Chapter 2

Hilbert and Sobolev spaces

”The unified character of mathematics lies in its very nature; indeed,mathematics is the foundation of all exact natural sciences.”David Hilbert (1862-1943)

Hilbert spaces, named after the German mathematician D. Hilbert (1862-1943), are complete infinite-dimensional spaces in which distances and angles can be measured. These spaces have a major impactin analysis and topology and will provide a convenient and proper setting for the functional analysis ofpartial di!erential equations. The notion of smoothness of mathematical solutions is often related tothe notion of continuity. We will see that a more interesting and stronger notion of smoothness is thatof di!erentiability (derivatives must be continuous). Moreover, spaces like Ck or C! have limitationswhen dealing with the solutions of partial di!erential equations of mathematical physics. The notionof weak derivative has emerged from the works of J. Leray (1906-1998) and S. Sobolev (1908-1989) todeal with non-continuous functions for which the derivatives exist almost everywhere. In a certain sense,functions that belong to Sobolev spaces represent a good compromise as they have some, but not toogreat, smoothness properties [Evans, 2002].

Contents2.1 Hilbert spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.2 Sobolev spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.1 Hilbert spaces

In this chapter, we will consider vector spaces on K, with the convention that the identity u = u, ifK = R and that it denotes the usual conjugation operator if K = C. We recall some basic definitionsrelated to the inner product in a vector space.

2.1.1 Definitions

Definition 2.1 1. A mapping f : E ! F between two vector spaces on K is said to be antilinear (orsemilinear) if

f(!x + y) = !f(x) + f(y) , "(x, y) # E $ E ,"! # K .

17

18 Chapter 2. Hilbert and Sobolev spaces

2. A sesquilinear form is a map f : E $E ! K that is linear on the first argument and antilinear onthe second:

"! # K ,"(x, y, z) # E3 , f(!x + y, z) = !f(x, y) + f(y, z)f(z, !x + y) = !f(z, x) + f(z, y) .

3. A mapping f : E $ E ! K is called an inner product on E if and only if it is sesquilinear and isa positive-definite hermitian form satisfying the following axioms:

(i) "(x, y) # E2 , f(y, x) = f(x, y) (hermitian)(ii) "x # E , f(x, x) # R+ , (nonnegativity)

(iii) "x # E , f(x, x) = 0 % x = 0 , (definite)(2.1)

Remark 2.1 In general and from now, the notation &·, ·' will be used to denote the inner product 1.

A (complex) vector space with an inner product satisfying (i)-(iii) is sometimes called a pre-Hilbert space.Inner product spaces have a naturally defined norm.

Theorem 2.1 (Cauchy-Schwarz inequality) Consider a inner product &·, ·' on E. The followinginequality holds for all x, y # E:

|&x, y'|2 ( &x, x' · &y, y' (2.2)

Furthermore, if the equality |&x, y'|2 = &x, x' · &y, y' holds, the vectors x and y are colinear (linearlydependent).

Proof. For ! # K, x, y # E, we have

&x) !y, x) !y' = &x, x) !y' ) !&y, x) !y'= &x, x' ) !&x, y' ) !&x, y'+ !!&y, y'= &x, x' ) 2*e(!&x, y') + |!|2&y, y' .

There exists " # R+ and # # R such that &x, y' = " exp (i#). For t # R, we set ! = t exp (i#). Then, we have:

P (t) = &x, x' ) 2t" + t2&y, y' + 0 , "t # R .

The second-order polynomial P has a negative or null discriminant. Hence, we deduce

"! = "2 ) &x, x' · &y, y' ( 0or |&x, y'|2 ( &x, x' · &y, y' .

In case of equality, the discriminant "! = 0. Then, either y = 0 and thus y = 0 · x or P has a double roott0 = &x, y' exp ()i#)/&y, y' which means that P (t0) = ,x) t0 exp(i#)y,2 and thus x = t0 exp(i#)y. !

Theorem 2.2 If &·, ·' is a inner product on E, the mapping x -! ,x, =!&x, x' is a norm on E.

The norm ,x, is thought as the length of the vector x. It is well defined by the nonnegativity axiom ofthe definition.

1This introduction to Hilbert spaces is inspired by lecture notes on Analysis from several sources, notably B. Perthame,Topologie et analyse di!erentielle, ENS, Paris and J. Saint-Raymond, Topologie et calcul di!erentiel, UPMC, Paris.

2.1. Hilbert spaces 19

Proof. From the definition of the inner product, we have &x, x' = 0 if and only if x = 0 and &!x, !x' = |!|2&x, x'.In other words, ,x, = 0 % x = 0 and ,!x, = |!| ,x,. Considering x, y # E, we have

,x + y,2 = &x + y, x + y' = &x, x'+ &y, y'+ &x, y'+ &y, x'= ,x,2 + ,y,2 + 2*e(&x, y')

and using Cauchy-Schwarz inequality,

,x + y,2 ( ,x,2 + ,y,2 + 2|&x, y'| ( ,x,2 + ,y,2 + 2!&x, x' · &y, y'

( ,x,2 + ,y,2 + 2,x, · ,y, = (,x,+ ,y,)2 .

hence, ,x + y, ( ,x,+ ,y,. !

Definition 2.2 Two non zero vectors x, y of a pre-Hilbert space E are orthogonal if their product iszero: &x, y' = 0. We will then note x.y.

Proposition 2.1 If two vectors x, y # E are orthogonal, we have

,x + y,2 = ,x,2 + ,y,2 (Pythagorean theorem) .

Theorem 2.3 If E is a pre-Hilbert space, the inner product is continuous from E $ E in K.

Proof. For all a, b, x, y # E, we have

&x, y' ) &a, b' = &x, y ) b'+ &x) a, b' = &a, y ) b'+ &x) a, b'+ &x) a, y ) b' ,

and according to Cauchy-Schwarz inequality

|&x, y' ) &a, b'| ( ,a, · ,y ) b,+ ,b, · ,x) a,+ ,x) a, · ,y ) b,

and for $ > 0 given, it is su#cient to take ,x) a, < min"

1,$

3,b,

#and ,y) b, < min

"$

3,

$

3,a,)

#to ensure that

all three terms are lesser than $/3 and thus that |&x, y' ) &a, b'| ( $. This proves the continuity. !

