of 26 /26
Chapter 2 Sobolev Spaces Sobolev spaces turn out often to be the proper setting in which to apply ideas of functional analysis to get information concerning par- tial diﬀerential equations. Here, we collect a few basic results about Sobolev spaces. A general reference to this topic is Adams , Gilbarg- Trudinger , or Evans . 2.1 H¨olderspaces Assume Ω R n is open and 0 1. Deﬁnition 2.1.1 (i) If u R is bounded and continuous, we write uC( ¯ Ω) := sup xΩ |u(x)|. (ii) The γ th -H¨ older semi-norm of u R is [u] C 0( ¯ Ω) := sup x,yΩ,x =y |u(x) u(y)| |x y| γ , and the γ th -H¨ older norm is uC 0( ¯ Ω) := uC( ¯ Ω) +[u] C 0( ¯ Ω) . (iii) The H¨ older space C k,γ ( ¯ Ω) 11
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• Chapter 2

Sobolev Spaces

Sobolev spaces turn out often to be the proper setting in which to

apply ideas of functional analysis to get information concerning par-

tial differential equations. Here, we collect a few basic results about

Sobolev spaces. A general reference to this topic is Adams , Gilbarg-

Trudinger , or Evans .

2.1 Holder spaces

Assume Rn is open and 0 < 1.

Definition 2.1.1 (i) If u : R is bounded and continuous, we

write

uC() := supx

|u(x)|.

(ii) The th-Holder semi-norm of u : R is

[u]C0,() := supx,y, x 6=y

{ |u(x) u(y)|

|x y|

}

,

and the th-Holder norm is

uC0, () := uC() + [u]C0, ().

(iii) The Holder space

Ck,()

11

• 12 CHAPTER 2. SOBOLEV SPACES

consists of all functions u Ck() for which the norm

uCk, () :=

||k

DuC() +

||=k

[Du]C0, () (2.1.1)

is finite.

Theorem 2.1.2 The space of functions Ck,() is a Banach space.

Proof. It is easy to verify that Ck,() is a norm (Exercise). It

suffices to check the completeness of Ck,().

1. Construction of limit functions.

Let {um}m=1 be a Cauchy sequence in C

k,(), i.e.,

um unCk,() 0, as m,n +.

Since any Cauchy sequence is a bounded sequence, so umCk, () is

uniformly bounded. Particularly, umC() is uniformly bounded, that

is, {um(x)} is a uniformly bounded sequence. Furthermore, the bound-

edness of DumC() or [Dum]C0, () implies the equicontinuity of the

sequence {um(x)}. The Arzela-Ascoli theorem implies there exists a

subsequence, still denoted by {um}m=1, which converges to a bounded

continuous function u(x) uniformly in . The same argument can be

repeated for any derivative of Dum of order up to and including k.

That is, there exist functions u C(), such that

Dum u, uniformly in , || k.

2. Du = u.

Deduce by induction, first let = (1, 0, , 0), since the series

u1(x) ++

j=1

(uj+1 uj) = limm

um+1(x) = u(x),

uniformly in and the series

Du1(x) +

+

j=1

(Duj+1 Duj) = lim

mDum+1(x) = u(x),

• 2.1. HOLDER SPACES 13

uniformly in . The theory of series implies that u(x) is differentiable

in and Du = u. The induction method shows that Du = u is

valid for all multi-index such that || k.

3. u Ck,().

It suffices to show that for any multi-index such that || = k,

there holds [Du]C0, () < +.

For any x, y , x 6= y, since Dum Du uniformly in , there

exists a N large enough, such that:

|DuN (z) Du(z)|

1

2|x y| , z .

Calculate:

|Du(x) Du(y)| |Du(x) DuN (x)| + |DuN (x) D

uN (y)|

+|DuN (y) Du(y)|

|x y| + |x y|DuNC0,()

(1 + supm

DumC0, ())|x y| ,

Thus the boundedness of DumC0,() implies [Du]C0,() < +.

