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**TABLE OF CONTENTS**

Epigraph ii

Acknowledgement iii

Dedication iv

Introduction 1

1 Spaces of Functions 5

1.1 Lp-spaces and some of its properties . . . . . . . . . . . . . . . . . . . . . . . 5

1.1.1 Basic Integration Results . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.1.2 Definition and basic properties . . . . . . . . . . . . . . . . . . . . . . 6

1.1.3 The Main properties of Lp(

) . . . . . . . . . . . . . . . . . . . . . . 7

1.1.4 Dual Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.1.5 Convolutions and Mollifiers . . . . . . . . . . . . . . . . . . . . . . . 11

1.1.6 Density of Cc(

) in Lp(

) . . . . . . . . . . . . . . . . . . . . . . . . 15

1.1.7 Density of D(

) in Lp(

). . . . . . . . . . . . . . . . . . . . . . . . . 18

1.2 Distribution Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

1.2.1 Test Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

1.2.2 Convergence in Function Spaces . . . . . . . . . . . . . . . . . . . . . 23

1.2.3 Continuity and Denseness on Dm(

) and D(

) . . . . . . . . . . . . 24

1.2.4 Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

1.2.5 The Support of a Distribution . . . . . . . . . . . . . . . . . . . . . . 26

1.2.6 Distributions with Compact Support . . . . . . . . . . . . . . . . . . 27

1.2.7 Convergence of Distributions . . . . . . . . . . . . . . . . . . . . . . . 29

1.2.8 Multiplication of Distributions . . . . . . . . . . . . . . . . . . . . . . 30

1.2.9 Differentiation of Distributions . . . . . . . . . . . . . . . . . . . . . 31

2 Sobolev spaces Wm;p 34

2.1 Definitions and main properties . . . . . . . . . . . . . . . . . . . . . . . . . 34

2.2 The Main Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

2.2.1 Approximation by smooth functions . . . . . . . . . . . . . . . . . . . 37

v

CONTENTS

2.2.2 Extension Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

2.2.3 Trace Theory. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

3 Variational Method 58

3.1 Optimization in Infinite Dimensional Spaces . . . . . . . . . . . . . . . . . . 58

3.1.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

3.1.2 Lower Semi Continuous Functions(lsc) . . . . . . . . . . . . . . . . . 59

3.1.3 Convex sets. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

3.1.4 Convex Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

3.1.5 Gateaux Differentiability . . . . . . . . . . . . . . . . . . . . . . . . . 63

3.1.6 Existence Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

3.1.7 Optimality Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . 65

3.2 Application to Elliptic Partial Differential Equation . . . . . . . . . . . . . . 67

Conclusion 71

Bibliography 72

vi

**CHAPTER ONE**

SPACES OF FUNCTIONS

In the following,

is a nonempty open subset of RN with the Lebesgue measure dx.

1.1 Lp-spaces and some of its properties

1.1.1 Basic Integration Results

Theorem 1.1.1.1 (Monotone Convergence Theorem) Let ffng be a nondecreasing sequence

of integrable functions such that:

sup

Z

fn dx < 1:

Then ffng converges pointwise to some function f. Futhermore f is integrable and

lim

n!+1

Z

jfn fj dx = 0:

Theorem 1.1.1.2 (Lebesgue Dominated Convergence Theorem) Let ffng be a sequence

of integrable functions such that:

(i) fn(x) ! f(x) a.e on

,

(ii) there exists a function g, integrable and jfn(x)j g(x) a.e on

.

Then f is integrable and

lim

n!+1

Z

jfn fj dx = 0:

Theorem 1.1.1.3 (Fatou Lemma) Let ffng be a sequence of integrable functions such that:

(i) 8 n; fn(x) 0 a.e on

,

(ii) sup

R

fn dx < 1.

For x 2

, set f(x) = lim infn fn(x). Then f is integrable and

Z

f dx lim inf

n

Z

fn dx:

5

+ Lp-spaces

1.1.2 Definition and basic properties

Definition Let 1 p < 1. We define:

(i) Lp(

) as the set of measurable functions f :

! R such that:

Z

jf(x)jp dx < +1

and

(ii) L1(

) as the set of measurable functions f :

! R such that:

ess sup jfj < +1

where

ess sup jfj = inf fK 0; jf(x)j K; a.e x 2

g

Definition We say that two functions f and g are equivalent if f = g almost everywhere.

Then we define Lp(

) spaces as the equivalent classes for this relation.

