
|x|
=
δi j|x|−
xi xj
|x|
|x|2
∈ Lp
(B(0,1))
whenever 1 p < n. Here
δi j =



1, i = j,
0, i = j,
is the Kronecker symbol. By the ACL characterization the coordinate functions of
u belong to W1,p
(B(0,1)) whenever 1 p < n.
Remark 1.44. We say that a closed set E ⊂ Ω to be removable for W1,p
(Ω), if
|E| = 0 and W1,p
(Ω E) = W1,p
(Ω) in the sense that every function in W1,p
(Ω E)

can be approximated by the restrictions of functions in C∞
(Ω). Theorem 1.41
implies the following removability theorem for W1,p
Ω): if H n−1
(E) = 0, then E
is removable for W1,p
(Ω). Observe, that if H n−1
(E) = 0, then E is contained in a
measure zero set of lines in a ﬁxed direction (equivalently the projection of E onto
a hyperplane also has H n−1
-measure zero).
This result is quite sharp. For example, let Ω = B(0,1) and E = {x ∈ B(0,1) :
x2 = 0}. Then 0 < H n−1
(E) < ∞, but E is not removable since, using Theorem 1.41
again, it is easy to see that the function which is 1 on the upper half-plane and 0
on the lower half-plane does not belong to W1,p
Ω). With a little more work we can
show that E = E ∩B(0, 1
2 ) is not removable for W1,p
(B(0,1)).

2Sobolev inequalities
The term Sobolev inequalities refers to a variety of inequalities involving functions
and their derivatives. As an example, we consider an inequality of the form
ˆ
Rn
|u|q
dx
1
q
c
ˆ
Rn
|Du|p
dx
1
p
(2.1)
for every u ∈ C∞
0 (Rn
), where constant 0 < c < ∞ and exponent 1 q < ∞ are
independent of u. By density of smooth functions in Sobolev spaces, see Theorem
1.18, we may conclude that (2.1) holds for functions in W1,p
(Rn
) as well. Let u ∈
C∞
0 (Rn
), u ≡ 0, 1 p < n and consider uλ(x) = u(λx) with λ > 0. Since u ∈ C∞
0 (Rn
),
it follows that (2.1) holds true for every uλ with λ > 0 with c and q independent of
λ. Thus
ˆ
Rn
|uλ|q
dx
1
q
c
ˆ
Rn
|Duλ|p
dx
1
p
for every λ > 0. By a change of variables y = λx, dx = 1
λn dy, we see that
ˆ
Rn
|uλ(x)|q
dx =
ˆ
Rn
|u(λx)|q
dx =
ˆ
Rn
|u(y)|q 1
λn
dy =
1
λn
ˆ
Rn
|u(x)|q
dx
and
ˆ
Rn
|Duλ(x)|p
dx =
ˆ
Rn
λp
|Du(λx)|p
dx
=
λp
λn
ˆ
Rn
|Du(y)|p
dy
=
λp
λn
ˆ
Rn
|Du(x)|p
dx.
Thus
1
λ
n
q
ˆ
Rn
|u|q
dx
1
q
c
λ
λ
n
p
ˆ
Rn
|Du|p
dx
1
p
37

CHAPTER 2. SOBOLEV INEQUALITIES 38
for every λ > 0, and equivalently,
u Lq(Rn) cλ
1− n
p + n
q Du Lp(Rn).
Since this inequality has to be independent of λ, we have
1−
n
p
+
n
q
= 0 ⇐⇒ q =
np
n− p
.
T H E M O R A L : There is only one possible exponent q for which inequality (2.1)
may hold true for all compactly supported smooth functions.
For 1 p < n, the Sobolev conjugate exponent of p is
p∗
=
np
n− p
.
Observe that
(1) p∗
> p,
(2) If p → n−, then p∗
→ ∞ and
(3) If p = 1, then p∗
= n
n−1 .
2.1 Gagliardo-Nirenberg-Sobolev inequal-
ity
The following generalized Hölder’s inequality will be useful for us.
Lemma 2.1. Let 1 p1,..., pk ∞ with 1
p1
+···+ 1
pk
= 1 and assume fi ∈ Lpi (Ω),
i = 1,...,k. Then
ˆ
Ω
|f1 ... fk|dx
k
i=1
fi Lpi (Ω).
Proof. Induction and Hölder’s inequality (exercise).
Sobolev proved the following theorem in the case p > 1 and Nirenberg and
Gagliardo in the case p = 1.
Theorem 2.2 (Gagliardo-Nirenberg-Sobolev). Let 1 p < n. There exists
c = c(n, p) such that
ˆ
Rn
|u|p∗
dx
1
p∗
c
ˆ
Rn
|Du|p
dx
1
p
for every u ∈ W1,p
(Rn
).

T H E M O R A L : The Sobolev-Gagliardo-Nirenberg inequality implies that
W1,p
(Rn
) ⊂ Lp∗
(Rn
), when 1 p < n. More precisely, W1,p
(Rn
) is continuously
imbedded in Lp∗
(Rn
), when 1 p < n. This is the Sobolev embedding theorem for
1 p < n.
Proof. (1) We start by proving the estimate for u ∈ C∞
0 (Rn
). By the fundamental
theorem of calculus
u(x1,...,xj,...,xn) =
ˆ xj
−∞
u
xj
(x1,...,tj,...,xn)dtj, j = 1,...,n.
This implies that
|u(x)|
ˆ
R
|Du(x1,...,tj,...,xn)|dtj, j = 1,...,n.
By taking product of the previous estimate for each j = 1,...,n, we obtain
|u(x)|
n
n−1
n
j=1
ˆ
R
|Du(x1,...,tj,...,xn)|dtj
1
n−1
.
We integrate with respect to x1 and then we use generalized Hölder’s inequality
for the product of (n−1) terms to obtain
ˆ
R
|u|
n
n−1 dx1
ˆ
R
|Du|dt1
1
n−1
ˆ
R
n
j=2
ˆ
R
|Du|dtj
1
n−1
dx1
ˆ
R
|Du|dt1
1
n−1 n
j=2
ˆ
R
ˆ
R
|Du|dx1 dtj
1
n−1
.
Next we integrate with respect to x2 and use again generalized Hölder’s inequality
ˆ
R
ˆ
R
|u|
n
n−1 dx1 dx2
ˆ
R
ˆ
R
|Du|dt1
1
n−1 n
j=2
ˆ
R
ˆ
R
|Du|dx1 dtj
1
n−1
dx2
=
ˆ
R
ˆ
R
|Du|dx1 dt2
1
n−1
·
ˆ
R
ˆ
R
|Du|dt1
1
n−1 n
j=3
ˆ
R
ˆ
R
|Du|dx1 dtj
1
n−1
dx2
ˆ
R
ˆ
R
|Du|dx1 dt2
1
n−1
·
ˆ
R
ˆ
R
|Du|dt1 dx2
1
n−1 n
j=3
ˆ
R
ˆ
R
ˆ
R
|Du|dx1 dx2 dtj
1
n−1
.
Then we integrate with respect to x3,...,xn and obtain
ˆ
Rn
|u|
n
n−1 dx
n
j=1
ˆ
R
...
ˆ
R
|Du|dx1 ... dtj ... dxn
1
n−1
=
ˆ
Rn
|Du|dx
n
n−1
.

This is the required inequality for p = 1.
If 1 1 is to be chosen later. Since γ > 1, we have v ∈ C1
(Rn
). Hölder’s
inequality implies
ˆ
Rn
|u|γ n
n−1 dx
n−1
n
ˆ
Rn
|D(|u|γ
)|dx
= γ
ˆ
Rn
|u|γ−1
|Du|dx
γ
ˆ
Rn
|u|
(γ−1)
p
p−1 dx
p−1
p
ˆ
Rn
|Du|p
dx
1
p
.
Now we choose γ so that |u| has the same power on both sides. Thus
γn
n−1
= (γ−1)
p
p −1
⇐⇒ γ =
p(n−1)
n− p
.
This gives
γn
n−1
=
p(n−1)
n− p
n
n−1
=
pn
n− p
= p∗
and consequently
ˆ
Rn
|u|p∗
dx
1
p∗
γ
ˆ
Rn
|Du|p
dx
1
p
.
This proves the claim for u ∈ C∞
0 (Rn
).
(2) Assume then that u ∈ W1,p
(Rn
). By Lemma 1.24 we have W1,p
(Rn
) =
W
1,p
0 (Rn
). Thus there exist ui ∈ C∞
0 (Rn
), i = 1,2,..., such that ui − u W1,p(Rn) →
0 as i → ∞. In particular ui − u Lp(Rn) → 0, as i → ∞. Thus there exists a
subsequence (ui) such that ui → u almost everywhere in Rn
and ui → u in Lp
(Rn
).
Claim: (ui) is a Cauchy sequence in Lp∗
(Rn
).
Reason. Since ui − uj ∈ C∞
0 (Rn
), we use the Sobolev-Gagliardo-Nirenberg inequal-
ity for compactly supported smooth functions and Minkowski’s inequality to
conclude that
ui − uj Lp∗
(Rn) c Dui − Duj Lp(Rn)
c Dui − Du Lp(Rn) + Du − Duj Lp(Rn) → 0.
Since Lp∗
(Rn
) is complete, there exists v ∈ Lp∗
(Rn
) such that ui → v in Lp∗
(Rn
) as
i → ∞.
Since ui → u almost everywhere in Rn
and ui → v in Lp∗
(Rn
), we have u = v
almost everywhere in Rn
. This implies that ui → u in Lp∗
(Rn
) and that u ∈ Lp∗
(Rn
).

Now we can apply Minkowski’s inequality and the Sobolev-Gagliardo-Nirenberg
inequality for compactly supported smooth functions to conclude that
u Lp∗
(Rn) u − ui Lp∗
(Rn) + ui Lp∗
(Rn)
u − ui Lp∗
(Rn) + c Dui Lp(Rn)
u − ui Lp∗
(Rn) + c Dui − Du Lp(Rn) + Du Lp(Rn)
→ c Du Lp(Rn),
since ui → u in Lp∗
(Rn
) and Dui → Du in Lp
(Rn
). This completes the proof.
Remarks 2.3:
(1) The Gagliardo-Nirenberg-Sobolev inequality shows that if u ∈ W1,p
(Rn
)with
1 p < n, then u ∈ Lp
(Rn
)∩ Lp∗
(Rn
), with p∗
> p.
(2) The Gagliardo-Nirenberg-Sobolev inequality shows that if u ∈ W1,p
(Rn
)
with 1 p < n and Du = 0 almost everywhere in Rn
, then u = 0 almost
everywhere in Rn
.
(3) The Sobolev-Gagliardo-Nirenberg inequality holds for Sobolev spaces with
zero boundary values in open subsets of Rn
by considering the zero exten-
sions. There exists c = c(n, p) > 0 such that
ˆ
Ω
|u|p∗
dx
1
p∗
c
ˆ
Ω
|Du|p
dx
1
p
for every u ∈ W
1,p
0 (Ω), 1 p < n. If |Ω| < ∞, by Hölder’s inequality
ˆ
Ω
|u|q
dx
1
q
ˆ
Ω
|u|p∗
dx
1
p∗
|Ω|
1− 1
p∗
c|Ω|
1− 1
p∗
ˆ
Ω
|Du|p
dx
1
p
whenever 1 q p∗
. Thus for sets with ﬁnite measure all exponents below
the Sobolev exponent will do.
(4) The Sobolev-Gagliardo-Nirenberg inequality shows that W
1,p
loc
(Rn
) ⊂ L
p∗
loc
(Rn
).
To see this, let Ω Rn
and choose a cutoff function η ∈ C∞
0 (Rn
) such that
η = 1 in Ω. Then ηu ∈ W
1,p
0 (Rn
) = W1,p
(Rn
) and ηu = u in Ω and
u Lp∗
(Ω) ηu Lp∗
(Rn) c D(ηu) Lp(Rn) < ∞.
(5) The Sobolev-Gagliardo-Nirenberg inequality holds for higher order Sobolev
spaces as well. Let k ∈ N, 1 p < n
k and p∗
=
np
n−kp . There exists c =
c(n, p,k) > 0 such that
ˆ
Rn
|u|p∗
dx
1
p∗
c
ˆ
Rn
|Dk
u|p
dx
1
p
for every u ∈ Wk,p
(Ω). Here |Dk
u|2
is the sum of squares of all kth order
partial derivatives of u (exercise).

Remark 2.4. When p = 1 the Sobolev-Gagliardo-Nirenberg inequality is related
to the isoperimetric inequality. Let Ω ⊂ Rn
be a bounded domain with smooth
boundary and set
uε(x) =



1, x ∈ Ω,
1−
dist(x,Ω)
ε
, 0 < dist(x,Ω) < ε,
0, dist(x,Ω) ε.
Note that u can been considered as an approximation of the characteristic function
of Ω. The Lipschitz constant of x → dist(x,Ω) is one so that the Lipschitz constant
of uε is ε−1
and thus this function belongs to W1,1
(Rn
), for example, by the ACL
characterization, see Theorem 1.41, we have
|Duε(x)|



1
ε
, 0 < dist(x,Ω) < ε,
0, otherwise.
The Sobolev-Gagliardo-Nirenberg inequality with p = 1 gives
|Ω|
n−1
n =
ˆ
Ω
|uε|
n
n−1 dx
n−1
n
ˆ
Rn
|uε|
n
n−1 dx
n−1
n
c
ˆ
Rn
|Duε|dx c
ˆ
{0<dist(x,Ω)<ε}
1
ε
dx
= c
|{x ∈ Rn : 0 < dist(x,Ω) < ε}|
ε
→ cHn−1
( Ω)
This implies
|Ω|
n
n−1 cH n−1
( Ω),
which is an isoperimetric inequality with the same constant c as in the Sobolev-
Gagliardo-Nirenberg inequality. According to he classical isoperimetric inequality,
if Ω ⊂ Rn
is a bounded domain with smooth boundary, then
|Ω|
n−1
n n−1
Ω
− 1
n
n H n−1
( Ω),
where H n−1
( Ω) stands for the (n − 1)-dimensional Hausdorff measure of the
boundary Ω. The isoperimetric inequality is equivalent with the statement that
among all smooth bounded domains with ﬁxed volume, balls have the least surface
area.
Conversely, the Sobolev-Gagliardo-Nirenberg inequality can be proved by the
isoperimetric inequality, but we shall not consider this argument here. From
these considerations it is relatively obvious that the best constant in the Sobolev-
Gagliardo-Nirenberg when p = 1 should be the isoperimetric constant n−1
Ω
− 1
n
n .
This also gives a geometric motivation for the Sobolev exponent in the case p = 1.

2.2 Sobolev-Poincaré inequalities
We begin with a Poincaré inequality for Sobolev functions with zero boundary
values in open subsets.
Theorem 2.5 (Poincaré). Assume that Ω ⊂ Rn
is bounded and 1 p < ∞. Then
there is a constant c = c(p) such that
ˆ
Ω
|u|p
dx cdiam(Ω)p
ˆ
Ω
|Du|p
dx
for every u ∈ W
1,p
0 (Ω).
T H E M O R A L : This shows that W
1,p
0 (Ω) ⊂ Lp
(Ω) when 1 p < ∞, if Ω ⊂ Rn
is bounded. The main difference compared to the Gagliardo-Nirenberg-Sobolev
inequality is that that this applies for the whole range 1 p < ∞ without the
Sobolev exponent.
Remark 2.6. The Poincaré inequality above also shows that of Du = 0 almost
everywhere, then u = 0 almost everywhere. For this it is essential that the
function belongs to the Sobolev space with zero boundary values.
0 (Ω). Let y = (y1,..., yn) ∈ Ω. Then
Ω ⊂
n
j=1
yj −diam(Ω), yj +diam(Ω) =
n
j=1
aj,bj ,
where aj = yj −diam(Ω) and bj = yj +diam(Ω), j = 1,...,n. As the proof of Theorem
2.2, we obtain
|u(x)|
ˆ bj
aj
|Du(x1,...,tj,...,xn)|dtj
(2diam(Ω))
1− 1
p
ˆ bj
aj
|Du(x1,...,tj,...,xn)|p
dtj
1
p
, j = 1,...,n.
The second inequality follows from Hölder’s inequality. Thus
ˆ
Ω
|u(x)|p
dx =
ˆ b1
a1
...
ˆ bn
an
|u(x)|p
dx1 ... dxn
(2diam(Ω))p−1
ˆ b1
a1
...
ˆ bn
an
ˆ b1
a1
|Du(t1,x2,...,xn)|p
dt1 dx1 ... dxn
(2diam(Ω))p
ˆ b1
a1
...
ˆ bn
an
|Du(t1,x2,...,xn)|p
dt1 ... dxn
= (2diam(Ω))p
ˆ
Ω
|Du(x)|p
dx.
(2) The case u ∈ W
1,p
0 (Ω) follows by approximation (exercise).

The Gagliardo-Nirenberg-Sobolev inequality in Theorem 2.2 and Poincaré’s
inequality in Theorem 2.5 do not hold for functions u ∈ W1,p
(Ω), at least when
Ω ⊂ Rn
is an open set |Ω| < ∞, since nonzero constant functions give obvious
counterexamples. However, there are several ways to obtain appropriate local
estimates also in this case.
Next we consider estimates in the case when Ω is a cube. Later we consider
similar estimates for balls. The set
Q = [a1,b1]×...×[an,bn], b1 − a1 = ... = bn − an
is a cube in Rn
. The side length of Q is
l(Q) = b1 − a1 = bj − aj, j = 1,...,n,
and
Q(x,l) = y ∈ Rn
: |yj − xj| l
2 , j = 1,...,n
is the cube with center x and sidelength l. Clearly,
|Q(x,l)| = ln
and diam(Q(x,l)) = nl
The integral average of f ∈ L1
loc
(Rn
) over cube Q(x,l) is denoted by
fQ(x,l) =
Q(x,l)
f dy =
1
|Q(x,l)|
ˆ
Q(x,l)
f (y)dy.
Same notation is used for integral averages over other sets as well.
Theorem 2.7 (Poincaré inequality on cubes). Let Ω be an open subset of Rn
.
Assume that u ∈ W
1,p
loc
(Ω) with 1 p < ∞. Then there is c = c(n, p) such that
Q(x,l)
|u − uQ(x,l)|p
dy
1
p
cl
Q(x,l)
|Du|p
dy
1
p
for every cube Q(x,l) Ω.
T H E M O R A L : The Poincaré inequality shows that if the gradient is small in
a cube, then the mean oscillation of the function is small in the same cube. In
particular, if the gradient is zero, then the function is constant.
(Ω). Let z, y ∈ Q = Q(x,l) = [a1,b1] × ··· ×
[an,bn]. Then
|u(z)− u(y)| |u(z)− u(z1,..., zn−1, yn)|+...+|u(z1, y2,..., yn)− u(y)|
n
j=1
ˆ bj
aj
|Du(z1,..., zj−1,t, yj+1,..., yn)|dt

By Hölder’s inequality and the elementary inequality (a1 +···+ an)p
np
(a
p
1 +
···+ a
p
n), ai 0, we obtain
|u(z)− u(y)|p
n
j=1
ˆ bj
aj
|Du(z1,..., zj−1,t, yj+1,..., yn)|dt
p


n
j=1
ˆ bj
aj
|Du(z1,..., zj−1,t, yj+1,..., yn)|p
dt
1
p
(bj − aj)
1− 1
p


p
np
lp−1
n
j=1
ˆ bj
aj
|Du(z1,..., zj−1,t, yj+1,..., yn)|p
dt.
By Hölder’s inequality and Fubini’s theorem
ˆ
Q
|u(z)− uQ|p
dz =
ˆ
Q Q
(u(z)− u(y))dy
p
dz
ˆ
Q Q
|u(z)− u(y)|dy
p
dz
ˆ
Q Q
|u(z)− u(y)|p
dz dy
np
lp−1
|Q|
n
j=1
ˆ
Q
ˆ
Q
ˆ bj
aj
|Du(z1,..., zj−1,t, yj+1,..., yn)|p
dtdydz
np
lp−1
|Q|
n
j=1
(bj − aj)
ˆ
Q
ˆ
Q
|Du(z)|p
dz dw
np+1
lp
ˆ
Q
|Du(z)|p
dz.
(2) The case u ∈ W
1,p
loc
(Ω) follows by approximation (exercise).
Theorem 2.8 (Sobolev-Poincaré inequality on cubes). Let Ω be an open sub-
set of Rn
. Assume that u ∈ W
1,p
loc
(Ω) with 1 p < n. Then there is c = c(n, p) such
that
Q(x,l)
|u − uQ(x,l)|p∗
dy
1
p∗
cl
Q(x,2l)
|Du|p
dy
1
p
for every cube Q(x,l) Ω
T H E M O R A L : The Sobolev-Poincaré inequality shows that W
1,p
loc
(Rn
) ⊂ L
p∗
loc
(Rn
),
when 1 p < n. This is a stronger version of the Poincaré inequality on cubes in
which we have the Sobolev exponent on the left-hand side.
Proof. Let η ∈ C∞
0 (Rn
) be a cutoff function such that
0 η 1, |Dη| c
l , suppη ⊂ Q(x,2l) and η = 1 in Q(x,l).
Notice that the constant c = c(n) does not depend on the cube. Then (u−uQ(x,l))η ∈
W1,p
(Rn
) and by the Gagliardo-Nirenberg-Sobolev inequality, see Theorem 2.2,

and the Leibniz rule, see Theorem 1.12 (5), we have
ˆ
Q(x,l)
|u − uQ(x,l)|p∗
dy
1
p∗
ˆ
Rn
|(u − uQ(x,l))η|p∗
dy
1
p∗
c
ˆ
Rn
D (u − uQ(x,l))η
p
dy
1
p
c
ˆ
Rn
ηp
|Du|p
dy
1
p
+ c
ˆ
Rn
|Dη|p
|u − uQ(x,l)|p
dy
1
p
c
ˆ
Q(x,2l)
|Du|p
dy
1
p
+
c
l
ˆ
Q(x,2l)
|u − uQ(x,l)|p
dy
1
p
.
By the Poincaré inequality on cubes, see Theorem 2.7, we obtain
ˆ
Q(x,2l)
|u − uQ(x,l)|p
dy
1
p
ˆ
Q(x,2l)
|u − uQ(x,2l)|p
dy
1
p
+
ˆ
Q(x,2l)
|uQ(x,2l) − uQ(x,l)|p
dy
1
p
cl
ˆ
Q(x,2l)
|Du|p
dy
1
p
+|uQ(x,2l) − uQ(x,l)||Q(x,2l)|
1
p .
By Hölder’s inequality and Poincaré inequality on cubes, see Theorem 2.7, we
have
|uQ(x,2l) − uQ(x,l)||Q(x,2l)|
1
p (2l)
n
p
Q(x,l)
|u − uQ(x,2l)|dy
(2l)
n
p
|Q(x,2l)|
|Q(x,l)| Q(x,2l)
|u − uQ(x,2l)|p
dy
1
p
cl
ˆ
Q(x,2l)
|Du|p
dy
1
p
.
By collecting the estimates above we obtain
ˆ
Q(x,l)
|u − uQ(x,l)|p∗
dy
1
p∗
c
ˆ
Q(x,2l)
|Du|p
dy
1
p
.
Remark 2.9. The Sobolev-Poincaré inequality also holds in the form
Q(x,l)
|u − uQ(x,l)|p∗
dy
1
p∗
cl
Q(x,l)
|Du|p
dy
1
p
.
Observe that there is the same cube on both sides. We shall return to this question
later.
Remark 2.10. In this remark we consider the case p = n.
(1) As Example 1.10 shows, functions in W1,n
(Rn
) are not necessarily bounded.