4. um u in Ck,().

It also suffices to show that for any multi-index such that || = k,

there holds

[Dum Du]C0, () 0, as m .

In fact, for any x, y , x 6= y, there holds

|Dum(x) Du(x) (Dum(y) D

u(y))|

lim supn

|Dum(x) Dun(x) (D

um(y) Dun(y))|

|x y| limn

[Dum Dun]C0, ().

Since {um}m=1 is a Cauchy sequence in C

k,(), [DumDu]C0,()

limn

[Dum Dun]C0,() 0 as m 0.

• 14 CHAPTER 2. SOBOLEV SPACES

2.2 Sobolev spaces

2.2.1 Lp spaces

The Lp() space consists of all the function u : R such that

|u|p dx

• 2.2. SOBOLEV SPACES 15

Lemma 2.2.2 (Young inequality with ) Let 1 < p, q < , 1p +1q = 1, then

ab ap

p+

1

q/pbq

q, a, b > 0.

Lemma 2.2.3 (Holder inequality) Let 1 p, q , 1p +1q = 1,

then if u Lp(), v Lq(), we have

|uv| dx uLp()vLq().

Proof. 1. In the cases where p = or q = , it is easy, because there

exists a subset , with || = ||, such that sup

|u| = uL()

or sup

|v| = vL().

2. In the cases where 1 < p, q < . By the homogeneity of the

inequality, we may assume that uLp() = vLq() = 1. Then the

Young inequality implies that

|uv| dx

1

p

|u|p dx+

1

q

|u|q dx = 1 = uLp()vLq().

By induction argument, it is easy to show the following Generalized

Holder inequality (Exercise):

Lemma 2.2.4 (Generalized Holder inequality) Let 1 p1, , pm

, 1p1 + +1

pm= 1, then if uk L

pk() for k = 1, ,m, we have

|u1 um| dx

m

k=1ukLpk ().

Lemma 2.2.5 (Minkowski inequality) Assume 1 p , u, v

Lp(), then

u+ vLp() uLp() + vLp().

• 16 CHAPTER 2. SOBOLEV SPACES

Proof. From the Holder inequality, we have

u+ vpLp() =

|u+ v|p dx

|u+ v|p1(|u| + |v|) dx

(

|u+ v|

(p1) pp1 dx

)p1

p(

(

|u|p dx)1/p + (

|v|p dx)1/p

)

= u+ vp1Lp()(uLp() + vLp()).

Lemma 2.2.6 (Interpolation inequality for Lp- norms) Assume 1

s r t , and 1r =s +

(1)t . Suppose also u L

s() Lt().

Then u Lr(), and

uLr() uLs()u

(1)Lt().

Proof. From the Holder inequality, we have

|u|r dx =

|u|r|u|(1)r dx

(

|u|r

sr dx

)rs(

|u|

(1)r t(1)r dx

)

(1)rt .

Theorem 2.2.7 (Completeness of Lp()) Lp() is a Banach space

if 1 p .

Proof. It is easy to check that Lp() is a norm (Exercise). It suffices

to verify the completeness of Lp()

1. Construction of the limit function.

Assume that 1 p < and let {um} be a Cauchy sequence in

Lp(). There is a subsequence {umj} of {um} such that

umj+1 umjp 1/2j , j = 1, 2 .

• 2.2. SOBOLEV SPACES 17

Let vm(x) =m

j=1|umj+1(x) umj (x)|, then

vmp

m

j=1

umj+1 umjp

m

j=1

1/2j < 1.

Putting v(x) = limm

vm(x), which may be infinite for some x, we obtain

by the Fatou Lemma that

|v(x)|p dx lim inf

m

|vm(x)|

p dx 1.

Hence v(x) < a.e. in and its absolutely convergence implies the

series

um1 +

j=1

(umj+1(x) umj (x)) (2.2.1)

converges to a limit u(x) a.e. in . Let u(x) = 0 whenever it is

undefined as the limit of series (2.2.1). Thus

limj

umj (x) = u(x) a.e. in .

