functional analysis week07 v2

Week 7: Distributions & Sobolev spacesDocument prepared by Anna Rozanova-Pierrat1

1 Lecture 7.2: Distributions

1.1 Topology on D()

Let be an open domain of Rn. We recall the notation D() = C0 ().

Definition 1 Let n N and = (1, 2, . . . , n) Nn. Then is called a multi-index. In

addition, we write|| = 1 + . . .+ n,

and for D()

=||

x11 x22 . . . x

nn

.

Lets equip D() with this topology:

Definition 2 V is an open set in D() iff it can be presented as an (finite or countable) unionjJUj of sets Uj of the following type:

Uj D() such that

for all K ( compact (in Rn!), and f Uj with supp f K (thus f D()!), there exist > 0and a multi-index , such that

{g D()| supp g K and x K, |f(x) g(x)| < } Uj .

A sequence (fk) of elements of D() converges in D() toward f D() iff

1. There exists K ( compact such that

k N, supp fk K,

2. For all multi-index fk(x) converges uniformly toward f on K.

Note that

it follows immediately that supp f K.

if in the space C l() we introduce the norm

fCl() = max||l

supx

|f(x)|,

1MAS, ECP

2 Week 7: Distributions & Sobolev spaces

then

Nn fk(x) f on K l N fk fCl(K) 0 for k +.

In the topology on D(), a linear operator A : D() D() is continuous iff

fk f in D() Afk Af in D().

The elements of D() are often called testing functions. D() is not a normable space (it is notpossible to define a norm).

Problem 1 Show that the following linear operators are continuous in D():

1. For a multi-index Nn

: D() D(), f 7 f,

2. For a fixed function C()

A : D() D(), f(x) 7 (x)f(x),

3. For = Rn and a fixed g D(Rn)

Sg : D() D(), f(x) 7 Sg(f)(x) = (f g)(x) =Rnf(x y)g(y)dy.

Example 1 An example of a function from D() is given in the PDF of Week 6 Remark 2 p.3 (seealso Fig. 1)

Lemma 1 For all domain Rn and all > 0 there exists a function C(Rn) such that (seeFig. 2 for =]a, b[ (a < b))

1. x Rn 0 (x) 1;

2. x (x) = 1;

3. x / 3 (x) = 0,

where = {x R

n \ | d(, x) < }.

Proof. Let 12 be the characteristic function of the set 2:

12 =

{1, x 20, x 6= 2.

Let us show that the function(x) =

Rn12(y)w(x y)dy

satisfies 1)-3). Here w(x) D(Rn) is the kernel defined in Definition 5 of Week 6 p.2 such that

x Rn w(x) 0,

suppw(x) = B(0),

Rn w(x)dx = 1.

Week 7: Distributions & Sobolev spaces 3

1 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 1

h = 14

h = 12

h = 12

wh(x)

x

Figure 1 Example of non trivial elements of D(] 1, 1[): wh(x) =1h(xh

), where (x) = e

11|x|2 if |x| < 1

and (x) = 0 if |x| 1.

1

a ba 3 a b+ 3b+

(x)

x

Figure 2 Example of D(]a, b[) from Lemma 1.

Consequently, (see Fig. 3) we have

(x) =2

w(x y)dy C(Rn);

0 (x) Rnw(x y)dy =

Rnw(z)dz = 1;

(x) =B(x)

12(y)w(x y)dy =

=

{ B(x)

w(x y)dy =Rn w(z)dz = 1, x ;

0, x / 3.


3

3

2

Figure 3 Example of supports of w(x y) for x and x / 3 for the proof of Lemma 1.

In addition, we know (see Week 6 Theorem 1 p.6) that D() is dense in Lp(). We have definedwhat is a Cauchy sequence in the metric spaces, but it can be also defined for the topological spaceswhich are not metrizable.

In particular, let us define what is a Cauchy sequence D():

Definition 3 Let k D() for k N. We say that (k) is a Cauchy sequence in D() if thereis some compact set K ( such that for all k N suppk K and such that

Nn l N (k m)Cl(K) 0 as k,m.

Thus we can prove:

Theorem 1 Let be a domain in Rn. D() is complete.