(2) Assume that u ∈ W1,n
(Rn
). The Poincaré inequality implies that
Q
|u(y)− uQ|dy
Q
|u(y)− uQ|n
dy
1
n
cl
Q
|Du(y)|n
dy
1
n
c Du Ln(Rn) < ∞
for every cube Q where c = c(n). Thus if u ∈ W1,n
(Rn
), then u is of bounded
mean oscillation, denoted by u ∈ BMO(Rn
), and
u ∗ = sup
Q⊂Rn Q
|u(y)− uQ|dy c Du Ln(Rn),
where c = c(n).
(3) Assume that u ∈ W1,n
(Rn
). The John-Nirenberg inequality for BMO func-
tions gives
Q
eγ|u(x)−uQ|
dx
c1γ u ∗
c2 −γ u ∗
+1
for every cube Q in Rn
with 0 < γ < c2
u ∗
, where c1 = c1(n) and c2 = c2(n).
By choosing γ = c2
2 u ∗
, we obtain
Q
e
c
|u(x)−uQ|
Du n dx
Q
ec
|u(x)−uQ|
u ∗ dx c
. In particular, this implies that u ∈ L
p
loc
(Rn
) for
every power p, with 1 p < ∞. This is the Sobolev embedding theorem in
the borderline case when p = n.
In fact, there is a stronger result called Trudinger’s inequality, which states
that for small enough c > 0, we have
Q
e
c
|u(x)−uQ|
Du n
n
n−1
dx c
, n 2, but we shall not discuss this issue here
T H E M O R A L : W1,n
(Rn
) ⊂ L
p
loc
(Rn
) for every p, with 1 p < ∞. This is the
Sobolev embedding theorem in the borderline case when p = n.
The next remark shows that it is possible to obtain a Poincaré inequality
on cubes without the integral average also for functions that do not have zero
boundary values. However, the functions have to vanish in a large subset.
Remark 2.11. Assume u ∈ W1,p
(Rn
) and u = 0 in a set A ⊂ Q(x,l) = Q satisfying
|A| γ|Q| for some 0 < γ 1.

This means that u = 0 in a large portion of Q. By the Poincaré inequality there
exists c = c(n.p) such that
Q
|u|p
dy
1
p
Q
|u − uQ|p
dy
1
p
+
Q
|uQ|p
dy
1
p
cl
Q
|Du|p
dy
1
p
+|uQ|,
where
|uQ| =
Q
u(y)dy
Q
χQA(y)|u(y)|dy
|Q A|
|Q|
1− 1
p
Q
|u(y)|p
dy
1
p
(1− c)
1− 1
p
Q
|u(y)|p
dy
1
p
.
Since 0 (1− c)
1− 1
p < 1, we may absorb the integral average to the left hand side
and obtain
(1−(1− c)
1− 1
p )
Q
|u|p
dy
1
p
cl
Q
|Du|p
dy
1
p
.
It follows that there exists c = c(n, p,γ) such that
Q
|u|p
dy
1
p
cl
Q
|Du|p
dy
1
p
.
A similar argument can be done with the Sobolev-Poincaré inequality on cubes
(exercise).
2.3 Morrey’s inequality
Let A ⊂ Rn
. A function u : A → R is Hölder continuous with exponent 0 < α 1, if
there exists a constant c such that
|u(x)− u(y)| c|x− y|α
for every x, y ∈ A. We deﬁne the space C0,α
(A) to be the space of all bounded
functions that are Hölder continuous with exponent α with the norm
u Cα(A) = sup
x∈A
|u(x)|+ sup
x,y∈A,x=y
|u(x)− u(y)|
|x− y|α
. (2.2)
Remarks 2.12:
(1) Every function that is Hölder continuous with exponent α > 1 in the whole
space is constant (exercise).

(2) There are Hölder continuous functions that are not differentiable at any
point. Thus Hölder continuity does not imply any differentiability proper-
ties.
(3) C0,α
(A) is a Banach space with the norm defined above (exercise).
(4) Every Hölder continuous function on A ⊂ Rn
can be extended to a Hölder
continuous function on Rn
with the same exponent and same constant.
Moreover, if A is bounded, we may assume that the Hölder continuous
extension to Rn
is bounded (exercise).
The next result shows that every function in W1,p
(Rn
) with p > n has a 1− n
p -
Hölder continuous representative up to a set of measure zero.
Theorem 2.13 (Morrey). Assume that u ∈ W1,p
(Rn
) with p > n. Then there is
c = c(n, p) > 0 such that
|u(z)− u(y)| c|z − y|
1− n
p Du Lp(Rn)
for almost every z, y ∈ Rn
.
Proof. (1) Assume first that u ∈ C∞
(Rn
)∩W1,p
(Rn
). Let x, y ∈ Q(x,l). Then
u(z)− u(y) =
ˆ 1
0 t
u(tz +(1− t)y) dt =
ˆ 1
0
Du(tz +(1− t)y)·(z − y)dt
and
|u(y)− uQ(x,l)| =
Q(x,l)
(u(z)− u(y))dz
=
Q(x,l)
ˆ 1
0
Du(tz +(1− t)y)·(z − y)dtdz
n
j=1
1
ln
ˆ
Q(x,l)
ˆ 1
0
u
xj
(tz +(1− t)y) |zj − yj|dtdz
n
j=1
1
ln−1
ˆ 1
0
ˆ
Q(x,l)
u
xj
(tz +(1− t)y) dz dt
=
n
j=1
1
ln−1
ˆ 1
0
1
tn
ˆ
Q(tx+(1−t)y,tl)
u
xj
(w) dwdt.
Here we used the fact that |zj − yj| l, Fubini’s theorem and finally the change of
variables w = tz+(1−t)y ⇐⇒ z = 1
t (w−(1−t)y), dz = 1
tn dw. By Hölder’s inequality
n
j=1
1
ln−1
ˆ 1
0
1
tn
ˆ
Q(tx+(1−t)y,tl)
u
xj
(w) dwdt
n
j=1
1
ln−1
ˆ 1
0
1
tn
ˆ
Q(tx+(1−t)y,tl)
u
xj
(w)
p
dw
1
p
|Q(tx+(1− t)y,tl)|
1
p dt
n Du Lp(Q(x,l))
l
n(1− 1
p )
ln−1
ˆ 1
0
t
n(1− 1
p )
tn
dt (Q(tx+(1− t)y,tl) ⊂ Q(x,l)))
=
np
p − n
l
1− n
p Du Lp(Q(x,l)).

Thus
|u(z)− u(y)| |u(z)− uQ(x,l)|+|uQ(x,l) − u(y)|
2
np
p − n
l
1− n
p Du Lp(Q(x,l)) (2.3)
for every z, y ∈ Q(x,l).
For every z, y ∈ Rn
, there exists a cube Q(x,l) z, y such that l = |z − y|. For
example, we may choose x =
z+y
2 . Thus
|u(z)− u(y)| c|z − y|
1− n
p Du Lp(Q(x,l)) c|z − y|
1− n
p Du Lp(Rn)
for every z, y ∈ Rn
.
(2) Assume then that u ∈ W1,p
(Rn
). Let uε be the standard mollification of u.
Then
|uε(z)− uε(y)| c|z − y|
1− n
p Duε Lp(Rn).
Now by Lemma 1.16 (2) and by Theorem 1.17, we obtain
|u(z)− u(y)| c|z − y|
1− n
p Du Lp(Rn).
when z and y are Lebesgue points of u. The claim follows from the fact that almost
every point of a locally integrable function is a Lebesgue point.
Remarks 2.14:
(1) Morrey’s inequality implies that u can be extended uniquely to Rn
as a
Hölder continuous function u such that
|u(x)− u(y)| c|x− y|
1− n
p Du Lp(Rn) for all x, y ∈ Rn
.
Reason. Let N be a set of zero measure such that Morrey’s inequality holds
for all points in Rn
N. Now for any x ∈ Rn
, choose a sequence of points
(xi) such that xi ∈ Rn
N, i = 1,2..., and xi → x as i → ∞. By Morrey’s
inequality (u(xi)) is a Cauchy sequence in R and thus we can define
u(x) = lim
i→∞
u(xi).
Now it is easy to check that u satisfies Morrey’s inequality in every pair of
points by considering sequences of points in Rn
N converging to the pair
of points.
(2) If u ∈ W1,p
(Rn
) with p > n, then u is essentially bounded.

Reason. Let y ∈ Q(x,1). Then Morrey’s and Hölder’s inequality imply
|u(z)| |u(z)− uQ(x,1)|+|uQ(x,1)|
Q(x,1)
|u(z)− u(y)|dy+
ˆ
Q(x,1)
|u(y)|dy
c Du Lp(Rn) +
ˆ
Q(x,1)
|u(y)|p
dy
1
p
c u W1,p(Rn)
for almost every z ∈ Rn
. Thus u L∞(Rn) c u W1,p(Rn).
This implies that
u
C
0,1− n
p (Rn)
c u W1,p(Rn), c = c(n, p),
where u is the Hölder continuous representative of u. Hence W1,p
(Rn
) is
continuously embedded in C
0,1− n
p (Rn
), when p > n.
(3) The proof of Theorem 2.13, see (2.3), shows that if Ω is an open subset of
Rn
and u ∈ W
1,p
loc
(Ω), p > n, then there is c = c(n, p) such that
|u(z)− u(y)| c|z − y|
1− n
p Du Lp(Q(x,l))
for every z, y ∈ Q(x,l), Q(x,l) Ω. This is a local version of Morrey’s
inequality.
T H E M O R A L : W1,p
(Rn
) ⊂ C
0,1− n
p (Rn
), when p > n. More precisely, W1,p
(Rn
) is
continuously embedded in C
0,1− n
p (Rn
), when p > n. This is the Sobolev embedding
theorem for p > n.
Deﬁnition 2.15. A function u : Rn
→ R is differentiable at x ∈ Rn
if there exists a
linear mapping L : Rn
→ R such that
lim
y→x
|u(y)− u(x)− L(x− y)|
|x− y|
= 0.
If such a linear mapping L exists at x, it is unique and we denote L = Du(x) and
call Du(x) the derivative of u at x.
Theorem 2.16. If u ∈ W
1,p
loc
(Rn
), n < p ∞, then u is differentiable almost every-
where and its derivative equals its weak derivative almost everywhere.
T H E M O R A L : By the ACL characterization, see Theorem 1.41, we know
that every function in W1,p
, 1 p ∞ has classical partial derivatives almost
everywhere. If p > n, then every function in W1,p
is also differentiable almost
everywhere.

Proof. Since W
1,∞
loc
(Rn
) ⊂ W
1,p
loc
(Rn
), we may assume n < p < ∞. By the Lebesgue
differentiation theorem
lim
l→0 Q(x,l)
|Du(z)− Du(x)|p
dz = 0
for almost every x ∈ Rn
. Let x be such a point and denote
v(y) = u(y)− u(x)− Du(x)·(y− x),
where y ∈ Q(x,l). Observe that v ∈ W
1,p
loc
(Rn
) with n < p < ∞. By (2.3) in the proof
of Morrey’s inequality, there is c = c(n, p) such that
|v(y)− v(x)| cl
Q(x,l)
|Dv(z)|p
dz
1
p
where l = |x− y|. Since v(x) = 0 and Dv = Du − Du(x), we obtain
|u(y)− u(x)− Du(x)·(y− x)|
|y− x|
c
Q(x,l)
|Du(z)− Du(x)|p
dz
1
p
→ 0
as y → x.
2.4 Lipschitz functions and W1,∞
Let A ⊂ Rn
and 0 L < ∞. A function f : A → R is called Lipschitz continuous with
constant L, or an L-Lipschitz function, if
|f (x)− f (y)| L|x− y|
for every x, y ∈ Rn
. Observe that a function is Lipschitz continuous if it is Hölder
continuous with exponent one. Moreover, C0,1
(A) is the space of all bounded
Lipschitz continuous functions with the norm (2.2).
Examples 2.17:
(1) For every y ∈ Rn
the function x → |x − y| is Lipschitz continuous with
constant one. Note that this function is not smooth.
(2) For every nonempty set A ⊂ Rn
the function x → dist(x, A) is Lipschitz
continuous with constant one. Note that this function is not smooth when
A = Rn
(exercise).
The next theorem describes the relation between Lipschitz functions and
Sobolev functions.
Theorem 2.18. A function u ∈ L1
loc
(Rn
) has a representative that is bounded and
Lipschitz continuous if and only if u ∈ W1,∞
(Rn
).

T H E M O R A L : The Sobolev embedding theorem for p > n shows that W1,p
(Rn
) ⊂
C
0,1− n
p (Rn
). In the limiting case p = ∞ we have W1,∞
(Rn
) = C0,1
(Rn
). This is the
Sobolev embedding theorem for p = ∞.
Proof. ⇐= Assume that u ∈ W1,∞
(Rn
). Then u ∈ L∞
(Rn
) and u ∈ W
1,p
loc
(Rn
) for
every p > n and thus by Remark 2.14 we may assume that u is a bounded
continuous function. Moreover, we may assume that the support of u is compact.
By Lemma 1.16 (3) and by Theorem 1.17, the standard molliﬁcation uε ∈ C∞
0 (Rn
)
for every ε > 0, uε → u uniformly in Rn
as ε → 0 and
Duε L∞(Rn) Du L∞(Rn)
for every ε > 0. Thus
|uε(x)− uε(y)| =
ˆ 1
0
Duε(tx+(1− t)y)·(x− y)dt
Duε L∞(Rn)|x− y|
Du L∞(Rn)|x− y|
. By letting ε → 0, we obtain
|u(x)− u(y)| Du L∞(Rn)|x− y|
.
=⇒ Assume that u is Lipschitz continuous. Then there exists L such that
|u(x)− u(y)| L|x− y|
. This implies that
D−h
j u(x) =
u(x− he j)− u(x)
h
L
for every x ∈ Rn
and h = 0. This means that
D−h
j u(x)
L∞(Rn)
L
for every h = 0 and thus
D−h
j u(x)
L2(Ω)
D−h
j u(x)
L∞(Rn)
|Ω|
1
2 L|Ω|
1
2 ,
where Ω ⊂ Rn
is bounded and open.
As in the proof of Theorem 1.38, the space L2
(Ω) is reﬂexive and thus there
exists gj ∈ L2
(Ω), j = 1,...,n, and a subsequence of hi → 0 such that D
−hi
j
u → gj
weakly in L2
(Ω). As in the proof of Theorem 1.38, we obtain
ˆ
Ω
u
ϕ
xj
dx =
ˆ
Ω
( lim
hi→0
D
hi
j
ϕ)u dx = lim
hi→0
ˆ
Ω
uD
hi
j
ϕdx
= lim
hi→0
ˆ
Ω
(D
−hi
j
u)ϕdx =
ˆ
Ω
gjϕdx

0 (Ω) and thus Dju = −gj, j = 1,...,n, in the weak sense.
Claim: Dju ∈ L∞
(Ω), j = 1,...,n,
Reason. Let fi = D
−hi
j
u, i = 1,2.... Since fi → Dju weakly in L2
(Ω), by Mazur’s
lemma, see theorem 1.31, there exists a subsequence (fil
) such that the convex
combitions
k
l=1
ai fil
→ Dju in Lp
(Ω) as k → ∞.
Observe that
k
l=1
ai fil
L∞(Ω)
k
l=1
ai D
−hil
j
u(x)
L∞(Ω)
L.
Since there exists a subsequence that converges almost everywhere, we conclude
that
Dju(x) L, j = 1,...,n,
for almost every x ∈ Ω.
This shows that Du ∈ L∞
(Ω), with Du L∞(Ω) L. As u is bounded, this
implies u ∈ W1,∞
(Ω) for all bounded subsets Ω ⊂ Rn
. Since the norm does not
depend on Ω, we conclude that u ∈ W1,∞
(Rn
).
A direct combination of Theorem 2.18 and Theorem 2.16 gives a proof for
Rademacher’s theorem.
Corollary 2.19 (Rademacher). Let f : Rn
→ R be locally Lipschitz continuous.
Then f is differentiable almost everywhere.
W A R N I N G : For an open subset Ω of Rn
, Morrey’s inequality and the charac-
terization of Lipschitz continuous functions holds only locally, that is, W1,p
(Ω) ⊂
C
0,1− n
p
loc
(Ω), when p > n and W1,∞
(Ω) ⊂ C
0,1
loc
(Ω).
Example 2.20. Let
Ω = {x ∈ R2
: 1 < |x| < 2}{(x1,0) ∈ R2
: x1 1} ⊂ R2
.
Then we can construct functions such that u ∈ W1,∞
(Ω), but u ∈ C0,α
(Ω), for
example, by deﬁning u(x) = θ, where θ is the argument of x in polar coordinates
with 0 < θ < 2π. Then u ∈ W1,∞
(Ω), but u is not Lipschitz continuous in Ω.
However, it is locally Lipschitz continuous in Ω.
2.5 Summary of the Sobolev embeddings
We summarize the results related to Sobolev embeddings below. Assume that Ω is
an open subset of Rn
.

1 p < n W1,p
(Rn
) ⊂ Lp∗
(Rn
), W
1,p
loc
(Ω) ⊂ L
p∗
loc
(Ω), p∗
=
np
n−p (Theorem 2.2
and Theorem 2.8).
p = n W1,n
(Rn
) ⊂ BMO(Rn
), W
1,n
loc
(Ω) ⊂ L
p
loc
(Ω) for every p, with 1 p < ∞
(Remark 2.10 (3)).
n < p < ∞ W1,p
(Rn
) ⊂ C
0,1− n
p (Rn
), W
1,p
loc
(Ω) ⊂ C
0,1− n
p
loc
(Ω) (Theorem 2.13).
p = ∞ W1,∞
(Rn
) = C0,1
(Rn
), W
1,∞
loc
(Ω) = C
0,1
loc
(Ω) (Theorem 2.18).
2.6 Direct methods in the calculus of vari-
ations
Sobolev space methods are important in existence results for PDEs. Assume that
Ω ⊂ Rn
is a bounded open set. Consider the Dirichlet problem



∆u = 0 in Ω,
u = g on Ω.
Let u ∈ C2
(Ω) be a classical solution to the Laplace equation
∆u =
n
j=1
2
u
x2
j
= 0
and let ϕ ∈ C∞
0 (Ω). An integration by parts gives
0 =
ˆ
Ω
ϕ∆u dx =
ˆ
Ω
ϕdivDu dx =
ˆ
Ω
n
j=1
2
u
x2
j
ϕdx
=
n
j=1
ˆ
Ω
2
u
x2
j
ϕdx =
n
j=1
ˆ
Ω
u
xj
ϕ
xj
dx =
ˆ
Ω
Du · Dϕdx
0 (Ω). Conversely, if u ∈ C2
(Ω) and
ˆ
Ω
Du · Dϕdx = 0 for every ϕ ∈ C∞
0 (Ω),
then by the computation above
ˆ
Ω
ϕ∆u dx = 0 for every ϕ ∈ C∞
0 (Ω).
By the fundamental lemma in the calculus of variations, see Corollary 1.5, we
conclude ∆u = 0 in Ω.
T H E M O R A L : Assume u ∈ C2
(Ω). Then ∆u = 0 in Ω if and only if
ˆ
Ω
Du · Dϕdx = 0 for every ϕ ∈ C∞
0 (Ω).
This gives a motivation to the deﬁnition below.

