## TABLE OF CONTENTS

Epigraph ii

Acknowledgement iii

Dedication iv

Introduction v

1 Spaces of Functions 1

1.1 Lp-spaces and some of its properties . . . . . . . . . . . . . . . 1

1.1.1 Basic Integration Results . . . . . . . . . . . . . . . . . . . 1

1.1.2 Definition and basic properties . . . . . . . . . . . . . . 2

1.1.3 The Main properties of Lp(

) . . . . . . . . . . . . . . . 4

1.1.4 Dual Space . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.1.5 Convolutions and Mollifiers . . . . . . . . . . . . . . . . . 9

1.1.6 Density of Cc(

) in Lp(

) . . . . . . . . . . . . . . . . . . 13

1.1.7 Density of D(

) in Lp(

). . . . . . . . . . . . . . . . . . . 16

1.2 Distribution Theory . . . . . . . . . . . . . . . . . . . . . . . . . . 21

1.2.1 Test Functions . . . . . . . . . . . . . . . . . . . . . . . . 21

1.2.2 Convergence in Function Spaces . . . . . . . . . . . . . . . 22

1.2.3 Continuity and Denseness on Dm(

) and D(

) . . . . . . 23

1.2.4 Distributions . . . . . . . . . . . . . . . . . . . . . . . . . 24

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

vii

2 Sobolev spaces Wm;p 35

2.1 Definitions and main properties . . . . . . . . . . . . . . . . . . . 35

2.2 The Main Theorems . . . . . . . . . . . . . . . . . . . . . . . . . 39

2.2.1 Approximation by smooth functions . . . . . . . . . . . . . 39

2.2.2 Extension Theorems . . . . . . . . . . . . . . . . . . . . . 48

2.2.3 Trace Theory. . . . . . . . . . . . . . . . . . . . . . . . . . 56

2.2.4 Sobolev Embedding . . . . . . . . . . . . . . . . . . . . . . 62

3 Variational Formulation of Some Linear Elliptic PDEs 67

3.1 Some abstract Variational Problems . . . . . . . . . . . . . . . . . 67

3.1.1 Riesz-Frechet Representation Theorem . . . . . . . . . . . 68

3.1.2 Lax-Milgram Representation Theorem . . . . . . . . . . . 69

3.2 Existence and uniqueness of weak solutions of some linear elliptic

PDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

3.2.1 Homogenious Dirichlet problem . . . . . . . . . . . . . . . 71

3.2.2 Homogenious Neumann problem . . . . . . . . . . . . . . . 73

Bibliography 78

## 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:

1

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:

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)

2

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 :

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:

3

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:

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):

4

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].

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.

5

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).

6

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.

7

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)

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 Z f = 0 a.e on

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

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

8

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).

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;

9

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;

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

10

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(xy)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 Z of functions ; > 0, is called a mollifier with kernel . Note that

RN

dx = 1.

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:

11

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(x y)(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:

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

12

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

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; ).

13

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.

14

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:

15

It follows that

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

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(RNZ ), 1 p < +1 and let f 2 Lp(RN). If

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

fg(x)fg(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)

16

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

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

17

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 kxzk < 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:

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

18

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.

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.

19

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

.

20

1.2 Distribution Theory

1.2.1 Test Functions

Let

be a nonempty open subset of RN.

Notations

If m 2 N, Cm(

) denotes the space of real-valued functions on

of class Cm and

C1(

) the space of those of class C1. By convention, C0(

) = C(

) the space

of continuous functions on

.

An element 2 Nn is called a multiindex. If = (1; ; n) is a multiindex,

we define the length of to be the sum jj = 1 + + n, and we put

! = 1! n!. We give Nn the product order: if ; 2 Nn , we write if

1 1; ; n n:

If 1 i n, we often use Di to denote @

@xi

. Then if is a multiindex, we write

D = D1

1 Dn

n =

Djj

@x1

1 @xn

n

:

The differential operator D is also denoted by @jj

x or @

x . By convention, D0

(the differential of order 0 with espect to any index) is the identity map. We see

that each operator D, where 2 Nn, acts on the space Cm(

), for jj m.

We recall the following classical result:

Proposition 1.2.1.1 (Leibniz’ formula) . Let u; v 2 Cm(

). For each multiindex

such that jj m,

D(uv) =

X

C

DuDv;

where

C

=

Yn

i=1

i!

i!(i i)!

=

!

!( )!