Remark 1.1.2.1 The space Lp(

) can be seen as a space of functions. We do however, need

to be careful sometimes. For example, saying that f 2 Lp(

) is continuous means that f is

equivalent to a continuous function. Now for f 2 Lp(

), we define:

kfkp =

Z

jf(x)jp dx

1

p

; 1 p < +1 (1.1)

kfk1 = ess sup jfj: (1.2)

Theorem 1.1.2.1 (Holder’s Inequality) . Let 1 p < +1, we define p0 by 1=p+1=p0 =

1: If f 2 Lp(

) and g 2 Lp0(

), then fg 2 L1(

) and

kfgk1 kfkpkgkp0 : (1.3)

Proof. The cases p = 1 and p0 = +1 are easy to prove. Now assume 1 < p < +1. We use

the following Young’s inequality: Let 1 < p < +1, a; b 0 then

ab

ap

p

+

bp0

p0 :

Assume that kfkp 6= 0 and kgkp0 6= 0 otherwise, nothing to do. Using Young’s inequality,

we have

jfj

kfkp

jgj

kgkp0

1

p

jfjp

kgkpp

+

1

p0

jgjp0

kfkp0

p

:

Thus Z

jfj

kfkp

jgj

kgkp0

dx

1

p

Z

jfjp

kfkpp

dx +

1

p0

Z

jgjp0

kgkp0

p

dx =

1

p

+

1

p0 = 1:

Hence Z

jfj jgj dx kfkp kgkp0 :

6

+ Lp-spaces

Theorem 1.1.2.2 (Minkowski’s Inequality) . If 1 p +1 and f; g 2 Lp(

) then

kf + gkp kfkp + kgkp: (1.4)

Proof. If f + g = 0 a:e, then the statement is trivial. Assume that f + g 6= 0 and p > 1

(the case p = 1 is easy to check). We evaluate as follows:

jf + gjp = jf + gjjf + gjp1 (jfj + jgj)jf + gjp1:

Integrating over

, we get

Z

jf + gjp dx

Z

(jfj + jgj)jf + gjp1 dx

=

Z

jfjjf + gjp1 dx +

Z

jgjjf + gjp1 dx

Using Holder’s inequality in the right hand side, we obtain

Z

jf + gjp dx (kfkp + kgkp)kf + gkp=q

p ;

from which it follows

kf + gkp kfkp + kgkp:

1.1.3 The Main properties of Lp(

)

Lp-Spaces are Banach

Theorem 1.1.3.1 The Lp-spaces are Banach for 1 p +1.

Proof.

Case1. Assume that p = 1. Let ffng be a Cauchy sequence in L1. Let k 1, there

exists Nk such that

kfm fnkp

1

k

8 n;m Nk:

There exists a set of measure zero Ak such that

jfm(x) fn(x)jp

1

k

8 x 2

Ak; 8 n;m Nk: (1.5)

Let A = [Ak (A is of measure zero) and forall x 2

A the sequence ffn(x)g is Cauchy in

R. Let fn(x) = limn fn(x) forall x 2

A. Letting m goes to +1 in (1.5), we obtain

jfn(x) f(x)jp

1

k

8 x 2

Ak; 8 n Nk:

Thus f 2 L1 and kfn fkp 1=k; 8 n Nk: So kfn fkp ! 0:

7

+ Lp-spaces

Case2. Assume that 1 p < +1. Let (fn)n1 be a Cauchy sequence in Lp(

), then

there exists a subsequence (fnk)k1 of (fn) such that:

kfnk+1 fnkkp

1

2k ; 8 k 1: (1.6)

To simplify the notations, let us replace fnk by fk so that:

kfk+1 fkkp

1

2k ; 8 k 1: (1.7)

Now set:

gn(x) =

Xn

k=1

jfk+1(x) fk(x)j:

It follows that:

kgnkp 1; 8 n 1:

Thus, from the monotone convergence theorem, gn(x) converge pointwise to some g(x) almost

every where and g 2 Lp: On the other hand we have: for all n;m 2

jfm(x) fn(x)j jfm(x) fm1(x)j + + jfn+1(x) fn(x)j g(x) gn1(x):

It follows that (fn(x)) is Cauchy in R and converges to some f(x) a.e. Letting m goes to

+1 leads to:

jf(x) fn(x)j g(x); 8 n 2:

Therefore f 2 Lp and by using dominate convergence theorem we have

kfn fkp ! 0:

We complete the proof by applying the following lemma

Lemma 1.1.3.1 Let E be a metric space and (xn) be a cauchy sequence in E. If (xn) has a

convergence subsequence, then it converges to the same limit.