Definition 2.21. A function u ∈ W1,2
(Ω) is a weak solution to ∆u = 0 in Ω, if
ˆ
Ω
Du · Dϕdx = 0
0 (Ω).
T H E M O R A L : There are second order derivatives in the definition of a classical
solution to the Laplace equation, but in the definition above is enough to assume
that only first order weak derivatives exist.
Th next lemma shows that, in the definition of a weak solution, the class of
test functions can be taken to be the Sobolev space with zero boundary values.
Lemma 2.22. If u ∈ W1,2
(Ω) is a weak solution to the Laplace equation, then
ˆ
Ω
Du · Dvdx = 0
for every v ∈ W
1,2
0 (Ω).
Proof. Let vi ∈ C∞
0 (Ω), i = 1,2,..., be such that vi → v in W1,p
(Ω). Then by the
Cauchy-Schwarz inequality and Hölder’s inequality, we have
ˆ
Ω
Du · Dvdx−
ˆ
Ω
Du · Dvi dx =
ˆ
Ω
Du ·(Dv− Dvi)dx
ˆ
Ω
|Du||Dv− Dvi|dx
ˆ
Ω
|Du|2
dx
1
2
ˆ
Ω
|Dv− Dvi|2
dx
1
2
→ 0
as i → ∞. Thus ˆ
Ω
Du · Dvdx = lim
i→∞
ˆ
Ω
Du · Dvi dx = 0.
Remark 2.23. Assume that Ω ⊂ Rn
is bounded and g ∈ W1,2
(Ω). If there exists a
weak solution u ∈ W1,2
(Ω) to the Dirichlet problem



∆u = 0 in Ω,
u − g ∈ W
1,2
0 (Ω),
then the solution is unique. Observe that the boundary values are taken in the
Sobolev sense.
Reason. Let u1 ∈ W1,2
(Ω), with u1 − g ∈ W
1,2
0 (Ω), and u2 ∈ W1,2
(Ω), with u2 − g ∈
W
1,2
0 (Ω), be solutions to the Dirichlet problem above. By Lemma 2.22
ˆ
Ω
Du1 · Dvdx = 0 and
ˆ
Ω
Du2 · Dvdx = 0

for every v ∈ W
1,2
0 (Ω) and thus
ˆ
Ω
(Du1 − Du2)· Dvdx = 0 for every v ∈ W
1,p
0 (Ω).
Since
u1 − u2 = (u1 − g)
∈W
1,2
0 (Ω)
−(u2 − g)
∈W
1,2
0 (Ω)
∈ W
1,2
0 (Ω),
we may choose v = u1 − u2 and conclude
ˆ
Ω
|Du1 − Du2|2
dx =
ˆ
Ω
(Du1 − Du2)·(Du1 − Du2)dx = 0.
This implies Du1 − Du2 = 0 almost everywhere in Ω. By the Poincaré inequality,
see Theorem 2.5, we have
ˆ
Ω
|u1 − u2|2
dx cdiam(Ω)2
ˆ
Ω
|Du1 − Du2|2
dx = 0.
This implies u1 − u2 = 0 ⇐⇒ u1 = u2 almost everywhere in Ω. This is a PDE
proof of uniqueness and in the proof of Theorem 2.26 we shall see a variational
argument for the same result.
Next we consider a variational approach to the Dirichlet problem for the
Laplace equation.
Deﬁnition 2.24. Assume that g ∈ W1,2
(Ω). A function u ∈ W1,2
(Ω) with u − g ∈
W
1,2
0 (Ω) is a minimizer of the variational integral
I(u) =
ˆ
Ω
|Du|2
dx
with boundary values g, if
ˆ
Ω
|Du|2
dx
ˆ
Ω
|Dv|2
dx
for every v ∈ W1,2
(Ω) with v− g ∈ W
1,2
0 (Ω).
T H E M O R A L : A minimizer u minimizes the variational integral I(u) in the
class of functions with given boundary values, that is,
ˆ
Ω
|Du|2
dx = inf
ˆ
Ω
|Dv|2
dx : v ∈ W1,2
(Ω), v− g ∈ W
1,2
0 (Ω) .
If there is a minimizer, then inﬁmum can be replaced by minimum.
Theorem 2.25. Assume that g ∈ W1,2
(Ω) and u ∈ W1,2
(Ω) with u − g ∈ W
1,2
0 (Ω).
Then ˆ
Ω
|Du|2
dx = inf
ˆ
Ω
|Dv|2
dx : v ∈ W1,2
(Ω), v− g ∈ W
1,p
0 (Ω)
if and only if u is a weak solution to the Dirichlet problem



∆u = 0 in Ω,
u − g ∈ W
1,2
0 (Ω).

T H E M O R A L : A function is a weak solution to the Dirichlet problem if and
only if it is a minimizer of the corresponding variational integral with the given
boundary values in the Sobolev sense.
Proof. =⇒ Assume that u ∈ W1,2
(Ω) is a minimizer with boundary values g ∈
W1,2
(Ω). We use the method of variations by Lagrange. Let ϕ ∈ C∞
0 (Ω) and ε ∈ R.
Then (u +εϕ)− g ∈ W
1,2
0 (Ω) and
ˆ
Ω
|D(u +εϕ)|2
dx =
ˆ
Ω
(Du +εDϕ)·(Du +εDϕ)dx
=
ˆ
Ω
|Du|2
dx+2ε
ˆ
Ω
Du · Dϕdx+ε2
ˆ
Ω
|Dϕ|2
dx
= i(ε).
Since u is a minimizer, i(ε) has minimum at ε = 0, which implies that i (0) = 0.
Clearly
i (ε) = 2
ˆ
Ω
Du · Dϕdx+2ε
ˆ
Ω
|Dϕ|2
dx
and thus
i (0) = 2
ˆ
Ω
Du · Dϕdx = 0.
This shows that ˆ
Ω
Du · Dϕdx = 0
0 (Ω).
⇐= Assume that u ∈ W1,2
(Ω) is a weak solution to ∆u = 0 with u− g ∈ W
1,2
0 (Ω)
and let v ∈ W1,2
1,2
0 (Ω). Then
ˆ
Ω
|Dv|2
dx =
ˆ
Ω
|D(v− u)+ Du|2
dx
=
ˆ
Ω
(D(v− u)+ Du)·(D(v− u)+ Du)dx
=
ˆ
Ω
|D(v− u)|2
dx+2
ˆ
Ω
D(v− u)· Du dx+
ˆ
Ω
|Du|2
dx.
Since
v− u = (v− g)
∈W
1,2
0 (Ω)
− (u − g)
∈W
1,2
0 (Ω)
∈ W
1,2
0 (Ω),
by Lemma 2.22 we have ˆ
Ω
Du · D(v− u)dx = 0
and thus
ˆ
Ω
|Dv|2
dx =
ˆ
Ω
|D(v− u)|2
dx+
ˆ
Ω
|Du|2
dx
ˆ
Ω
|Du|2
dx
for every v ∈ W1,2
1,2
0 (Ω). Thus u is a minimizer.

Next we give an existence proof using the direct methods in the calculus
variations. This means that, instead of the PDE, the argument uses the variational
integral.
Theorem 2.26. Assume that Ω is a bounded open subset of Rn
. Then for every
g ∈ W1,2
(Ω) there exists a unique minimizer u ∈ W1,2
(Ω) with u − g ∈ W
1,2
0 (Ω),
which satisfies
ˆ
Ω
|Du|2
dx = inf
ˆ
Ω
|Dv|2
dx : v ∈ W1,2
(Ω), v− g ∈ W
1,2
0 (Ω) .
T H E M O R A L : The Dirichlet problem for the Laplace equation has a unique
solution with Sobolev boundary values in any bounded open set.
W A R N I N G : It is not clear whether the solution to the variational problem
attains the boundary values pointwise.
Proof. (1) Since I(u) 0, in particular, it is bounded from below in W1,2
(Ω) and
since u is a minimizer, g ∈ W1,2
(Ω) and g − g = 0 ∈ W
1,2
0 (Ω), we note that
0 m = inf
ˆ
Ω
|Du|2
dx : u ∈ W1,2
(Ω), u − g ∈ W
1,2
0 (Ω)
ˆ
Ω
|Dg|2
dx < ∞.
The definition of infimum then implies that there exists a minimizing sequence
ui ∈ W1,2
(Ω) with ui − g ∈ W
1,2
0 (Ω), i = 1,2,..., such that
lim
i→∞
ˆ
Ω
|Dui|2
dx = m.
The existence of the limit implies the sequence (I(ui)) is bounded. Thus there
exists a constant M < ∞ such that
I(ui) =
ˆ
Ω
|Dui|2
dx M for every i = 1,2,...,
(2) By the Poincaré inequality, see Theorem 2.5, we obtain
ˆ
Ω
|ui − g|2
dx+
ˆ
Ω
|D(ui − g)|2
dx
cdiam(Ω)2
ˆ
Ω
|D(ui − g)|2
dx+
ˆ
Ω
|D(ui − g)|2
dx
(cdiam(Ω)2
+1)
ˆ
Ω
|Dui − Dg|2
dx
(cdiam(Ω)2
+1) 2
ˆ
Ω
|Dui|2
dx+2
ˆ
Ω
|Dg|2
dx
c(diam(Ω)2
+1) M +
ˆ
Ω
|Dg|2
dx < ∞
for every i = 1,2,... This shows that (ui − g) is a bounded sequence in W
1,2
0 (Ω).

(3) By reﬂexivity of W
1,2
0 (Ω), see Theorem 1.33, there is a subsequence (uik
−g)
and a function u ∈ W1,2
(Ω), with u−g ∈ W
1,2
0 (Ω), such that uik
→ u weakly in L2
(Ω)
and
uik
xj
→ u
xj
, j = 1,...,n, weakly in L2
(Ω) as k → ∞. By lower semicontinuity of
L2
-norm with respect to weak convergence, see (1.3), we have
ˆ
Ω
|Du|2
dx liminf
k→∞
ˆ
Ω
|Duik
|2
dx = lim
i→∞
ˆ
Ω
|Dui|2
dx.
Since u ∈ W1,2
(Ω), with u − g ∈ W
1,2
0 (Ω), we have
m
ˆ
Ω
|Du|2
dx lim
i→∞
ˆ
Ω
|Dui|2
dx = m
which implies ˆ
Ω
|Du|2
dx = m.
Thus u is a minimizer.
(4) To show uniqueness, let u1 ∈ W1,2
1,2
0 (Ω) and u2 ∈
W1,2
1,2
0 (Ω) be minimizers of I(u) with the same boundary
function g ∈ W1,2
(Ω). Assume that u1 = u2, that is, |{x ∈ Ω : u1(x) = u2(x)}| > 0. By
the Poincaré inequality, see Theorem 2.5, we have
0 <
ˆ
Ω
|u1 − u2|2
dx cdiam(Ω)2
ˆ
Ω
|Du1 − Du2|2
dx
and thus |{x ∈ Ω : Du1(x) = Du2(x)}| > 0. Let v = u1+u2
2 . Then v ∈ W1,2
(Ω) and
v− g =
1
2
(u1 − g)
∈W
1,2
0 (Ω)
+
1
2
(u2 − g)
∈W
1,2
0 (Ω)
∈ W
1,2
0 (Ω).
By strict convexity of ξ → |ξ|2
we conclude that
|Dv|2
<
1
2
|Du1|2
+
1
2
|Du2|2
on {x ∈ Ω : Du1(x) = Du2(x)}.
Since |{x ∈ Ω : Du1(x) = Du2(x)}| > 0 and using the fact that both u1 and u2 are
minimizers, we obtain
ˆ
Ω
|Dv|2
dx <
1
2
ˆ
Ω
|Du1|2
dx+
1
2
ˆ
Ω
|Du2|2
dx =
1
2
m+
1
2
m = m.
Thus I(v) < m. This is a contradiction with the fact that u1 and u2 are minimiz-
ers.
Remarks 2.27:
(1) This approach generalizes to other variational integrals as well. Indeed,
the proof above is based on the following steps:
(a) Choose a minimizing sequence.

(b) Use coercivity
ui W1,2(Ω) → ∞ =⇒ I(ui) → ∞.
to show that the minimizing sequence is bounded in the Sobolev space.
(c) Use reflexivity to show that there is a weakly converging subsequence.
(d) Use lower semicontinuity of the variational integral to show that the
limit is a minimizer.
(e) Use strict convexity of the variational integral to show uniqueness.
(2) If we consider C2
(Ω) instead of W1,2
(Ω) in the Dirichlet problem above,
then we end up having the following problems. If we equip C2
(Ω) with the
supremum norm
u C2(Ω) = u L∞(Ω) + Du L∞(Ω) + D2
u L∞(Ω),
where D2
u is the Hessian matrix of second order partial derivatives, then
the variational integral is not coercive nor the space is reflexive. Indeed,
when n 2 it is possible to construct a sequence of functions for which
the supremum tends to infinity, but the L2
norm of the gradients tends to
zero. The variationall integral is not coersive even when n = 1. If we try to
obtain coercivity and reflexivity in C2
(Ω) by changing norm to u W1,2(Ω)
then we lose completeness, since the limit functions are not necessarily
in C2
(Ω). The Sobolev space seems to have all desirable properties for
existence of solutions to PDEs.

3Maximal function approach to
Sobolev spaces
We recall the definition of the maximal function.
Definition 3.1. The centered Hardy-Littlewood maximal function M f : Rn
→
[0,∞] of f ∈ L1
loc
(Rn
) is
M f (x) = sup
r>0
1
|B(x,r)|
ˆ
B(x,r)
|f (y)|dy,
where B(x,r) = {y ∈ Rn : |y− x| < r} is the open ball with the radius r > 0 and the
center x ∈ Rn
.
T H E M O R A L : The maximal function gives the maximal integral average of
the absolute value of the function on balls centered at a point.
Note that the Lebesgue differentiation theorem implies
|f (x)| = lim
r→0
1
|B(x,r)|
ˆ
B(x,r)
|f (y)|dy
sup
r>0
1
|B(x,r)|
ˆ
B(x,r)
|f (y)|dy = M f (x)
.
We are interested in behaviour of the maximal operator in Lp
-spaces and begin
with a relatively obvious result.
Lemma 3.2. If f ∈ L∞
(Rn
), then M f ∈ L∞
(Rn
) and M f L∞(Rn) f L∞(Rn).
T H E M O R A L : If the original function is essentially bounded, then the maximal
function is essentially bounded and thus finite almost everywhere. Intuitively this
62

CHAPTER 3. MAXIMAL FUNCTION APPROACH TO SOBOLEV SPACES 63
is clear, since the integral averages cannot be larger than the essential supremum
of the function. Another way to state this is that M : L∞
(Rn
) → L∞
(Rn
) is a
bounded operator.
Proof. For every x ∈ Rn
and r > 0 we have
1
|B(x,r)|
ˆ
B(x,r)
|f (y)|dy
1
|B(x,r)|
f L∞(Rn)|B(x,r)| = f L∞(Rn).
By taking supremum over r > 0, we have M f (x) f L∞(Rn) for every x ∈ Rn
and
thus M f L∞(Rn) f L∞(Rn).
The following maximal function theorem was ﬁrst proved by Hardy and Little-
wood in the one-dimensional case and by Wiener in higher dimensions.
Theorem 3.3 (Hardy-Littlewood-Wiener).
(1) If f ∈ L1
(Rn
), there exists c = c(n) such that
|{x ∈ Rn
: M f (x) > λ}|
c
λ
f L1(Rn) for every λ > 0. (3.1)
(2) If f ∈ Lp
(Rn
), 1 < p ∞, then M f ∈ Lp
(Rn
) and there exists c = c(n, p) such
that
M f Lp(Rn) c f Lp(Rn). (3.2)
T H E M O R A L : The ﬁrst assertion states that the Hardy-Littlewood max-
imal operator maps L1
(Rn
) to weak L1
(Rn
) and the second claim shows that
M : Lp
(Rn
) → Lp
(Rn
) is a bounded operator for p > 1.
W A R N I N G : f ∈ L1
(Rn
) does not imply that M f ∈ L1
(Rn
) and thus the Hardy-
Littlewood maximal operator is not bounded in L1
(Rn
). In this case we only have
the weak type estimate.
3.1 Representation formulas and Riesz po-
tentials
We begin with considering the one-dimensional case. If u ∈ C1
0(R), there exists an
interval [a,b] ⊂ R such that u(x) = 0 for every x ∈ R[a,b]. By the fundamental
theorem of calculus,
u(x) = u(a)+
ˆ x
a
u (y)dy =
ˆ x
−∞
u (y)dy, (3.3)
since u(a) = 0. On the other hand,
0 = u(b) = u(x)+
ˆ b
x
u (y)dy = u(x)+
ˆ ∞
x
u (y)dy,

so that
u(x) = −
ˆ ∞
x
u (y)dy. (3.4)
Equalities (3.3) and (3.4) imply
2u(x) =
ˆ x
−∞
u (y)dy−
ˆ ∞
x
u (y)dy
=
ˆ x
−∞
u (y)(x− y)
|x− y|
dy+
ˆ ∞
x
u (y)(x− y)
|x− y|
dy
=
ˆ
R
u (y)(x− y)
|x− y|
dy,
from which it follows that
u(x) =
1
2
ˆ
R
u (y)(x− y)
|x− y|
dy for every x ∈ R.
Next we extend the fundamental theorem of calculus to Rn
.
Lemma 3.4 (Representation formula). If u ∈ C1
0(Rn
), then
u(x) =
1
ωn−1
ˆ
Rn
Du(y)·(x− y)
|x− y|n
dy for every x ∈ Rn
,
where ωn−1 = nΩn is the (n−1)-dimensional measure of B(0,1).
T H E M O R A L : This is a representation formula for a compactly supported
smooth function in terms of its gradient. A function can be integrated back from
its derivative using this formula.
Proof. If x ∈ Rn
and e ∈ B(0,1), by the fundamental theorem of calculus
u(x) = −
ˆ ∞
0 t
(u(x+ te))dt = −
ˆ ∞
0
Du(x+ te)· e dt.

By the Fubini theorem
ωn−1u(x) = u(x)
ˆ
B(0,1)
1dS(e)
= −
ˆ
B(0,1)
ˆ ∞
0
Du(x+ te)· e dtdS(e)
= −
ˆ ∞
0
ˆ
B(0,1)
Du(x+ te)· e dS(e)dt (Fubini)
= −
ˆ ∞
0
ˆ
B(0,t)
Du(x+ y)·
y
t
1
tn−1
dS(y)dt
(y = te, dS(e) = t1−n
dS(y))
= −
ˆ ∞
0
ˆ
B(0,t)
Du(x+ y)·
y
|y|n
dS(y)dt
= −
ˆ
Rn
Du(x+ y)· y
|y|n
dy (polar coordinates)
= −
ˆ
Rn
Du(z)·(z − x)
|z − x|n
dz (z = x+ y, dy = dz)
=
ˆ
Rn
Du(y)·(x− y)
|x− y|n
dy.
Remark 3.5. By the Cauchy-Schwarz inequality and Lemma 3.4, we have
|u(x)| =
1
ωn−1
ˆ
Rn
Du(y)·(x− y)
|x− y|n
dy
1
ωn−1
ˆ
Rn
|Du(y)||x− y|
|x− y|n
dy
=
1
ωn−1
ˆ
Rn
|Du(y)|
|x− y|n−1
dy
=
1
ωn−1
I1(|Du|)(x),
where Iα f , 0 < α < n, is the Riesz potential
Iα f (x) =
ˆ
Rn
f (y)
|x− y|n−α
dy.
T H E M O R A L : This gives a useful pointwise bound for a compactly supported
smooth function in terms of the Riesz potential of the gradient.
Remark 3.6. A similar estimate holds almost everywhere if u ∈ W1,p
(Rn
) or u ∈
W
1,p
0 (Ω) (exercise).
We begin with a technical lemma for the Riesz potential for α = 1.
Lemma 3.7. If E ⊂ Rn
is a measurable set with |E| < ∞, then
ˆ
E
1
|x− y|n−1
dy c(n)|E|
1
n .

Proof. Let B = B(x,r) be a ball with |B| = |E|. Then |E B| = |B E| and thus
ˆ
EB
1
|x− y|n−1
dy |E B|
1
rn−1
= |B E|
1
rn−1
ˆ
BE
1
|x− y|n−1
dy.
This implies
ˆ
E
1
|x− y|n−1
dy =
ˆ
EB
1
|x− y|n−1
dy+
ˆ
E∩B
1
|x− y|n−1
dy
ˆ
BE
1
|x− y|n−1
dy+
ˆ
E∩B
1
|x− y|n−1
dy
=
ˆ
B
1
|x− y|n−1
dy
= c(n)r = c(n)|B|
1
n = c(n)||E|
1
n .
Lemma 3.8. Assume that |Ω| < ∞ and 1 p < ∞. Then
I1(|f |χΩ) Lp(Ω) c(n, p)|Ω|
1
n f Lp(Ω).
T H E M O R A L : If |Ω| < ∞, then I1 : Lp
(Ω) → Lp
(Ω) is a bounded operator for
1 p < ∞.
Proof. If p > 1, Hölder’s inequality and Lemma 3.7 give
ˆ
Ω
|f (y)|
|x− y|n−1
dy =
ˆ
Ω
|f (y)|
|x− y|
1
p (n−1)
1
|x− y|
1
p
(n−1)
dy
ˆ
Ω
|f (y)|p
|x− y|n−1
dy
1
p
ˆ
Ω
1
|x− y|n−1
dy
1
p
c|Ω|
1
np
ˆ
Ω
|f (y)|p
|x− y|n−1
dy
1
p
= c|Ω|
p−1
np
ˆ
Ω
|f (y)|p
|x− y|n−1
dy
1
p
.
For p = 1, the inequality above is clear. Thus by Fubini’s theorem and Lemma 3.7,
we have
ˆ
Ω
|I1(|f |χΩ)(x)|p
dx c|Ω|
p−1
n
ˆ
Ω
ˆ
Ω
|f (y)|p
|x− y|n−1
dydx
c|Ω|
p−1
n |Ω|
1
n
ˆ
Ω
|f (y)|p
dy.
Next we show that the Riesz potential can be bounded by the Hardy-Littlewood
maximal function. We shall do this for the general α although α = 1 will be most
important for us.
Lemma 3.9. If 0 < α < n, there exists c = c(n,α), such that
ˆ
B(x,r)
|f (y)|
|x− y|n−α
dy crα
M f (x)
for every x ∈ Rn
and r > 0.