:

+ We denote by Dm(

) the space of functions of class Cm having compact

support in

. In particular, D0(

) = Cc(

). Clearly, m0 m implies Dm0(

)

Dm(

). Now we set

D(

) =

\

m2N

Dm(

);

Thus D(

) is the space of functions of class C1(

) having compact support in

; such functions are called test functions on

.

21

Finally, if K is a compact subset of

, we denote by DK(

) the space of functions

of class C1 having support contained in K.

DK(

) =

\

m2N

Dm

K(

)

thus

D(

) =

[

K2K(

)

DK(

)

where K(

) is the set of compact subsets of

.

Clearly, a function in Dm(

) or D(

), when extended with the 0 value outside

becomes an element of Dm(RN) or D(RN), respectively. Thus Dm(

) and D(

)

can be considered as subspaces of Dm(RN) and D(RN). We will often make this

identification. Conversely, an element ‘ 2 Dm(RN) or D(RN) belong to all the

spaces Dm(

) or D(

) such that supp(‘)

.

1.2.2 Convergence in Function Spaces

Convergence in Dm

K(

) and DK(

). Let K be a compact subset of

. We say

that a sequence (‘n) in Dm

K(

) converges to ‘ 2 Dm

K(

), if for every multiindex

such that jj m, the sequence (D’n) converges uniformly to D’. An

analogous definition applies with the replacement of Dm

K(

) by DK(

) where

now there is no restriction on the multiindex 2 Nn.

The convergence thus defined on Dm

K(

) clearly corresponds to the convergence

in the norm k k(m) defined on Dm

K(

) by:

k’k(m) =

X

jjm

kD ‘k1;

where k k1 denote the uniform norm. In contrast, no norm on DK(

) yields the

notion of convergence we have defined in that space.

Convergence in Dm(

) and D(

). We say that a sequence (‘n) in Dm(

)

converges to ‘ in Dm(

) if the followings are satisfied:

(i) there exists a compact subset K of

such that

supp(‘) K and supp(‘n) K for all n;

(ii) the sequence (‘n) converge to ‘ in Dm

K(

).

An analogous definition applies with the replacement of Dm(

) and Dm

K(

) by

D(

) and DK(

).

22

Convergence in Cm(

) and C1(

). We say that a sequence (fn) in Cm(

)

converge to f 2 Cm(

), if for every multiindex such that jj m and for every

compact K in

, the sequence (Dfn) converges to (Df) uniformly on K. An

analogous definition applies with the replacement of Cm(

) by C1(

), where

now there is no restriction on the multiindex . For m = 0, the convergence in

C0(

) = C(

) thus defined coincides with the uniform convergence on compact

subsets.

Remark 1.2.2.1 The definitions of convergence of sequences just made extend immediately

to families (‘), where runs over a subset in R and ! 0; 0 2

[1;+1].

We will see that it is possible to give the spaces DK(

), Cm(

) and C1(

)

complete metric structures for which convergence of sequences coincides with the

notions just defined. In contrast, one can show that the convergence we have

defined in Dm(

) and D(

) cannot come from a metric structure.

In fact, the only topological notions that we will use in connection with these

function spaces are continuity and denseness, and these notions, in the case of

metric spaces, can always be expressed in terms of sequences.

1.2.3 Continuity and Denseness on Dm(

) and D(

)

A subset C of Dm(

) or D(

) will be called dense in Dm(

) or D(

), if for

every ‘ in Dm(

) or D(

), there exists a sequence (‘n) in C converging to ‘ in

Dm(

) or D(

).

A function F on Dm(

) or D(

) and taking values in a metric space or in

one of the spaces just introduced will be called continuous, if for every sequence

(‘n) in Dm(

) or D(

) that converges to ‘ in Dm(

) or D(

), the sequence

(F(‘n)) converges to F(‘) in the space considered.

For example, the Canonical Injection from Dm(

) to Cm(

) is continuous. This

means simply that every sequence in Dm(

) that converges in Dm(

) also converges

in Cm(

) to the same limit.

Proposition 1.2.3.1 For every m 2 N, the space D(

) is dense in Dm(

). In

particular, D(

) is dense in Cc(

).

Lemma 1.2.3.1 Let

be an open subet of RN. For n 2 N, define

Kn = fx 2 RN j kxk n and d(x;

c)

1

n

g

where d is the usual distance in RN. Then

1. Each Kn is a compact subset of

and Kn Kn+1.

23

2.

=

1[

n=1

Kn =

1[

n=2

Kn.