The preceding proof contains a result which is interesting enough to be stated separetely:

Theorem 1.1.3.2 (Convergence criteria for Lp functions) Let 1 p < +1. Let (fn)

and f in Lp(

) such that (fn) converges to f in Lp(

). Then there exists a subsequence

(fnk) of (fn) and h 2 Lp(

) such that fnk(x) ! f(x) for a.e, x 2

and fnk(x) h(x), a.e

x 2

.

Remark 1.1.3.1 It is in general not true that the entire sequence itself converge pointwise to

the limit f, without some futher conditions holding.

Example 1.1.3.1 Let X = [0; 1], and consider the subintervals

h

0;

1

2

i

;

h1

2

; 1

i

;

h

0;

1

3

i

;

h1

3

;

2

3

i

;

h2

3

; 1

i

;

h

0;

1

4

i

;

h1

4

;

2

4

i

;

h2

4

;

3

4

i

;

h3

4

; 1

i

;

h

1;

1

5

i

;

Let fn denote the indicator function of the nth interval of the above sequence. Then kfnkp !

0, but fn(x) does not converge for any x 2 [0; 1].

8

+ Lp-spaces

Example 1.1.3.2 Let

= R, and for n 2 N, set fn = X[n; n + 1]. Then fn(x) ! 0 as

n ! 1; but kfnkp = 1 for p 2 [0;1). Thus fn converge pointwise but not in norm.

Theorem 1.1.3.3 Let 1 p < 1. Let ffng be a sequence in Lp such that fn(x) ! f(x)

a.e. If

lim

n

kfnk = kfk

then ffng converges to f in norm.

Theorem 1.1.3.4 The Lp spaces are reflexive for 1 < p < 1.

Proof. For 2 p < 1. We have the follwing first Clarkson inequality:

f + g

2

p

p

+

f g

2

p

p

1

2

kfkpp

+ kgkpp

; 8 f; g 2 Lp:

For 1 < p 2, we have the second Clarkson inequality:

f + g

2

p0

p

+

f g

2

p0

p

h1

2

kfkpp

+

1

2

kgkpp

i1=(p1)

; 8 f; g 2 Lp:

Using the Clarkson inequalities, we prove that Lp is uniformly convex for 1 < p < 1. So it

is reflexive by Milman-Pettis Theorem

Theorem 1.1.3.5 Let 1 p < 1. Then Lp is separable.

Proof. Let (i)i2I be the family of N-cubes of RN of the form =

YN

k=1

]ak; bk[ where

ak; bk 2 Q and

. Let E be the Q-vector space spanned by the functions Xi .

Claim: E is a countable dense subspace of Lp.

Remark 1.1.3.2 L1 is not separable. To establish this, we need the following:

Lemma 1.1.3.2 Let E be a banach space. We assume that there exists a familly (Oi)i2I

such that:

(i) For all i 2 I Oi is a nonempty open subset of E;

(ii) Oi \ Oj = ; if i 6= j;

(iii) I is uncountable.

Then E is not separable.

Now we apply this lemma for L1 as follows:

For all a 2

, let ra such that 0 < ra < d(a;

c). Set fa = XB(a;ra) and

Oa = ff 2 L1 j kf fak1 <

1

2

g:

One can check that the family (Oa)a2

satisfies (i), (ii) and (iii).

9

+ Lp-spaces

1.1.4 Dual Space

Theorem 1.1.4.1 (Riesz representation theorem.) Let 1 < p < +1 and let 2 (Lp)0.

Then there exists a unique g 2 (Lp)0 such that:

h; fi =

Z

g f dx; 8 f 2 Lp(

):

Futhermore

kk(L1)0 = kgk1:

Proof. Let 1 < p < +1 and let p0 such that 1=p + 1=p0 = 1. For g 2 Lp0(

), we define

Tg : Lp(

) ! R; hTg; fi =

Z

f g dx:

Using Holder’s inequality, we observe that Tg is well defined, linear and

jhTg; fij kgkp0kfkp:

Thus

kTgk(Lp)0 kgkp0 :

In fact we have kTgk(Lp)0 = kgkp0 . This follows by choosing f = jgjp02g.

