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1.1. Innitely smooth functions 5

Denition 1.1 The function f : R d R m is said to be di ff erentiable at x if there exists an element f (x) L(R d , R m ) such that

f (x + h) f (x) f (x)h = o( h ) , h 0 .

Lemma 1.1 If f : R is di ff erentiable at x0, then f has a partial derivative with respect to x j at x0

and f x j

(x0) = f (x0)e j , j {1, . . . , d } ,

where (e1, . . . , e d ) denotes the canonical basis of R d . Moreover, we have

f (x0)v = f (x0), v , v R d ,

the vector f (x0) =d

j =1

f x j

(x0)e j is the gradient of f at x0,

Proposition 1.1 Suppose I is an open interval of R and f is continuous in I and di ff erentiable except at point x0 I . If x I and lim

x x 0f (x) = a R , then f (x0) exists and we have f (x0) = a.

Denition 1.2 Consider an open subset R d and k N . A function f C k ( ) if one of the following conditions holds:

(i) k = 0 and f is continuous on ,

(ii) k 1 and f has partial derivatives f x j

, j = 1 , . . . , d that are in C k 1( ).

Moreover, f C ( ) if f C k ( ) for all k N .

Lemma 1.2 (Schwarz) If f C 2( ) then 2f

x i x j=

2f x j x i

, i, j {1, . . . , d} .

Notice that if f C k ( , this lemma refers to the property of interchanging the order of taking partialderivatives of a function. The matrix of second-order partial derivatives of f is called the Hessian matrix.In most cases, this matrix is a symmetric matrix.

Theorem 1.1 (Taylors formula) Let be an open subset of R d , m N and f C m +1 ( ). Consider x, y such that x + ty , for all t [0, 1]. Then, we have:

f (x + y) =| | m

y

! f (x) + ( m + 1)

| |= m +1

y

! 1

0(1 t)m f (x + ty) dt . (1.4)

Corollary 1.1 Let be an open subset of R d , f C m +1 ( ) and K a compact subset of . Consider x, y K such that [x, x + y] K . Then, there exists a constant C = C (K,m,f ) such that

f (x + y) | | m

y

! f (x) C y m +1 . (1.5)

Corollary 1.2 If f C k (B ) where B = {x R d , x < 1}, then there exists f 1, . . . , f d C k 1(B ) such that:

f j (0) = j f (0)

1 + | | , sup

x B| f j | sup

x B| j f | , j = 1 , . . . , d f (x) f (0) =

d

j =1x j f j (x) .

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6 Chapter 1. Distributions on open sets of R d

1.1.2 Existence of C functions

Let R d be an open set. We recall that the set of innitely smooth functions is dened as:

C ( ) =k 0

C k ( ) .

Denition 1.3 Given u C 0( ), the support of u, denoted as Supp( u), is dened as the closure of the set {x , u(x) = 0 } in . It is the smallest closed subset of such that u = 0 in \ Supp( u).Notice that if x Supp( u), then there exists a sequence ( xn )n N such that u(xn ) = 0 for all n Nand such that lim n xn = x.

Denition 1.4 If k N {+ }, the space C k0 ( ) is composed of all functions u C k ( ) having a compact subset of as support. The elements of C 0 ( ), hereafter denoted by D( ), are called testfunctions .

It is common to nd the notations C 0 ( ) or C c ( ) instead of the symbol D( ). Every functionu C k0 ( ) can be extended to a function of C k0 (R d ). Thus, C k0 ( ) can be seen as a subspace of C k0 (R d ).In this respect, given an open set R d , the set C k0 ( ) can be dened as the set of elements u C k0 (R d )

for which Supp( u) .The space D( ) is not empty. Indeed, we have the following result.Lemma 1.3 There exists a function u D(R d ) such that u(0) > 0 and u(x) 0, for all x R d .

Proof. Consider the function f C (R ) dened as:

f (x) = 0 x 0exp( 1/x ) x > 0

Then, the functionu(x) = f (1 x 2)

satises the assumptions. By translation and scaling, we show that for every r > 0, the function

x ux x0

ris positive on R , strictly positive at x0 and its support is the ball of radius r centered at x0 . The existence of such function u allows to prove a classical result.