Proof. Let (k) be a Cauchy sequence in D() (see Definition 3). Since

l N (C l(K), Cl(K)) is a Banach space,

it follows that there exists a function C(K) such that

Nn l N (k )Cl(K) 0 as k .

But if for all k N suppk K, then supp K also. Hence Cc () = D() and k in

D().

Problem 2 Prove that D() is dense in C():

f C() (k), k D() such that k f (k ) in C().


1.2 Dual space of D(). Distributions

Definition 4 Let D() be the dual space of D(): space of linear continuous functionals on D().Elements D() of are called distributions.

Example 2 (Regular distributions) For f in L1loc() define Tf : D() R by

Tf() =< Tf , >=f(x)(x)dx.

Tf is linear and continuous: Tf D(). Indeed, as supp is a compact in , we have for all l N

| < Tf , > | =

f(x)(x)dx

supp

|f(x)| |(x)|dx Cl(supp)

supp

|f(x)|dx.

As f L1loc() and supp is a compact in , we havesupp

|f(x)|dx =< f2, >

i.e.,

f1 = f2 in D().

The converse is also true thanks to Lemma 2 of Week 6 p.7:

if f1, f2 L1loc() and f1 = f2 in D

() f1 = f2 a.e. in .

Example 3 (Dirac) Define T : D() R by

T () =< T, >= (0).

T is linear and continuous: T D().

T is called a Dirac (sometimes a Dirac delta function). This distribution is usually denoted by .It is not in L1loc() (prove it!).

To describe the space D(), we give (without the proof) the following Theorem:

Theorem 2

T D() bounded ( M = M() ]0,[ and m = m() N such that

D() | < T, > | MCm().


Definition 5 Let T D() be a distribution and g C(). We define the operator

gT : T D() 7 gT D()

of the product gT by D() < gT, >=< T, g > .

(Since g C() and D() then g D().)

Definition 6 Let T D() be a distribution.

We define T : D() R, the derivative of the distribution T , by

T () = T ().

(Since D(), then D(), and since D() is a linear vector space, T D(). Conse-quently, T D(), i.e. a linear continuous functional on D().)

Therefore, we define the operator D : D() D() by

D(), < T , >= < T, > .

This is consistent with the derivative of functions thanks to the integration by parts: since has acompact support in , then (x) = 0 for all x and the boundary term in the formula of theintegration by parts is equal to zero.

Example 4 We consider = R. Let us consider the derivative of the Heaviside function:

(x) =

{0, x < 0,1, x 0.

Clearly, L1loc(R) and thus D(R) with

D(R) < , >=R(x)(x)dx.

Let us calculate :

D(R) < , >= < , >= R(x)(x)dx =

+0

(x)dx = (x)|+0

= 0 + (0) =< , > .

Finally, D(R) < , >=< , >,

which means that = in the sense of distributions.

More generally,

Definition 7 Let be a domain in Rn. Let T D() be a distribution and let Nn be amulti-index.

We define the derivative operator D : D() D() by

D() < DT, >= (1)|| < T, > .


Remark 2 If f C() then all its derivatives f (in the usual, or classical sense) are equal tothe corresponding derivatives of f in the sense of distributions:

As f C() then f L1loc() and thus f D(). We summarize:

C() ( D().

Therefore, by integration by parts,

D() < Df, >= (1)||f(x)(x)dx =

f(x)(x)dx =< f, > .

Example 5 Let T D(R3) be a distribution and let = (1, 4, 5). By Definition 7, the operatorD(1,4,5) : D(R3) R is defined by

D(R3), < D(1,4,5)T, >=< T, (1,4,5) >

Corollary 1 1. All distributions are infinitely derivable.

2. All convergent series in D() are infinitely derivable term by term:

if (uk)kN ( D() such that

kN

uk = S D(), then

NnkN

Duk = DS in D().

Proof. The first statement follows from the definition of a destribution: since D(), then forall Nn D(), therefore, for all T D() the distribution DT is correctely defined forall Nn.

The second point follows from

S = limm

mk=1

uk in D(), i.e. D() < S, >= lim

m,

D() Nn < DS, >= (1)|| < S, >= (1)|| limm

= limm

< D(mk=1

uk), >= limm

= .

Remark 3 For all Rn, it holds (prove it!)