2. u Lp() and um u in Lp().

For any > 0, there exists N such that if m,n N , then um

unp < . Hence by the Fatou Lemma again, we have

|um(x) u(x)|

p dx =

lim

j|um(x) umj (x)|

p dx

lim infj

|um(x) umj (x)|

p dx

p.

If m N , thus u = (u um) + um Lp(), and um up 0 as

m .

3. In the case where p = , let {um} be a Cauchy sequence in L(),

then there exists a subset A , with |A| = 0, such that if x \A,

then for n,m = 1, , there holds

|um(x)| um, |um(x) un(x)| um un.

That is, um(x) is uniformly bounded in \ A, and um(x) is a Cauchy

sequence for any x \A, thus um(x) converges uniformly on \A to

• 18 CHAPTER 2. SOBOLEV SPACES

a bounded function u(x). Set u(x) = 0 if x A. We have u L()

and um u 0 as m .

Theorem 2.2.8 (Riesz Representation Theorem for Lp()) Let

1 < p

• 2.2. SOBOLEV SPACES 19

Motivation for definition of weak derivatives. Assume we are

given a function u C1(). Then if C0 (), we see from the

integration by parts that

uxi dx =

uxidx (i = 1, , n). (2.2.2)

More general now, if k is a positive integer, u Ck(), and =

(1, , n) is a multi-index of order || = 1 + + n = k, then

u(D) dx = (1)||

(Du)dx. (2.2.3)

We next examine that formula (2.2.3) is valid for u Ck(), and

ask whether some variant of it might still be true even if u is not k times

continuously differentiable. Note that the left-hand side of (2.2.3) make

sense if u is only locally summable: the problem is rather that if u is

not Ck, then the expression Du on the right-hand side of (2.2.3) has

no obvious meaning. We overcome this difficulty by asking if there

exists a locally summable function v for which formula (2.2.3) is valid,

with v replacing Du:

Definition 2.2.13 Suppose u, v L1loc(), and is a multi-index. We

say that v is the th-weak partial derivative of u, written

Du = v,

provided that

uDdx = (1)||

v dx (2.2.4)

for all test functions C0 ().

Lemma 2.2.14 (Uniqueness of weak derivative) The weak partial

derivative of u, if it exists, is uniquely defined up to a set of measure

zero.

Example 2.2.15 Let n = 1, = (0, 2), and

u(x) =

{

x, if 0 < x 1,

1, if 1 x < 2.

• 20 CHAPTER 2. SOBOLEV SPACES

Define

v(x) =

{

1, if 0 < x 1,

0, if 1 x < 2.

From the definition 2.2.13, u = v. (Exercise)

Example 2.2.16 Let n = 1, = (0, 2), and

u(x) =

{

x, if 0 < x 1,

2, if 1 x < 2.

We assert u does not exist in the weak sense. (Exercise)

2.2.3 Definition of Sobolev spaces

Fix 1 p and let k be a nonnegative integer. We define now cer-

tain function spaces, whose members have weak derivatives of various

orders lying in various Lp spaces.

Definition 2.2.17 i) The Sobolev space

W k,p()

consists of all locally summable functions u : R such that

for each multi-index with || k, Du exists in the weak sense

and belongs to Lp().

ii) If u W k,p(), we define its norm to be

uW k,p() :=

(

||k

|Du|p dx

)1/p, (1 p

• 2.2. SOBOLEV SPACES 21

ii) We write

um u in Wk,ploc ()

to mean

um u in Wk,p(U),

for each U .

Definition 2.2.19 We denote by

W k,p0 ()

the closure of C0 () in Wk,p().

Remark 2.2.20 (i) If p = 2, we usually write

Hk() = W k,2(), Hk0 () = Wk,20 (), (k = 0, 1, ).