Theorem 1.2 If f, g C 0( ) and if the following identity holds

fudx = gudx, u D( ) ,then, f = g.

Proof. Let consider h = f g, then

hudx = 0 , u D(

) .If h is a complex valued function, we will consider the real and imaginary parts separately. Hence, consider h is areal valued function and that the previous equality holds for all u D( ), u being a real valued function. If thereexists x0 such that h(x0) = 0 then we select u such that u(x0) > 0 with support in a neighborhood of x0 and suchthat uh has constant sign. This obviously is in contradiction with the assumption, thus h 0 in .

Lemma 1.4 There exists an increasing function C (R ) such that

(x) = 0 x 01 x 1

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8 Chapter 1. Distributions on open sets of R d

Corollary 1.3 Let K 1, K 2 be two disjoined compact subsets of the open set R d . Then, there exists a function u D( ) such that

u(x) = 1 x K 1 1 x K 2

and such that |u(x)| 1 for all x

.

Proof. Consider U 1 and U 2 two open sets of such that

K 1 U 1 , K 2 U 2 , U 1 U 2 = .

From the previous proposition, we know that there exists u1 , u2 D( ) such that:

u i 1 , in K i , ui D(U i ) , 0 ui (x) 1 , i {1, 2} .

The function u dened as: u(x) = u1(x) u2(x), x satises the desired properties.

1.1.3 Partition of unity

Partitions of unity are useful because they often allow one to extend local constructions to the wholespace.

Proposition 1.4 Let K be a compact subset of R d and let consider an open cover (U j ) j =1 ,...,N of K .Then, there exists compacts sets (K j ) j =1 ,...,N such that K j U j , for all j = 1 , . . . , N and

K =N

j =1

K j . (1.6)

Proof. For every x K , consider rx > 0 such that B(x, r x ) x U j

U j . Hence, we have K x K

B (x, r x ) and

thus there exists x1 , . . . , x M K such that K M

i =1

B (x i , r x i ). We pose

K j = K B (x i ,r x i ) U j

B (x i , r x i ) .

Then by denition, K j is a compact set included in K such that K j U j . Let consider x K . There exists

i {1, . . . , M } such that x B(x i , r x i ). Moreover, there exists j0 {1, . . . , N } such that xi U j 0 , thusB (x i , r x i ) U j 0 . We conclude that x K j 0 .

Theorem 1.3 (Partition of unity) Consider a compact subset K of R d and open sets U j of R d . Sup-

pose we have K N

j =1U j , then there exists a collection (u j ) j =1 ,...,N such that u j D(U j ), 0 u j 1 for

all j = 1 , . . . , N and such that N

j =1u j = 1 in the neighborhood of K .

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1.1. Innitely smooth functions 9

Proof. We know that there exists compact sets ( K j ) j {1,...,N } such that K j U j , for every j = 1, . . . , N and

such that K =N

j =1K j . Moreover, from the previous proposition, we deduce that for every j = 1 , . . . , N , there

exists vj D(U j ) such that vj (x) [0, 1], for all x R d and vj (x) = 1 if x K j . Consider the open set

V = x N

j =1

U j ,N

j =1

vj (x) > 0 .

We have then K V . Hence, there exists D(V ) such that (x) [0, 1] for x R d and 1 on an open setW such that K W V . Considering the function

u i = vj

(1 ) + N k =1 vk,

we notice that u i D(U j ) as the denominator is strictly positive on V and is equal to 1 outside V . Since 1 on

the set W , the relation implies thatN

j =1u i 1 on W .

Denition 1.5 Under the hypothesis of Theorem 1.3, the collection (u j ) j =1 ,...,N is called a partition of unity subordinate to the open cover (U j ) j =1 ,...,N of K .

1.1.4 Lebesgue integration

We briey recall some important results about Lebesgue integration and L p spaces and we refer thereader to Appendix D for more details and results.

If is an open subset of R d and p 1, we denote by L p( ) the space of p-power integrable functionswith values in R . The space L p( ) endowed with the norm

f Lp

( ) = |f (x) | p

dx

1

p

, f L p

(

) , (1.7)

is a Banach space. If p 1, we denote by L ploc ( ) the space of locally integrable functions f such thatf L p(K ), for every compact K .