Now we define the map

T : Lp0

! (Lp)0; by T(g) = Tg 8 g 2 Lp0

:

We have to prove that T is onto. For this, let E = T(Lp0). We have to show that E is closed

and dense in (Lp). E is closed by using the fact that kTgk = kgkp0 and Lp0 is Banach. For

density we will show that if L 2 (Lp)00 and L = 0 on E then L = 0 on (Lp)0. Since Lp is

reflexive, we identify (Lp)00 to Lp through the canonical embeding. Thus there exists f 2 Lp

such that hL; i = h; fi, for all 2 (Lp)0. So L = 0 on E leads hTg; fi = 0 for all g 2 Lp0

and this implys that f = 0 so L is.

Theorem 1.1.4.2 (Dual space of L1). Let 2 (L1)0, then there exists a unique g 2 L1

such that

h; fi =

Z

g f dx; 8 f 2 Lp(

):

and

kk(L1)0 = kgk1:

Remark 1.1.4.1 The spaces L1(

) and L1(

) are not reflexive.

Indeed assume that L1 is reflexive and let

open such that assume that 0 2

. Let

fn = nXB(0;1=n), where n =

B(0; 1=n)

1

so that kfnk1 = 1: For n large enough, we

have B(0; 1=n)

. By reflexivity, ffng has a weakly convergence subsequence fnk to some

function f in L1(

). Thus

Z

fnk’ dx !

Z

f’ dx; 8 ‘ 2 L1(

): (1.8)

10

+ Lp-spaces

So for ‘ 2 Cc(

f0g), we have

Z

fnk’ dx = 0 for k large enough. By (1.8) it follows that

Z

f’ dx = 0; 8′ 2 Cc(

f0g):

Thus f = 0 a.e on

. On the other hand, taking ‘ 1 in (1.8) leads to

Z

f dx = 1.

Contradiction. So L1(

) is not reflexive.

Since a Banach space is reflexive if and only if its dual E0 is reflexive, then L1(

) is not

reflexive.

Remark 1.1.4.2 Since (L1)0 = L1, then from Banach-Alaogulu theorem any bounded sequence

in L1(

) has a w-convergence subsequence.

Proposition 1.1.4.1 There exists a linear continuous forms on L1(

) such that there is no

g 2 L1(

) such that

hT; fi =

Z

g f dx; 8 f 2 L1(

):

Proof. Let

an open subset of Rn such that 0 2

. Let

0 : Cc(

) ! R; h0; ‘i = ‘(0):

0 is a linear continuous form on (Cc(

); k k1). So by the Hann-Banach extension

theorem, 0 can be extended to a continuous linear form on L1(

), say . We summarize

the main properties of the Lp spaces as follows:

Completeness Reflexivity Separability Dual Space

Lp; 1 < p < 1 yes yes yes Lp0 ; 1=p + 1=p0 = 1

L1 yes no yes L1

L1 yes no no Contains strctly L1

1.1.5 Convolutions and Mollifiers

Two usefull theorems

Let

1 RN,

2 RN open subsets of RN and F :

1

2 ! R be a measurable function.

Theorem 1.1.5.1 (Tonelli) Assume that

Z

2

jF(x; y)j dy < 1 a:e x 2

1

and Z

1

Z

2

jF(x; y)j dy

dx < 1:

Then F 2 L1(

1

2).

11

+ Lp-spaces

Theorem 1.1.5.2 (Fubini) Assume that F 2 L1(

1

2).

Then for a.e x 2

1

F(x; ) 2 L1(

2) and

Z

2

F(; y) dy 2 L1(

1):

Similarly, for a.e y 2

2

F(; y) 2 L1(

1) and

Z

1

F(x; ) dx 2 L1(

2):

Futhermore, we have

Z

1

Z

2

F(x; y) dxdy =

Z

2

Z

1

F(x; y) dx

dy =

Z

1

Z

2

F(x; y) dy

dx:

Definition Let f and g be measurable functions on RN. We define the convolution product

f g of f and g by:

f g(x) =

Z

RN

f(x y)g(y) dy

for those x, if any, for which the integral converges.

Theorem 1.1.5.3 (Minkowski’s Inequality) . Let 1 p < +1 and let (X;A; dx) and

(Y; B; dy) be -finite measure spaces. Let F be a measurable function on the product space

X Y . Then

Z

X

Z

Y

F(x; y) dy

p

dx

1

p

Z

Y

Z

X

jF(x; y)jp dx

1

p

dy;

in the sense that the integral on the left hand side exists if the one on the right hand side is

finite, and in this case the inequality holds. Note that the inequality may also be writen as:

Z

Y

F(; y) dy

p

Z

Y

kF(; y)kp dy:

Theorem 1.1.5.4 Let 1 p +1. If f 2 L1(RN) and g 2 Lp(RN) then

f g(x) =

Z

RN

f(x y)g(y) dy

exists for almost all x and defines a function f g 2 Lp(RN). Moreover

kf gkp kfk1kgkp:

Proof.