Proof. Let x ∈ Rn
and denote Ai = B(x,r2−i
), i = 0,1,2,.... Then
ˆ
B(x,r)
|f (y)|
|x− y|n−α
dy =
∞
i=0
ˆ
AiAi+1
|f (y)|
|x− y|n−α
dy
∞
i=0
r
2i+1
α−n ˆ
Ai
|f (y)|dy
= Ωn
∞
i=0
1
2
α−n
r
2i
α
1
Ωn
r
2i
−n ˆ
Ai
|f (y)|dy
= Ωn
∞
i=0
1
2
α−n
r
2i
α
1
|Ai|
ˆ
Ai
|f (y)|dy
cM f (x)rα
∞
i=0
1
2α
i
= crα
M f (x).
Theorem 3.10 (Sobolev inequality for Riesz potentials). Assume that α >
0, p > 1 and αp < n. Then there exists c = c(n, p,α), such that for every f ∈ Lp
(Rn
)
we have
Iα f Lp∗
(Rn) c f Lp(Rn), p∗
=
pn
n−αp
.
T H E M O R A L : This is the Sobolev inequality for the Riesz potentials. Observe
that of α = 1, then p∗
is the Sobolev conjugate of p.
Proof. If f = 0 almost everywhere, the claim is clear. Thus we may assume that
f = 0 on a set of positive measure and thus M f > 0 everywhere. By Hölder’s
inequality
ˆ
RnB(x,r)
|f (y)|
|x− y|n−α
dy
ˆ
RnB(x,r)
|f (y)|p
dy
1
p
ˆ
RnB(x,r)
|x− y|(α−n)p
dy
1
p
,
where
ˆ
RnB(x,r)
|x− y|(α−n)p
dy =
ˆ ∞
r
ˆ
B(x,ρ)
|x− y|(α−n)p
dS(y)dρ
=
ˆ ∞
r
ρ(α−n)p
ˆ
B(x,ρ)
1dS(y)
=ωn−1 ρn−1
dρ
= ωn−1
ˆ ∞
r
ρ(α−n)p +n−1
dρ
=
ωn−1
(n−α)p − n
rn−(n−α)p
.
The exponent can be written in the form
n−(n−α)p = n−(n−α)
p
p −1
=
αp − n
p −1
,

and thus ˆ
RnB(x,r)
|f (y)|
|x− y|n−α
dy cr
α− n
p f Lp(Rn).
Lemma 3.9 implies
|Iα f (x)|
ˆ
Rn
|f (y)|
|x− y|n−α
dy
=
ˆ
B(x,r)
|f (y)|
|x− y|n−α
dy+
ˆ
RnB(x,r)
|f (y)|
|x− y|n−α
dy
c rα
M f (x)+ r
α− n
p f Lp(Rn) .
By choosing
r =
M f (x)
f Lp(Rn)
−
p
n
,
we obtain
|Iα f (x)| cM f (x)1−
αp
n f
αp
n
Lp(Rn)
. (3.5)
By raising both sides to the power p∗
=
np
n−αp , we have
|Iα f (x)|p∗
cM f (x)p
f
αp
n p∗
Lp(Rn)
The maximal function theorem, see (3.2), implies
ˆ
Rn
|Iα f (x)|p∗
dy c f
αp
n p∗
Lp(Rn)
ˆ
Rn
(M f (x))p
dx
= c f
αp
n p∗
Lp(Rn)
M f
p
Lp(Rn)
c f
αp
n p∗
Lp(Rn)
f
p
Lp(Rn)
and thus
Iα f Lp∗
(Rn) c f
αp
n +
p
p∗
Lp(Rn)
= c f Lp(Rn).
Remark 3.11. From the proof of the previous theorem we also obtain a weak type
estimate when p = 1. Indeed, by (3.5) with p = 1, there exists c = c(n,α) such that
|Iα f (x)| cM f (x)1− α
n f
α
n
L1(Rn)
and thus the maximal function theorem with p = 1, see (3.1), implies
|{x ∈ Rn
: |Iα f (x)| > t}| x ∈ Rn
: cM f (x)
n−α
n f
α
n
L1(Rn)
> t
x ∈ Rn
: M f (x) > ct
n
n−α f
− α
n · n
n−α
L1(Rn)
ct− n
n−α f
α
n−α
L1(Rn)
f L1(Rn)
= ct− n
n−α f
n
n−α
L1(Rn)
for every t > 0. This also implies that
|{x ∈ Rn
: |Iα f (x)| t}| ct− n
n−α f
n
n−α
L1(Rn)
for every t > 0.

This gives a second proof for the Sobolev-Gagliardo-Nirenberg inequality, see
Theorem 2.2.
Corollary 3.12 (Sobolev-Gagliardo-Nirenberg inequality). If 1 p < n, there
exists a constant c = c(n, p) such that
u Lp∗
(Rn) c Du Lp(Rn), p∗
=
np
n− p
,
for every u ∈ C1
0(Rn
).
T H E M O R A L : The Sobolev-Gagliardo-Nirenberg inequality is a consequence
of pointwise estimates for the function in terms of the Riesz potential of the
gradient and the Sobolev inequality for the Riesz potentials.
Proof. 1 < p < ∞ By Remark 3.5
|u(x)|
1
ωn−1
I1(|Du|)(x) for every x ∈ Rn
,
Thus Theorem 3.10 with α = 1 gives
u Lp∗
(Rn) c I1(|Du|) Lp∗
(Rn) c Du Lp(Rn).
p = 1 Let
Aj = {x ∈ Rn
: 2j
< |u(x)| 2j+1
}, j ∈ Z,
and let ϕ : R → R, ϕ(t) = max{0,min{t,1}}, be an auxiliary function. For j ∈ Z deﬁne
uj : Rn
→ [0,1],
uj(x) = ϕ(21−j
|u(x)|−1) =



0, |u(x)| 2j−1
,
21−j
|u(x)|−1, 2j−1
< |u(x)| 2j
,
1, |u(x)| > 2j
.
Lemma 1.25 implies uj ∈ W1,1
(Rn
), j ∈ Z. Observe that Duj = 0 almost everywhere
in Rn
Aj−1, j ∈ Z. Then
|Aj| |{x ∈ Rn
: |u(x)| > 2j
}|
= |{x ∈ Rn
: uj(x) = 1}| (|u(x)| > 2j
=⇒ 21−j
|u(x)|−1 > 1)
{x ∈ Rn
: I1(|Duj|)(x) ωn−1} (Remark 3.5)
c
ˆ
Rn
|Duj(x)|dx
n
n−1
(Remark 3.11)
= c
ˆ
A j−1
|Duj(x)|dx
n
n−1
c
ˆ
A j−1
ϕ (21−j
|u(x)|−1)21−j
|Du(x)|dx
n
n−1
= c2−j n
n−1
ˆ
A j−1
|Du(x)|dx
n
n−1
.

By summing over j ∈ Z, we obtain
ˆ
Rn
|u(x)|
n
n−1 dx =
j∈Z
ˆ
A j
|u(x)|
n
n−1 dx
j∈Z
2(j+1) n
n−1 |Aj|
c
j∈Z
ˆ
A j−1
|Du(x)|dx
n
n−1
c
j∈Z
ˆ
A j−1
|Du(x)|dx
n
n−1
= c
ˆ
Rn
|Du(x)|dx
n
n−1
.
In the last equality we used the fact that the sets Aj, j ∈ Z, are pairwise disjoint.
Remark 3.13. The Sobolev-Gagliardo-Nirenberg inequality for u ∈ W1,p
(Rn
) fol-
lows from Corollary 3.12 by using the fact that C1
0(Rn
) is dense in W1,p
(Rn
),
1 p < n.
3.2 Sobolev-Poincaré inequalities
Next we consider Sobolev-Poincaré inequalities in balls, compare with Theorem
2.7 and Theorem 2.8 for the corresponding estimates over cubes.
First we study the one-dimensional case. Assume that u ∈ C1
(R) and let
y, z ∈ B(x,r) = (x− r,x+ r). By the fundamental theorem of calculus
u(z)− u(y) =
ˆ y
z
u (t)dt.
Thus
|u(z)− u(y)|
ˆ y
z
|u (t)|dt
ˆ x+r
x−r
|u (t)|dt =
ˆ
B(x,r)
|u (t)|dt
and
|u(z)− uB(x,r)| = u(z)−
B(x,r)
u(y)dy
=
B(x,r)
u(z)dy−
B(x,r)
u(y)dy
B(x,r)
|u(z)− u(y)|dy
ˆ
B(x,r)
|u (y)|dy.
This is a pointwise estimate of the oscillation of the function. Next we generalize
this to Rn
.
Lemma 3.14. Let u ∈ C1
(Rn
) and B(x,r) ⊂ Rn
. There exists c = c(n) such that
u(z)− uB(x,r) c
ˆ
B(x,r)
|Du(y)|
|z − y|n−1
dy

for every z ∈ B(x,r).
T H E M O R A L : This is a pointwise estimate for the oscillation of the function
in terms of the Riesz potential of the gradient.
Proof. For any y, z ∈ B(x,r), we have
u(y)− u(z) =
ˆ 1
0 t
(u(ty+(1− t)z))dt =
ˆ 1
0
Du(ty+(1− t)z)·(y− z)dt.
By the Cauchy-Schwarz inequality
|u(y)− u(z)| |y− z|
ˆ 1
0
|Du(ty+(1− t)z)|dt.
Let ρ > 0. In the next display, we make a change of variables
w = ty+(1− t)z ⇐⇒ y =
1
t
(w−(1− t)z), dS(y) = t1−n
dS(w).
Then we have |w−z| = t|y−z| and tn−1
= |z−w|
ρ
n−1
, where ρ = |y−z|. We arrive at
ˆ
B(x,r)∩ B(z,ρ)
|u(y)− u(z)|dS(y)
ρ
ˆ 1
0
ˆ
B(x,r)∩ B(z,ρ)
|Du(ty+(1− t)z)|dS(y)dt
= ρ
ˆ 1
0
1
tn−1
ˆ
B(x,r)∩ B(z,tρ)
|Du(w)|dS(w)dt
= ρn
ˆ 1
0
ˆ
B(x,r)∩ B(z,tρ)
|Du(w)|
|z − w|n−1
dS(w)dt
= ρn−1
ˆ ρ
0
ˆ
B(x,r)∩ B(z,s)
|Du(w)|
|z − w|n−1
dS(w)ds (s = tρ, dt =
1
ρ
ds)
= ρn−1
ˆ
B(x,r)∩B(z,ρ)
|Du(w)|
|z − w|n−1
dw. (polar coordinates)
Since B(x,r) ⊂ B(z,2r), an integration in polar coordinates gives
u(z)− uB(x,r)
B(x,r)
|u(z)− u(y)|dy
=
1
|B(x,r)|
ˆ 2r
0
ˆ
B(x,r)∩ B(z,ρ)
|u(y)− u(z)|dS(y)dρ
1
|B(x,r)|
ˆ 2r
0
ρn−1
ˆ
B(x,r)∩B(z,ρ)
|Du(y)|
|z − y|n−1
dydρ
1
|B(x,r)|
ˆ 2r
0
ρn−1
dρ
ˆ
B(x,r)
|Du(y)|
|z − y|n−1
dy
= c(n)
ˆ
B(x,r)
|Du(y)|
|z − y|n−1
dy.

Remarks 3.15:
(1) Assume that u ∈ C1
(Rn
). By Lemma 3.14 and Lemma 3.9, we have
|u(z)− uB(x,r)| c
ˆ
B(x,r)
|Du(y)|
|z − y|n−1
dy
= cI1(|Du|χB(x,r))(z)
c
ˆ
B(z,2r)
|Du(y)|χB(x,r)(y)
|z − y|n−1
dy
crM(|Du|χB(x,r))(z),
for every z ∈ B(x,r).
Next we show that the corresponding inequalities hold true almost every-
where if u ∈ W
1,p
loc
(Rn
), 1 p < ∞. Since C∞
(B(x,r)) is dense in W1,p
(B(x,r)),
there exists a sequence ui ∈ C∞
(B(x,r)), i = 1,2,..., such that ui → u in
W1,p
(B(x,r)) as i → ∞. By passing to a subsequence, if necessary, we
obtain an exceptional set N1 ⊂ Rn
with |N1| = 0 such that
lim
i→∞
ui(z) = u(z) < ∞
for every z ∈ B(x,r) N1. By linearity of the Riesz potential and by Lemma
3.8, we have
I1(|Dui|χB(x,r))− I1(|Du|χB(x,r)) Lp(B(x,r))
= I1((|Dui|−|Du|)χB(x,r)) Lp(B(x,r))
c|B(x,r)|
1
n |Dui|−|Du| Lp(B(x,r)),
which implies that
I1(|Dui|χB(x,r)) → I1(|Du|χB(x,r)) in Lp
(B(x,r)) as i → ∞.
By passing to a subsequence, if necessary, we obtain an exceptional set
N2 ⊂ B(x,r) with |N2| = 0 such that
lim
i→∞
I1(|Dui|χB(x,r))(z) = I1(|Du|χB(x,r))(z) < ∞
for every z ∈ B(x,r) N2. Thus
|u(z)− uB(x,r)| = lim
i→∞
|ui(z)−(ui)B(x,r)|
c lim
i→∞
I1(|Dui|χB(x,r))(z)
crM(|Du|χB(x,r))(z),
for every z ∈ B(x,r)(N1 ∪ N2).

(2) By Lemma 3.14 and (3.5), we have
|u(z)− uB(x,r)| c
ˆ
B(x,r)
|Du(y)|
|z − y|n−1
dy
cM(|Du|χB(x,r))(z)1−
p
n |Du|χB(x,r)
p
n
Lp(Rn)
for every z ∈ B(x,r). The corresponding inequalities hold true almost
everywhere if u ∈ W
1,p
loc
(Rn
), 1 p < ∞.
This gives a proof for the Sobolev-Poincaré inequality on balls, see Theorem
2.8 for the correspoding statement for cubes. Maximal function arguments can be
used for cubes as well.
Theorem 3.16 (Sobolev-Poincaré inequality on balls). Assume that u ∈ W1,p
(Rn
)
and let 1 < p < n. There exists c = c(n, p) such that
B(x,r)
|u − uB(x,r)|p∗
dy
1
p∗
cr
B(x,r)
|Du|p
dy
1
p
for every B(x,r) ⊂ Rn
.
T H E M O R A L : The Sobolev-Poincaré inequality is a consequence of pointwise
estimates for the oscillation of the function in terms of the Riesz potential of the
gradient and the Sobolev inequality for the Riesz potentials.
Proof. By Remark 3.15, we have
|u(y)− uB(x,r)| cI1(|Du|χB(x,r))(y)
for almost every y ∈ B(x,r). Thus Theorem 3.10 implies
ˆ
B(x,r)
|u − uB(x,r)|p∗
dy
1
p∗
c
ˆ
Rn
I1(|Du|χB(x,r))p∗
dy
1
p∗
c
ˆ
Rn
(|Du|χB(x,r))p
dy
1
p
= c
ˆ
B(x,r)
|Du|p
dy
1
p
.
A similar argument can be used to prove a counterpart of Theorem 2.7 as well.
Theorem 3.17 (Poincaré inequality on balls). Assume that u ∈ W1,p
(Rn
)and
let 1 < p < ∞. There exists c = c(n, p) such that
B(x,r)
|u − uB(x,r)|p
dy
1
p
cr
B(x,r)
|Du|p
dy
1
p
.

Proof. By Remark 3.15, we have
|u(y)− uB(x,r)| crM(|Du|χB(x,r))(y)
for almost every y ∈ B(x,r). The maximal function theorem with p > 1, see (3.2),
implies
ˆ
B(x,r)
|u − uB(x,r)|p
dy crp
ˆ
Rn
M(|Du|χB(x,r))p
dy
crp
ˆ
Rn
(|Du|χB(x,r))p
dy
= crp
ˆ
B(x,r)
|Du|p
dy.
The maximal function approach to Sobolev-Poincaré inequalities is more in-
volved in the case p = 1, since then we only have a weak type estimate. However,
it is possible to consider that case as well, but this requires a different proof. We
begin with two rather technical lemmas.
Lemma 3.18. Assume that E ⊂ Rn
is a measurable set and that f : E → [0,∞] is
a measurable function for which
{x ∈ E : f (x) = 0}
|E|
2
.
Then for every a ∈ R and λ > 0, we have
{x ∈ E : f (x) > λ} x ∈ E : |f (x)− a| >
λ
2
.
Proof. Assume ﬁrst that |a| λ
2 . If x ∈ E with f (x) > λ, then
|f (x)− a| >
λ
2
f (x)−|a| >
λ
2
.
Thus {x ∈ E : f (x) > λ} ⊂ x ∈ E : |f (x)− a| > λ
2 and
{x ∈ E : f (x) > λ} x ∈ E : |f (x)− a| >
λ
2
.
Assume then that |a| > λ
2 . If x ∈ E with f (x) = 0, then
|f (x)− a| = |a| >
λ
2
.
Thus
{x ∈ E : f (x) = 0} ⊂ x ∈ E : f (x) >
λ
2
.
If |E| = ∞, then by assumption
{x ∈ E : f (x) = 0}
|E|
2
= ∞

and thus x ∈ E : f (x) λ
2 = ∞. On the other hand, if |E| < ∞, then
{x ∈ E : f (x) > λ} |E|− {x ∈ E : f (x) = 0}
|{x ∈ E : f (x) = 0}|
x ∈ E : |f (x)− a| >
λ
2
.
This completes the proof.
Lemma 3.19. Assume that u ∈ C0,1
(Rn
), that is, u is a bounded Lipschitz contin-
uous function in Rn
, and let B(x,r) be a ball in Rn
. Then there exists λ0 ∈ R for
which
{y ∈ B(x,r) : u(y) λ0}
|B(x,r)|
2
and {y ∈ B(x,r) : u(y) λ0}
|B(x,r)|
2
.
Proof. Denote Eλ = {y ∈ B(x,r) : u(y) λ}, λ ∈ R, and set
λ0 = sup λ ∈ R : |Eλ|
|B(x,r)|
2
.
Note that |λ0| u L∞(Rn) < ∞. Thus there exists an increasing sequence of real
numbers (λi) such that λi → λ0 and
|Eλi
|
|B(x,r)|
2
for every i = 1,2,....
Since Eλ0 = ∞
i=1
Eλi
, Eλ1 ⊃ Eλ2 ⊃ ... and |Eλi
| |B(x,r)| < ∞, we conclude that
|Eλ0 | = lim
i→∞
|Eλi
|
|B(x,r)|
2
.
This shows that
{y ∈ B(x,r) : u(y) λ0}
|B(x,r)|
2
.
A similar argument shows the other claim (exercise).
The next result is Theorem 3.16 with p = 1.
Theorem 3.20. Assume that u ∈ W
1,1
loc
(Rn
). There exists c = c(n) such that
B(x,r)
|u − uB(x,r)|
n
n−1 dy
n−1
n
cr
B(x,r)
|Du|dy
.
Proof. By Lemma 3.19 there is a number λ0 ∈ R for which
{y ∈ B(x,r) : u(y) λ0}
|B(x,r)|
2
and {y ∈ B(x,r) : u(y) λ0}
|B(x,r)|
2
.
Denote
v+ = max{u −λ0,0} and v− = −min{u −λ0,0}.

Both of these functions belong to W
1,1
loc
(Rn
). For the rest of the proof v 0 denotes
either v+ or v−. All statements are valid in both cases.
Let
Aj = {y ∈ B(x,r) : 2j
< v(y) 2j+1
}, j ∈ Z,
and let ϕ : R → R, ϕ(t) = max{0,min{t,1}}, be an auxiliary function. We deﬁne
vj : B(x,r) → [0,1],
vj(y) = ϕ(21−j
v(y)−1), j ∈ Z.
Lemma 1.25 implies vj ∈ W1,1
(B(x,r)), j ∈ Z. By Remark 3.15 (2) with p = 1, we
have
|vj(y)−(vj)B(x,r)|
n
n−1 cM(|Dvj|χB(x,r))(y) |Dvj|χB(x,r)
1
n−1
L1(Rn)
.
Lemma 3.18 with λ = 1
2 and a = (vj)B(x,r) gives
|Aj| |{y ∈ B(x,r) : v(y) > 2j
}|
y ∈ B(x,r) : vj(y) >
1
2
y ∈ B(x,r) : |vj(y)−(vj)B(x,r)| >
1
4
y ∈ Rn
: M(|Dvj|χB(x,r))(y) c |Dvj|χB(x,r)
1
1−n
L1(Rn)
.
The last term can be estimated using the weak type estimate for the maximal
function, see (3.1), and the fact that
|Dvj| = 21−j
|Dv|χA j−1
almost everywhere in B(x,r). Thus we arrive at
y ∈ Rn
:M(|Dvj|χB(x,r))(y) c |Dvj|χB(x,r)
1
1−n
L1(Rn)
c |Dvj|χB(x,r)
1
n−1
L1(Rn)
ˆ
Rn
|Dvj(y)|χB(x,r)(y)dy
= c |Dvj|χB(x,r)
n
n−1
L1(Rn)
c2−
jn
n−1 |Dv|χA j−1∩B(x,r)
n
n−1
L1(Rn)
.
Combining the above estimates for |Aj|, we obtain
ˆ
B(x,r)
v(y)
n
n−1 dy =
j∈Z
ˆ
A j
v(y)
n
n−1 dy =
j∈Z
2
(j+1)n
n−1 |Aj|
c
j∈Z
2
(j+1)n
n−1 2−
jn
n−1 |Dv|χA j−1∩B(x,r)
n
n−1
L1(Rn)
c
j∈Z
|Dv|χA j−1∩B(x,r)
n
n−1
L1(Rn)
c |Du|χB(x,r)
n
n−1
L1(Rn)
.