Proposition 2.2 If E is a pre-Hilbert space, the inner product of two vectors x, y # E is given by:

&x, y' =14

$,x + y,2 + i,x + iy,2 ) ,x) y,2 ) i,x) iy,2

%.

Proof. We have, for k = 0, . . . , 3

,x + iky,2 = &x, x'+ &y, y'+ i"k&x, y'+ ik&y, x'ik,x + iky,2 = ik,x,2 + ik,y,2 + &x, y'+ ()1)k&y, x'

3&

k=0

ik,x + iky,2 = ,x,2(1 + i + i2 + i3) + ,y,2(1 + i + i2 + i3)

+ &x, y'(1 + 1 + 1 + 1) + &y, x'(1) 1 + 1) 1) = 4&x, y' .

and this proves the desired equality. !

Theorem 2.4 If x, y, z are three vectors of a pre-Hilbert space E and u =x + y

2denotes the midpoint

of the segment [x, y], we have

12

$,x,2 + ,y,2

%=

''''x + y

2

'''' +''''x) y

2

'''' (parallelogram law)

,z ) x,2 + ,z ) y,2 = 2,z ) u,2 +12,x) y,2 .

20 Chapter 2. Hilbert and Sobolev spaces

Definition 2.3 A pre-Hilbert space E is a Hilbert space if and only if it is a complete normed space(i.e. a Banach space) under the norm associated with the inner product.

Remark 2.2 Actually, the hypothesis of completness is weak, it is always possible to complete a spaceK endowed with a inner product. The ”completed” norm is then associated with the inner product.

Example 2.1 1. On Rd, the inner product is usually denoted x · y and is defined by:

x · y =d&

i=1

xi yi . (2.3)

2. The vector space %2(C) of the square integrable sequences:

%2(C) =

((xn)n"N # CN , s.t.

&

n"N|xn|2 < /

)

endowed with the inner product defined by:

&x, y' =!&

n=0

xnyn (2.4)

is a Hilbert space. It plays an important role as it is isomorphic to all separable Hilbert spaces.

Proof. If x is in %2 so will be !x. For a, b # C, we have |a + b|2 ( 2(|a|2 + |b|2), then for x, y # %2(C):

#&

n=0

|xn + yn|2 ( 2

* #&

n=0

|xn|2 +#&

n=0

|yn|2+

< +/ ,

hence, x + y # %2(C).The Cauchy-Schwarz inequality shows that, for every m # N:

m&

n=0

|xnyn| (*

m&

n=0

|xn|2+1/2 *

m&

n=0

|yn|2+1/2

(* #&

n=0

|xn|2+1/2 * #&

n=0

|yn|2+1/2

< +/ ,

thus yielding the convergence of the series of general term (xnyn) and allowing to define &x, y'. It is easy toshow that the formula (2.4) defines a inner product on %2 and that the associated norm is defined by:

,x, =

* #&

n=0

|xn|2+1/2

.

The completness of the space %2(C) is established via the convergence of Cauchy sequences and is left as anexercise to the reader. !

3. The integration theory gives a Hilbert space: the space L2($, dµ) of the square-measurable integrablefunctions, for the inner product:

&f, g' =,

!f(x)g(x)dµ(x) ,

for any measure dµ(x), for example the Lebesgue measure dx.

2.1. Hilbert spaces 21

2.1.2 Projection on a closed convex

We recall that a set C is said to be convex if, for all x, y # C and t # [0, 1], tx + (1) ty) # C.

Theorem 2.5 Suppose E is a pre-Hilbert space and A is a nonempty complete convex subset of E. Then,for all x # E, there exists a unique point y # A, called the orthogonal projection of x onto A, such that

d(x,A) = infz"A

,x) z, = ,x) y, .

This point is often denoted as PA(x) and is characterized by the property:

"z # A , *e(&x) y, z ) y') ( 0 .

Proof. If x # A, then y = x is the only point that minimizes the distance to x. Moreover, we have &x)x, z)x' = 0for every z # A. If for y # A we have *e(&x) y, z ) y') ( 0 for all z # A, then ,x) y,2 = *e(&x) y, x) y') ( 0,thus x = y.On the other hand, if x /# A, we denote & the distance from x to A. For every n # N, we pose

Cn = {z # A , ,z ) x,2 ( &2 + 2"2n} .

Hence, the sequence (Cn) is a decreasing sequence of nonempty closed subsets of the complete space A. It issu#cient to show that the diameter of Cn vanishes to conclude that the intersection of the (Cn) is a singleton {y}.Let z, w be two points in Cn. Since A is convex, u = z+w

2 is in A, hence we deduce that ,x ) u, + &. Accordingto theorem 2.4, we can write

2&2 +12,z ) w,2 ( 2,x) u,2 +

12,z ) w,2 = ,x) z,2 + ,x) w,2

( &2 + 2"2n + &2 + 2"2n

and we deduce that ,z ) w,2 ( 4.2"2n, i.e. ,z ) w, ( 21"n and finally that diam(Cn) ( 21"n. Now, we haveestablished the existence of a point y # A such that

-n$N Cn = {y}. The point y is the unique point in A such

that ,x) y, = d(x,A).Let consider z # A. Since A is convex, we have zt := y + t(z ) y) # A for all t # [0, 1]. Hence,

&2 ( ,x) zt,2 = &x) y ) t(z ) y), x) y ) t(z ) y)'= ,x) y,2 + t2,z ) y,2 ) 2t*e(&x) y, z ) y')= &2 + t2,z ) y,2 ) 2t*e(&x) y, z ) y')

We deduce that 2*e(&x) y, z ) y') ( t,z ) y,2, and since t can be considered as arbitrarily small, we obtain that*e(&x) y, z ) y') ( 0.If a point y! # A satisfies this relation for every z # A, we have for z = y,

0 + *e(&x) y!, y ) y!') = *e(&x) y, y ) y!') + *e(&y ) y!, y ) y!')= ,y ) y!,2 )*e(&x) y, y! ) y') + ,y ) y!,2

since y is the projection of x onto A. This leads to conclude that ,y ) y!, = 0 thus y = y!. !The projection PA is continuous and 1-Lipschitz continuous as stated now.