Case1. If p = +1, we have

Z

RN

jf(x y)g(y)j dy kgk1

Z

RN

jf(x y)j dy = kgk1kfk1;

12

+ Lp-spaces

by invariance of Lebesgue’s measure under translation. Thus f g(x) exists a.e and

jf g(x)j kgk1kfk1; a.e x 2 RN:

So f g 2 L1(

) and

kf gk1 kfk1kgk1:

Case2. For p = 1, let

F(x; y) = f(x y)g(y):

For almost every y 2 RN, we have

Z

RN

jF(x; y)j dx = jg(y)j

Z

RN

jf(x y)j dx = kfk1jg(y)j < 1

and Z

RN

Z

RN

jF(x; y)j dx

dy = kfk1kgk1 < 1:

Using Tonelli’s Theorem, we have F 2 L1(RN RN). By Fubini’s Theorem, we obtain

Z

RN

jF(x; y)j dy < 1 a.e x 2 RN

and Z

RN

Z

RN

jF(x; y)j dy

dx kfk1kgk1:

So

kf gk1 kfk1kgk1:

Case3. For 1 < p < +1, let q be the conjugate exponent of p. From Case2., we know that

for a.e x 2 RN fixed, y 7! jf(xy)jjg(y)jp is integrable or equivalently y 7! jf(xy)j1=pjg(y)j

is in Lp(RN). Since y 7! jf(x y)jq is in Lq(RN), we have from Holder’s inequality that

jf(x y)jjg(y)j = jf(x y)jq jf(x y)j1=pjg(y)j 2 L1(RN)

and

jf(x y)jjg(y)j

Z

RN

jf(x y)jjg(y)jp dy

1=p

kfk1=q

1

i.e

jf g(x)jp (jfj jgjp)(x) kfkp=q

1 :

Using again case2. we have

f g 2 Lp(

) and kf gkp kfk1kgkp:

Definition Let 2 L1(RN) such that

Z

RN

(x) dx = 1. Let (x) =

1

N (

x

). The family

of functions ; > 0, is called a mollifier with kernel . Note that

Z

RN

dx = 1.

13

+ Lp-spaces

Definition If f is a function on RN and a 2 RN, we define the translation of f by a, af

as follow:

af(x) = f(x a)

Proposition 1.1.5.1 Let be a mollifier, 1 p < +1 and f 2 Lp(RN). Then for each

> 0

kf fkp

Z

RN

kyf fkpj(y)j dy: (1.9)

Proof. Since

Z

RN

(x) dx = 1 we have

f (x) f(x) =

Z

RN

[f(x y) f(x)](y) dy:

by Minkowski’s inequality (1.1.5.3)

kf fkp =

Z

RN

Z

RN

[f(x y) f(x)](y) dy

p

dx

1

p

Z

RN

Z

RN

jf(x y) f(x)jpj(y)j dx

1

p

dy

=

Z

RN

kyf fkpj(y)j dy:

Corollary 1.1.5.1 If is such that

Z

RN

(x) dx = 0 then

kf kp

Z

RN

kyf fkpj(y)j dy:

Theorem 1.1.5.5 Assume that 0. Let f be a bounded continuous function on RN.

Then f is continous on RN for each > 0 and for each x 2 RN we have

lim

!0+

f (x) = f(x):

Proof. Let > 0, we have

f (x) =

Z

RN

f(x y)(y) dy =

Z

RN

f(x y)(y) dy:

Let M be the bound on the absolute value of f. Then jf(x y)(y)j M(y) a.e. Since

2 L1(RN) and the function x ! f(xy)(y) is continuous a.e y 2 RN then by Lebesgue

dominated convergence theorem f is continuous.

