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 + gjp?1 (jfj + jgj)jf + gjp?1:
Integrating over
, we get
Z

jf + gjp dx
Z

(jfj + jgj)jf + gjp?1 dx
=
Z

jfjjf + gjp?1 dx +
Z

jgjjf + gjp?1 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) ? fm?1(x)j + + jfn+1(x) ? fn(x)j g(x) ? gn?1(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=(p?1)
; 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 = jgjp0?2g.
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(x?y)jjg(y)jp is integrable or equivalently
y 7! jf(x?y)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 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 k?xf ? ?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) =
(
e?1=t if t > 0;
0 if t 0 :
(1.11)
16
Since for any integer k, lim
t!0
1
tk e?1
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 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:
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 ? N?1) 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
D?uDv;
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
ne?nx2
; 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
e?nx2
‘(x) dx =
Z
R
e?y2
‘(
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
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