Cours de Francis Clarke

Francis Clarke

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Niveau: Supérieur, Master
Chapter 5 Banach spaces Analyse Master 1 Cours de Francis Clarke (2011) 5.1 Completeness of normed spaces A normed space X is said to be a Banach space if its metric topology is complete: every Cauchy sequence xi in X (that is, one that satisfies lim i, j?∞ xi ? x j = 0) admits a limit in X : there exists a point x ? X such that xi? x? 0. Informally, the absence of such a point x would mean that the space has a hole where x should be. For purposes of minimization, one of our principal themes, it is clear that the existence of minimizers is imperiled by such voids. Consider, for example, the vector space Q of rational numbers. The minimization of the function (x2?2)2 does not admit a solution over Q, as the Greek mathematicians of antiquity were able to prove. The existence of solutions to minimization problems is not the only compelling reason to require the completeness of a normed space, as we shall see. The property is essential for such tools as uniform boundedness, variational principles, and weak compactness. The reader is asked to observe that completeness of a normed space is invariant under equivalent norms, and that a closed subspace of a Banach space is itself a Banach space. It is easy to see as well that the Cartesian product of two Banach spaces is a Banach space.

• since acp


• minimization principles


• lim n?∞


• banach spaces


• banach space


• zn ?


• theorem readily


• every cauchy sequence

Sujets

• Clarke
• Banach space

Informations

• Publié par: cidur
• Nombre de lectures: 47
• Langue: English

Extrait

Chapter 5

Banach spaces

Analyse Master 1

Cours de Francis Clarke (2011)

5.1 Completeness of normed spaces

A normed spaceXis said to be aBanach spaceif its metric topology iscomplete: every Cauchy sequencex i inX(that is, one that satisfies lim i,j→∞ ￿x i −x j ￿=0) admits a limit inX: there exists a pointx∈Xsuch that￿x i −x￿ →0. Informally, the absence of such a pointxmean that the space has a hole wherewould xshould be. For purposes of minimization, one of our principal themes, it is clear that the existence of minimizers is imperiled by such voids. Consider, for example, the vector spaceQof rational numbers. The minimization of the function(x 2 −2) 2 does not admit a solution overQas the Greek mathematicians of antiquity were, able to prove.

The existence of solutions to minimization problems is not the only compelling reason to require the completeness of a normed space, as we shall see. The property is essential for such tools as uniform boundedness, variational principles, and weak compactness.

The reader is asked to observe that completeness of a normed space is invariant under equivalent norms, and that a closed subspace of a Banach space is itself a Banach space. It is easy to see as well that the Cartesian product of two Banach spaces is a Banach space. A less evident fact is that the dual space is always a Banach space.

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5.1 Theorem.The dualX ∗ of a normed spaceXis a Banach space.

Proof.It is understood, of course, thatX ∗ is equipped with its usual dual norm ￿∙￿ ∗ . The issue here is to show that a given Cauchy sequenceζ n inX ∗ admits a limitζ∈X ∗ . For eachx∈X, it follows that ￿ ζ n ,x  is a Cauchy sequence inR, since  ￿ ζ n ,x  − ￿ ζ m ,x   ￿  ζ n −ζ m  ∗ ￿x￿. SinceRis complete, the sequence converges to a limit, denoted ￿ ζ,x  . The function ζso defined is a linear functional; we now show that it is continuous. Fixx∈B, together withNsuch thatm,nN=⇒  ζ m −ζ n  ∗ ￿1.Then, for m,nN, we have:  ￿ ζ n ,x   ￿  ￿ ζ n −ζ m ,x   +  ￿ ζ m ,x   ￿1+  ζ m  ∗ . Lettingn→∞, we deduce  ￿ ζ,x   ￿1+  ζ m  ∗ , which implies thatζ∈X ∗ . To conclude, we must confirm that  ζ n −ζ  ∗ →0. Letε>0. There existsN ε such thatm,nN ε =⇒  ζ n −ζ m  ∗ <ε. FixnN ε . Then, for anyx∈Bandm>N ε , we have  ￿ ζ n −ζ,x   ￿  ￿ ζ n −ζ m ,x   +  ￿ ζ m −ζ,x   ￿ε+  ￿ ζ m −ζ,x   . Now ￿ ζ m −ζ,x  →0 asm→∞, whence  ￿ ζ n −ζ,x   ￿ε. Sincexis an arbitrary point inB, this reveals  ζ n −ζ  ∗ ￿ε∀nN ε . The proof of the theorem readily adapts to prove