Since |u −λ0| = v+ + v−, we obtain
B(x,r)
|u − uB(x,r)|
n
n−1 dy
n−1
n
2
B(x,r)
|u −λ0|
n
n−1 dy
n−1
n
2
B(x,r)
v+(y)
n
n−1 dy
n−1
n
+2
B(x,r)
v−(y)
n
n−1 dy
n−1
n
c |Du|χB(x,r) L1(Rn)
= c
ˆ
B(x,r)
|Du(y)|dy.
T H E M O R A L : The proof shows that in this case a weak type estimate implies
a strong type estimate. Observe carefully, that this does not hold in general. The
reason why this works here is that we consider gradients, which have the property
that they vanish on the set where the function itself is constant.
Next we give a maximal function proof for Morrey’s inequality, see Theorem
2.13 and Remark 2.14 (3).
Theorem 3.21 (Morrey’s inequality). Assume that u ∈ C1
(Rn
) and let n < p <
∞. There exists c = c(n, p) such that
|u(y)− u(z)| cr
B(x,r)
|Du|p
dw
1
p
and y, z ∈ B(x,r).
Proof. By Lemma 3.14
|u(y)− u(z)| |u(y)− uB(x,r)|+|uB(x,r) − u(z)|
c
ˆ
B(x,r)
|Du(w)|
|y− w|n−1
dw+ c
ˆ
B(x,r)
|Du(w)|
|z − w|n−1
dw
for every y, z ∈ B(x,r). Hölder’s inequality gives
ˆ
B(x,r)
|Du(w)|
|y− w|n−1
dw
ˆ
B(x,r)
|Du|p
dw
1
p
ˆ
B(x,r)
|y− w|(1−n)p
dw
1
p
,
where
ˆ
B(x,r)
|y− w|(1−n)p
dw
ˆ
B(y,2r)
|y− w|(1−n)p
dw
=
ˆ 2r
0
ˆ
B(y,ρ)
ρ(1−n)p
dS(w)dρ
= ωn−1
ˆ 2r
0
ρ(1−n)p +n−1
dρ = crn−(n−1)p
.
Since
(n−(n−1)p ) 1
p = 1− n
p ,

we have
ˆ
B(x,r)
|Du(w)|
|y− w|n−1
dw cr
1− n
p
ˆ
B(x,r)
|Du|p
dw
1
p
.
The same argument applies to the other integral as well, so that
|u(y)− u(z)| cr
1− n
p
ˆ
B(x,r)
|Du|p
dw
1
p
.
3.3 Sobolev inequalities on domains
In this section we study open sets Ω ⊂ Rn
for which the Sobolev-Poincaré inequality
ˆ
Ω
|u − uΩ|p∗
dy
1
p∗
c(p,n,Ω)
ˆ
Ω
|Du|p
dy
1
p
, 1 p < n, p∗
=
np
n− p
,
holds true for all u ∈ C∞
(Ω). We already know that this inequality holds if Ω
is a ball, but are there other sets for which it holds true as well? We begin by
introducing an appropriate class of domains.
Definition 3.22. A bounded open set Ω ⊂ Rn
is a John domain, if there is cJ 1
and a point x0 ∈ Ω so that every point x ∈ Ω can be joined to x0 by a path γ : [0,1] →
Ω such that γ(0) = x, γ(1) = x0 and
dist(γ(t), Ω) c−1
J |x−γ(t)|
for every t ∈ [0,1].
T H E M O R A L : In a John domain every point can be connected to the distin-
guished point with a curve that is relatively far from the boundary.
Remarks 3.23:
(1) A bounded and connected open set Ω ⊂ Rn
satisfies the interior cone condi-
tion, if there exists a bounded cone
C = {x ∈ Rn
: x2
1 +···+ x2
n−1 ax2
, 0 xn b}
such that every point of Ω is a vertex of a cone congruent to C and entirely
contained in Ω. Every domain with interior cone condition is a John
domain (exercise). Roughly speaking the main difference between the
interior cone condition and a John domain is that rigid cones are replaced
by twisted cones.
(2) The collection of John domains is relatively large. For example, a domain
whose boundary is von Koch snowflake is a John domain.

Theorem 3.24. If Ω ⊂ Rn
is a John domain and 1 p < n, then
ˆ
Ω
|u − uΩ|p∗
dx
1
p∗
c(p,n, cJ)
ˆ
Ω
|Du|p
dx
1
p
, 1 < p < n,
for all u ∈ C∞
(Ω).
T H E M O R A L : The Sobolev-Poincaré inequality holds for many other sets than
balls as well.
W A R N I N G : A rooms and passages example shows that the Sobolev-Poincaré
inequality does not hold for all sets.
Proof. Let x0 ∈ Ω be the distinguished point in the John domain. Deonte B0 =
B(x0,r0), r0 = 1
4 dist(x0, Ω). We show that there is a constant M = M(cJ,n) such
that for every x ∈ Ω there is a chain of balls Bi = B(xi,ri) ⊂ Ω, i = 1,2,..., with the
properties
(1) |Bi ∪Bi+1| M|Bi ∩Bi+1|, i = 1,2,...,
(2) dist(x,Bi) Mri, ri → 0, xi → x as i → ∞ and
(3) no point of Ω belongs to more than M balls Bi.
To construct the chain, first assume that x is far from x0, say x ∈ ΩB(x,2r0). Let
γ be a John path that connects x to x0. All balls on the chain are centered on
γ. We construct the balls recursively. We have already defined B0. Assume that
B0,...,Bi have been constructed. Starting from the center xi of Bi we move along
γ towards x until we leave Bi for the last time. Let xi+1 be the point on γ where
this happens and define
Bi+1 = B(xi+1,ri+1), ri+1 =
1
4cJ
|x− xi+1|.
By construction Bi ⊂ Ω. Property (1) and dist(x,Bi) Mri in (2) follow from
the fact that the consecutive balls have comparable radii and that the radii are
comparable to the distances of the centers of the balls to x.
To prove (3) assume that y ∈ Bi1 ∩ ··· ∩ Bik
. Observe that the radii of Bi j
,
j = 1,...,k, are comparable to |x− y|. By construction, if i j < im, the the center of
Bim does not belong to Bi j
. This implies that the distances between the centers of
Bi j
are comparable to |x−y|. The number of points in Rn
with pairwise comparable
distances is bounded, that is, if z1,..., zm ∈ Rn
satisfy
r
c
< dist(zi, zj) < cr for i = j,
then m N = N(c,n). Thus k is bounded by a constant depending only on n and
cJ. This implies (3). Property (3) implies , ri → 0, xi → x as i → ∞.
The case x ∈ B(x,2r0) is left as an exercise.

Since
uBi
=
Bi
u(y)dy → u(x)
for every x ∈ Ω as i → ∞, we obtain
|u(x)− uB0 |
∞
i=0
|uBi
− uBi+1
|
∞
i=0
|uBi
− uBi∩Bi+1
|+|uBi∩Bi+1
− uBi+1
|
∞
i=0
|Bi|
|Bi ∩Bi+1| Bi
|u − uBi
|dy+
|Bi+1|
|Bi ∩Bi+1| Bi+1
|u − uBi+1
|dy
c
∞
i=0 Bi
|u − uBi
|dy (property (1))
c
∞
i=0
ri
Bi
|Du|dy (Poincaré inequality, see Theorem 3.17)
= c
∞
i=0
ˆ
Bi
|Du|
rn−1
i
dy.
Property (2) implies |x− y| cri for every y ∈ Bi and
1
rn−1
i
c
|x− y|n−1
for every y ∈ Bi.
Thus
|u(x)− uB0 | c
∞
i=0
ˆ
Bi
|Du(y)|
|x− y|n−1
dy c
ˆ
Ω
|Du(y)|
|x− y|n−1
dy.
The last inequality follows from (3). We observe that
|u(x)− uΩ| |u(x)− uB0 |+|uB0 − uΩ|,
where by Lemma 3.7 we have
|uB0 − uΩ|
1
|Ω|
ˆ
Ω
|u(x)− uB0 |dx
c
1
|Ω|
ˆ
Ω
ˆ
Ω
|Du(y)|
|x− y|n−1
dxdy
= c
1
|Ω|
ˆ
Ω
|Du(y)|
ˆ
Ω
1
|x− y|n−1
dx dy
c|Ω|−1+ 1
n
ˆ
Ω
|Du(y)|dy.
By the John condition we have
c|Ω|
1
n dist(x0, Ω) c−1
J |x− x0|
and by taking supremum over x ∈ Ω we obtain
diamΩ c(n, cJ)|Ω|
1
n

and thus
|Ω|− n−1
n
c
|x− y|n−1
for every y ∈ Ω.
This implies
|uB0 − uΩ| c
ˆ
Ω
|Du(y)|
|x− y|n−1
dy
and thus
|u(x)− uΩ| c
ˆ
Ω
|Du(y)|
|x− y|n−1
dy = cI1(|Du|χΩ)(x).
Theorem 3.10 implies
ˆ
B(x,r)
|u(x)− uΩ|p∗
dx
1
p∗
c
ˆ
Rn
|I1(|Du(x)|χΩ(x))|p∗
dx
1
p∗
c
ˆ
Rn
(|Du(x)|χΩ(x))p
dx
1
p
= c
ˆ
Ω
|Du(x)|p
dx
1
p
.
3.4 A maximal function characterization
of Sobolev spaces
Similar argument as in the proof of Sobolev-Poincaré inequality gives the following
pointwise estimate.
Theorem 3.25. Assume that u ∈ C1
(Rn
). There exists a constant c = c(n) > 0 such
that
|u(x)− u(y)| c|x− y|(M|Du|(x)+ M|Du|(y))
.
Proof. Let x, y ∈ Rn
. Then x, y ∈ B(x,2|x− y|) and B(x,2|x− y|) ⊂ B(y,4|x− y|). By
Remark 3.15 we obtain
|u(x)− u(y)| |u(x)− uB(x,2|x−y|)|+|uB(x,2|x−y|) − u(y)|
c|x− y|(M|Du|(x)+ M|Du|(y)).
Remarks 3.26:
(1) If |Du| ∈ Lp
(Rn
), 1 < p ∞, then by (3.2) we have M|Du| ∈ Lp
(Rn
).
(2) If |Du| ∈ L1
(Rn
), then by (3.1) we have M|Du| < ∞ almost everywhere.
(3) If |Du| ∈ L∞
(Rn
), then M|Du| M|Du| L∞(Rn) Du L∞(Rn) everywhere.
Thus
|u(x)− u(y)| c Du L∞(Rn)|x− y|
. In other words, u is Lipschitz continuous.

Theorem 3.27. Assume that u ∈ W1,p
(Rn
), 1 < p < ∞. There exists c = c(n) and a
set N ⊂ Rn
with |N| = 0 such that
N.
Proof. C∞
0 (Rn
) is dense in W1,p
(Rn
) by Lemma 1.24. Thus there exists a sequence
ui ∈ C∞
0 (Rn
), i = 1,2,..., such that ui → u in W1,p
(Rn
) as i → ∞. By passing to a
subsequence, if necessary, we obtain an exceptional set N1 ⊂ Rn
with |N1| = 0 such
that
lim
i→∞
ui(x) = u(x) < ∞
for every x ∈ Rn
N1. By the sublinearity of the maximal operator and the maximal
function theorem
M|Dui|− M|Du| Lp(Rn) M(|Dui|−|Du|) Lp(Rn)
c |Dui|−|Du|| Lp(Rn)
c Dui − Du Lp(Rn)
which implies that M|Dui| → M|Du| in Lp
(Rn
) as i → ∞. By passing to a sub-
sequence, if necessary, we obtain an exceptional set N2 ⊂ Rn
with |N2| = 0 such
that
lim
i→∞
M|Dui|(x) = M|Du|(x) < ∞
for every x ∈ Rn
N2. By Theorem 3.25
|u(x)− u(y)| = lim
i→∞
|ui(x)− ui(y)|
c|x− y| lim
i→∞
(M|Dui|(x)+ M|Dui|(y))
c|x− y|(M|Du|(x)+ M|Du|(y))
for every x ∈ Rn
(N1 ∪ N2).
Remark 3.28. Compare the proof above to Remark 3.15, which shows that the
result holds for u ∈ W1,p
(Rn
), 1 p ∞.
The following deﬁnition motivated by Theorem 3.25.
Deﬁnition 3.29. Assume that 1 < p < ∞ and let u ∈ Lp
(Rn
). For a measurable
function g : Rn
→ [0,∞] we denote g ∈ D(u) if there exists an exceptional set
N ⊂ Rn
such that |N| = 0 and
|u(x)− u(y)| |x− y|(g(x)+ g(y)) (3.6)
N. We say that u ∈ Lp
(Rn
) belongs to the Hajłasz-Sobolev
space M1,p
(Rn
), if there exists g ∈ Lp
(Rn
) with g ∈ D(u). This space is endowed
with the norm
u M1,p(Rn) = u Lp(Rn) + inf
g∈D(u)
g Lp(Rn).

T H E M O R A L : The space M1,p
(Rn
) is defined through the pointwise inequality
(3.6).
Theorem 3.30. Assume that 1 < p < ∞. Then M1,p
(Rn
) = W1,p
(Rn
) and the asso-
ciate norms are equivalent, that is, there exists c such that
1
c
u W1,p(Rn) u M1,p(Rn) c u W1,p(Rn)
for every measurable function u that belongs to M1,p
(Rn
) = W1,p
(Rn
).
T H E M O R A L : This is a pointwise characterization of Sobolev spaces. This can
be used as a definition of the first order Sobolev spaces on metric measure spaces.
Proof. ⊃ Assume that u ∈ W1,p
(Rn
). By Theorem 3.27 there exists c = c(n) and a
set N ⊂ Rn
with |N| = 0 such that
N. Thus g = cM|Du| ∈ D(u) ∩ Lp
(Rn
) and by the maximal
function theorem
u M1,p(Rn) = u Lp(Rn) + inf
g∈D(u)
g Lp(Rn)
u Lp(Rn) + cM|Du| Lp(Rn)
u Lp(Rn) + c Du Lp(Rn)
c u W1,p(Rn),
where c = c(n, p).
⊂ Assume then that u ∈ M1,p
(Rn
). Then u ∈ Lp
(Rn
) and there exists g ∈ Lp
(Rn
)
with g ∈ D(u). Then
|u(x+ h)− u(x)| |h|(g(x+ h)+ g(x))
for almost every x, h ∈ Rn
and thus
ˆ
Rn
|u(x+ h)− u(x)|p
dx |h|p
ˆ
Rn
(g(x+ h)+ g(x))p
dx
2p
|h|p
ˆ
Rn
(g(x+ h)p
+ g(x)p
)dx
2p+1
g
p
Lp(Rn)
|h|p
.
By the characterization of the Sobolev space with the integrated difference quo-
tients, see Theorem 1.38, we conclude u ∈ W1,p
(Rn
) and
u W1,p(Rn) c u Lp(Rn) + c g Lp(Rn).
The inequality u W1,p(Rn) c u M1,p(Rn) follows by taking infimum over all g ∈
D(u)∩ Lp
(Rn
).

Remark 3.31. The pointwise characterization of Sobolev spaces in Theorem 3.30 is
very useful in studying properties of Sobolev spaces. For example, if u ∈ M1,p
(Rn
)
and g ∈ D(u)∩ Lp
(Rn
), then by the triangle inequality
|u(x)|−|u(y)| |u(x)− u(y)| |x− y|(g(x)+ g(y))
Thus g ∈ D(|u|)∩ Lp
(Rn
) and consequently |u| ∈ M1,p
(Rn
).
The pointwise characterization of Sobolev spaces in Theorem 3.30 can be used
to show a similar result as Theorem 1.35.
Lemma 3.32. The function u belongs to W1,p
(Rn
) if and only if u ∈ Lp
(Rn
) and
there are functions ui ∈ Lp
(Rn
), i = 1,2,..., such that ui → u almost everywhere
and gi ∈ D(ui)∩ Lp
(Rn
) such that gi → g almost everywhere for some g ∈ Lp
(Rn
).
Proof. If u ∈ W1,p
(Rn
), then the claim of the lemma is clear. To see the converse,
suppose that u, g ∈ Lp
(Rn
), gi ∈ D(ui)∩Lp
(Rn
) and ui → u almost everywhere and
gi → g almost everywhere. Then
|ui(x)− ui(y)| |x− y| gi(x)+ gi(y) (3.7)
for all x, y ∈ Rn
Fi with |Fi| = 0, i = 1,2,... Let A ⊂ Rn
be such that ui(x) → u(x)
and gi(x) → g(x) for all x ∈ Rn
A and |A| = 0. Write F = A ∪ ∞
i=1
Fi. Then |F| = 0.
Let x, y ∈ Rn
F, x = y. From (3.7) we obtain
|u(x)− u(y)| |x− y| g(x)+ g(y)
and thus g ∈ D(u)∩ Lp
(Rn
). This completes the proof.
3.5 Pointwise estimates
In this section we revisit pointwise inequalities for Sobolev functions.
Deﬁnition 3.33. Let 0 < β < ∞ and R > 0. The fractional sharp maximal function
of a locally integrable function f is deﬁned by
f #
β,R(x) = sup
0<r<R
r−β
B(x,r)
|f − fB(x,r)|dy,
If R = ∞ we simply write f #
β
(x).
T H E M O R A L : The fractional sharp maximal function controls the mean
oscillation of the function instead of the average of the function as in the Hardy-
Littlewood maximal function.
Next we prove a more general pointwise inequality than in Theorem 3.27.

Lemma 3.34. Suppose that f is locally integrable and let 0 < β < ∞. Then there
is c = c(β,n) and a set E with |E| = 0 such that
|f (x)− f (y)| c|x− y|β
f #
β,4|x−y|(x)+ f #
β,4|x−y|(y) (3.8)
E.
T H E M O R A L : This is a pointwise inequality for a function without the gradi-
ent.
Proof. Let E be the complement of the set of Lebesgue points of f . By Lebesgue’s
theorem |E| = 0. Fix x ∈ Rn
E, 0 < r < ∞ and denote Bi = B(x,2−i
r), i = 0,1,...
Then
|f (x)− fB(x,r)|
∞
i=0
|fBi+1
− fBi
|
∞
i=0
|Bi|
|Bi+1| Bi
|f − fBi
|dy
c
∞
i=0
(2−i
r)β
(2−i
r)−β
Bi
|f − fBi
|dy
crβ
f #
β,r(x).
Let y ∈ B(x,r) E. Then B(x,r) ⊂ B(y,2r) and we obtain
|f (y)− fB(x,r)| |f (y)− fB(y,2r)|+|fB(y,2r) − fB(x,r)|
crβ
f #
β,2r(y)+
B(x,r)
|f − fB(y,2r)|dz
crβ
f #
β,2r(y)+ c
B(y,2r)
|f − fB(y,2r)|dz
crβ
f #
β,2r(y).
Let x, y ∈ Rn
E, x = y and r = 2|x− y|. Then x, y ∈ B(x,r) and hence
|f (x)− f (y)| |f (x)− fB(x,r)|+|f (y)− fB(x,r)|
c|x− y|β
f #
β,4|x−y|(x)+ f #
β,4|x−y|(y) .
Remark 3.35. Lemma 3.34 gives a Campanato type characterization for Hölder
continuity. Assume that f ∈ L1
loc
(Rn
) and let 0 < β 1. By Lemma 3.34 there exists
a set E ⊂ Rn
with |E| = 0 such that
|f (x)− f (y)| c(n,β)|x− y|β
f #
β (x)+ f #
β (y)
E. If f #
β
∈ L∞
(Rn
), then f can be redeﬁned on a set of measure
zero so that the function is Hölder continuous in Rn
with exponent β. On the other

hand, if f ∈ C0,β
(Rn
), then
|f (y)− fB(x,r)| = f (y)−
B(x,r)
f (z)dz
B(x,r)
|f (y)− f (z)|dz crβ
for every y ∈ B(x,r). Thus
f #
β,R(x) = sup
0<r<R
r−β
B(x,r)
|f (y)− fB(x,r)|dy c
for every x ∈ Rn
and this implies that f #
β
∈ L∞
(Rn
). Thus f can be redeﬁned on a
set of measure zero so that the function is Hölder continuous with exponent β if
and only if f #
β
∈ L∞
(Rn
).
Deﬁnition 3.36. Let 0 α < n and R > 0. The fractional maximal function of
f ∈ L1
loc
(Rn
) is
Mα,R f (x) = sup
0<r<R
rα
B(x,r)
|f |dy,
For R = ∞, we write Mα,∞ = Mα. If α = 0, we obtain the Hardy–Littlewood
maximal function and we write M0 = M.
If u ∈ W
1,1
loc
(Rn
), then by the Poincaré inequality with p = 1, see Theorem 3.20,
there is c = c(n) such that
B(x,r)
|u − uB(x,r)|dy cr
B(x,r)
|Du|dy
for every ball B(x,r) ⊂ Rn
. It follows that
rα−1
B(x,r)
|u − uB(x,r)|dy crα
B(x,r)
|Du|dy
and consequently
u#
1−α,R(x) cMα,R|Du|(x)
for every x ∈ Rn
and R > 0. Thus we have proved the following useful inequality.
Corollary 3.37. Let u ∈ W
1,1
loc
(Rn
) and 0 α < 1. Then there is c = c(n,α) and a
set E ⊂ Rn
with |E| = 0 such that
|u(x)− u(y)| c|x− y|1−α
Mα,4|x−y||Du|(x)+ Mα,4|x−y||Du|(y)
E.
The next result shows that this gives a characterization of W1,p
(Rn
) for 1 <
p ∞.
Theorem 3.38. Let 1 < p < ∞. Then the following four conditions are equivalent.

from which we conclude that
uh − u Lp(Rn) |h|( gh Lp(Rn) + g Lp(Rn)) = 2|h| g Lp(Rn).
The claim follows from this.
Remark 3.39. It can be shown that u ∈ W1,1
(Rn
) if and only if u ∈ L1
(Rn
) and there
is a nonnegative function g ∈ L1
(Rn
) and σ 1 such that
|u(x)− u(y)| |x− y| Mσ|x−y| g(x)+ Mσ|x−y| g(y)
E with |E| = 0. Moreover, if this inequality holds, then |Du|
c(n,σ)g almost everywhere.
3.6 Approximation by Lipschitz functions
Smooth functions in C∞
(Ω) and C∞
0 (Ω) are often used as canonical test functions
in mathematical analysis. However, in many occasions smooth functions can be
replaced by a more flexible class of Lipschitz functions. One highly useful property
of Lipschitz functions, not shared by the smooth functions, is that the pointwise
minimum and maximum over L-Lipschitz functions are still L-Lipschitz. The
same is in fact true also for pointwise infimum and supremum of L-Lipschitz
functions, if these are finite at a single point. In particular, it follows that if
u : A → R is an L-Lipschitz function, then the truncations max{u, c} and min{u, c}
with c ∈ R are L-Lipschitz.
Theorem 3.40 (McShane). Assume that A ⊂ Rn
, 0 L < ∞ and that f : A → R
is an L-Lipschitz function. There exists an L-Lipschitz function f ∗ : Rn
→ R such
that f ∗
(x) = f (x) for every x ∈ A.
T H E M O R A L : Every Lipschitz continuous function defined on a subset A of
Rn
can be extended as a Lipschitz continuous function to the whole Rn
.
Proof. Define f ∗ : Rn
→ R,
f ∗
(x) = inf f (a)+ L|x− a| : a ∈ A .
We claim that f ∗
(b) = f (b) for every b ∈ A. To see this we observe that
f (b)− f (a) |f (b)− f (a)| L|b − a|,
which implies f (b) f (a)+ L|b − a| for every a ∈ A. By taking infimum over a ∈ A
we obtain f (b) f ∗
(b). On the other hand, by the definition f ∗
(b) f (b) for every
b ∈ A. Thus f ∗
(b) = f (b) for every b ∈ A.