Now, fix x 2 RN. Since

Z

(y) dy = 1 we have:

f (x) f(x) =

Z

RN

[f(x y) f(x)](y) dy:

14

+ Lp-spaces

Let > 0. By the continuity of f at x, there is > 0, such that

jf(x y) f(x)j

2

; for jyj < :

Since Z

jyj

(y) dy =

Z

jyj

(y) dy ! 0; as ! 0

then there exists 0 > 0 such that

Z

jyj

(y) dy <

4M

; for < 0

It follows that for all such > 0, we can write the integral as a sum over jyj < and jyj

and get

jf (x) f(x)j

2

+

2

= :

1.1.6 Density of Cc(

) in Lp(

)

Proposition 1.1.6.1 Let

be an open subset of RN. Let (Uj)j2J be a collection of open

subsets of

with union U. Let E U. If E \ Uj is a set of Lebesgue measure 0 for each

j 2 J then E has measure 0.

Proof. Let Q be the countable set consisting of all open balls in RN with rational radius

and rational center coordinates. Then for each j 2 J

Uj =

[

fB j B 2 Q; B Ujg

so E is a countable union of sets of measure 0 of the form E \ B.

Note that it is important that be Uj to be open.

Now let f 2 L1(

). Then by the proposition above there exists a largest open subset U

of

on which f is 0 almost everywhere, just take the union of open sets on which f vanishes.

Definition The complement of U is called the support of f in

and is denoted by supp(f).

Proposition 1.1.6.2 If f :

! R is continuous then the support of f in

is the closure of

fx 2

j f(x) 6= 0g

Definition If

is an open subset of RN, we denote by Cc(

) the set of continuous functions

on RN with compact support in

. We denote by D(

) the set of infinitely continuously

differentiable functions with compact support in

15

+ Lp-spaces

Let : RN ! R defined by

(x) =

8><

>:

c(1 kxk) if kxk 1;

0 if kxk > 1 :

(1.10)

where the constant c is chosen so that

Z

RN

(x) dx = 1. Then is a continuous mollifier

and moreover supp () is the -Ball B0(0; ).

Lemma 1.1.6.1 (Uryshon.) Let

be an open subset of RN and K

be a compact set.

Then there exists 2 Cc(

) such that 0 1 and = 1 on some neighborhood of K.

Proof. Let be a continuous mollifier as above and let L be the closed -neighborhood of

K, that is

L = fx 2 RN; j dist(x;K) g

where =

1

3

dist(K; @

). Let

(x) = XL (x) =

Z

RN

XL(x y)(y) dy =

Z

L

(x y) dy

For 0 < < , we have 2 C(

), has it support in the closed 2-neighborhood of K and

so has compact support in

, 0 1 and = 1 on the ( )-neighborhood of K.

Theorem 1.1.6.1 (Density of Cc(

) in Lp(

) ) . Let

be an open subset of RN and let

1 p < +1. Then Cc(

) is dense in Lp(

).

Proof. We denote the Lebesgue measure of measurable set B by m(B). Since the simple

functions are dense in Lp(

) for finite p, it suffices to show that we can approximate the

characteristic function XA of a measurable set A of finite measure by function in Cc(

). Let

> 0. By the regularity of Lebesgue measure there exits a compact set K A and an open

set U, A U such that m(U K) < p. From Uryshon’s Lemma, there is 2 Cc(U) such

that 0 1 and = 1 on K. We have jXA j XU XK and so

kXA kp m(U K)

1

p < :

Remark 1.1.6.1 If 1 p < 1, Theorem 1.1.6.1 says that Cc(

) is dense in Lp(

), and Theorem

1.1.3.1 shows that Lp(

) is complete. Thus Lp(

) is the completion of the metric space

which is obtained by endowing C0(

) with the Lp-metric.

Of course, every metric space S has a completion S whose elements may be viewed abstractly

as equivalent classes of Cauchy sequence in S. The important point in the present situation is

that the various Lp-completion of Cc(

) again turn out to be spaces of functions on

.

The case p = +1 differs from the cases p < 1. The L1-completion of Cc(

) is not L1(

),

but is C0(

), the spaces of all continuous functions on

which vanish at infinity.

16

+ Lp-spaces

Definition A function f :

! R is said to vanish at infinity if for every > 0, there exists

a compact set K

such that jf(x)j < for all x not in K.

We denote by C0(

),the class of all continuous functions on

which vanish at infinity.

It is clear that Cc(

) C0(

).

Theorem 1.1.6.2 C0(

) is the completion of Cc(

), relative to the metric defined by the

supremum norm:

kfk1 = sup

x2

jf(x)j:

Proof. An elementary verification shows that C0(

) satisfies the axioms of a metric space

if the distance between f and g is taken to be kf gk1. We have to show that (i) Cc(

) is

dense in C0(

) and (ii) C0 is complete.