The letter H is used, since - as we will see - Hk() is a Hilbert

space. Note that H0() = L2().

(ii) We henceforth identity functions in W k,p() which agree almost

everywhere.

Example 2.2.21 Let = B(0, 1), the open unit ball in Rn, and

u(x) = |x| (x , x 6= 0), > 0.

Then u W 1,p() if and only if 0 < ( + 1)p < n. In particular

u 6W 1,p() for each p n. (Exercise)

Example 2.2.22 Let {rk}k=1 be a countable, dense subset of =

B(0, 1) Rn. Set

u(x) =

k=1

1

2k|x rk|

(x ).

Then u W 1,p() if and only if ( + 1)p < n. If 0 < ( + 1)p < n,

we see that u belongs to W 1,p() and yet is unbounded on each open

subset of . (Exercise)

Theorem 2.2.23 (Properties of weak derivatives) Assume u, v

W k,p(), || k. Then

• 22 CHAPTER 2. SOBOLEV SPACES

(i) Du W k||,p() and D(Du) = D(Du) = D+u for all

multi-indices , with |+ | k;

(ii) For each , R, u + v W k,p() and D(u + v) =

Du+ Dv, || k;

(iii) If is an open subset of , then u W k,p();

(iv) If C0 (), then u Wk,p() and

D(u) =

(

)

DDu (Leibniz formula),

where

(

)

=!

!( )!.

Theorem 2.2.24 (Sobolev spaces as function spaces) For each k

= 0, 1, and 1 p , Sobolev space W k,p() is a Banach space;

W k,p() is reflexive if and only if 1 < p < ; For 1 p < , the

subspace W k,pC() is dense in W k,p(), thus W k,p() is separable;

Moreover, W k,2() is a Hilbert space with scalar product

(u, v)W k,2 =

k

DuDv dx.

Proof. 1. It is easy to check that W k,p() is a norm (Exercise).

Next to show that W k,p() is complete.

2. Construction of limit functions.

Assume {um}m=1 is a Cauchy sequence in W

k,p(). Then for each

|| k, {Dum}m=1 is a Cauchy sequence in L

p(). Since Lp() is

complete, there exist functions u Lp(), such that

Dum u, in Lp(),

for each || k. In particular,

um u(0, ,0) =: u in Lp().

3. u W k,p() and Du = u, || k.

• 2.2. SOBOLEV SPACES 23

In fact, let C0 (), then

uD dx = lim

m

umD

dx

= limm

(1)||

(Dum) dx

=

u dx.

4. um u in Wk,p().

Since Dum u = Du in Lp() for all || k, it follows that

um u in Wk,p().

5. Regard W k,p() as a closed subspace of a Cartesian product space

of spaces Lp(): Let N = N(n, k) =

0||k

1 be the number of multi-

indexes satisfying 0 || k. For 1 p , let LpN =N

j=1Lp(),

the product norm of u = (u1, , uN ) in LpN being given by

uLpN=

(

N

j=1uj

pp

)1/p, if 1 p

• 24 CHAPTER 2. SOBOLEV SPACES

p and n, such that

uLp (Rn) CDuLp(Rn), (2.2.5)

for all u C10 (Rn).

Proof. 1. First assume p = 1.

Since u has compact support, for each i = 1, 2 , n and x Rn

we have

u(x) =

xi

uxi(x1, , xi1, yi, xi+1, , xn) dyi,

and so

|u(x)|

+

|Du(x1, , xi1, yi, xi+1, , xn)| dyi, i = 1, 2 , n.

Consequently

|u(x)|n

n1 n

i=1

(

+

|Du(x1, , xi1, yi, xi+1, , xn)| dyi

)1

n1.