Then we claim that f ∗
is L-Lipschitz in Rn
. Let x, y ∈ Rn
. Then
f ∗
(x) = inf f (a)+ L|x− a| : a ∈ A
inf f (a)+ L(|y− a|+|x− y|) : a ∈ A
inf f (a)+ L|y− a| : a ∈ A + L|x− y|
= f ∗
(y)+ L|x− y|.
By switching the roles of x and y, we arrive at f ∗
(y) f ∗
(x)+L|x− y|. This implies
that −L|x− y| f ∗
(x)− f ∗
(y) L|x− y|.
Remark 3.41. The function f∗ : Rn
→ R,
f∗(x) = sup f (a)− L|x− a| : a ∈ A .
is an L-Lipschitz extension of f as well. We can see, that f ∗
is the largest and f∗
the smallest L-Lipschitz extensions of f .
Since C∞
0 (Rn
) is dense in W1,p
(Rn
), also compactly supported Lipschitz func-
tions are dense in W1,p
(Rn
). By Theorem 3.27, we give a quantitative density
result for Lipschitz functions in W1,p
(Rn
). The main difference of the following
result to the standard molliﬁcation approximation uε → u as ε → 0 is that the
value of the function is not changed in a good set {x ∈ Rn : uε(x) = u(x)} and there
is an estimate for the measure of the bad set {x ∈ Rn : uε(x) = u(x)}.
(Rn
), 1 0 there
exists a Lipschitz continuous function uε : Rn
→ R such that
(1) |{x ∈ Rn : uε(x) = u(x)}| < ε and
(2) u − uε W1,p(Rn) < ε.
Proof. Let Eλ = {x ∈ Rn : M|Du|(x) λ}, λ > 0. We show that u is cλ-Lipschitz in
Eλ. By Theorem 3.27
|u(x)− u(y)| c|x− y|(M|Du|(x)+ M|Du|(y)) cλ|x− y|
for almost every x, y ∈ Eλ. The McShane extension theorem allows us to ﬁnd a
cλ-Lipschitz extension vλ : Rn
→ R. We truncate vλ and obtain a 2cλ-Lipschitz
function
uλ = max{−λ,min{vλ,λ}}.
Observe that |uλ| λ in Rn
and uλ = u almost everywhere in Eλ.
(1) We consider measure of the set
Rn
Eλ = {x ∈ Rn
: M|Du|(x) > λ}.
There exits c = c(n, p) such that
λp
|{x ∈ Rn
: M|Du|(x)| > λ}| c
ˆ
{x∈Rn:|Du(x)|> λ
2 }
|Du(x)|p
dx → 0

as λ → ∞, since Du ∈ Lp
(Rn
). This follows by choosing f = |Du| in the following
general fact for the Hardy-Littlewood maximal function.
Claim: If f ∈ Lp
(Rn
), then there exists c = c(n, p) such that
|{x ∈ Rn
: M f (x) > λ}|
c
λ
ˆ
{x∈Rn:|f (x)|> λ
2 }
|f (x)|p
dx, λ > 0.
Reason. Let f = f1 + f2, where f1 = f χ{|f |> λ
2 } and f2 = f χ{|f | λ
2 }. Then
ˆ
Rn
|f1(x)|dx =
ˆ
{x∈Rn:|f (x)|> λ
2 }
|f1(x)|p
|f1(x)|1−p
dx
λ
2
1−p
f
p
Lp(Rn)
< ∞.
This shows that f1 ∈ L1
(Rn
). On the other hand, |f2(x)| λ
2 for every x ∈ Rn
,
which implies f2 L∞(Rn)
λ
2 and f2 ∈ L∞
(Rn
). Thus every Lp
function can be
represented as a sum of an L1
function and an L∞
function. By Lemma 3.2, we
have
M f2 L∞(Rn) f2 L∞(Rn)
λ
2
From this we conclude using sublinearity of the maximal operator that
M f (x) = M(f1 + f2)(x) M f1(x)+ M f2(x) M f1(x)+
λ
2
for every x ∈ Rn
and thus M f (x) > λ implies M f1(x) > λ
2 . It follows that
|{x ∈ Rn
: M f (x) > λ}| x ∈ Rn
: M f1(x) >
λ
2
for every λ > 0.
p = 1 By the maximal function theorem on L1
(Rn
), see (3.1), we have
x ∈ Rn
: M f1(x) >
λ
2
c
λ
f1 L1(Rn) =
c
λ
ˆ
{x∈Rn:|f (x)|> λ
2 }
|f (x)|dx
for every λ > 0.
1 1, see (3.2), we have
x ∈ Rn
: M f1(x) >
λ
2
2
λ
p ˆ
Rn
(M f1(x))p
dx
c
λp
ˆ
Rn
|f1(x)|p
dx
=
c
λp
ˆ
{x∈Rn:|f (x)|> λ
2 }
|f (x)|dx
for every λ > 0.
Thus we conclude that
λp
|Rn
Eλ| λp
|{x ∈ Rn
: M|Du|(x)| > λ}|
c
ˆ
{x∈Rn:|Du(x)|> λ
2 }
|Du(x)|p
dx

and consequently λp
|Rn
Eλ| → 0 and |Rn
Eλ| → 0 as λ → ∞.
(2) Next we prove an estimate for u − uε W1,p(Rn). Since uλ = u in Eλ and
|uλ| λ in Rn
, we have
uλ − u
p
Lp(Rn)
=
ˆ
RnEλ
|uλ − u|p
dx
2p
ˆ
RnEλ
|uλ|p
dx+
ˆ
RnEλ
|u|p
dx
2p
λp
|Rn
Eλ|+
ˆ
RnEλ
|u|p
dx → 0
as λ → ∞.
To prove the corresponding estimate for the gradients, we note that
D(uλ − u) = χRnEλ
D(uλ − u) = χRnEλ
Duλ −χRnEλ
Du
almost everywhere. Recall that uλ is cλ-Lipschitz and thus |Duλ| cλ almost
everywhere.
D(uλ − u)
p
Lp(Rn)
=
ˆ
RnEλ
|D(uλ − u)|p
dx
2p
ˆ
RnEλ
|Duλ|p
dx+
ˆ
RnEλ
|Du|p
dx
2p
(2cλ)p
|Rn
Eλ|+
ˆ
RnEλ
|Du|p
dx → 0
as λ → ∞. Thus u − uλ W1,p(Rn) → 0 as λ → ∞. Observe that
{x ∈ Rn
: u(x) = uλ(x)} ⊂ Ω Eλ,
with |Rn
Eλ| → 0 as λ → ∞. This proves the claims.
Remark 3.43. Let Eλ = {x ∈ Rn : M|Du|(x) λ}, λ > 0. Let Qi, i = 1,2,... be a
Whitney decomposition of Rn
Eλ with the following properties: each Qi is open,
cubes Qi, i = 1,2,..., are disjoint, Rn
Eλ = ∪∞
i=1
Qi, 4Qi ⊂ Rn
Eλ, i = 1,2,...,
∞
i=1
χ2Qi
N < ∞,
and
c1 dist(Qi,Eλ) diam(Qi) c2 dist(Qi,Eλ)
for some constants c1 and c2.
Then we construct a partition of unity associated with the covering 2Qi,
i = 1,2,... This can be done in two steps. First, let ϕi ∈ C∞
0 (2Qi) be such that
0 ϕi 1, ϕi = 1 in Qi and
|Dϕi|
c
diam(Qi)
,

for i = 1,2,... Then we define
φi(x) =
ϕi(x)
∞
j=1 ϕj(x)
for every i = 1,2,.... Observe that the sum is over finitely many terms only since
ϕi ∈ C∞
0 (2Qi) and the cubes 2Qi, i = 1,2,..., are of bounded overlap. The functions
φi have the property
∞
i=1
φi(x) = χRnEλ
(x)
for every x ∈ Rn
.
Then we define the function uλ by
uλ(x) =



u(x), x ∈ Eλ,
∞
i=1 φi(x)u2Qi
, x ∈ Rn
Eλ.
The function uλ is a Whitney type extension of u|Eλ
to the set Rn
Eλ.
First we claim that
uλ W1,p(RnEλ) c u W1,p(RnEλ). (3.9)
Since the cubes 2Qi, i = 1,2,..., are of bounded overlap, we have
ˆ
RnEλ
|uλ|p
dx =
ˆ
RnEλ
∞
i=1
φi(x)u2Qi
p
dx c
∞
i=1
ˆ
2Qi
|u2Qi
|p
dx
c
∞
i=1
|2Qi|
2Qi
|u|p
dx c
ˆ
RnEλ
|u|p
dx.
Then we consider an estimate for the gradient. We recall that
Φ(x) =
∞
i=1
φi(x) = 1
for every x ∈ Rn
Eλ. Since the cubes 2Qi, i = 1,2,..., are of bounded overlap, we
see that Φ ∈ C∞
(Rn
Eλ) and
DjΦ(x) =
∞
i=1
Djφi(x) = 0, j = 1,2,...,n,
for every x ∈ Rn
Eλ. Hence we obtain
|Djuλ(x)| =
∞
i=1
Djφi(x)u2Qi
=
∞
i=1
Djφi(x)(u(x)− u2Qi
)
c
∞
i=1
diam(Qi)−1
|u(x)− u2Qi
|χ2Qi
(x)
and consequently
|Djuλ(x)| c
∞
i=1
diam(Qi)−p
|u(x)− u2Qi
|p
χ2Qi
(x).

Here we again used the fact that the cubes 2Qi, i = 1,2,..., are of bounded overlap.
This implies that for every j = 1,2,...,n,
ˆ
RnEλ
|Djuλ|dx c
ˆ
RnEλ
∞
i=1
diam(Qi)−p
|u − u2Qi
|p
χ2Qi
dx
∞
i=1
ˆ
2Qi
diam(Qi)−p
|u − u2Qi
|p
dx
c
∞
i=1
ˆ
2Qi
|Du|p
dx c
ˆ
RnEλ
|Du|p
dx.
Then we show that uλ ∈ W1,p
(Rn
). We know that uλ ∈ W1,p
(Rn
Eλ) and that it
is Lipschitz continuous in Rn
(exercise). Moreover u ∈ W1,p
(Rn
) and u = uλ in Eλ
by (i). This implies that w = u − uλ ∈ W1,p
(Rn
Eλ). and that w = 0 in Eλ. By the
ACL-property, u is absolutely continuous on almost every line segment parallel to
the coordinate axes. Take any such line. Now w is absolutely continuous on the
part of the line segment which intersects Rn
Eλ. On the other hand w = 0 in the
complement of Eλ. Hence the continuity of w in the line segment implies that w
is absolutely continuous on the whole line segment.
We have
u − uλ W1,p(Rn) = u − uλ W1,p(Eλ)
u W1,p(Eλ) + uλ W1,p(Eλ) c u W1,p(Eλ).
3.7 Maximal operator on Sobolev spaces
Assume that u is Lipschitz continuous with constant L, that is
|uh(y)− u(y)| = |u(y+ h)− u(y)| L|h|
for every y,h ∈ Rn
, where we denote uh(y) = u(y+ h). Since the maximal function
commutes with translations and the maximal operator is sublinear, we have
|(Mu)h(x)− Mu(x)| = |M(uh)(x)− Mu(x)|
M(uh − u)(x)
= sup
r>0
1
|B(x,r)|
ˆ
B(x,r)
|uh(y)− u(y)|dy
L|h|.
This means that the maximal function is Lipschitz continuous with the same
constant as the original function if Mu is not identically inﬁnity. Observe, that
this proof applies to Hölder continuous functions as well.
Next we show that the Hardy-Littlewood maximal operator is bounded in
Sobolev spaces.

Theorem 3.44. Let 1 1, is a bounded operator. Thus
the maximal operator is not only bounded on Lp
(Rn
) but also on W1,p
(Rn
) for p > 1.
Proof. The proof is based on the characterization of W1,p
(Rn
) by integrated dif-
ference quotients, see Theorem 1.38. By the maximal function theorem with
1 < p < ∞, see (3.2), we have Mu ∈ Lp
(Rn
) and
(Mu)h − Mu Lp(Rn) = M(uh)− Mu Lp(Rn)
M(uh − u) Lp(Rn)
c uh − u Lp(Rn)
c Du Lp(Rn)|h|
for every h ∈ Rn
. Theorem 1.38 gives Mu ∈ W1,p
(Rn
) with
DMu Lp(Rn) c Du Lp(Rn).
Thus by the maximal function theorem
Mu W1,p(Rn) = Mu
p
Lp(Rn)
+ DMu
p
Lp(Rn)
1
p
Mu Lp(Rn) + DMu Lp(Rn)
c u Lp(Rn) + Du Lp(Rn)
c u W1,p(Rn).
A more careful analysis gives a pointwise estimate for the partial derivatives.
Theorem 3.45. Let 1 < p < ∞. If u ∈ W1,p
(Rn
), then Mu ∈ W1,p
(Rn
) and
|Dj Mu| M(Dju), j = 1,2,...,n, (3.11)
.
T H E M O R A L : Differentiation commutes with a linear operator. The sublinear
maximal operator semicommutes with differentiation.
Proof. If χB(0,r) is the characteristic function of B(0,r) and
χr =
χB(0,r)
|B(0,r)|
,

then
1
|B(x,r)|
ˆ
B(x,r)
|u(y)|dy =
1
|B(0,r)|
ˆ
B(0,r)
|u(x− y)|dy
=
1
|B(0,r)|
ˆ
Rn
χB(0,r)|u(x− y)|dy
= (|u|∗χr)(x),
where ∗ denotes the convolution. Now |u|∗χr ∈ W1,p
(Rn
) and by Theorem 1.17 (1)
Dj(|u|∗χr) = χr ∗ Dj|u|, j = 1,2,...,n,
.
Let rm, m = 1,2,..., be an enumeration of positive rationals. Since u is locally
integrable, we may restrict ourselves to the positive rational radii in the deﬁnition
of the maximal function. Hence
Mu(x) = sup
m
(|u|∗χrm )(x).
We deﬁne functions vk : Rn
→ R, k = 1,2,..., by
vk(x) = max
1 m k
(|u|∗χrm )(x).
Now (vk) is an increasing sequence of functions in W1,p
(Rn
), which converges to
Mu pointwise and
|Djvk| max
1 m k
Dj(|u|∗χrm )
= max
1 m k
χrm ∗ Dj|u|
M(Dj|u|) = M(Dju), j = 1,2,...,n,
. Here we also used Remark 1.27 and the fact that by
Theorem 1.26
Dj|u| = |Dju|, j = 1,2,...,n,
almost everywhere. Thus
Dvk Lp(Rn)
n
j=1
Djvk Lp(Rn)
n
j=1
M(Dju) Lp(Rn)
and the maximal function theorem implies
vk W1,p(Rn) Mu Lp(Rn) +
n
j=1
M(Dju) Lp(Rn)
c u Lp(Rn) + c
n
j=1
Dju Lp(Rn) c < ∞

for every k = 1,2,... Hence (vk) is a bounded sequence in W1,p
(Rn
) which converges
to Mu pointwise. Theorem 1.35 implies Mu ∈ W1,p
(Rn
), vk → Mu weakly in Lp
(Rn
)
and Djvk → Dj Mu weakly in Lp
(Rn
).
Next we prove the pointwise estimate for the gradient. By Mazur’s lemma, see
Theorem 1.31, there is a subsequence of (vk), still denoted by (vk), such that the
convex combinations
wl =
l
k=1
akDjvk → Dj Mu, j = 1,...,n,
in Lp
(Rn
) as l → ∞. There is a subsequence of (wl) which converges almost
everywhere to Dj Mu. Thus we have
|wl|
l
k=1
ak Djvk
l
k=1
akM(Dju) = M(Dju)
for every l = 1,2,... and finally
|Dj Mu| = lim
i→∞
|wl| M(Dju), j = 1,...,n,
. This completes the proof.
Remarks 3.46:
(1) Estimate (3.10) also follows from (3.11). To see this, we may use the
maximal function theorem, see (3.2), and (3.11) to obtain
Mu W1,p(Rn) Mu Lp(Rn) + DMu Lp(Rn)
c u Lp(Rn) + M|Du| Lp(Rn)
c u W1,p(Rn),
where c is the constant in (3.2).
(2) If u ∈ W1,∞
(Rn
), then a slight modification of our proof shows that Mu
belongs to W1,∞
(Rn
). Moreover,
Mu W1,∞(Rn) = Mu L∞(Rn) + DMu L∞(Rn)
u L∞(Rn) + M|Du| L∞(Rn)
u W1,∞(Rn).
Hence in this case the maximal operator is bounded with constant one.
Recall, that after a redefinition on a set of measure zero u ∈ W1,∞
(Rn
) is a
bounded and Lipschitz continuous function, see Theorem 2.18.

4Pointwise behaviour of Sobolev
functions
In this chapter we study fine properties of Sobolev functions. By definition, Sobolev
functions are defined only up to Lebesgue measure zero and thus it is not always
clear how to use their pointwise properties to give meaning, for example, to
boundary values.
4.1 Sobolev capacity
Capacities are needed to understand pointwise behavior of Sobolev functions.
They also play an important role in studies of solutions of partial differential
equations.
Definition 4.1. For 1 < p < ∞, the Sobolev p-capacity of a set E ⊂ Rn
is defined
by
capp(E) = inf
u∈A (E)
u
p
W1,p(Rn)
= inf
u∈A (E)
u
p
Lp(Rn)
+ Du
p
Lp(Rn)
= inf
u∈A (E)
ˆ
Rn
|u|p
+|Du|p
dx,
where A (E) = u ∈ W1,p
(Rn
) : u 1 on a neighbourhood of E . If A (E) = , we set
capp(E) = ∞. Functions in A (E) are called admissible functions for E.
T H E M O R A L : Capacity measures the size of exceptional sets for Sobolev
functions. Lebesgue measure is the natural measure for functions in Lp
(Rn
) and
the Sobolev p-capacity is the natural outer measure for functions in W1,p
(Rn
).
97

CHAPTER 4. POINTWISE BEHAVIOUR OF SOBOLEV FUNCTIONS 98
Remark 4.2. In the definition of capacity we can restrict ourselves to the admissi-
ble functions u for which 0 u 1. Thus
capp(E) = inf
u∈A (E)
u
p
W1,p(Rn)
,
where A (E) = u ∈ W1,p
(Rn
) : 0 u 1, u = 1 on a neighbourhood of E .
Reason. (1) Since A (E) ⊂ A (E), we have
capp(E) inf
u∈A (E)
u
p
W1,p(Rn)
.
(2) For the reverse inequality, let ε > 0 and let u ∈ A (E) such that
u
p
W1,p(Rn)
capp(E)+ε.
Then v = max{0,min{|u|,1}} ∈ A (E), |v| |u| and by Remark 1.27 we have |Dv|
|Du| almost everywhere. Thus
inf
u∈A (E)
u
p
W1,p(Rn)
v
p
W1,p(Rn)
u
p
W1,p(Rn)
capp(E)+ε
and by letting ε → 0 we obtain
inf
u∈A (E)
u
p
W1,p(Rn)
capp(E).
Remarks 4.3:
(1) There are several alternative definitions for capacity and, in general, it
does not matter which one we choose. For example, when 1 < p < n, we
may consider the defintion
capp(E) = inf
ˆ
Rn
|Du|p
dx,
where the infimum is taken over all u ∈ Lp∗
(Rn
) with |Du| ∈ Lp
(Rn
), u 0
and u 1 on a neighbourhood of E. Some estimates and arguments may
become more transparent with this definition, but we stick to our original
definition.
(2) The definition of Sobolev capacity applies also for p = 1, but we shall not
discuss this case here.
The Sobolev p-capacity enjoys many desirable properties, one of the most
important of which says that it is an outer measure.
Theorem 4.4. The Sobolev p-capacity is an outer measure, that is,
(1) capp( ) = 0,
(2) if E1 ⊂ E2, then capp(E1) capp(E2) and

(3) capp
∞
i=1
Ei
∞
i=1 capp(Ei) whenever Ei ⊂ Rn
, i = 1,2,....
T H E M O R A L : Capacity is an outer measure, but measure theory is useless
since there are very few measurable sets.
Proof. (1) Clearly capp( ) = 0.
(2) A (E2) ⊂ A (E1) implies capp(E1) capp(E2) .
(3) Let ε > 0. We may assume that ∞
i=1 capp(Ei) < ∞. Choose ui ∈ A (Ei) so
that
ui
p
W1,p(Rn)
capp(Ei)+ε2−i
, i = 1,2,....
Claim: v = supi ui is admissible for ∞
i=1
Ei.
Reason. First we show that v ∈ W1,p
(Rn
). Let
vk = max
1 i k
ui, k = 1,2,....
Then (vk) is an increasing sequence such that vk → v pointwise as k → ∞. More-
over
|vk| = | max
1 i k
ui| |sup
i
ui| = |v|, k = 1,2,...,
and by Remark 1.27
|Dvk| max
1 i k
|Dui| sup
i
|Dui|, k = 1,2,....
We show that (vk) is a a bounded sequence in W1,p
(Rn
). To conclude this, we
observe that
vk
p
W1,p(Rn)
=
ˆ
Rn
|vk|p
dx+
ˆ
Rn
|Dvk|p
dx
ˆ
Rn
sup
i
|ui|p
dx+
ˆ
Rn
sup
i
|Dui|p
dx
ˆ
Rn
∞
i=1
|ui|p
dx+
ˆ
Rn
∞
i=1
|Dui|p
dx
=
∞
i=1
ˆ
Rn
|ui|p
dx+
ˆ
Rn
|Dui|p
dx
∞
i=1
(capp(Ei)+ε2−i
)
∞
i=1
capp(Ei)+ε < ∞, k = 1,2,....
Since vk → v almost everywhere, by weak compactness of Sobolev spaces, see
Theorem 1.35, we conclude that v ∈ W1,p
(Rn
). Since ui ∈ A (Ei), there exists an
open set Oi ⊃ Ei such that ui 1 on Oi for every i = 1,2,.... It follows that
v = supi ui 1 on ∞
i=1
Oi, which is a neighbourhood of ∞
i=1
Ei.