Proposition 2.3 Consider a nonempty complete convex subset A of E. Then, For all x, y # E, we have

,PA(x)) PA(y), ( ,x) y, .

22 Chapter 2. Hilbert and Sobolev spaces

y

x

z

A

Figure 2.1: Projection onto a convex closed subset.

Proof. Let denote a = PA(x) and b = PA(y). The characterization of the projection (Theorem 2.5) allows towrite

0 ( &a) x, b) a' 0 ( &b) y, a) b' .

By adding these relations, we find

0 ( &a) x) b + y, b) a' = ),b) a,2 + &y ) x, b) a' ,

hence,b) a,2 ( &y ) x, b) a' ( ,x) y, ,b) a, ,

and the result follows. !

Proposition 2.4 If A is a subset of a pre-Hilbert space E, the orthogonal of A, denoted A# is the setof vectors x # E orthogonal to each element of A:

A#def= {x # E , s.t. "y # A , x.y} .

The set A# is a closed subspace of E.

Proof. For every y # A, y% = {x, &x, y' = 0} is the kernel of the continuous linear form: x -! &x, y', thus it is aclosed subspace of E. This yields to conclude that:

A% =.

y$A

y%

is a closed subspace of E. !

Theorem 2.6 If F is a closed subspace of the Hilbert space E, there exists for every x # E a uniquecouple (y, z) with y # F , z # F# such that x = y + z. We have then ,y, ( ,x, and ,z, ( ,x,.

Proposition 2.5 Let F be a closed subspace of the Hilbert space E. Then, F## = F .

Theorem 2.7 Let X be a subspace of the Hilbert space E. Then, X is dense if and only if X# = {0}.

2.1. Hilbert spaces 23

Proof. If X is dense and if y # X%, we have X 0 y%. Since y% is closed, it contains the closure of X, i.e.E; in particular y # y%, thus y = 0. This shows that X% = {0}. Conversely, if X is not dense, its closure is aclosed vector subspace F distinct from E and every vector z # F% is in X%. Having F% = {0} would imply thatF = F% = {0} = E, wich is in contradiction with the assumption. Hence, X% 1= {0}. !

2.1.3 Hilbert bases, Bessel-Parseval identities

Recall that a subspace E is separable if it contains a countable dense subset in E, i.e. a set with acountable number of elements whose closure is the space itself.

Definition 2.4 Consider E a separable Hilbert space of infinite dimension. A Hilbert basis or orthonor-mal basis is a sequence (en)n"N of elements of E that is complete (i.e. the linear span of (en)n"N is densein E) and such that:

&ei, ej' = &i,j , with &i,j the Kronecker symbol . (2.5)

Remark 2.3 A Hilbert basis is not an algebraic basis: the elements are linearly independent but do notform a spanning set of E.

Example 2.2 Consider the space %2(N). The sequence (en)n"N of elements of %2(N) defined by en(k) =&n,k is an Hilbert basis of %2(N).

Theorem 2.8 In any separable Hilbert space E, there exists Hilbert bases.

Proof. Consider a complete countable set (an)n$N. We can assume that the family (an) is linearly independent.We use the Gram-Schmidt orthogonalization procedure. To this end, we pose

e1 =a1

,a1,.

Suppose the terms (ej)1&j&n satisfying the relation (2.5) have been defined and that

Span{a1, . . . , an} = Span{e1, . . . , en} .

We pose

en+1 =fn+1

,fn+1,with fn+1 = an+1 )

n&

j=1

&an+1, ej'ej .

The relation (2.5) is immediately satisfied and the result follows. !

Definition 2.5 Let (En)n"N be a countable set of closed subspaces of E. The space E is said to be theHilbertian sum of (en) and we denote E =

/n"N EN if:

(i) the elements ei of the set (en) are mutually orthogonal:

&x, y' = 0 , "x # en , v # em , n 1= m ,

(ii) the vector space generated by the span of (en) is dense in E.

Theorem 2.9 Let E be a Hilbertian sum of (en)n"N. For any x # E, we denote xn = Pen(x) theorthogonal projection of x onto en. Then, we have

(i) x =&

n"Nxn (convergent series),

24 Chapter 2. Hilbert and Sobolev spaces

(ii) moreover, ,x,2 =&

n"N,xn,2, (Parseval identity).

Conversely, given a sequence (xn)n"N # En such that&

n"N,xn,2 < /, then the series x =

&

n"Nxn is

convergent and xn = PEn(x) for all n + 1.

Remark 2.4 This result is fundamental, for at least two reasons:

(i) it indicates that every separable Hilbert space with a Hilbert basis (en)n"N is isomorphic and iso-metric to the space %2(N): there exists a bijective linear map ' : E ! %2(N), x -! (&x, en')n"N;

(ii) it gives also a practical way of getting the best approximation in a finite dimensional subspace byusing an orthonormal basis. This will reveal extremely useful in (least-squares) polynomial approx-imations, orthogonal polynomials, Fourier series, etc.

2.1.4 Duality and Riesz theorem

Consider a Hilbert space E on K. Recall that a continuous linear form on E is a linear mappingf : E ! K, continuous on E.

Definition 2.6 The vector space of all continuous linear forms on E is called the topological dual of Eand is denoted E$.

Theorem 2.10 (Riesz-Frechet) For every continuous linear form f on the Hilbert space E, there existsa unique y # E such that for all x # E the following identity holds f(x) = &x, y'. Moreover, we have,y, = ,f,.The mapping from E to E$ that associates to y the linear form x -! &x, y' is an antilinear isomorphismfrom E to E$ (i.e. E and E$ are isometrically anti-isomorphic).