3. For all compact K in

, there exists N 1 such that K KN.

Proposition 1.2.3.2 The space D(

) is dense in C1(

) and in Cm(

) for every

m 2 N

Proof. Let Kn be a sequence of compact subsets of

exhausting

(as above).

Then there exists for each n, an element ‘n 2 D(

) such that

0 ‘n 1; ‘n = 1 on Kn; supp(‘n) Kn+1:

Now let f 2 C1(

), we have (f’n) 2 D(

) for every n 2 N. If K is a compact

subset of

, there is an integer N such that K KN. Thus for every n N

and for every 2 Nn, we have D(f’n) = Df on K. By the definition of

convergence in C1(

), we deduce that (f’n) converge to f in C1(

).

1.2.4 Distributions

Definition A distribution on

is a continuous linear mapping T from D(

)

into R. The set of all distributions is denoted by D0(

).

Remark 1.2.4.1 By Linearity, to show that T is continuous, it is enough to show

that, if ‘n ! 0 in D(

) then (T; ‘n) ! 0 in R.

Theorem 1.2.4.1 Let T be a linear mapping from D(

) into R. Then T is a

distribution if and only if, for any compact set K in

, there exists an integer

nK 2 N and a positive constant CK such that:

j(T; ‘)j CK

X

jjnK

sup

K

jD'(x)j; 8 ‘ 2 DK(

):

Definition If nK can be chosen independent of K, then the smallest n with this

property is called the order of the distribution T.

Example 1.2.4.1 (Distribution given by a locally integrable function) Let

f 2 L1(

; loc), then f gives a distribution Tf defined by:

(Tf ; ‘) =

Z

f(x) ‘(x) dx; 8 ‘ 2 D(

):

24

The linearity of Tf follows from the linearity of integral. Now let K be a compact

subset of

and let ‘ 2 D(

) with supp(‘) K then we have

j(Tf ; ‘)j

Z

K

jf(x)j dx

sup

K

j'(x)j

;

so T is a distribution of order 0.

We define the maping f ! Tf . It is linear and one to one. In fact let f 2 L1(

; loc)

such that Tf = 0, Then by using Dubois-Reymond’s lemma, we show that f = 0 a.e

in

. From now, we can identify L1(

; loc) as a subset of D0(

).

Example 1.2.4.2 (Dirac distribution) .

Let x0 2

. We denote by x0 the linear form defined on D(

) by

(x0 ; ‘) = ‘(x0); 8 ‘ 2 D(

):

Let K be a compact subset of

. Since j(x0 ; ‘)j sup

K

j'(x)j for all ‘ 2 D(

)

with supp(‘) K, then x0 is a distribution.

Example 1.2.4.3 (The distribution Principal Value of 1=x) .

Consider the function x 7! 1=x from R to R. This function is clearly not locally

integrable on R but it is on R. We will see how we can extend to R the distribution

defined by this function on R.

Proposition 1.2.4.1 For every ‘ 2 D(R), the limit

(pv(1=x); ‘) = lim

!0+

Z

jxj

‘(x)

x

dx

exists. The linear form pv(1=x) thus defined a distribution of order 1 on R, and

is an extension to R of the distribution [1=x] 2 D0(R).

We call pv(1=x) the principal value of 1=x.

Example 1.2.4.4 (The distribution Finite part of 1=x2. ) Let ‘ 2 D(

),

we call the distribution Finite part, the distribution denoted by fp(1=x2) and defined

by:

(fp(1=x2); ‘) = lim

!0+

Z

jxj

‘(x)

x2 dx 2

‘(0)

8 ‘ 2 D(R):

Proposition 1.2.4.2 Let T be a distribution on

such that every point x 2

has an open neighborhood Vx such that (T; ‘) = 0 for all ‘ 2 D(Vx). Then T = 0.

25

Proof. Let ‘ 2 D(

), we will show that (T; ‘) = 0. Let K = supp(‘) and let

x 2 K then by hypothesis, there exists an open neighborhood Vx of x such that

(T; ‘) = 0 for all ‘ 2 D(Vx). Since K is compact, there exists x1; ; xN 2 K

such that

K

N[

i=1

Vxi

Let 1: ; N be a partition of unity associated to this open cover of K. Then

‘ =

XN

i=1

i’

Since supp(i’) Vxi then (T; i’) = 0 and so is (T; ‘).