To prove (i), let f 2 C0(

) and > 0, there exists a compact set K

such that jf(x)j <

outside K. Uryshon’s lemma gives us that there exists a function ‘ 2 C0(

) such that

0 ‘ 1 and ‘(x) = 1 on K. Put h = ‘f. Then h 2 Cc(

) and kf hk1 < .

To prove (ii), let ffng be a Cauchy sequence in C0(

). Using the definition of Cauchy

sequence and supremum norm, we can assume that ffng converges uniformly. Then its

pointwise limit function f is continuous. Given > 0, there exists an N so that kfN fk1 <

=2 and there exists a compact set K so that jfN(x)j < =2 outside K. Hence jf(x)j <

outside K, and we have proved that f vanishes at infinity. Thus C0(

) is complete.

Proposition 1.1.6.3 (Continuity of Translation in Lp(

)) . Let 1 p < +1 and f 2

Lp(RN). Let : RN ! Lp(RN) be the map defined by

(y) = yf; 8 y 2 RN:

Then is uniformly continuous on RN.

Proof. Let > 0. By density choose g 2 Cc such that kf gkp <

3

. Let y; z 2 RN and

v = y z, then

k(y) (z)kp = kyf zfkp kyf ygkp + kyg zgkp + kzg zfkp

2

3

+ kyg zgkp

2

3

+ kvg gkp

by translation invariance of Lebesgue measure. Since g has compact support, then the

support of vg stays in a fixed compact set K for kvk 1. Since g is bounded we have

jvgj M XK:

It follows that

jvg gjp (2M)pXK 2 L1(RN); for kvk 1

17

+ Lp-spaces

since g is continuous we have vg ! g as v ! 0 pointwise. By the dominate convergence

theorem

Z

RN

jvg gjp dx ! 0 as v ! 0. Thus there exists > 0 such that 0 < < 1 and

kvg gkp <

1

3

; if kvk < :

Hence, the uniform continuity of follows.

Theorem 1.1.6.3 Let 2 L1(RN), 1 p < +1 and let f 2 Lp(RN). If

Z

RN

dx = 1 then

f ! f in Lp(RN) as ! 0. If

Z

RN

dx = 0 then f ! 0 in Lp(RN) as ! 0

Proof. for the first case, we know that:

kf fkp

Z

RN

kyf fkpj(y)j dy:

The integrand is bounded by 2kfkpjj 2 L1(RN) and goes to 0 as ! 0 by continuity of the

translation. Thus the Lebesgue dominated convergence theorem yields the desired result.

Corollary 1.1.6.1 Let 1 < p < +1 and q 1 such that 1=p + 1=q = 1. If f 2 Lp(RN) and

g 2 Lq(RN) then f g is uniformly continuous.

Proof. We have

f g(x)f g(z) =

Z

RN

(f(x y) f(z y)) g(y)dy =

Z

RN

(xf(y) zf(y)) g(y) dy

therefore by using Holder’s inequality we have

jf g(x) f g(z)j kxf zfkpkgkq:

We conclude by using the fact that the translation is uniformly continuous.

1.1.7 Density of D(

) in Lp(

).

One important application of the convolution product is regularization of functions, that is,

the approximation of functions by smooth functions. Let

u(t) =

(

e1=t if t > 0;

0 if t 0 :

(1.11)

Since for any integer k, lim

t!0

1

tk e1

t = 0, then u 2 C1(R). Let (x) = cu(1 kxk2); x 2 RN.

Then 2 C1(RN) and (x) = 0 if kxk 1. Moreover, for a suitable choice of the constant

c we have (x) 0 and

Z

RN

(x) dx = 1. Let

(x) =

1

N (

x

):

Then

18

+ Lp-spaces

1. 2 C1(RN),

2. supp() = B0(0; ) = fx 2 RN j kxk g,

3. (x) 0;

4.

Z

RN

(x) dx = 1:

Any family () satisfying these four properties is called Friedrichs’s mollifier.