Integrate this inequality with respect to x1: +

|u(x)|

nn1 dx1

+

n

i=1

(

+

|Du(x1, , xi1, yi, xi+1, , xn)| dyi

)1

n1dx1

=(

+

|Du(y1, x2 , , xn)| dy1

)1

n1

+

n

i=2

(

+

|Du(x1, , xi1, yi, xi+1, , xn)| dyi

)1

n1dx1

(

+

|Du| dy1

)1

n1( n

i=2

+

+

|Du| dx1dyi

)1

n1,

(2.2.6)

the last inequality follows from the Generalized Holder inequality.

Now integrate (2.2.6) with respect to x2:

+

+

|u(x)|

nn1 dx1 dx2

(

+

+

|Du| dx1dy2

)1

n1

+

n

i=1,i6=2I

1n1

i dx2,

• 2.2. SOBOLEV SPACES 25

where

I1 =

+

|Du| dy1, Ii =

+

+

|Du| dx1dyi, i = 3, , n.

Applying once more the Generalized Holder inequality, we find

+

+

|u(x)|

nn1 dx1 dx2

(

+

+

|Du| dx1dy2

)1

n1(

+

+

|Du| dy1dx2

)1

n1

n

i=3

(

+

+

+

|Du| dx1dx2dyi

)1

n1.

We continue by integrating with respect to x3, , xn, eventually to

find

Rn|u(x)|

nn1 dx

n

i=1

(

+

+

|Du| dx1 dxn

)1

n1

=(

Rn|Du| dx

)n

n1.

(2.2.7)

This is conclusion for p = 1.

2. Consider now the case that 1 < p < n. We apply (2.2.7) to v := |u| ,

where > 1 is to be selected. Then

(

Rn|u(x)|

nn1 dx

)n1

n

Rn|D|u| | dx =

Rn|u|1|Du| dx

(

Rn|u|(1)

pp1 dx

)p1

p(

Rn|Du|p

)1p .

(2.2.8)

We choose so thatn

n 1= ( 1)

p

p 1. That is, we get

=p(n 1)

n p> 1,

in which casen

n 1= ( 1)

p

p 1=

np

n p= p. Thus, (2.2.8)

becomes (2.2.5).

• 26 CHAPTER 2. SOBOLEV SPACES

Remark 2.2.26 The exponent p =np

n pis called critical Sobolev

exponent, because it is the unique exponent q = p to make the following

inequality:

uLq(Rn) CDuLp(Rn)

be true for all u C10(Rn).

Proof. For any u C10 (Rn) and > 0, define the scaled function

u(x) = u(x). Direct calculations show that

uLq(Rn) = n

q uLq(Rn)

and

DuLp(Rn) = 1n

p DuLp(Rn).

If the inequality in remark is true for all u C10 (Rn), replacing u by

u, there holds

uLq(Rn) C1n

p+ n

q DuLp(Rn).

If 1 np +nq 6= 0, upon sending to either 0 or , we will get C = .

Theorem 2.2.27 (Poincare inequality for W 1,p0 ()) Assume is

a bounded, open subset of Rn, Suppose u W 1,p0 () for some 1 p n implies (n 1) pp1 < n; so that

B(x,1)

dy

|x y|(n1) p

p1

• 2.2. SOBOLEV SPACES 29

Similarly, there holds

W|u(y) u(z)| dz Cr1

np DuLp(Rn).

Substituting this estimate and (2.2.15) into (2.2.14) yields

|u(x) u(y)| Cr1n

p DuLp(Rn) = C|x y|1n

p DuLp(Rn).

Thus

[u]C0, 1n/p(Rn) = supx 6=y

{ |u(x) u(y)|

|x y|1n/p}

CDuLp(Rn).

This inequality and (2.2.13) complete the proof.

2.2.5 Embedding theorems and Trace theorems

Definition 2.2.30 Let (X, X), (Y, Y ) be Banach spaces. X is

(continuously) embedded into Y , denoted X Y , if there exists an

injective linear map i : X Y and a constant C such that

i(x)Y CxX , x X.

X is compactly embedded into Y , denoted X Y , if i maps

bounded subsets of X into precompact subsets of Y .