We conclude that
capp
∞
i=1
Ei v
p
W1,p(Rn)
∞
i=1
ui
p
W1,p(Rn)
∞
i=1
capp(Ei)+ε.
The claim follows by letting ε → 0.
Remark 4.5. The Sobolev p-capacity is outer regular, that is,
capp(E) = inf{capp(O) : E ⊂ O, O open}.
Reason. (1) By monotonicity,
capp(E) inf{capp(O) : E ⊂ O, O open}.
(2) To see the inequality into the other direction, let ε > 0 and take u ∈ A (E)
such that
u
p
W1,p(Rn)
capp(E)+ε.
Since u ∈ A (E) there is an open set O containing E such that u 1 on O, which
implies
capp(O) u
p
W1,p(Rn)
capp(E)+ε.
T H E M O R A L : Capacity of a set is completely determined by capacities of open
sets containing the set. The same applies to the Lebesgue outer measure.
4.2 Capacity and measure
We are mainly interested in the sets of vanishing capacity, since they are in some
sense exceptional sets in the theory Sobolev spaces. Our ﬁrst result is rather
immediate.
Lemma 4.6. |E| capp(E) for every E ⊂ Rn
.
T H E M O R A L : Sets of capacity zero are of measure zero. Thus capacity is a
ﬁner measure than Lebesgue measure.
Proof. If capp(E) = ∞, there is nothing to prove. Thus we may assume that
capp(E) < ∞. Let ε > 0 and take u ∈ A (E) such that
u
p
W1,p(Rn)
capp(E)+ε.
There is an open O ⊃ E such that u 1 in O and thus
|E| |O|
ˆ
O
|u|p
dx u
p
Lp(Rn)
u
p
W1,p(Rn)
capp(E)+ε.

Remark 4.7. Lemma 4.6 shows that capp(B(x,r)) > 0 for every x ∈ Rn
, r > 0. This
implies that capacity is nontrivial in the sense that every nonempty open set has
positive capacity.
Lemma 4.8. Let x ∈ Rn
and 0 < r 1. Then there exists c = c(n, p) such that
capp(B(x,r)) crn−p
T H E M O R A L : For the Lebesgue measure of a ball we have |B(x,r)| crn
, but
for the Sobolev capacity of a ball we have capp(B(x,r)) crn−p
. Thus the natural
scaling dimension for capacity is n− p. Observe, that the dimension for capacity
is smaller than n−1.
Proof. Define a cutoff function
u(y) =



1, y ∈ B(x,r),
2−
|y−x|
r , y ∈ B(x,2r)B(x,r),
0, y ∈ Rn
B(x,2r).
Observe that 0 u 1, u is 1
r -Lipschitz and |Du| 1
r almost everywhere. Thus
u ∈ A (B(x,r)) and
capp(B(x,r))
ˆ
B(x,2r)
|u(y)|p
dy+
ˆ
B(x,2r)
|Du(y)|p
dy
(1+ r−p
)|B(x,2r)| (r−p
+ r−p
)|B(x,2r)|
= 2r−p
|B(x,2r)| = crn−p
,
with c = c(n, p)
Remarks 4.9:
(1) Lemma 4.8 shows that every bounded set has finite capacity. Thus there
are plenty of sets with finite capacity.
Reason. Assume that E ⊂ Rn
is bounded. Then E ⊂ B(0,r) for some r,
0 < r < ∞, and
capp(E) capp(B(0,r)) crn−p
< ∞.
(2) Lemma 4.8 implies that capp({x}) = 0 for every x ∈ Rn
when 1 < p < n.
Reason.
capp({x}) capp(B(x,r)) crn−p
, 0 < r 1.
The claim follows by letting r → 0.

Remark 4.10. Let x ∈ Rn
and 0 < r 1
2 . Then there exists c = c(n) such that
capn(B(x,r)) c log
1
r
1−n
.
Reason. Use the test function
u(y) =



log 1
r
−1
log 1
|x−y| , y ∈ B(x,1)B(x,r),
1, y ∈ B(x,r),
0, y ∈ Rn
B(x,1).
This implies that capp({x}) = 0 for every x ∈ Rn
when p = n (exercise).
We have shown that a point has zero capacity when 1 n.
Lemma 4.11. If p > n, then capp({x}) > 0 for every x ∈ Rn
.
T H E M O R A L : For p > n every set containing at least one point has a positive
capacity. Thus there are no nontrivial sets of capacity zero. In practice this means
that capacity is a useful tool only when p n.
Proof. Let x ∈ Rn
and assume that u ∈ A ({x}). Then there exists 0 < r 1 such that
u(y) 1 on B(x,r). Take a cutoff function η ∈ C∞
0 (B(x,2)) such that 0 η 1, η = 1
in B(x,r) and |Dη| 2. By Morrey’s inequality, see Theorem 2.13, there exists
c = c(n, p) > 0 such that
|(ηu)(y)−(ηu)(z)| c|y− z|
1− n
p D(ηu) Lp(Rn)
for almost every y, z ∈ Rn
. Choose y ∈ B(x,r) and z ∈ B(x,4) B(x,2) so that
(ηu)(y) 1 and (ηu)(z) = 0. Then 1 |y− z| 5 and thus
ˆ
B(x,2)
|D(ηu)(y)|p
dy = D(ηu)
p
Lp(Rn)
c|y− z|n−p
5n−p
|(ηu)(y)−(ηu)(z)|p
1
c > 0.
On the other hand
ˆ
B(x,2)
|D(ηu)(y)|p
dy 2p
ˆ
B(x,2)
|Dη(y)u(y)|p
dy+
ˆ
B(x,2)
|η(y)Du(y)|p
dy
= 2p


ˆ
B(x,2)
|Dη(y)|p
2p
|u(y)|p
dy+
ˆ
B(x,2)
|η(y)|p
1
|Du(y)|p
dy


4p
ˆ
B(x,2)
|u(y)|p
dy+
ˆ
B(x,2)
|Du(y)|p
dy
4p
u
p
W1,p(Ω)
.
This shows that there exists c = c(n, p) > 0 such that u
p
W1,p(Ω)
c > 0 for every
u ∈ A ({x}) and thus capp({x}) c > 0.

In order to study the connection between capacity and measure, we need to
consider lower dimensional measures than the Lebesgue measure. We recall the
definition of Hausdorff measures.
Definition 4.12. Let E ⊂ Rn
and s 0. For 0 < δ ∞ we set
H s
δ (E) = inf
∞
i=1
rs
i
: E ⊂
∞
i=1
B(xi,ri), ri δ .
The (spherical) s-Hausdorff measure of E is
H s
(E) = lim
δ→0
H s
δ (E) = sup
δ>0
H s
δ (E).
The Hausdorff dimension of E is
inf s : H s
(E) = 0 = sup s : H s
(E) = ∞ .
T H E M O R A L : The Hausdorff measure is the natural s-dimensional measure
up to scaling and the Hausdorff dimension is the measure theoretic dimension of
the set. Observe that the dimension can be any nonnegative real number less or
equal than the dimension of the space.
We begin by proving a useful measure theoretic lemma. In the proof we need
some tools from measure and integration theory and real analysis.
Lemma 4.13. Assume that 0 < s < n, f ∈ L1
loc
(Rn
) and let
E = x ∈ Rn
: limsup
r→0
1
rs
ˆ
B(x,r)
|f |dy > 0 .
Then H s
(E) = 0.
T H E M O R A L : Roughly speaking the lemma above says that the set where
a locally integrable function blows up rapidly is of the corresponding Hausdorff
measure zero.
Proof. (1) Assume first that f ∈ L1
(Rn
).
(2) By the Lebesgue differentiation theorem
lim
r→0 B(x,r)
|f (y)|dy = |f (x)| < ∞,
. If x is a Lebesgue point of |f |, then
limsup
r→0
1
rs
ˆ
B(x,r)
|f (y)|dy = climsup
r→0
rn−s
B(x,r)
|f (y)|dy = 0.
This shows that all Lebesgue points of |f | belong to the complement of E and thus
|E| = 0.

(3) Let ε > 0 and
Eε = x ∈ Rn
: limsup
r→0
1
rs
ˆ
B(x,r)
|f |dy > ε .
Since Eε ⊂ E and |E| = 0, we have |Eε| = 0.
Claim: H s
(Eε) = 0 for every ε > 0.
Reason. Let 0 < δ < 1. For every x ∈ Eε there exists rx with 0 < rx δ such that
1
rs
x
ˆ
B(x,rx)
|f |dy > ε.
By the Vitali covering theorem, there exists a subfamily of countably many pair-
wise disjoint balls B(xi,ri), i = 1,2,..., such that
Eε ⊂
∞
i=1
B(xi,5ri).
This gives
H s
5δ(Eε)
∞
i=1
(5ri)s 5s
ε
∞
i=1
ˆ
B(xi,ri)
|f |dy =
5s
ε
ˆ
∞
i=1
B(xi,ri)
|f |dy.
By disjointness of the balls
∞
i=1
B(x,ri) =
∞
i=1
|B(xi,ri)| = c
∞
i=1
rn
i
c
∞
i=1
rn
i
εrs
i
ˆ
B(xi,ri)
|f |dy
c
δn−s
ε
ˆ
Rn
|f |dy → 0 as δ → 0.
By absolute continuity of integral
ˆ
∞
i=1
B(xi,ri)
|f |dy → 0 as δ → 0.
Thus
H s
(Eε) = lim
δ→0
H s
5δ(Eε)
5s
ε
lim
δ→0
ˆ
∞
i=1
B(xi,ri)
|f |dy = 0.
This shows that H s
(Eε) = 0 for every ε > 0.
(4) By subadditivity of the Hausdorff measure
H s
(E) = H s
∞
k=1
E 1
k
∞
k=1
H s
(E 1
k
) = 0.
This shows that H s
(E) = 0.

(5) Assume then that f ∈ L1
loc
(Rn
). Then
H s
(E) = H s
x ∈ Rn
: limsup
r→0
1
rs
ˆ
B(x,r)
|f |dy > 0
= H s
∞
k=1
x ∈ Rn
: limsup
r→0
1
rs
ˆ
B(x,r)
|f χB(0,k)|dy > 0
∞
k=1
H s
x ∈ Rn
: limsup
r→0
1
rs
ˆ
B(x,r)
|f χB(0,k)|dy > 0 = 0.
Next we compare capacity to the Hausdorff measure.
Theorem 4.14. Assume that 1 < p < n. Then there exists c = c(n, p) such that
capp(E) cH n−p
(E) for every E ⊂ Rn
.
T H E M O R A L : Capacity is smaller than (n−p)-dimensional Hausdorff measure.
In particular, H n−p
(E) = 0 implies capp(E) = 0.
Proof. Let B(xi,ri), i = 1,2,..., be any covering of E such that the radii satisfy
ri δ. Subadditivity implies
capp(E)
∞
i=1
capp(B(xi,ri)) c
∞
i=1
r
n−p
i
.
By taking the inﬁmum over all coverings by such balls and observing that H s
δ
(E)
H s
(E) we obtain
capp(E) cH
n−p
δ
(E) cH n−p
(E).
We next consider the converse of the previous theorem. We prove that sets of
p-capacity zero have Hausdorff dimension at most n− p.
Theorem 4.15. Assume that 1 n− p.
Proof. (1) Let E ⊂ Rn
be such that capp(E) = 0. Then for every i = 1,2,..., there is
ui ∈ A (E) such that ui
p
W1,p(Rn)
2−i
. Deﬁne u = ∞
i=1 ui.
Claim: u ∈ A (E).
Reason. Let vk = k
i=1 ui, k = 1,2,.... Then vk ∈ W1,p
(Rn
) and
vk W1,p(Rn) =
k
i=1
ui
W1,p(Rn)
k
i=1
ui W1,p(Rn)
∞
i=1
ui W1,p(Rn)
∞
i=1
2
− i
p < ∞.
Thus (vk) is a bounded sequence in W1,p
(Rn
). Since 0 ui 1, we observe that
(vk) is an increasing sequence and thus vk → u almost everywhere. Theorem 1.35
implies u ∈ W1,p
(Rn
). Moreover, u 1 almost everywhere on a neighbourhood of E
which shows that u ∈ A (E).

(2) Claim:
limsup
r→0 B(x,r)
u dy = ∞ for every x ∈ E. (4.1)
Reason. Let m ∈ N and x ∈ E. Then for r > 0 small enough B(x,r) is contained in
an intersection of open sets Oi, i = 1,...,m, with the property that ui = 1 almost
everywhere on Oi. This implies that u = ∞
i=1 ui m almost everywhere in B(x,r)
and thus
B(x,r)
u dy m.
This proves the claim.
T H E M O R A L : This gives a method to construct a function that blows up on
any set of zero capacity.
(3) Claim: If s > n− p, then
limsup
r→0
1
rs
ˆ
B(x,r)
|Du|p
dy = ∞ for every x ∈ E.
Reason. Let x ∈ E and, for a contradiction, assume that
limsup
r→0
1
rs
ˆ
B(x,r)
|Du|p
dy < ∞.
Then there exists c < ∞ such that
limsup
r→0
1
rs
ˆ
B(x,r)
|Du|p
dy c.
The we choose R > 0 so small that
ˆ
B(x,r)
|Du|p
dy crs
for every 0 < r R. Denote Bi = B(x,2−i
R), i = 1,2,.... Then by Hölder’s inequality
and the Poincaré inequality, see Theorem 3.17, we have
|uBi+1
− uBi
|
Bi+1
|u − uBi
|dy
|Bi|
|Bi+1| Bi
|u − uBi
|dy
c
Bi
|u − uBi
|p
dy
1
p
c2−i
R
Bi
|Du|p
dy
1
p
c(2−i
R)
p−n+s
p .
For k > j, we obtain
|uBk
− uBj
|
k−1
i=j
|uBi+1
− uBi
| c
k−1
i=j
(2−i
R)
p−n+s
p

and thus (uBi
) is a Cauchy sequence when s > n − p. This contradicts (4.1) and
thus the claim holds true.
(4) Thus
E ⊂ x ∈ Rn
:
1
rs
ˆ
B(x,r)
|Du|p
dy = ∞
⊂ x ∈ Rn
:
1
rs
ˆ
B(x,r)
|Du|p
dy > 0 .
Lemma 4.13 implies
H s
(E) H s
x ∈ Rn
:
1
rs
ˆ
B(x,r)
|Du|p
dy > 0 = 0.
This shows that H s
(E) = 0 whenever n− p < s < n. The claim follows from this,
since H s
(E) = 0 implies H t
(E) = 0 for every t s.
Remark 4.16. It can be shown that even H n−p
(E) < ∞, 1 < p < n, implies capp(E) =
0.
4.3 Quasicontinuity
In this section we study fine properties of Sobolev functions. It turns out that
Sobolev functions are defined up to a set of capacity zero.
Definition 4.17. We say that a property holds p-quasieverywhere, if it holds
except for a set of p-capacity zero.
T H E M O R A L : Qusieverywhere is a capacitary version of almost everywhere.
Recall that by Meyers-Serrin theorem 1.18 W1,p
(Rn
)∩C(Rn
) is dense in W1,p
(Rn
)
for 1 p < ∞ and, by Theorem 1.13, the Sobolev space W1,p
(Rn
) is complete. The
next result gives a way to find a quasieverywhere converging subsequence.
Theorem 4.18. Assume that ui ∈ W1,p
(Rn
)∩C(Rn
), i = 1,2,..., and that (ui) is a
Cauchy sequence in W1,p
(Rn
). Then there is a subsequence of (ui) that converges
pointwise p-quasieverywhere in Rn
. Moreover, the convergence is uniform outside
a set of arbitrarily small p-capacity.
T H E M O R A L : This is a Sobolev space version of the result that for every
Cauchy sequence in Lp
(Rn
), there is a subsequence that converges pointwise
almost everywhere. The claim concerning uniform convergence is a Sobolev space
version of Egorov’s theorem.

Proof. There exists a subsequence of (ui), which we still denote by (ui), such that
∞
i=1
2ip
ui − ui+1
p
W1,p(Rn)
< ∞.
For i = 1,2,..., denote
Ei = x ∈ Rn
: |ui(x)− ui+1(x)| > 2−i
and Fj =
∞
i=j
Ei.
By continuity 2i
(ui − ui+1) ∈ A (Ei) and thus
capp(Ei) 2ip
ui − ui+1
p
W1,p(Rn)
.
By subadditivity we obtain
capp(Fj)
∞
i=j
capp(Ei)
∞
i=j
2ip
ui − ui+1
p
W1,p(Rn)
.
Thus
capp
∞
j=1
Fj lim
j→∞
capp(Fj) (
∞
j=1
Fj ⊂ Fj, Fj+1 ⊂ Fj, j = 1,2,...)
lim
j→∞
∞
i=j
2ip
ui − ui+1
p
W1,p(Rn)
= 0.
Here we used the fact that the tail of a convergent series tends to zero. We observe
that (ui) converges pointwise in Rn
∞
j=1
Fj. Moreover,
|ul(x)− uk(x)|
k−1
i=l
|ui(x)− ui+1(x)|
k−1
i=l
2−i
21−l
for every x ∈ Rn
Fj for every k > l > j, which shows that the convergence is
uniform in Rn
Fj.
Definition 4.19. A function u : Rn
→ [−∞,∞] is p-quasicontinuous in Rn
if for
every ε > 0 there is a set E such that capp(E) < ε and the restriction of u to Rn
E,
denoted by u|RnE, is continuous.
Remark 4.20. By outer regularity, see Remark 4.5, we may assume that E is open
in the definition above.
The next result shows that a Sobolev function has a quasicontinous represen-
tative.
Corollary 4.21. For each u ∈ W1,p
(Rn
) there is a p-quasicontinuous function
v ∈ W1,p
(Rn
) such that u = v almost everywhere in Rn
.
T H E M O R A L : Every Lp
function is defined almost everywhere, but every
W1,p
function is defined quasieverywhere.

Proof. By Theorem 1.18, for every function u ∈ W1,p
(Rn
), there are functions
ui ∈ W1,p
(Rn
) ∩ C(Rn
), i = 1,2,..., such that ui → u in W1,p
(Rn
) as i → ∞. By
Theorem 4.18 there exists a subsequence that converges uniformly outside a set
of arbitrarily small capacity. Uniform convergence implies continuity of the limit
function and thus the limit function is continuous outside a set of arbitrarily small
p-capacity. This completes the proof.
Next we show that the quasicontinuous representative given by Corollary 4.21
is essentially unique. We begin with a useful observation.
Remarks 4.22:
(1) If G ⊂ Rn
is open and E ⊂ Rn
with |E| = 0, then capp(G) = capp(G E).
Reason. Monotonicity implies capp(G) capp(G E).
Let ε > 0 and let u ∈ A (G E) be such that
u
p
W1,p(Rn)
capp(G E)+ε.
Then there exists an open O ⊂ Rn
with (G E) ⊂ O and u 1 almost every-
where in O. Since O ∪G is open G ⊂ (O ∪G) and u 1 almost everywhere
in O ∪ (G E), and almost everywhere in O ∪ G since |E| = 0, we have
u ∈ A (G).
capp(G) u
p
W1,p(Rn)
capp(G E)+ε.
By letting ε → 0, we obtain capp(G) capp(G E).
(2) For any open G ⊂ Rn
we have |G| = 0 ⇐⇒ capp(G) = 0.
Reason. =⇒ If |G| = 0, then (1) implies
capp(G) = capp(G G) = capp( ) = 0.
⇐= If capp(G) = 0, then Lemma 4.6 implies |G| capp(G) = 0.
W A R N I N G : It is not true in general that capacity and measure have
the same zero sets.
Theorem 4.23. Assume that u and v are p−quasicontinuous functions on Rn
. If
u = v almost everywhere in Rn
, then u = v p-quasieverywhere in Rn
.
T H E M O R A L : Quasicontinuous representatives of Sobolev functions are
unique.
Proof. Let ε > 0 and choose open G ⊂ Rn
such that capp(G) < ε and that the
restrictions of u and v to Rn
G are continuous. Thus {x ∈ Rn
G : u(x) = v(x)} is
open in the relative topology on Rn
G, that is, there exists open U ⊂ Rn
with
U G = {x ∈ Rn
G : u(x) = v(x)}

and
|U G| = |{x ∈ Rn
G : u(x) = v(x)}| = 0.
Moreover,
{x ∈ Rn
: u(x) = v(x)} ⊂ G ∪{x ∈ Rn
G : u(x) = v(x)} = G ∪U.
Remark 4.22 (1) with G and E replaced by U ∪G and U G, respectively, implies
capp({x ∈ Rn
: u(x) = v(x)}) capp(G ∪U) = capp(G) < ε.
Remarks 4.24:
(1) The same proof gives the following local result: Assume that u and v are
p−quasicontinuous on an open set O ⊂ Rn
. If u = v almost everywhere in
O, then u = v p-quasieverywhere in O.
(2) Observe that if u and v are p−quasicontinuous and u v almost ev-
erywhere in an open set O, then max{u − v,0} = 0 almost everywhere
in O and max{u − v,0} is p−quasicontinuous. Then Theorem 4.23 im-
plies max{u − v,0} = 0 p-quasieverywhere in O and consequently u v
p-quasieverywhere in O.
(3) The previous theorem enables us to define the trace of a Sobolev function
to an arbitrary set. If u ∈ W1,p
(Rn
) and E ⊂ Rn
, then the trace of u to E is
the restriction to E of any p−quasicontinuous representative of u. This
definition is useful only if capp(E) > 0.
4.4 Lebesgue points of Sobolev functions
By the maximal function theorem with p = 1, see (3.1), there exists c = c(n) such
that
|{x ∈ Rn
: M f (x) > λ}|
c
λ
f L1(Rn)
for every λ > 0. By Chebyshev’s inequality and the maximal function theorem
with 1 λ}|
1
λp
M f
p
Lp(Rn)
c
λp
f
p
Lp(Rn)
for every λ > 0. Thus the Hardy-Littlewood maximal function satisfies weak type
estimates with respect to Lebesgue measure for functions in Lp
(Rn
). Next we
consider capacitary weak type estimates for functions in W1,p
(Rn
).
(Rn
), 1 λ}
c
λp
u
p
W1,p(Rn)
.
for every λ > 0.