Proof. Let denote fy the linear form x -! &x, y', for every y # E. According to the definition of the innerproduct, the mapping y -! fy is antilinear from E to E!. Cauchy-Schwarz inequality allows to write

|fy(x)| = |&x, y'| ( ,x, · ,y,

thus ,fy, ( ,y,. Moreover, since fy(y) = &y, y' = ,y,2, we have ,fy, + ,y,. Hence, ,fy, = ,y,.We have to show that every continuous linear form f on E is of the form fy for a certain y. If f = 0, the trivialsetting y = 0 is appropriate. If f 1= 0, the kernel H of f is a closed hyperplane of E. Let a /# H. We pose ( = f(a)and if we denote b the orthogonal projection of a onto H, the vector y = !

'a"b'2 (a ) b) is orthogonal to H sincea) b is. In particular, &b, a) b' = 0, this &a, a) b' = &a) b, a) b' = ,a) b,2. Moreover,

&a, y' =(

,a) b,2 &a, a) b' = ( .

This yields fy(a) = f(a). Hence, f and fy coincide on a and on K since they both vanishes. Since every vector ofE can be written as the sum of a vector of H and of a vector colinear to a, f and fy coincide on E: f = fy. !

Remark 2.5 The identification of E and E$ may be erroneous and misleading, when considering theinner product of E = L2. For example, let consider a dense subspace X = L2 (R; (1 + |t|) dt) 0 L2(R),endowed with the inner product:

&x, y' =,

R(1 + |t|)x(t)y(t) dt .

A linear form f # L2(R)$ is also an element of X $. The identification of f to an element f $ # L2(R) doesnot define a linear form on X and the identity f(y) = &f $, y' makes no sense on E. Here, we must write

X 0 E = E$ 0 X $ .

2.1. Hilbert spaces 25

Definition 2.7 Let (xn)n"N be a sequence of elements in a Hilbert space E and consider a point x # E.The sequence (xn)n"N is said to converge weakly to x and we note (xn)n"N ) x if and only if

"y # E , limn%!

&y, xn' = &y, x' .

Any weakly convergent sequence is bounded. In addition, if (xn)n"N ) x then

,x, ( lim infn%!

,xn, .

2.1.5 Lax-Milgram and Stampacchia theorems

We consider a Hilbert space E and we will give two important results for bilinear forms on Hilbert spaces.

Definition 2.8 Let a(·, ·) : E $ E ! R be a bilinear form. It is said to be

(i) continuous if there exists a constant C > 0 such that

|a(x, y)| ( C ,x, ,y, , "x, y # E ,

(ii) (-elliptic (or coercive) if there exists a constant ( > 0 such that

a(x, x) + (,x,2 , "x # E .

Theorem 2.11 (Lax-Milgram) Suppose a(·, ·) : E$E ! R is a continuous and coercive bilinear form.For every linear form l # E$, there exists a unique x # E such that:

a(x, y) = l(y) , "y # E .

Moreover, if a(·, ·) is symmetric, then x # E is characterized by:

12a(x, x)) l(x) = min

y"E

"12a(y, y)) l(y)

#.

Proof.

(i) (Existence and uniqueness of x) Riesz-Frechet Theorem 2.10 allows to define a linear mapping A : E ! Eby:

a(x, y) = &A(x), y' , and ,A(x), ( C,x, .

Similarly, we associate with l # E! an element f # E as l(y) = &f, y'. Hence, we have to solve A(x) = f or

x) rA(x) + rf = x , in E .

It is su#cient to prove that the mapping T : E ! E

T (x) = x) rA(x) + rf

is a strict contraction for r > 0 su#ciently small. This can be achieved using the Banach fixed-point theoremand is left as an exercise. From the existence and uniqueness of a fixed point of T we deduce the solution xto the problem.

26 Chapter 2. Hilbert and Sobolev spaces

(ii) (variational principle) If the bilinear form a(·, ·) is symmetric, we write:

12a(x, x)) 1

2a(y, y)) l(x) y) =

12a(x, x)) 1

2a(y, y)) a(x, x) y)

= )12a(x, x)) 1

2a(y, y) + a(x, y)

= )12a(x) y, x) y) ( 0 .

!

Corollary 2.1 Let A # L(E,E) (i.e. a continuous linear mapping) such that &A(x), x' + *,x,2, for allx # E with * > 0. Then, A # Isom(E,E).

An extension of Lax-Milgram Theorem is given by the following theorem.

Theorem 2.12 (Stampacchia) Under the assumptions of Theorem 2.11, let C be a closed nonemptyconvex set. Then, for all l # E$, there exists a unique x # C such that

a(x, y ) x) + l(y ) x) , "y # C .

Moreover, if the bilinear form a(·, ·) is symmetric, then x is characterized by

x # C ,12a(x, x)) l(x) = min

y"C

"12a(x, y)) l(y)

#.

As we will see in Chapter 3, Lax-Milgram Theorem has an interesting application in the resolution ofpartial di!erential equations of the type:

01112

1113

)"u(x) = )d&

i=1

+2u(x)+x2

i

= f(x) # L2($) , "x # $ ,

u(x) = 0 "x # +$ ,

where $ denotes a open subset of Rd and +$ its boundary. Indeed, this problem can be reduced tosolving:

a(u, v) =,

!2u(x) ·2v(x) dx =

,

!f(x)v(x) dx , "v # E ,

and the di#culty is then to select the appropriate Hilbert space E. And this is where and why Sobolevspaces appeared.

2.2 Sobolev spaces

We are considering functions for which all the derivatives, in the distributional sense, belongs to L2 spaceand such that most of the classical derivation rules can be applied. Before turning to Sobolev spaces, weintroduce Holder spaces.

2.2. Sobolev spaces 27

2.2.1 Holder spaces

Suppose $ 0 Rd is an open set and 0 < , ( 1. Recall that k-Lipschitz continuous functions f : $ ! Rsatisfy by definition the following estimate:

|f(x)) f(y)| ( k,x) y, , "x, y # $ , k # R+ .

This relation provides a uniform modulus of continuity. It is often useful to consider functions f satisfying

|f(x)) f(y)| ( k,x) y,! , "x, y # $ .

Such functions are said to be Holder continuous with exponent , # R+.

Definition 2.9 Consider f : $ ! R.

(i) if f is bounded and continuous, we write

,f,C0(!)def= sup

x"!|f(x)| .