Theorem 2.2.31 (Exercise)

(i) For Rn with Lebesgue measure Ln()

• 30 CHAPTER 2. SOBOLEV SPACES

(ii) If 0 m < k np < m + 1, then Wk,p() Cm,(), for

0 km np ; the embedding is compact, if < kmnp .

Proof. 1. The embedding of the case where kp < n.

Assume that kp < n and u W k,p(). Then Du Lp() for all

|| = k. The Gagliardo-Nirenberg-Sobolev inequality implies

DuLp1 () CuW k,p(), for || = k 1, p1 =np

n p,

so u W k1,p1(). Similarly, we find u W k2,p2(), where p2 =np1n p1

=np

n 2p. Continuing, we eventually discover after k steps

that W 0,pk() = Lpk(), for pk =npn1n pn1

=np

n kp, and

uLpk () CuW k,p().

Since is bounded, thus for any 1 q p = npnkp , Wk,p() Lq().

2. The compactness of embedding.

That is to show that if {um}m=1 is a bounded sequence in W

k,p(),

there exists a subsequence {umj}j=1 which converges in L

q(). For

simplicity, we just prove the case k = 1. By induction argument, one

get the proof of general case.

By extension argument, without loss of generality assume that

= Rn and the functions {um}m=1 all have compact support in some

bounded open set U Rn, moreover,

supm

umW k,p(U) 0,m = 1, 2, ),

where :=1

n(x

) denotes the usual mollifier and satisfies

C0 (Rn), 0,

Rn(x) dx = 1.

A concrete example of is

(x) :=

Cexp(1

|x|2 1), if |x| < 1,

0, if |x| 1,

• 2.2. SOBOLEV SPACES 31

and constant C > 0 is selected so that

Rn(x) dx = 1. We may

suppose that the functions {um}m=1 all have support in U as well.

3. Claim

um um in Lq(U) as 0, uniformly in m. (2.2.17)

To prove this claim, we first note that if um is smooth, then

um(x) um(x) =

B(0,1)(y)(um(x y) um(x)) dy

=

B(0,1)(y)

1

0

d

dtum(x ty) dtdy

=

B(0,1)(y)

1

0Dum(x ty) y dtdy.

Thus

U|um(x) um(x)| dx

B(0,1)(y)

1

0

U|Dum(x ty)| dxdtdy

U|Dum(z)| dz.

By approximation this estimate holds if um W1,p(U). Hence

um umL1(U) DumL1(U) DumLp(U),

the latter inequality holds since U is bounded. Owing to (2.2.16) we

thereby discover

um um in L1(U) as 0, uniformly in m. (2.2.18)

But then since 1 q < p, we using the Interpolation inequality for

Lp-norm that

um umLq(U) um um

L1(U)u

m um

1Lp(U)

,

where1

q= +

(1 )

p, 0 < 1. Consequently, (2.2.16) and the

Gagliardo-Nirenberg-Sobolev inequality imply

um umLq(U) Cum um

L1(U);

• 32 CHAPTER 2. SOBOLEV SPACES

whence assertion (2.2.17) follows from (2.2.18).

4. Next we claim

for each fixed > 0, the sequence {um}m=1

is uniformly bounded and equi-continuous.(2.2.19)

Indeed, if x Rn, then

|um(x)|

B(x,)(x y)|um(y)|dy

L(Rn)umL1(U) C

n 0 so

small that

um umLq(U) /2, (2.2.21)

for m = 1, 2, .

We now observe that since functions {um}m=1, and thus functions

{um}m=1, have support in some fixed bounded set U R

n, we may

use (2.2.19) and the Arzela-Ascoli theorem, to obtain a subsequence

{umj}j=1 {u

m}

m=1 which converges uniformly on U . In particular

therefore

lim supj,k

umj umk

Lq(U) = 0. (2.2.22)

Then (2.2.21) and (2.2.22) imply (2.2.20).