Lp() L1loc().

Therefore functions in Lp() are regular distributions.

Thus, we can now differentiate any function in Lp! But the derivative may (or may not) be a functionin Lp. When it is the case, we have a Sobolev space, we will see this in the next section.


2 Lecture 7.3: Wm,p Spaces

2.1 Definition and main properties

Let be a domain in Rn (not necessarily bounded).

Definition 8 Let m N and p [1,]. The Sobolev Space Wm,p() is defined by

Wm,p() = {f Lp()| Df Lp() for any || m}

(derivative in the distributional sense) with a norm

fm,p = fWm,p() = (

0||m

|Df |p)

1p = (

0||m

DfpLp())1p for p 0 N N : m, k N um ukWm,p() < .

The inequality um ukWm,p() < means that0||m

Dum DukLp() < .

Then for 0 || m the sequence (Dui) is a Cauchy sequence in Lp(). Since Lp() is complete,

there exist functions v Lp() and v Lp() for 1 || m such that

ui v in Lp() and Dui v in L

p() (1 || m).

To finish the proof, we need to show that

Dv = v for all 1 || m,

from where we could conclude that v Wm,p() and thus (ui) converge to v Wm,p() in Wm,p().

As Lp() L1loc() and so ui determines a regular distribution Tui (or itself can be considered as aregular distribution):

D() < ui, >=ui(x)(x)dx.


For any D() we have

| < ui, > < v, > | |ui(x) v(x)||(x)|dx Lp()ui vLp(),

where we have applied the Hlder inequality (see Week 2 p.26 Definition 35 and Proposition 12).Hence < ui, >< v, > for every D() as i.

Similarly, D() < Dui, >< v, > i.

It follows that

D() < v, >= limi

< Dui, >= limi

(1)|| < ui, >= (1)|| < v, > .

Thus Dv = v for all 1 || m, from where v Wm,p() and ui vWm,p() 0 for i.

Remark 5 If classical partial derivatives exist and are continuous then they coincide with the dis-tributional partial derivative. Thus the set

S = {f Cm()|fWm,p()


Example 7 The Sobolev space

W 1,2(R) = {f L2(R)| f L2(R)},

endowed with the norm

fW 1,2(R) = (R|f |2 +

R|f |2)

12

is a Hilbert space with the inner product

< f, g >=Rfg +

Rf g.

Theorem 4 Wm,p() is separable if 1 p


2.2 Sobolev spaces Wm,p() for bounded domains with a regular bound-

ary

Definition 9 Let be a bounded domain of Rn. Its boundary is locally Lipschitz if

x neighborhood Ux : Ux is the graph of a Lipschitz continuous function f.

Remark 6 Here by a Lipschitz continuous function we understand f : Rn1 R which is continuousand there exist M > 0 such that

x, y Rn1 |f(x) f(y)| Mx yRn1.

Definition 10 The boundary is called a regular boundary of the class Ck (k 1, seeFig. 4):

x0 f(x) Ck(Ux0) such that {x Ux0 | f(x) = 0} = Ux0 , df |Ux0 6= 0.

Ux0

x0 f(x) = 0

Figure 4 Example of a regular boundary from Definition 10.

If is a bounded domain with a regular boundary (at least locally Lipschitz), then we have followingresults:

Theorem 5 Let be a bounded domain with a locally Lipschitz boundary. Then it holds

1. C() is dense in Wm,p().

2. We can extend f W 1,p() from to a bigger domain ( ( ) without loss of regularity:f W 1,p() and f = f on .

3 Lecture 7.4: Hm Spaces

3.1 Definition and properties

Definition 11 Let be a domain in Rn and m N. Define Hm() = Wm,2().

For f and g in Hm(), we define a inner product

< f, g >Hm()=

0||m

D fDg =< f, g >L2() +

m||=1

< Df,Dg >L2(), (4)


which is associated with the norm

fm = fHm() = (||m

|Df |2)

12 = (

||m

f2L2())12 .

Corollary 2 Hm() with the inner product (4) is a separable Hilbert space.

Proposition 3 1. Canonic injection of Hm() in L2() is continuous:

v Hm() vHm() CvL2().

2. Hm() is dense in L2().

Proof.