(ii) the ,th-Holder seminorm of f is:

[f ]C0,!(!)def= sup

x,y"!x &=y

"|f(x)) f(y)|,x) y,!

#.

and the ,th-Holder norm of f is:

,f,C0,!(!)def= ,f,C0(!) + [f ]C0,!(!) .

Definition 2.10 The Holder space Ck,!($) consists of all functions f # Ck($) for which the norm

,f,Ck,!(!)def=

&

|"|'k

,+"f,C0(!) +&

|"|=k

[+"f ]C0,!(!) ,

is finite.

Theorem 2.13 The space of functions Ck,!($) is a Banach space.

2.2.2 Sobolev spaces of integer order

Let $ 0 Rd and let v be a function of L2($), it can be identified to a distribution on $ as a functionof L1

loc($), also denoted as v and we can define its derivatives +xiv, 1 ( i ( n as distributions on $. Ingeneral, +xiv is not an element of L2($). Hence, we introduce

Definition 2.11 We call Sobolev space of order 1 on $ the space

H1($) =4v # L2($) , +xiv # L2($) , 1 ( i ( d

5.

The space H1 is endowed with the norm associated to the inner product:

&u, v'1,! =,

!(uv +

d&

i=1

+xiu +xiv) dx ,

and we note the corresponding norm:

,v,1,! =6&v, v'1,! =

",

!|u|2 dx +

,

!|2u|2 dx

#1/2

.

28 Chapter 2. Hilbert and Sobolev spaces

And we have the generalization of such spaces.

Definition 2.12 Let m # N. A function v # L2($) belongs to the Sobolev space of order m, denotedHm($), if all the derivatives of v up to order m, in the distributional sense, belong to L2($). Byconvention, we note H0($) = L2($).

Theorem 2.14 The spaces Hm($), m + 0 endowed with the following inner product are Hilbert spaces:

&u, v'm,! =&

|"|'m

,

!+"u(x) +"v(x) dx , (2.6)

with the associated norm,u,m,! =

7 &

|"|'m

,+"u,2L2(!) .

Proof. The expression (2.6) defines an inner product. It is su#cient to show that the space is complete for theassociated norm. Let consider a Cauchy sequence (un)n$N for this norm. We have

limj,k(#

&

|!|&m

,+!uj ) +!uk,2L2(!) = 0 ,

which means that, for each ( such that |(| ( m, the sequence (+!uj) is a Cauchy sequence in L2($). Since L2($)is complete, there exists v! such that +!uj ! v! in L2($). Since the convergence in L2 implies the convergencein the distributional sense, we have uj ! v0. Hence, we have +!uj ! +!u0, in the distributional sense and thus+!v0 = v! # L2($) and consequently v0 # Hm($). Moreover, we have

,v0 ) uj,2m,! =&

|!|&m

,v! ) +!uj,2L2(!) ! 0 ,

and thus uj ! v0 for the Hm($) norm. !

Definition 2.13 More generally, we can define for every 1 ( p ( / and for every m # N, m + 1, theSobolev spaces as:

Wm,p =8

u # Lp($) , +"u # Lp($) ,"( # Nd , |(| ( m9

,

endowed with the norm:

,u,W 1,p =

:

;&

|"|'m

,

!|+"u|p dx

<

=1/p

.

All these spaces are Banach spaces. However, we will consider here the spaces Wm,2($) = Hm($).

Remark 2.6 The inclusion Hm($) 0 L2($) is continuous. Moreover, since D($) is dense in L2($) (cf.Chapter 1) and D($) 0 Hm($), we have that Hm($) is dense in L2($).

Example 2.3 To apprehend the concept of regularity for function in Hm($), we consider the followingexample2. Consider, for every ( # R the function defined by:

u"(x) =>

x" exp ()x) x > 0)()x)" exp(x) x ( 0

According to the definition of Sobolev spaces, we can establish that2This example is taken from the lecture notes of A. Munnier, Espaces de Sobolev et introduction aux EDP, Institut ELie

Cartan, Nancy (2006).

2.2. Sobolev spaces 29

22 Chap. 2: Les espaces de Sobolev

ce qui conclut la demonstration. !

Si l’on a en general une bonne intuition de ce qu’est la regularite d’une fonction u declasse Ck, il est moins aise de bien comprendre ce que signifie u ! H1(R) et encoremoins u ! H1/2(R) ou bien u ! H!1/2(R). Afin d’illustrer ce nouveau concept deregularite, donnons une nouvelle serie d’exemples :

Exemple 2.2 On considere, pour tout ! ! R, la fonction impaire :

u!(x) =

!x!e!x si x > 0,

"("x)!ex si x # 0.

Les graphes de ces fonctions sont donnes ci-dessous pour certaines valeurs de !.Remarquer que la fonction u! est d’autant plus reguliere en x = 0 que ! est grand.Par exemple, la fonction u! est continue en 0 si et seulement si ! > 0.

4

2

2

1

0

-2

0

-4

-1-2

Fig. 2.1 – Graphe de la fonction u! avec ! = "1/2. La fonction presente uneasymptote verticale en x = 0.

0

1

2

0,5

-2

0

-0,5

-4

-1

4

Fig. 2.2 – Graphe de la fonction u! avec ! = 0. La fonction est discontinue en 0mais sans asymptote.

22 Chap. 2: Les espaces de Sobolev

ce qui conclut la demonstration. !

Si l’on a en general une bonne intuition de ce qu’est la regularite d’une fonction u declasse Ck, il est moins aise de bien comprendre ce que signifie u ! H1(R) et encoremoins u ! H1/2(R) ou bien u ! H!1/2(R). Afin d’illustrer ce nouveau concept deregularite, donnons une nouvelle serie d’exemples :

Exemple 2.2 On considere, pour tout ! ! R, la fonction impaire :

u!(x) =

!x!e!x si x > 0,

"("x)!ex si x # 0.

Les graphes de ces fonctions sont donnes ci-dessous pour certaines valeurs de !.Remarquer que la fonction u! est d’autant plus reguliere en x = 0 que ! est grand.Par exemple, la fonction u! est continue en 0 si et seulement si ! > 0.