Theorem 1.1.7.1 Let be a Friedrichs’s mollifier. If f 2 L1(RN; loc) the convolution

f (x) =

Z

RN

f(x y)(y) dy =

Z

RN

f(x y)(y) dy

exists for each x 2 RN. Moreover

1. f 2 C1(RN),

2. supp(f ) supp(f) + B0(0; ),

3. if 1 p < +1 and f 2 Lp(RN), then f ! f in Lp(RN), as ! 0. In fact we

have kf fkp sup

kyk

kyf fkp

4. If K, the set of continuity points of f is compact, then f ! f uniformly on K as

! 0

Proof. The convolution exists for each x because the mollifier has compact support. Note

that f (x) =

Z

RN

(x y)f(y) dy implies f 2 C1(RN) by standard results on

differentiating under the integral sign (since has compact support). The second is obvious

and the third follows from

kf fkp

Z

RN

kfyf fkp(y) dy supkykkyf fkp

Assume that K the set of continuity points of f is compact. Then f is uniformly continuous

on K and this shows a little bit more: let > 0, then there exists > 0 such that if x 2 K,

z 2 RN and kx zk < then it follows that jf(x) f(z)j < . Note that we do not require

z to be in K. Now

f (x) f(x) =

Z

kyk

(f(x y) f(x)) (y) dy:

Hence if 0 < < then

jf (x) f(x)j

Z

kyk

jf(x y) f(x)j (y) dy

Z

RN

(y) dy =

for each x 2 K:

19

+ Lp-spaces

Corollary 1.1.7.1 Let

be an open subset of RN and K is a compact subset of RN then

there exits 2 D(

) with 0 1 and = 1 on K.

Proof. Let =

1

3

dist(K; @

). Let L be the closed -neighbborhood of K, that is:

L := fx 2 RN j dist(x;K) g

Let f be the characteristic function of L and let 0 < < . then = f 2 C1(RN) has

its support in the closed 2-neighborhood of K and so has compact support in

. We have

that 0 1 and = 1 on the ( )-neighborhood of K.

Theorem 1.1.7.2 (Density of D(

) in Lp(

)) . Let

be an open subset of RN and let

1 p < +1. Then D(

) is dense in Lp(

)

Proof. Let f 2 Lp(

) and let > 0. By theorem (1.1.6.1) there exits g 2 Cc(

) such that

kf gkp <

2

. Let > 0 and define g = g . Then g 2 C1(

) and g ! g in Lp(

).

Moreover

supp(g) supp(g) + B0(0; )

Since g ! g in Lp(

) as ! 0, there exists > 0 such that kg gkp <

2

for < . Now

let < min(; dist(supp(g); @

)). Then g 2 D(

) and kf gkp <

Theorem 1.1.7.3 (Partition of unity) . Let

be an open subset of RN and let (Uj)j2J

be a locally finite open cover of

such that each Uj has compact closure in

. Then there

exsits j such that

j 2 D(Uj); j 0 and

X

j2J

j(x) = 1; 8 x 2

Proof. There exists an open covering (wj) of

such that wj Uj U

j for all j 2 Uj .

Choose j 2 D(Uj) such that 0 j 1 and j = 1 on wj . The sum

(x) =

X

j2J

j(x)

is locally finite and bounded below by 1. Thus 2 C1(

) and 1 . take j =

j

The j are called a smooth partion of unity subordinate to the locally finite open cover

(Uj).

Theorem 1.1.7.4 (Finite partition of unity) . Let K be a compact subset of RN and let

(Uj)j=1; ;N be a finite open cover of K. Then there exists functions j 2 D(Uj) such that

j 0 and

XN

j=1

j = 1

in a neighborhood of K.

20

+ Lp-spaces

Proof. For x 2 K, let Vx be an open neighborhood of x such that Vx is a compact subset

of Uj with x 2 Uj . Since K is compact there exsit a finite set x1; ; xm in K such that

K

m[

k=1

Vxk :

For each j let Kj be the union of those Vxk which are contained in Uj . Then Kj is compact,

Kj Uj and

K K1 [ [ KN

By corollary1.1.7.1, we may choose j 2 D(Uj), 0 j 1 in a neighborhood of Kj and

j = 1 on Kj . Finally let

1 = 1

2 = (1 1) 2

3 = (1 1)(1 2) 3

N = (1 1)(1 2) (1 N1) N

We have, j 0 and

1 + 2 + + N = 1 (1 1)(1 2) (1 N):

For each x 2 K there is j so that j(x) = 1. Thus 1 + 2 + + N = 1 on K To obtain

the equality on a neighborhood of K, we would enlarge K a bit.

Theorem 1.1.7.5 (Dubois-Reymond) . Let

be an open subset of RN. If f 2 L1(

; loc)

and

Z

f(x)(x) = 0 for each 2 D(

) then f = 0 a.e in

.