1.2.5 The Support of a Distribution

Definition Let T be a distribution on

, an open of nullity of T is an open

subset U of

such that (T; ‘) = 0 for all ‘ 2 D(U).

Proposition 1.2.5.1 Any distribution T has a largest open of nullity

0. Its

complement is called the support of T and denoted by supp(T).

Proof. Let U be the collection of opens of nullity of T, and let

0 =

S

U2U U

be there union. It suffices to show that

0 is it self an open of nullity of T.

Take ‘ 2 D(

0). By compactness of the support of ‘, their exists a finite

collection of opens sets U1; ; UN whose union contains the support of ‘. Let

i; i = 1; ;N be a partion of unity associated to this open cover of supp(‘).

It follows that

‘ =

XN

i=1

‘ i:

Since each ‘ i is supported in the open of nullity Ui, this implies that

(T; ‘) =

XN

i=1

(T; ‘ ) = 0:

This proves that

0 is indeed an open of nullity of T, and by construction it is

the largest of such open sets.

Somes consequences of the definition:

1. x0 =2 supp(T) if and only if there exists an open neighborhood Vx0 of x0

such that

(T; ‘) = 0; 8 ‘ 2 D(Vx0);

26

2. x0 2 supp(T) if and only for all open neighborhood Vx0 of x0, there exists

‘ 2 D(Vx0) such that (T; ‘) 6= 0.

Proposition 1.2.5.2 Let T be a distribution on

and ‘ 2 D(

) such that

supp (T) \ supp(‘) = ;:

Then (T; ‘) = 0.

1.2.6 Distributions with Compact Support

Theorem 1.2.6.1 Let T be a distribution on

. A necessary and sufficient condition

for the support of T to be compact is that T has an extension to a continuous

linear form on C1(

). The extension is then unique.

Proof. Suppose first that the support of T is compact. Then there exists a

compact K in

whose interior contains the support of T. It follows from corollary

1.1.7.1 that there exists 2 D(

) such that 0 1 and (x) = 1 on K. We

then set, for f 2 C1(

),

( T; f) = (T; f): (1.12)

It is clear that this does define a linear form T on C1(

). On the other hand, if

‘ 2 D(

), we have

supp (‘ ‘)

K

supp (T);

this implies that

supp (‘ ‘) \ supp (T) = ;:

So by proposition 1.2.5.2, it follows that

( T; ‘) = (T; ‘):

Thus T is an extension of T to C1(

).

Finally, if (fn) is a sequence in C1(

) that converges to 0 in C1(

), then from

the definitions and Leibniz’s formula the sequence (fn) converges to 0 in D(

),

so that

lim

n!+1

( T; fn) = lim

n!+1

(T; fn) = 0

This proves that T is continuous on C1(

). Since D(

) is dense in C1(

), the

extension is unique.

For the converse, assume that T can be extended to a continuous linear form

T on C1(

). Let (Kn) be an exhausting sequence of compact subsets of

. If

the support of T is not compact, then there exists for each n 2 N, an element

‘n 2 D(

) such that

supp(‘n)

Kn and (T; ‘n) 6= 0:

27

Put

n =

‘n

(T; ‘n)

;

so we have

(T; n) = 1; 8 n 2 N:

Now we will show that the series

Xn

n=1

n converges in C1(

). To this end, let

K be a compact subset of

, then there exists N 2 N such that K KN. But

for n > N, we have n = 0 on KN, and so on K, the sum

X1

n=0

n reduces to a

finite sum on K, and this holds for every compact subset K on

. So the sum

converges in C1(

). By the continuity of T, it follows that the series

X1

n=0

(T; n)

converges, contradicting the fact that (T; n) = 1.

Remark 1.2.6.1 The restriction to D(

) of a continuous linear form on C1(

) is a

distribution on

(since a sequence in D(

) that converges in D(

) also converges in

C1(

)), and by the preceding theorem this distribution has compact support. Thus

we can identify the space of distributions having compact support with the space of

continuous linear forms on C1(

) denoted by C1(

)0.