4

2

2

1

0

-2

0

-4

-1-2

Fig. 2.1 – Graphe de la fonction u! avec ! = "1/2. La fonction presente uneasymptote verticale en x = 0.

0

1

2

0,5

-2

0

-0,5

-4

-1

4

Fig. 2.2 – Graphe de la fonction u! avec ! = 0. La fonction est discontinue en 0mais sans asymptote.

( = )1/2 ( = 02.1 Definitions et premieres proprietes 23

0,2

00

-0,2

-2

-0,4

-4 2

0,4

4

Fig. 2.3 – Graphe de la fonction u! avec ! = 1/2. La fonction est continue en x = 0mais non derivable.

0

-0,2

-2

0,2

-0,1

0,1

-4

-0,3

2

0

4

0,3

Fig. 2.4 – Graphe de la fonction u! avec ! = 1. La fonction est derivable en x = 0.

Par un calcul assez simple et en utilisant la definition (D1) des espaces de So-bolev, on etablit que

– Si ! > m (m ! N) alors u! ! Cm(R).– Si ! > m " 1/2 (m ! N) alors u! ! Hm(R). En particulier si 1/2 < ! <

1, u! ! H1(R) et u! ! C(R) mais u!!/ C1(R). En utilisant maintenant ladefinition (D2) des espaces de Sobolev on peut donner un resultat plus precisfaisant intervenir la transformee de Fourier de u!. On montre dans l’annexe B,paragraphe B.1 que, pour tout ! > "1 :

!u!(") = "i

#2##

!(! + 1)(1 + "2)(!+1)/2

sin((! + 1) arctan "), $ " ! R, (2.1)

ou ! designe la fonction Gamma d’Euler. On en deduit l’equivalence suivante,pour tout ! > "1 et s ! R :

u! ! Hs(R) % s < ! +12.

La definition des espaces de Sobolev avec la transformee de Fourier nous permetegalement des a present de resoudre certaines edp dans les espaces Hs. Consideronsl’exercice suivant :

Exercice 2.1 En utilisant la transformee de Fourier, montrer que pour tout T !S !(RN ), l’edp suivante :

""u + u = T dans S !(RN ),

2.1 Definitions et premieres proprietes 23

0,2

00

-0,2

-2

-0,4

-4 2

0,4

4

Fig. 2.3 – Graphe de la fonction u! avec ! = 1/2. La fonction est continue en x = 0mais non derivable.

0

-0,2

-2

0,2

-0,1

0,1

-4

-0,3

2

0

4

0,3

Fig. 2.4 – Graphe de la fonction u! avec ! = 1. La fonction est derivable en x = 0.

Par un calcul assez simple et en utilisant la definition (D1) des espaces de So-bolev, on etablit que

– Si ! > m (m ! N) alors u! ! Cm(R).– Si ! > m " 1/2 (m ! N) alors u! ! Hm(R). En particulier si 1/2 < ! <

1, u! ! H1(R) et u! ! C(R) mais u!!/ C1(R). En utilisant maintenant ladefinition (D2) des espaces de Sobolev on peut donner un resultat plus precisfaisant intervenir la transformee de Fourier de u!. On montre dans l’annexe B,paragraphe B.1 que, pour tout ! > "1 :

!u!(") = "i

#2##

!(! + 1)(1 + "2)(!+1)/2

sin((! + 1) arctan "), $ " ! R, (2.1)

ou ! designe la fonction Gamma d’Euler. On en deduit l’equivalence suivante,pour tout ! > "1 et s ! R :

u! ! Hs(R) % s < ! +12.

La definition des espaces de Sobolev avec la transformee de Fourier nous permetegalement des a present de resoudre certaines edp dans les espaces Hs. Consideronsl’exercice suivant :

Exercice 2.1 En utilisant la transformee de Fourier, montrer que pour tout T !S !(RN ), l’edp suivante :

""u + u = T dans S !(RN ),

( = 1/2 ( = 1

Figure 2.2: Graphs of the function u" for di!erent values of the parameter (.

• if ( > m, m # N then u" # Cm(R)

• if ( > m) 12, m # N, then u" # Hm(R);

if12

< ( < 1, u" # H1(R) and u" # C0(R) but u" /# C1(R).

The figure 2.2 shows the graphs of this function for di!erent values of (.

Theorem 2.15 Suppose $ is an open set of class C1 and its boundary +$ is bounded (or $ = Rd+).

Then, for every m # N, there exists a linear extension operator P : Hm(Omega) ! Hm(Rd) such thatfor every u # Hm($) we have:

1. Pu|! = u,

2. ,Pu,Hm(Rd) ( C ,u,Hm(!), where C depends upon $ only.

Proof. The proof is established in the particular case where N = 1, m = 1 and $ is a subset of R+. Let consider$ =]0,+/[ and # H1($). We define

(Pu)(x) = u)(x) =>

u(x) x + 0u()x) x < 0

and we pose

v(x) =>

u!(x) x > 0)u!()x) x < 0 .

It is easy to see that v # L2(R). To conclude, we have simply to show that (Pu)! = v in D!(R). Indeed, underthis assumption, we deduce that Pu # H1(R) and that ,Pu,H1(R) ( 2,u,H1(!). To prove that the hypothesis(Pu)! = v in D!(R) holds, we introduce the sequence (-k) of functions of C#(R) defined for every k # N) by

-k(t) = -(kt) , t # R , k # N) ,

30 Chapter 2. Hilbert and Sobolev spaces

where - # C#(R) is a given function such that (Figure 2.3):

-(t) =>

0 t < 12

1 t > 1

Let consider v # D(R). We have

B.1 Demonstrations des resultats du chapitre 2 67

Il su!t ensuite de remarquer que

(1 + i!)!+1 =!"

1 + !2ei arctan "#!+1

= (1 + !2)(!+1)/2ei(!+1) arctan ",

pour obtenir que :

$u!(!) = !i

"2""

"(# + 1)(1 + !2)(!+1)/2

sin((# + 1) arctan !).

!