Lemma 1.5 If f L ploc ( ) then, there exists a largest open set V in such that f |V = 0 almost everywhere in V .

Denition 1.6 Consider R d and f L1loc ( ). We call essential support of f the closed set \ V where V is the largest open V such that f |V = 0 almost everywhere in V .

Theorem 1.4 (Lebesgue dominated convergence) Let (f n )n N be a sequence of functions of L1( ),

with an open set of Rd. Suppose that

(i) f n (x) f (x) almost everywhere on ,

(ii) there exists a function g L1( ) such that for every n, |f n (x)| g(x) almost everywhere in .

Then, f L1( ) and limn f n f L 1 ( ) = 0 .

Theorem 1.5 (Lebesgue) Let be an open set of R d , f : R p R and k N . Suppose that

(i) the function f x : R dened by f x ( ) = f (x, ) belongs to C k ( ) for every xed x R p,

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10 Chapter 1. Distributions on open sets of R d

(ii) for all N n such that | | k we have

sup

| f x ( ) | w (x) ,

where w L1(R p).

Then, the function g( ) = R p f (x, ) dx is a function of C k ( ) and for every R we have: g( ) = R p f (x, ) dx , N n | | k .

Consider 1 R n 1 , 2 R n 2 two open sets and a measurable function F : 1 2 C . We havetwo classical results.

Theorem 1.6 (Tonelli) Suppose that

2

|F (x, y)| dy < for almost every x 1 and that

1 dx 2 |F (x, y )| dy < .Then F L1( 1 2).

Theorem 1.7 (Fubini) Suppose F L1( 1 2). Then, we have F (x, y ) L1( 2) for almost every

x 1 and 2 F (x, y) dy L1( 1). Similarly, we have F (x, y) L1( 1) for almost every y 2 and 1 F (x, y) dx L1( 2). Furthermore, we have:

1 dx 2 F (x, y) dy = 2 dy 1 F (x, y) dx = 1 2 F (x, y ) dxdy .

To conclude this section, we give the following density result.

Theorem 1.8 (i) If p 1, then the space C 00 ( ) is dense in L p( ).

(ii) If 1 p < , the space D( ) is dense in L p( ).

1.2 The concept of distribution

1.2.1 Main denitions

Let be an open subset of R d and let K denote a compact of . In this section, we call test function on any function of D( ) and we denote by DK ( ) the set of test functions with support in K . Furthermore,we like to introduce the following notations: given ( , ) N d N d , x R d , and u C ( ) :

| | def =

d

j =1

j , ! def =

d

j =1

j ! , x def =

d

j =1x j j ,

u def =

d

j =1

j j u .

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1.2. The concept of distribution 13

Remark 1.2 This result has a physical meaning. Indeed, if the functions n are considered as mass densities, the limit (in a certain sense) must be seen as a ponctual mass associated with the origin.

Denition 1.10 We call principal value of the function 1/x and we denote it pv 1x

the distribution dened by:

pv 1x

, u = 12

u(x) u( x)

x dx , u D(R ) .

Notice that this formula denes a distribution since we have, for every test function u D[ R,R ],

pv 1x

, u R u L .

1.2.2 Operations on distributions

We dene now several usual operations on distributions that are already familiar for classical smoothfunctions, for example test functions.

Denition 1.11 (Restriction) Suppose and are two open subsets in Rd

such that and consider a distribution T on . We dene the restriction of T to the open set , denoted by T | , as

u D() , T | , u def = T, u .

Furthermore, if f L1loc ( ) we have:

u D( ) , f | (x)u(x) dx = f | (x)u(x) dx .This denition coincide with the notion of restriction for functions.

Remark 1.3 The restriction of a distribution to a set is only dened for an open set.

If T 1 and T 2 are distributions of order m on Rd

, it is obvious that T 1 + T 2 and aT 1 are distributions of the same order for all a R . However, the product T 1T 2 of two distributions is not a distribution. Forexample, if we were to dene T 1T 2, u = T 1, u T 2, u , this is not linear in u.

1.2.3 Di ff erentiation of distributions

One of the strongest advantage of the distribution theory is that the di ff erentiation operation is alwaysdened and is continuous. Indeed, consider a function f C 1(R d ) and u D(R d ), then:

f x i

, u = R d f x i (x)u(x) dx ,=

R d

u

x i(x)f (x) dx = f,

u

x i.