Proposition 1.2.6.1 Every distribution T with compact support in

has finite

order. More precisely, there exists an integer m 2 N and a constant C0 0 such

that

j(T; ‘)j C0k’k(m); 8 ‘ 2 D(

):

Proof. Let K be the support of T and let K1;K2 be compact subsets of

such

that

K K1 K1 K2 K2

:

Then by theorem 1.2.4.1, there exists an integer m 2 N and a constant C 0

such that

j(T; ‘)j Ck’k(m); 8 ‘ 2 DK2(

):

By corollary 1.1.7.1, there exists 2 D(

) such that 0 1, = 1 on K1

and supp( ) K2. If ‘ 2 D(

) then ‘ 2 DK2(

) and

supp (‘ ‘ )

K1

K:

Since K is the support of T, then

(T; ‘ ‘ ) = 0;

so there exists a positive constant C depending only on C0, m and such that

j(T; ‘)j = j(T; ‘ )j C0k’ k(m) Ck’k(m):

The last inequality being a consequence of Leibniz’ formula.

28

Remark 1.2.6.2 One can deduce from the preceding result that if, T is a distribution

with compact support, there exists an integer m 2 N such that T extends to a

continuous linear form on Cm(

) and this extension is unique.

1.2.7 Convergence of Distributions

We assume that

is an open subset of RN.

Definition Let (Tn)n2N be a sequence of distributions in

. We say that (Tn)

converges to the distribution T if

lim

n!+1

(Tn; ‘) = (T; ‘); for all ‘ 2 D(

):

Theorem 1.2.7.1

1. Let 1 p +1. If fn; f 2 Lp(

) with fn ! f in Lp(

) then fn ! f in

D0(

).

2. The pointwise convergence does not imply the convergence in D0(

).

Proof.

1. Let q such that 1=p + 1=q = 1. Then by Holder’s inequality we have:

j(fn; ‘) (f; ‘)j = j(fn f; ‘)j

Z

jfn fj j’j dx kfn fkpk’kq ! 0:

2. Let (fn) be the sequence of functions defined by

fn(x) =

p

nenx2

; x 6= 0;

then fn(x) ! 0 for all x 6= 0 but fn !

p

0 in D0(

). In fact let ‘ 2 D(

), by

Lebesgue dominated convergence therorem we have

(fn; ‘) =

p

n

Z

R

enx2

‘(x) dx =

Z

R

ey2

‘(

y

p

n

) dy !

p

‘(0) =

p

(0; ‘)

Examples

Example 1.2.7.1 Let (Tn)n2N, be a sequence of distributions on R defined by:

Tn(x) = sin(nx):

Let ‘ 2 D(R), we have

(Tn; ‘) =

Z

R

sin(nx)'(x) dx =

1

n

Z

R

cos(nx)’0(x) dx ! 0:

So Tn converge to 0 in D0(R).

29

Example 1.2.7.2 For : > 0 define

v(x) =

8>><

>>:

1

if x 2 [0; ];

0 if x =2 [0; ] :

and w(x) =

8>><

>>:

1

2

if x 2 [; ];

0 if x =2 [; ] :

Then we have:

v ! 0 in D0(R) as ! 0 and w ! 0 in D0(R) as ! 0:

1.2.8 Multiplication of Distributions

Now, we define the product of distribution by a smooth function. The definition

arises from the following lemma.

Lemma 1.2.8.1 Let 2 C1(

). The map ‘ ! ‘ from D(

) to D(

) is linear

and continuous. In other words if (‘n) is a sequence in D(

) converging to ‘ in

D(

) then the sequence (‘n) converges to ‘ in D(

).

Definition If T 2 D0(

) and 2 C1(

), the product distribution T on

is

defined by setting:

(T; ‘) = (T; ‘); 8 ‘ 2 D(

):

The fact that T defines a distribution follows from the preceding lemma.

The definition immediately implies that if 2 C1(

), the linear map T ! T

from D0(

) to D0(

) is continuous in the sense that, if (Tn) converge to T in

D0(

) then (Tn) converge to (T) in D0(

).

Proposition 1.2.8.1 Let T 2 D0(

) and 2 C1(

), then we have

supp(T) supp() \ supp(T):

Proof. Let ‘ 2 D(

). If supp(‘)

-supp(), then ‘ = 0, so (T; ‘) = 0. It

follows that

-supp() is contained in

-supp(T), so supp(T) supp().

Now if If supp(‘)

-supp(T), then supp(‘) supp(‘)

supp(T), which

implies that (T; ‘) = 0. Therefore

supp(T) is contained in

supp(T) so

supp(T) supp(T).

The inclusion in the proposition may be strict. For example, if T = is the

dirac distribution in RN, and 2 C1(RN) is such that (0) = 0 and 0 2 sup()

(say (x) = x), then T = (0) = 0 and the support of T is empty, whereas

supp()\supp(T) = f0g.

30

Proposition

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