Demonstration de la proposition 2.5 : Supposons dans un premier temps que# = Q+ (ou # = RN

+ , le raisonnement etant exactement le meme). On prolongealors u par reflexion sur Q tout entier, c’est a dire que l’on pose :

u!(x", xN ) :=

%u(x", xN ) si xN > 0,

u(x",!xN ) si xN < 0,

ou x = (x", xN ) # RN#1$R. Remarquer que u! etant une fonction de L2(Q), il su!tde la definir presque partout sur Q (en particulier, on n’a pas besoin de connaıtreu! sur Q0 qui est de mesure nulle). Soit $ # D(Q). On a, pour tout i = 1, . . . , N !1et en posant d(x", xN ) := dx = dx" % dxN :

&

Qu!(x)

%$

%xi(x) dx =

&

Q+

u(x", xN )%$

%xi(x", xN )d(x", xN ) +

&

Q!

u(x",!xN )%$

%xi(x", xN )d(x", xN ) =

&

Q+

u(x", xN )'

%$

%xi(x", xN ) +

%$

%xi(x",!xN )

(d(x", xN ).

Posant &(x", xN ) := $(x", xN ) + $(x",!xN ), on obtient que :&

Qu!(x)

%$

%xi(x) dx =

&

Q+

u(x)%&

%xi(x) dx, & i = 1, . . . , N ! 1.

Remarquer que &, bien que C$, n’est pas a support compact dans Q+ et ne peutdonc pas etre utilisee comme fonction test. Soit ' et 'k les fonctions de C$(R)definies pour tout k # N! par :

'(t) :=

%0 si t < 1/2,

1 si t > 1,

et 'k(t) := '(kt).

0,80,6

0,4

0,20

0,8

1,41,21

0,6

0,4

1

0,2

0

Fig. B.2 – Graphe de la fonction '.Figure 2.3: Graph of the - function.

, +#

0u)(x)v!(x) dx =

, +#

0u(x).!(x) dx ,

where .(x) = v(x)) v()x). Since -k(x) # D(0,/) we have, +#

0u(-k.)! dx =

, +#

0u!(-k.) dx . (2.7)

However, we have the inequalities (-k.)!(x) = -k(x).!(x) + k-!(kx).(x) and????, +#

0ku(x)-!(kx).(x) dx

???? ( kMC

, 1/k

0x|u(x)| dx ( MC

, 1/k

0|u(x)| dx ,

with the constants C = supt$[0,1] |-((t)| and .(x) ( M |x|, hence

limk(#

????, +#

0ku(x)-!(kx) dx

???? = 0 .

Thus, we can deduce from the previous relations:, +#

0u.! dx = )

, +#

0u!./, dx = )

,

Ru!v dx .

From the Relation (2.7) and the previous relation, we deduce that (Pu)! = v in D!(R). Now, we consider a boundedinterval, say I =]0, 1[. We define - # C#(R) such that -(x) # [0, 1], for all x # R and

-(x) =>

1 x < 14

0 x > 34

.

If u # H1(I), we define u :]0,+/[! R as:

u(x) =>

u(x) 0 < x < 10 x > 1 .

2.2. Sobolev spaces 31

Thus, -u # H1(]O,+/[) and (-u)! = -!u + -u!. The function u # H1(I) can be written as

u = -u + (1) -)u .

At first, the function -u is extended to ]0,+/[ to the function -u # H1(]0,+/[) and then extended to R byreflection. In this manner, we obtain a function v1 # H1(R) that extends -u and such that

,v1,L2(R) ( 2 ,u,L2(I) , ,v1,H1(R) ( C ,u,H1(I) ,

where the constant C depends on ,-!,L! . Similarly, we extend the function (1)-)u to obtain a function v2 # H1(R)such that

,v2,L2(R) ( 2 ,u,L2(I) , ,v2,H1(R) ( C ,u,H1(I) ,

Finally, Pu = v1 + v2 gives the result. !We can show that this result still holds if the boundary +$ is only piecewise C1 continuous. Nonetheless,the following counter-example shows that for some open subsets $, smooth functions on $ cannot beextended in a regular manner.

Example 2.4 Consider the domain $ defined as ]) 1, 2[$]) 1, 1[ without the segment ]0, 2[$0. On $,we consider the function u defined as:

u(x, y) =

02

3

exp()1/x2) x > 0, y > 0)exp()1/x2) x > 0, y < 00 in the other cases

We can show that u # C!($) and +"u # L!($) for all ( # N2. Hence, this yields u # Hm($) for evereym # N. However, the function u cannot be extended to a regular function on all R2.

The previous result has been extended to Sobolev spaces Wm,p.

Corollary 2.2 (Stein) Consider a Lipschitz domain $. There exists a linear extension operator Pbounded of Wm,p($) in Wm,p(Rd) such that for all u # Wm,p($):

1. Pu|! = u,

2. ,Pu,W m,p(Rd) ( C,u,W m,p(!), where C depends on $.

Using the extension operator introduced above, we deduce the following result.

Lemma 2.1 If $ is of class C1 with +$ bounded (or if $ = Rd+), then for every u # H1($) there exists

a sequence of functions (un)n"N 0 D(Rd) such that

,un|! ) u,H1(!) ! 0 , n !/ .

This lemma states that only the restrictions of the functions of D(Rd) are dense in H1($) (and not thefunctions of D($)). Actually, in general D($) is not dense in H1($).

Theorem 2.16 (Meyer-Serrin) Let $ be an open subset of Rd. Then, the space C!($) is dense inHm($) for every m # N.

Definition 2.14 We denote by H10 ($) the closure of D($) in H1($). By extension, we note Hm

0 ($) theclosure of D($) in Hm($) (for the norm , · ,Hm(Rd)).

Theorem 2.17 The space D(Rd) is dense in H1(Rd), i.e.

H10 (Rd) = H1(Rd) .

32 Chapter 2. Hilbert and Sobolev spaces

Proof. The proof is obtained by truncation and regularization. It can be found in [Raviart-Thomas, 1998]. !

Theorem 2.18 If v # H10 ($), the function v, extension of v by 0 in Rd\$ is a function of H1(Rd).