1. It is a direct corollary of the definition of the norms in Hm and in L2.

2. Firstly, we notice thatC() ( Hm() ( L2().

Secondly, C() is dense in L2() by the norm L2 (it is easy to see that C is dense in

(C0, ), which is dense in L2). Thus, Hm() is dense in L2().

3.2 Dual space (Hm())

Since Hm() is a Hilbert space, the Riesz representation theorem yields

(Hm()) = Hm(),

where elements of (Hm()) are linear continuous forms defined by the inner product of Hm(). Itmeans that to f Hm() corresponds the following element of (Hm()):

u Hm() 7< f, u >Hm()=||m

Df Du.

By definition of Hm, we see thatHm() ( L2(),

where L2() is also a Hilbert space for < f, g >L2()= fg. This gives, if we identify L

2() withL2(), and (Hm()) with Hm(), that

Hm() = (Hm()) ( L2() = L2().

Definition 12 By tradition, L2() is always identified with L2() by Riesz Theorem, but what iscalled by the dual space of Hm, is the space

(Hm()) = { all linear continuous forms on Hm() defined by the inner product of L2()}.

For instance, for a fixed f L2() the application

T : u Hm()< f, u >L2() R

is a linear continuous form on L2() and also on Hm(). Thus T (Hm()). We define :L2() (Hm()) by

f L2() 7 (f) = [u Hm()< f, u > R] = T (Hm()).

Let us prove the following properties of :


Lemma 2 1. (f)(Hm) fL2

2. is injective

3. (L2) is dense in (Hm)

Proof.

1. (f)(Hm) fL2:

We have for all u Hm

| < f, u >L2 | fL2uL2 fL2uHm,

from where

(f)(Hm) = supu 6=0

| < f, u >L2 |

uHm fL2.

2. is injective:

By definition of ,u Hm, f L2 (f) =< f, u >L2 .

Therefore,Ker = {f L2| u Hm < f, u >L2= 0}.

As Hm is dense in L2 (see Proposition 3), it follows that if f Ker then

u L2 < f, u >L2= 0 f = 0.

3. (L2) is dense in (Hm) :

Let us apply Corollary 6 of Week 3 p. 12: By Theorem 3 of Week 3 p. 8, (Hm) is a Banachspace. Clearly, (L2) = Range() is a subspace of (Hm). We also notice that Hm is dense inL2 and Hm (Hm) L2.

If (L2) is not dense in (Hm), then, by Corollary 6 of Week 3 p. 12,

u Hm u 6= 0 such that f L2 < u, f >L2= 0,

which is not possible (u = 0).

Therefore, using , we prove the inclusion of L2 = (L2) in (Hm):

Hm L2 = (L2) (Hm).

L2 is called the pivot space. From now on, we will make this choice.

3.3 Trace operator

In what follows, we suppose that is a bounded domain of Rn and its boundary is locallyLipschitz, writing that is regular or sufficiently smooth.

We note (see L.C. Evans Partial differential equations p. 252 for the proof)

Proposition 4 For a regular of a bounded domain , C() is dense in Hm().


For a regular of the class C1 of a bounded domain , it is possible to define a boundary measureand define the space L2():

L2() = {v measurable on ||v(x)|2dx 0 N N such that m, k N um ukH1() 0 N N such that m, k N um| uk|L2() C()um ukH1() < .

Thus (um|) is a Cauchy sequence in L2(). The space L2() is complete, consequently

v L2() such that um| v for m in L2().

Uniqueness of v:

Let (uk) be a sequence in C() such that

(uk) 6= (um) and

{uk u in H

1(),uk| v in L

2().

Then, using (5) and the triangle inequality, we obtain

v vL2() = v um| + um| u| + u| uk| + uk| vL2()

v um|L2() + um| u|L2() + u| uk|L2() + uk| vL2()

v um|L2() + uk| vL2() + C()(um uH1() + u ukH1()) 0 for m, k .

Hence, v vL2() = 0 and thus v = v. We conclude that for all u H1() our construction allows

to define an unique element of L2(), which we call the trace of u.


3.4 Spaces Hm0 ()

Remark 7 In what following we write for functions in Sobolev spaces understanding the deriva-tive in the sense of distributions.

Definition 13 Define Hm0 () as the closure of D() in Hm().