• 2.2. SOBOLEV SPACES 33

6. We next employ assertion (2.2.20) with = 1,1

2,1

3, and use

the standard diagonal argument to extract a subsequence {uml}l=1 of

{um}m=1 such that

lim supl,k

uml umkLq(U) = 0.

Conclusion (i) of theorem is proved by steps 1 - 6. Next to prove

conclusion (ii) of theorem.

7. Assume that 0 m < k np < m+ 1. Then as above we see

u W kl,r(), (2.2.23)

for r =np

n lpprovided that lp < n. We choose the integer l0 = [

n

p].

Consequently, (2.2.23) holds for r =np

n pl0> n. Hence (2.2.23) and

Morrey inequality imply that Du C0,1nr () for all || k l0 1.

Observe also that 1n

r= 1

n

p+ l0 = [

n

p]+1

n

p. Thus u Cm, ()

with m = k l0 1 = k [n

p] 1, = 1

n

r= [

n

p] + 1

n

p, and the

stated estimate follows easily.

8. The Arzela-Ascoli theorem implies the compactness of embedding

W k,p() Cm,() provided that < = k m np .

Denote (u) =

u dx = average of u over .

Theorem 2.2.33 (Poincare inequality for W 1,p()) Let Rn

be bounded and connected, with C1 boundary . Assume 1 p .

Then there exists a constant C, depending only on p, n and , such

that

u (u)Lp() CDuLp(), (2.2.24)

for all u W 1,p(Rn).

Proof. Argue by contradiction. If the stated estimate is false, that is,

for each integer k = 1, 2, , there would exist a function uk W1,p()

such that

uk (uk)Lp() > kDukLp(). (2.2.25)

• 34 CHAPTER 2. SOBOLEV SPACES

Normalize by defining

vk :=uk (uk)

uk (uk)Lp(), k = 1, 2, .

Then, there holds

(vk) = 0, vkLp() = 1, (2.2.26)

and from (2.2.25), we have

DvkLp() 0 there

exists a C = C(, n) such that for u H1(B1) with

{x B1;u = 0}

B1

,

there holds

B1

u2 C

B1

|Du|2.

Proof. Suppose not. Then there exists a sequence {um} H1(B1)

such that

{x B1;um = 0}

B1

,

and

B1

u2m = 1,

B1

|Dum|2 0 as m .

Hence we may assume um u0 H1(B1) strongly in L

2(B1) and

weakly in H1(B1). Then limm

(

|Dum|2 |D(um u0)|

2)

= |Du0|2 and

|Du0|2 lim

m|Dum|

2 = 0. Thus u0 is a nonzero constant. So

0 = limm

B1

|um u0|2 lim

m

{um=0}|um u0|

2

|u0|2 inf

m

{um = 0}

> 0.

Definition 2.2.36 For a domain with Ck-boundary = , k

N, 1 < p

• 36 CHAPTER 2. SOBOLEV SPACES

Theorem 2.2.37 (Trace Theorem) Assume is a domain with Ck-

boundary = , k N, 1 < p < . Then W k1p,p() is a Banach

space.

Theorem 2.2.38 (Extension Theorem) For any domain with Ck-

boundary = , k N, 1 < p < , there exists a continuous linear

extension operator ext : W k1p,p() W k,p() such that

(

ext(u))

=

u, for all u W k1p,p().

Theorem 2.2.39 Suppose is a domain with Ck-boundary =

, k N, 1 < p < . Then W k,p() Wk 1

p,p() W k1,p()

and both embeddings are compact.

EXERCISES

1. Prove The Young inequality with - Lemma 2.2.2.

2. Prove the generalized Holder inequality - Lemma 2.2.4.

3. Complete Example 2.2.15.

4. Complete Example 2.2.16.

5. Complete Example 2.2.21.

6. Complete Example 2.2.22.

7. Verify Ck,(), Lp() and W k,p() are norms.

8. Prove Theorem 2.2.31.