Proof. If v # D($), the function v, extension of v by 0 in Rd\$, belongs to D(Rd). Moreover, we have

,v,1,Rd = ,v,1,! .

Hence, the mapping v -! v is continuous from D($), endowed with the norm induced by , · ,1,!, in H1(Rd). Itcan be extended by continuity into a linear mapping v -! v from H1

0 ($) to H1(Rd) and we have

v =>

v a.e. in $0 a.e. in Rd\$

The result follows. !

Definition 2.15 For k # N, the Sobolev spaces W(k,p($) are defined as dual spaces (W(k,q0 ($))$, where

q is conjugate to p: 1p + 1

q = 1. Their elements are distributions:

W(k,p($) =

02

3u # D$($) , u =&

|"|'k

+"u" , for some u" # Lp($)

@A

B .

Naturally, W(k,p($) is a Banach space with the norm:

,u,W!k,p(!) = supv"W k,q(!))v)

Wk,q(!)&=0

|&u, v'|,v,W k,q(!)

.

For any integer k, +" is a bounded operator from W k,p to W k(|"|,p.

2.2.3 Poincare’s inequality

Poincare’s inequality is a fundamental tool for solving partial di!erential equations.

Theorem 2.19 (Poincare’s inequality) If $ is bounded, there exists a constant C = C($) > 0 suchthat the following inequality holds:

"v # H10 ($) , ,v,0,! ( C($)

:

;d&

j=1

,+xjv,20,!

<

=1/2

. (2.8)

Proof. Let consider a function v # D($) and let v be the extension of v by 0 outside $. We verify easily thatv # D(Rd). Moreover, the support of v in included in $ that is bounded in at least one direction, say xd. Thus,we have

Supp(v) 0 $ 0 {x = (x!, xd) , x! # Rd"1 , a < xd < b} .

We write, since v(·, a) = 0:

v(x!, xd) =, xd

a+xd v(x!, t) dt , a ( xd ( b .

Cauchy-Schwarz inequality leads us to write

|v(x!, xd)|2 ( (xd ) a), xd

a|+xd v(x!, t)|2 dt ( (b) a)

, b

a|+xd v(x!, t)|2 dt .

2.2. Sobolev spaces 33

Integrating with respect to the variable x! in Rd"1 yields, +#

"#|v(x!, xd)|2 dx! ( (b) a),+xd v,2L2(Rd) .

Integrating with respect to the variable xd between a and b leads to

,v,2L2(!) ( (b) a)2,+xdv,2L2(!) ( (b) a)2|v|21,! ,

and the result follows with C = (b) a). !Poincare’s inequality is also known under the following form:

Proposition 2.6 Consider an open subset $ bounded in one direction. Then, there exists a constantC = C($) > 0 (depending only upon $) such that:

"v # H10 ($) , ,v,L2(!) ( C($) ,2v,L2(!) .

Corollary 2.3 Consider an open subset of $ 0 Rd, bounded in one direction. Then,

,u,H10

= ,2u,L2

defines a norm on H10 ($) that is equivalent to the usual norm of H1($).

2.2.4 Trace of a function

When a function u # L2($), where $ 0 Rd, it is not possible to consider its restriction to a zero-measureset since the functions in L2($) are defined except to a zero-measure set. However, the functions inSobolev spaces are more regular than L2 functions. We will now consider the restriction of the functionto +$, called the trace of the function on the boundary of the domain.

Let consider first the half-space

$ = Rd+ = {(x$, xd) # Rd(1 $ R , xd > 0}

for which we have+$ = {(x$, 0) , x$ # Rd(1} ,

that will be identified to Rd(1.

Theorem 2.20 (Trace) There exists a continuous linear map, called the trace operator and denoted:

,0 : H1(Rd+) ! H1/2(Rd(1) ,

that extends the usual restriction mapping of the continuous functions. This mapping is surjective andits kernel is Ker(,0) = H1

0 (Rd+).

Hence, functions in H10 (Rd

+) vanishe on the boundary of the domain. then, we extend this notion toLipschitz domains.

Theorem 2.21 Consider $ an open set of class C1. Then, there exists a ontinuous linear operator,called trace operator and denoted ,0 of H1($) in L2(+$) that coincide with the usual restriction operatorfor continuous functions. Its kernel is Ker(,0) = H1

0 ($).

34 Chapter 2. Hilbert and Sobolev spaces

Definition 2.16 Consider an open subset of class Cm, m # N* and ,0 the trace operator introduced inthe previous theorem. We define the space:

Hm(1/2(+$) = ,0(Hm($)) ,

endowed with the norm:,u,Hm!1/2(#!) = inf

v"!!1O ({u})

,v,Hm(!) .

We denote by H(1($) the dual space of H10 ($) endowed with the norm

,u,H!1 = sup)v)H1

0=1&u, v'H!1,H1

0.

Hence, a function u # L2($) can be identified with an element of H(1($) by using the linear form definedby the inner product in L2:

,u,H!1 = sup)v)H1

0=1

,

!uv .

We have then the following inclusion rule

H10 ($) 0 L2($) 0 H(1($) .

Remark 2.7 (i) The trace operator is not defined for L2($) functions.

(ii) Thanks to the density of D($) in H1($), we can establish the Green formula for the functions ofH1($):

"u, v # H1($) ,

,

!u +xiv dx = )

,

!v +xiu dx +

,

#!uvni d/ , (2.9)

where ni denotes the ith component of n.

We have defined H(1/2(+$) as the dual space of H1/2(+$). This space plays a role in the definition of

the trace of the normal derivative. Let denote ,1 the linear operator that asociates to u the function+u

+ndefined on +$:

"u # H2($) , ,1u = 2u · n .

We show that ,Au # L2(+$) for any function u # H2($) and we have the Green formula for the Laplacian:

"u # H2($) ,"v # H1($),

!"uv dx = )

,

!2u ·2v dx +

,

#!

+u

+nv d/ , (2.10)

where+u

+n=

d&

i=1

,0(+xiu)ni .

And we close this section by enouncing the following result without giving a formal proof.

Theorem 2.22 Suppose u # H1($) such that "u # L2($). Then, ,1u defines an element of H(1/2($).