We admit this characterization of Hm0 ():

Theorem 6 For all domains in Rn

1. u Hm0 () (um) ( D() such that um u in H1(),

2. Hm0 () is a Hilbert space with the inner product of Hm(),

3. Hm0 () ( Hm() (if = Rn then Hm0 () = H

m()),

4. If u Hm() and v D(), then vu Hm0 (),

5. H10 () = {f H1()| f = 0 on },

6. Hm0 () = {f Hm()| for || < m,Df = 0 on }.

Theorem 7 (Poincar inequality) Let be a bounded domain of Rn.

Then there exists a constant C = C() > 0 depending only on , such that for all u in H10 (),

uL2() C()uL2(), (6)

where

uL2() =n

k=1

uxkL2()

.

Proof. Since D() is dense in H10 (), for all u H10 () there exists a sequence of functions

vk D() which converges toward u in H10 ():

vk uH1() 0 for k .

Firstly we show (6) for vk D(). After it, since the convergence by the norm of H1() implies the

convergence of vk uL2() and (vk u)L2() toward 0, we obtain (6) for u H10 () taking the

limit for k by the norm of H1.

Let be a parallelepiped such that ( and, without loss a generality, we suppose that

= {x Rn|0 < xi < di, i = 1, . . . , n} and d1 = mini=1,...,n

di.

As for all k supp vk ( , we extend vk from to by 0.

For all vk D() we have

vk(x1, . . . , xn) = x10

vky1

(y1, x2, . . . , xn)dy1. (7)

Introducing the following notations x = (x2, . . . , xn), = {x Rn1|0 < xi < di i = 2, . . . , n},

we take the square of (7) and integrate it over :

v2kdx =

d10

dx1

( x10

vky1

(y1, x)dy1

)2dx. (8)


Using the 1D Cauchy-Schwartz inequality

( x10

1 vky1

(y1, x)dy1

)2 x10

12dy1

x10

(vky1

)2dy1 x1

d10

(vky1

)2dy1,

we obtain v2kdx

d10

dx1

x1

d10

(vky1

)2dy1

dx = d21

2

(vky1

)2dx.

Since vk 0 out of , we conclude that

v2kdx

d212

(vky1

)2dx,

from where taking the limit by the norm of H1, we find (6).

Remark 8 The Poincar inequality holds true for domains bounded at least in one direction. Butthe result is not true in H1(). u = 1 everywhere on would not work!

Theorem 8 When the open set is bounded, the semi-norm | |m is a norm on Hm0 which is

equivalent to m.

Proof. Instead of the general result, let us prove the equivalence of the norm

uH10 () =

(u2 + |u|2)dx (9)

and

|u|1 =

|u|2dx = uL2(). (10)

We need to show that there exist constants C1 > 0 and C2 > 0, such that

C2|u|1 uH10 () C1|u|1.

Thanks to the Poincar inequality and the positivity of the norm, we find that indeed

u2L2() u2L2()

+ u2L2() (C2() + 1)u2L2(),

from where it is sufficient to take C2 = 1 and C1 =C()2 + 1.

3.5 Greens formula and integration by parts

Definition 14 If is C1, then along is defined the outward pointing unit normal vectorfield = (1, . . . , n). The unit normal at any point x0 is (x0).

We denote by the normal derivative of u

u

=< ,u >Rn .


Theorem 9 Let be sufficiently smooth. Let be the unit outward normal to . Let {ei} be thecanonical basis of Rn. For any u and v in H1() we have

u(iv) =

(iu)v +

uv cos(, ei)

In particular, there is useful formula which can be considered as a classical integration by parts:

uv =

u

v

uv.

4 Lecture 7.5: Sobolev embeddings

Sobolevs embedding theorems give the fundamental properties of the Sobolev spaces.

These properties are connected with the degree of smoothness that can be expected in a Sobolevspace.

The larger the product mp ... the smoother the function!

The critical value is the space dimension n. If mp > n then all functions inWm,p are continuous(usual disclaimer: we are dealing with classes).

Proposition 6 For any domain in Rn, we have

Wm,p() Lq() for 1/q = 1/pm/n if mp < n

andWm,p() Lq() for any q [1,[ if mp = n.