5.2 Corollary.IfYis a Banach space, thenL c (X,Y)is a Banach space.

We remark that any space that is isometric to a dual space is necessarily a Banach space, in view of the following fact:

5.3 Exercise.A normed space which is isometric to a Banach space is itself a Ba- nach space.

5.4 Exercise.We return to the spaces of sequences defined in Example 1.5. a) Show that￿ 1 is a Banach space. b) Use Theorem 5.1 and Exer. 5.3 to prove that￿ p (1<p￿∞)is a Banach space, and deduce that the spacescandc 0 are Banach spaces. c) Show that￿ c ∞ is not a Banach space. 

We proceed now to pass in review some of the well-known Banach spaces that will be of use to us.

5.1 Completeness of normed spaces

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Spaces of continuous functions.We have met the normed spaceC(K)in Example 1.3. Iff i is a Cauchy sequence inC(K), then, for eachx∈K, the sequencef i (x)is Cauchy inR. If we denote its limit byf(x), then an argument very similar to the proof of Theorem 5.1 shows thatf∈C(K)and  f i −f  C(K) →0. Thus,C(K)is a Banach space.

We observed earlier in Example 1.1 that whenK= [0,1], another norm on the vec- tor spaceC[0,1]is provided by  f  1 =  01  f(t)  dt. This norm is not complete, however: 5.5 Exercise.Show that the sequencef i (t) =  min  2t,1  i is Cauchy inL 1 (0,1) relative to the norm￿∙￿ 1 , and thatf i converges pointwise and inL 1 (0,1)to a dis- continuous function.

When the compact metric spaceKis replaced by a normed spaceX, we can restrict attention to bounded functions, and in so doing, obtain a Banach space:

5.6 Proposition.LetXbe a normed space andYbe a Banach space. Then the vector spaceC b (X,Y)of bounded continuous functionsg:X→Yis a Banach space when equipped with the norm

 g  C b (X,Y) = x su ∈ p X  g(x)  Y

Once again, the proof can be patterned after that of Theorem 5.1.We turn now to Lipschitz functions.

5.7 Proposition.LetXbe a normed space andYbe a Banach space. The vector spaceLip b (X,Y)of bounded, Lipschitz functionsϕ:X→Yequipped with the norm  ϕ Lip b ϕ  C b (X,Y) +sup  ϕ(x)−ϕ(u)  Y  (X,Y) =  x,u∈X  x−u  X x=u is a Banach space.

Proof.It is easy to see that we do have a normed space; we verify that it is complete. Letϕ n be a Cauchy sequence in Lip b (X,Y). Thenϕ n is also a Cauchy sequence in C b (X,Y), which is complete. In consequence, there existsϕ∈C b (X,Y)such that  ϕ n −ϕ  C b (X,Y) →0. Note that the functionsϕ n have a common Lipschitz constant, which we denote byL. Passing to the limit in the inequality  ϕ n (x)−ϕ n (u)  Y ￿L  x−u  X , we see thatϕis Lipschitz. Thusϕ∈Lip b (X,Y).