This result is still valid if we set f to be a distribution. More precisely,

Proposition 1.6 Let T D ( ) be a distribution on an open set R d . The linear form T ( j ) on D( )dened by:

T ( j ) , u def = T,

u x j

= T, x j u ,

denes a distribution on . Furthermore, if (T n )n N is a sequence in D ( ) converging toward T , then the sequence (T ( j )n )n N converges toward T ( j ) .

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14 Chapter 1. Distributions on open sets of R d

Proof. Since T is a distribution, for every compact K in , there exists a constant C and an integer N suchthat

v DK , | T, v | C sup| | N

v L .

For every u DK , we apply this to v = x j u DK . Hence, we have

u DK , | T ( j ) , u | = | T, x j u | C sup| | N +1

u L .

Thus, T ( j ) is a distribution on . Moreover, for every function u D( ) we have:

limn

T ( j )n , u = limn T n , x j u = T, x j u = T ( j ) , u .

the results follows.

Denition 1.12 Suppose T D ( ). Then, the distribution T ( j ) is called the partial derivative of T with respect to the variable x j and is denoted x j T and is such that:

x j T, u = T, x j u , u D( ) . (1.10)

Notice that if f is a diff erentiable function, the derivative of the distribution associated with f coincidewith the usual derivative of f .

Example 1.2 (i) The Heaviside step function H is the characteristic function of R + . We have x H = 0 in D (R ). Indeed, for any test function u D(R ), we have:

x H, u =dH dx

, u = H, u = +

u (x)H (x) dx

= +

0u (x) dx = u(0) since u(+ ) = 0

= 0, u .

(1.11)

Hence, the derivative of H in the sens of distributions is the Dirac mass dened hereabove.

(ii) Since the function f (x) = log |x | is in the space L1loc (R ), it denes an element of D (R ). Indeed,

the derivative of f in the distributional sense is the distribution pv 1x

introduced previously:

df dx

= pv 1x

.

According to the denition of the derivative in D (R ), we have for all u D(R ):

df

dx, u =

Rlog(|x |)u (x) dx .

We observe that:

|x |> log(|x |)u (x) dx = [u( ) u()]log |x |> u (x) dx ,and thus, we deduce that

R log(|x |)u (x) dx = limn |x |> log(|x |)u (x) dx = pv 1x

, u .

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1.2. The concept of distribution 15

The next results establishes a relation between the derivative and the primitive, in the distributionalsense.

Theorem 1.11 Consider a distribution T on an open interval I R . If its derivative, in the distribu-tional sense, T is a function of L1loc (I ), then T is a continuous function and we have, for all a I :

T (x) = xa T (x)dx + C .

Lemma 1.6 Suppose I is an open interval of R . Then,

1. the distributions on I for which the identity T = 0 holds, are the constant functions.

2. for any T 2 D (I ), there exists T 1 D (I ) such that T 1 = T 2.

3. for a I and f L1loc (I ), we have (in the distributional sense):

ddx xa f (y)dy = f (x) .

Proof. Let consider D(I ), such that I (x) dx = 1, and u D(I ). Then, the function(x) = u(x) I u(x) dx (x) D(I ) and I (x) dx = 0 .

This implies the existence of a unique function v D(I ) such that v = . Hence, there exists a unique functionv D(I ) such that

v = u

I u(x) dx . (1.12)

Now, let consider T D (I ) such that T = 0. We have

T, u = I u(x) dx T, + T, v .The term T, v = T , v is equal to zero and we obtain by denoting C the constant term T, :

T, u = C I u(x) dx .Thus, T is equal to a constant C .To nd a primitive T 1 of T 2 , we pose

T 1 , u = T 2 , v ,where v is the unique function of D(I ) associated with u via the relation (1.12). It is easy to show the linearityand continuity of T 1 and thus we have

T 1 , u = T 2 , u , u D(I ) ,

and thus we have proved that T 1 = T 2 .

Lemma 1.7 Consider T D (R d ) and suppose that x j T = 0 for all i = 1 , . . . , d . Then, T is a constant (function).

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