Definition 15 Let X and Y be two normed vector spaces. A linear application A : X Y iscompact if from any bounded sequence (xk) of elements of X one can extract a subsequence (xkl)such that its image (Axkl) is strongly convergent in Y .

If there exists a compact linear application from X to Y , we note X Y

Theorem 10 Let be a bounded domain in Rn with locally Lipschitz boundary . Then

1. Wm,p() Lr() for any r [1, q[, where 1/q = 1/pm/n provided mp < n.

2. Wm,p() Lq() for any q [1,[ provided mp = n.

3. Wm,p() C() if mp > n.

Example 8 Let be a a bounded domain of R2 (n = 2) with a regular boundary and let takeH1() = W 1,2().

As 1 2 = 2, therefore W 1,2() Lq() for any q [1,[.

Thus, if ukH1() is bounded then, one can extract a subsequence in L2() that converges strongly

in L2(). The unit ball of H1() is compact in L2() but not in H1().

Let us define the inclusion operator

Definition 16 The operator i(u) = u which associates u H1() with the same u as an element ofL2() is called the operator of inclusion of H1() in L2().


Instead to prove Theorem 10, we prove the particular case

Theorem 11 Let be a bounded domain in Rn with locally Lipschitz boundary . Then theinclusion operator i : H1() L2() is linear, bounded and compact.

Proof. Continuity

We have

i(u)L2() uH1().

Compactness

Let (um) is a bounded sequence in H1():

m umH1() C.

Thus, sincei(u)L2() uH1() C,

the sequence (i(um)) is also bounded in L2(). To simplify the notations, we will simply write um

for i(um) L2.

Since L2() is a separable Hilbert space, all bounded sets in L2() are weakly compact in L2():

(umk) (um) such that u L2() : umk u L

2().

As (L2()) = L2(), by the Riesz theorem,

umk u L2() v L2()

(umk u)vdx 0.

We also notice that umk u L2() implies that (umk) is a Cauchy sequence in L

2. In addition, ifwe take v = 1, then

(um uk)dx 0 for k,m +.

On the other hand, we can extend um from to a parallelepiped , containing , such that

u =

{u, x 0, x \

and u H1().

We chose this parallelepiped in the way as

( , = {x|0 < xi < di}.

Let

= Mi=1i, where i = nk=1[ai, ai +

dkN].

For this it is possible to prove the following inequality for all u H1():

u2dx

Mi=1

1

|i|

(iudx

)2+n

2

nk=1

(dkN

)2 (u

xk

)2dx. (11)

Thus, for two members of the subsequence umk with sufficiently large p and k, we have

upuq2L2() upuq

2L2()

Nni=1

1

|i|

(i(up uq)dx

)2+

n

2N2

nk=1

d2k

upxk uqxk

2

L2()

L2([0,1])=< g, limnun >L2([0,1])=< g, u >L2([0,1])= f(u).

Let us prove that system (12)(13) can be written in the variational form as

u H a(x, u) = f(u).

We multiply x + x = g by u H and integrate over [0, 1]:

[0,1]

(xu+ xu)d =[0,1]

gud.

We can integrate by parts in the usual way:

[0,1]

(xu+ xu)d = xu|t=1t=0 +[0,1]

(xu + xu)d.

By the choice of our boundary conditions (13), we have

xu|t=1t=0 = x(1)u(1) + x(0)u(0) = 0.

Therefore,

u H[0,1]

(xu + xu)d =[0,1]

gud.

By Riesz Theorem, or more generally the Lax-Milgram Lemma, we conclude that system (12)(13)has a unique solution x in H .

This solution is called a weak solution.

Remark 9 Once we have a weak solution of a problem, we can study its regularity.

Lecture 7.2: DistributionsTopology on D()Dual space of D(). Distributions

Lecture 7.3: Wm,p SpacesDefinition and main propertiesSobolev spaces Wm,p() for bounded domains with a regular boundary

Lecture 7.4: Hm SpacesDefinition and propertiesDual space (Hm())'Trace operatorSpaces Hm0()Green's formula and integration by parts

Lecture 7.5: Sobolev embeddingsApplication of the Lax Milgram Lemma and of the Riesz representation theorem

functional analysis week07 v2

Documents