T H E M O R A L : This is a capacitary version of weak type estimates for the
Hardy-Littlewood maximal function.
Proof. Denote Eλ = {x ∈ Rn : Mu(x) > λ}. Then Eλ is open and by Theorem 3.44
Mu ∈ W1,p
(Rn
). Thus
Mu
λ
∈ A (Eλ).
Since the maximal operator is bounded on W1,p
(Rn
), see (3.10), we obtain
capp Eλ
Mu
λ
p
W1,p(Rn)
=
1
λp
Mu
p
W1,p(Rn)
c
λp
u
p
W1,p(Rn)
.
This weak type inequality can be used in studying the pointwise behaviour of
Sobolev functions. We recall that x ∈ Rn
is a Lebesgue point for u ∈ L1
loc
(Rn
) if the
limit
u∗
(x) = lim
r→0 B(x,r)
u(y)dy
exists and
lim
r→0 B(x,r)
|u(y)− u∗
(x)|dy = 0.
The Lebesgue differentiation theorem states that almost all points are Lebesgue
points for a locally integrable function. If a function belongs to W1,p
(Rn
), then
using the capacitary weak type estimate, see Theorem 4.25, we shall prove
that it has Lebesgue points p-quasieverywhere. Moreover, we show that the
p-quasicontinuous representative given by Corollary 4.21 is u∗
.
We begin by proving a measure theoretic result, which is analogous to Lemma
4.13.
Lemma 4.26. Let 1 0 .
Then capp(E) = 0.
T H E M O R A L : Roughly speaking the lemma above says that the set where an
Lp
function blows up rapidly is of capacity zero. The main difference compared to
Lemma 4.13 is that the size of the set is measured by capacity instead of Hausdorff
measure.
Proof. The argument is similar to the proof of Lemma 4.13, but we reproduce it
here.
(1) By the Lebesgue differentiation theorem
lim
r→0 B(x,r)
|f (y)|p
dy = |f (x)|p
< ∞,

. If x is a Lebesgue point of |f |p
, then
limsup
r→0
rp
B(x,r)
|f (y)|p
dy = 0.
This shows that all Lebesgue points of |f |p
belong to the complement of E and
thus |E| = 0.
(2) Let ε > 0 and
Eε = x ∈ Rn
: limsup
r→0
rp
B(x,r)
|f |p
dy > ε .
Since Eε ⊂ E and |E| = 0, we have |Eε| = 0. We show that capp(Eε) = 0 for every
ε > 0, then the claim follows by subadditivity. Let 0 < δ < 1
5 . For every x ∈ Eε there
is rx with 0 < rx δ such that
r
p
x
B(x,rx)
|f |p
dy > ε.
By the Vitali covering theorem, there exists a subfamily of countably many pair-
wise disjoint balls B(xi,ri), i = 1,2,..., such that
Eε ⊂
∞
i=1
B(xi,5ri).
By subadditivity of the capacity and Lemma 4.8 we have
capp(Eε)
∞
i=1
capp(B(xi,5ri)) c
∞
i=1
r
n−p
i
c
ε
∞
i=1
ˆ
B(xi,ri)
|f |p
dy =
c
ε
ˆ
∞
i=1
B(xi,ri)
|f (y)|p
dy.
Here c = c(n, p). Finally we observe that by the disjointness of the balls
∞
i=1
B(x,ri) =
∞
i=1
|B(xi,ri)|
∞
i=1
r
p
i
ε
ˆ
B(xi,ri)
|f |p
dy
δp
ε
ˆ
Rn
|f |p
dy → 0
as δ → 0. By absolute continuity of integral
ˆ
∞
i=1
B(xi,ri)
|f |p
dy → 0
as δ → 0. Thus
capp(Eε)
c
ε
ˆ
∞
i=1
B(xi,ri)
|f |p
dy → 0
as δ → 0, which implies that capp(Eε) = 0 for every ε > 0.
Now we are ready for a version of the Lebesgue differentiation theorem for
Sobolev functions.

(Rn
) with 1 < p < ∞. Then there exists
E ⊂ Rn
such that capp(E) = 0 and
lim
r→0 B(x,r)
u(y)dy = u∗
(x)
exists for every x ∈ Rn
E. Moreover
lim
r→0 B(x,r)
|u(y)− u∗
(x)|dy = 0
for every x ∈ Rn
E and the function u∗
is the p-quasicontinuous representative
of u.
T H E M O R A L : A function in W1,p
(Rn
) with 1 < p < ∞ has Lebesgue points
p-quasieverywhere. Moreover, the p-quasicontinuous representative is obtained
as a limit of integral averages.
Proof. (1) By Theorem 1.18 there exist ui ∈ C∞
(Rn
)∩W1,p
(Rn
) such that
u − ui
p
W1,p(Rn)
2−i(p+1)
, i = 1,2,....
Denote
Ei = {x ∈ Rn
: M(u − ui)(x) > 2−i
}, i = 1,2,....
By Theorem 4.25 there exists c = c(n, p) such that
capp(Ei) c2ip
u − ui
p
W1,p(Rn)
c2−i
, i = 1,2,....
Clearly
|ui(x)− uB(x,r)|
B(x,r)
|ui(x)− u(y)|dy
B(x,r)
|ui(x)− ui(y)|dy+
B(x,r)
|ui(y)− u(y)|dy,
which implies that
limsup
r→0
|ui(x)− uB(x,r)|
limsup
r→0 B(x,r)
|ui(x)− ui(y)|dy+limsup
r→0 B(x,r)
|ui(y)− u(y)|dy
M(ui − u)(x) 2−i
,
for every x ∈ Rn
Ei. Here we used the fact that
limsup
r→0 B(x,r)
|ui(x)− ui(y)|dy = 0, i = 1,2,...,
since ui is continuous and
B(x,r)
|ui(y)− u(y)|dy M(ui − u)(x) for every r > 0.

whenever x ∈ Rn
(E ∪ F). Finally we observe that
capp(E ∪ F) capp(E)+capp(F) = 0.
(3) Let ε > 0 and choose k large enough so that capp(Fk) < ε
2 . Then by outer
regularity of the capacity, see Remark 4.5, there is an open set O containing Fk so
that capp(O) < ε. Since (ui) converges uniformly to u∗
on Rn
O we conclude that
u∗
|XO is continuous. Thus u∗
is p-quasicontinuous.
4.5 Sobolev spaces with zero boundary
values
In this section we return to Sobolev spaces with zero boundary values started in
Section 1.9. Assume that Ω is an open subset of Rn
and 1 p < ∞. Recall that
W
1,p
0 (Ω) with 1 p < ∞ is the closure of C∞
0 (Ω) with respect to the Sobolev norm,
see Defintion 1.20. Using pointwise properties of Sobolev functions we discuss the
definition of W
1,p
0 (Ω).
The first result is a W
1,p
0 (Ω) version of Corollary 4.21 which states that for
every u ∈ W1,p
(Rn
) there is a p-quasicontinuous function v ∈ W1,p
(Rn
) such that
u = v almost everywhere in Rn
.
Theorem 4.28. If u ∈ W
1,p
0 (Ω), there exists a p-quasicontinuous function v ∈
W1,p
(Rn
) such that u = v almost everywhere in Ω and v = 0 p-quasieverywhere in
Rn
Ω.
T H E M O R A L : Quasicontinuous functions in Sobolev spaces with zero bound-
ary values are zero quasieverywhere in the complement.
Proof. Since u ∈ W
1,p
0 (Ω), there exist ui ∈ C∞
0 (Ω), i = 1,2,..., such that ui → u in
W1,p
(Ω) as i → ∞. Since (ui) is a Cauchy sequence in W1,p
(Rn
), by Theorem 4.18
it has a subsequence of (ui) that converges pointwise p-quasieverywhere in Rn
to a function v ∈ W1,p
(Rn
). Moreover, the convergence is uniform outside a set
of arbitrary small p-capacity and, as in Corollary 4.21, the limit function v is
p-quasicontinuous.
Theorem 4.29. If u ∈ W1,p
(Rn
) is p-quasicontinuous and u = 0 p-quasieverywhere
in Rn
Ω, then u ∈ W
1,p
0 (Ω).
T H E M O R A L : Quasicontinuous functions in a Sobolev space on the whole
space which are zero quasieverywhere in the complement belong to the Sobolev
space with zero boundary values. In particular, continuous functions in a Sobolev
space on the whole space which are zero everywhere in the complement belong to
the Sobolev space with zero boundary values.

Proof. (1) We show that u can be approximated by W1,p
(Rn
) functions with com-
pact support in Ω. If we can construct such a sequence for u+ = max{u,0}, then we
can do it for u− = −min{u,0}, and we obtain the result for u = u+ + u−. Thus we
may assume that u 0. By Theorem 1.24 we may assume that u has a compact
support in Rn
and by considering truncations min{u,λ}, λ > 0, we may assume
that u is bounded (exercise).
(2) Let δ > 0 and let O ⊂ Rn
be an open set such that capp(O) < δ and the
restriction of u to Rn
O is continuous. Denote
E = {x ∈ Rn
Ω : u(x) = 0}.
By assumption capp(E) = 0. Let v ∈ A (O ∪ E) such that 0 v 1 and
v
p
W1,p(Rn)
< δ,
see Remark 4.2. Then v = 1 in an open set G containing O ∪ E. Deﬁne
uε(x) = max{u(x)−ε,0}, 0 < ε < 1.
Let x ∈ ΩG. Since u(x) = 0 and the restriction of u to Rn
G is continuous, there
exists rx > 0 such that uε = 0 in B(x,rx)G. Thus (1− v)uε = 0 in B(x,rx)∪G for
every x ∈ ΩG. This shows that (1− v)uε is zero in a neighbourhood of Rn
Ω,
which implies that (1− v)uε is compactly supported in Ω. Lemma 1.23 implies
(1− v)uε ∈ W
1,p
0 (Ω). We show that this kind of functions converge to u in W1,p
(Rn
).
(3) Since
uε =



u −ε in {x ∈ Rn : u(x) ε},
0 in {x ∈ Rn : u(x) ε},
by Remark 1.27 we have
Duε =



Du almost everywhere in {x ∈ Rn : u(x) ε},
0 almost everywhere in {x ∈ Rn : u(x) ε}.
Thus
u −(1− v)uε W1,p(Rn) u − uε W1,p(Rn) + vuε W1,p(Rn).
Using the facts that u − uε ε and supp(u − uε) ⊂ suppu, we obtain
u − uε W1,p(Rn) u − uε Lp(Rn) + Du − Duε Lp(Rn)
ε χsuppu Lp(Rn) + χ{0<u ε}Du Lp(Rn) → 0
as ε → 0. Observe that, by the dominated convergence theorem, we have
lim
ε→0
χ{0<u ε}Du Lp(Rn) = lim
ε→0
ˆ
Rn
χ{0<u ε}|Du|p
dx
1
p
=
ˆ
Rn
lim
ε→0
χ{0<u ε}|Du|p
dx
1
p
= 0,

where χ{0<u ε}|Du|p
|Du|p
∈ L1
(Rn
) may be used as an integrable majorant. On
the other hand,
vuε W1,p(Rn) vuε Lp(Rn) + D(vuε) Lp(Rn)
vuε Lp(Rn) + uεDv Lp(Rn) + vDuε Lp(Rn)
uv Lp(Rn) + uDv Lp(Rn) + vDuε Lp(Rn)
u L∞(Rn) v Lp(Rn) + u L∞(Rn) Dv Lp(Rn) + vDu Lp(Rn)
2 u L∞(Rn) v W1,p(Rn) + vDu Lp(Rn)
2δ
1
p u L∞(Rn) + vDu Lp(Rn).
Since v = vδ → 0 in Lp
(Rn
) as δ → 0, there is a subsequence (δi) for which vi =
vδi
→ 0 almost everywhere as i → ∞. By the dominated convergence theorem, we
have
lim
i→∞
viDu Lp(Rn) = lim
i→∞
ˆ
Rn
|vi|p
|Du|p
dx
1
p
=
ˆ
Rn
( lim
i→∞
|vi|p
)|Du|p
dx
1
p
= 0,
where |vi|p
|Du|p
|Du|p
, so that |Du|p
∈ L1
(Rn
) may be used as an integrable
majorant. Thus we conclude that
lim
i→∞
viuε W1,p(Rn) lim
i→∞
2δ
1
p
i
u L∞(Rn) + viDu Lp(Rn) = 0.
Thus
u −(1− vi)uε W1,p(Rn) → 0
as ε → 0 and i → ∞. Since
(1− vi)uε ∈ W
1,p
0 (Ω) and (1− vi)uε → u in W1,p
(Rn
)
as ε → 0 and i → ∞, we conclude that u ∈ W
1,p
0 (Ω).
Remark 4.30. If u ∈ W1,p
(Rn
) is continuous and zero everywhere in Rn
Ω, then
u ∈ W
1,p
0 (Ω).
We obtain a very useful characterization of Sobolev spaces with zero boundary
values on an arbitrary open set by combining the last two theorems.
Corollary 4.31. u ∈ W
1,p
0 (Ω) if and only if there exists a p-quasicontinuous
function u∗
∈ W1,p
(Rn
) such that u∗
= u almost everywhere in Ω and u = 0 p-
quasieverywhere in Rn
Ω.
T H E M O R A L : Quasicontinuous functions in Sobolev spaces with zero bound-
ary values are precisely functions in the Sobolev space on the whole space which

are zero quasieverywhere in the complement. This result can be used to show that
a given function belongs to the Sobolev space with zero boundary values without
constructing an approximating sequence of compactly supported smooth functions.
There is also a characterization of Sobolev spaces with zero boundary values
using Lebesgue points for Sobolev functions.
Theorem 4.32. Assume that Ω ⊂ Rn
is an open set and u ∈ W1,p
(Rn
) with 1 < p <
∞. Then u ∈ W
1,p
0 (Ω) if and only if
lim
r→0 B(x,r)
u(y)dy = 0
for p-quasievery x ∈ Rn
Ω.
T H E M O R A L : A function in the Sobolev space on the whole space belongs to
the Sobolev space with zero boundary values if and only if the limit of integral
averages is zero quasieverywhere in the complement.
Proof. =⇒ If u ∈ W
1,p
0 (Ω), then by Theorem 4.28 there exists a p-quasicontinuous
function u∗
∈ W1,p
(Rn
) such that u∗
= u almost everywhere in Ω and u∗
= 0
p-quasieverywhere in Rn
Ω. Theorem 4.27 shows that the limit
u∗
(x) = lim
r→0 B(x,r)
u(y)dy
exists p-quasieverywhere and that the function u∗
is a p-quasicontinuous repre-
sentative of u. This shows that
lim
r→0 B(x,r)
u(y)dy = u∗
(x) = 0
Ω.
⇐= Assume then that u ∈ W1,p
(Rn
) and
lim
r→0 B(x,r)
u(y)dy = 0
Ω. Theorem 4.27 shows that the limit
u∗
(x) = lim
r→0 B(x,r)
u(y)dy
exists p-quasieverywhere and that the function u∗
is a p-quasicontinuous repre-
sentative of u. We conclude that u∗
(x) = 0 for p-quasievery x ∈ Rn
Ω.
Example 4.33. Let Ω = B(0,1){0} and u : Ω → R, u(x) = 1−|x|. Then u ∈ W
1,p
0 (Ω)
for 1 n.

T H E M O R A L : A function that belongs to the Sobolev space with zero boundary
values does not have to be zero at every point of the boundary.
Remark 4.34. Theorem 4.32 gives a practical tool to show that a function belongs
to a Sobolev space with zero boundary values. For example, the following claims
follow from Theorem 4.32 (exercise).
(1) Assume that u ∈ W1,p
(Ω) has a compact support, then u ∈ W
1,p
0 (Ω).
(2) Assume that u ∈ W
1,p
0 (Ω). Then |u| ∈ W
1,p
0 (Ω).
1,p
0 (Ω). If v ∈ W1,p
(Ω) and 0 v u almost everywhere
in Ω, then v ∈ W
1,p
0 (Ω).
1,p
0 (Ω). If v ∈ W1,p
(Ω) and |v| |u| almost everywhere
in Ω K, where K is a compact subset of Ω, then v ∈ W
1,p
0 (Ω).
Let E ⊂ Ω be a relatively closed set, that is, there exists a closed F ⊂ Rn
such
that E = Ω∩ F, with |E| = 0. It is clear that W
1,p
0 (Ω E) ⊂ W
1,p
0 (Ω). By
W
1,p
0 (Ω E) = W
1,p
0 (Ω)
we mean that every u ∈ W
1,p
0 (Ω) can be approximated by functions in C∞
0 (Ω E)
or in W
1,p
0 (Ω E).
Theorem 4.35. Assume that E is a closed subset of Ω. Then W
1,p
0 (Ω) = W
1,p
0 (Ω
E) if and only if capp(E) = 0.
Proof. ⇐= Assume capp(E) = 0. Lemma 4.6 implies |E| = 0 so that it is reasonable
to ask whether W
1,p
0 (Ω) = W
1,p
0 (Ω E) when we consider functions deﬁned up to a
set of measure.
It is clear that W
1,p
0 (Ω E) ⊂ W
1,p
0 (Ω). To see reverse inclusion, let ui ∈ C∞
0 (Ω),
i = 1,2,..., be such that ui → u in W1,p
(Ω) as i → ∞. Since capp(E) = 0 there are
vj ∈ A (E), j = 1,2,..., be such that vj W1,p(Rn) → 0 as j → ∞. Then (1− vj)ui ∈
W1,p
(Ω) and, since vj = 1 in a neighbourhood of E, supp(1 − vj)ui is a compact
subset of ΩE for every i, j = 1,2,.... Lemma 1.23 implies (1−vj)ui ∈ W
1,p
0 (ΩE),
i, j = 1,2,....
Moreover, we have
u −(1− vj)ui W1,p(Ω) u − ui W1,p(Ω) + vjui W1,p(Ω),
where u − ui W1,p(Ω) → 0 as i → ∞ and
vjui W1,p(Ω) vjui Lp(Ω) + D(vjui) Lp(Ω)
ui L∞(Ω) vj Lp(Ω) + vjDui Lp(Ω) + uiDvj Lp(Ω)
ui L∞(Ω) vj Lp(Ω) + vjDui Lp(Ω) + ui L∞(Ω) Dvj Lp(Ω)
2 ui L∞(Ω) vj W1,p(Ω) + vjDui Lp(Ω).

Since vj → 0 in Lp
(Ω) as j → ∞, there is a subsequence, still denoted by (vj),
for which vj → 0 almost everywhere as j → ∞. By the dominated convergence
theorem, we have
lim
j→∞
vjDui Lp(Ω) = lim
j→∞
ˆ
Ω
|vj|p
|Dui|p
dx
1
p
=
ˆ
Ω
( lim
j→∞
|vj|p
)|Dui|p
dx
1
p
= 0.
Observe that |vj|p
|Dui|p
|Dui|p
for j = 1,2,..., so that |Dui|p
∈ L1
(Ω) may be
used as an integrable majorant. Thus
u −(1− vj)ui W1,p(Ω) → 0 as i, j → ∞.
Since
(1− vj)ui ∈ W
1,p
0 (Ω) and (1− vj)ui → u in W1,p
(Ω E)
as i, j → ∞, we conclude that u ∈ W
1,p
0 (Ω E).
=⇒ Let x0 ∈ Ω and let i0 ∈ N be large enough that
dist(x0,Rn
Ω) >
1
i0
.
Deﬁne
Ωi = x ∈ Ω : dist(x,Rn
Ω) >
1
i
∩B(x0, i), i = i0, i0 +1,....
Observe that Ωi Ωi+1 ··· Ω and Ω = ∞
i=i0
Ωi. Let ui : Rn
→ R,
ui(x) = dist(x,Rn
Ω2i).
Then ui is Lipschitz continuous, ui ∈ W
1,p
0 (Ω) and ui(x) 1
2i for every x ∈ E ∩Ωi,
i = 1,2,.... Since W
1,p
0 (Ω) = W
1,p
0 (Ω E) we have ui ∈ W
1,p
0 (Ω E), i = 1,2,....
Fix i and let vj ∈ C∞
0 (Ω E), j = 1,2,..., such that vj → ui in W1,p
(Ω E) as
j → ∞. Since 3i(ui − vj) 1 in a neigbourhood of E ∩Ωi. Thus
capp(E ∩Ωi) 3i(ui − vj)
p
W1,p(ΩE)
= (3i)p
ui − vj
p
W1,p(ΩE)
→ 0 as j → ∞.
Thus capp(E ∩Ωi) = 0, i = 1,2..., and by subadditivity
capp(E) = capp
∞
i=1
(E ∩Ωi)
∞
i=1
capp(E ∩Ωi) = 0.

THE END

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