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 + 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:
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) ? fm?1(x)j + + jfn+1(x) ? fn(x)j g(x) ? gn?1(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=(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).
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 = 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.
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(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
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(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:
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 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)
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
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 ? 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.
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

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