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We complete the proof by showing that  ϕ n −ϕ  Lip b (X,Y) →0, which amounts to showing that lim sup   ϕ n −ϕ  (x)−  ϕ n −ϕ  (u)  Y =0. n→∞ x,u  x−u  X

The left side of this desired equality coincides with lim sup lim   ϕ n −ϕ m  (x)−  ϕ n −ϕ m  (u)  Y n→∞ x,u m →∞  x−u  X ￿lim lim sup   ϕ n −ϕ m  (x)−  ϕ n −ϕ m  (u)  Y n→∞m→∞ x,u  x−u  X

￿ n li → m ∞m li → m ∞  ϕ n −ϕ m  Lip b (X,Y) =0,

since the sequenceϕ n is Cauchy in Lip b (X,Y).



A space of differentiable functions.The following space will be used later in connection with minimization principles.

5.8 Proposition.LetXbe a normed space. The vector spaceC b 1,1 (X)of bounded continuously differentiable functionsg:X→Rwhose derivativeg  is bounded and Lipschitz is a Banach space when equipped with the norm   g  C b 1,1 (X) =  g  C b (X,R) +  g  Lip b (X,X ∗ ) .

t P h r a o t o it f. eltlg,ydrnicoAce.etplomsci ewodAniagntdeatthti,ivise g h n n b ar a o e emsdaevs c a e m inquenh p y ac s e e ; C i a t u i cC a b 1 t , t 1 e ( r X o ) gni h y e f . i T r h e e vf nt sequenceg n is Cauchy in the Banach spaceC b (X,R), so there existsg∈C b (X,R) such that  g n −g  C b (X,R) = x s ∈ u X p  g n (x)−g(x)  →0. It also follows that the sequenceg  n is Cauchy in the space Lip b (X,X ∗ ), which is complete by the preceding result; thus, there also existsϕ∈Lip b (X,X ∗ )such that  g  n −ϕ  Lip b (X,X ∗ ) →0. We now claim thatg  exists and coincides withϕ. Note that the functionsg  n have a common Lipschitz constantL. Letxandube distinct points inX. Then g(u)−g(x)− ￿ ϕ(x),u−x  = n li → m ∞ g n (u)−g n (x)− ￿ g  n (x),u−x  = n li → m ∞ ￿ g  n (z n )−g  n (x),u−x  ,

5.1 Completeness of normed spaces

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for somez n ∈  x,u  , by the Mean Value theorem. We deduce  g(u)−g(x)− ￿ ϕ(x),u−x   n→∞ L  z n −x   u−x  ￿L  u−x  2 . ￿lim This estimate reveals thatg  =ϕ, and it follows that  g n −g  C 1,1 (X) →0. b

Lebesgue spaces and Sobolev spaces.It is shown in courses on integration that the Lebesgue spaceL p (Ω) (1￿p￿∞)is complete 1 . SinceAC p [a,b](see Example 1.12) is isometric toR×L p (a,b), it follows that it too is a Banach space (by Exer. 5.3).

We introduce now a class of spaces that may be thought of as extensions of abso- lutely continuous functions to several dimensions, and which play a central role in the multiple integral calculus of variations, and in partial differential equa- tions. Letube an element ofL p (Ω,R), whereΩis a nonempty open subset of R n . Thenuis said to admit aweak derivativeinL p if there exists an element  g 1 ,g 2 , . . . ,g n  ∈L p (Ω,R n )such that, for each indexi,  u(x) ∂ ∂x ϕ i (x)dx=−  Ω g i (x)ϕ(x)dx∀ϕ∈C c ∞ (Ω,R). Ω Here,C c ∞ (Ω,R)refers to the functionsϕ:Ω→Rwhich admit partial derivatives of all orders, and which have compact support inΩ. We learn in integration thatC c ∞ (Ω,R)is a dense subset ofL p (Ω); this implies that the weak derivativeg=  g 1 ,g 2 , . . . ,g n  ofuis unique when it exists. It is clear from the integration by parts formula that ifuhappens to be continuously differentiable, then its weak derivative is precisely its gradient. This justifies, by extension, the use of the notationDufor the weak derivativegofu. We also writeD i ufor the function g i . It is easy to see that if two functionsuandvadmit the weak derivativesDuandDv, and ifc,kare scalars, thencu+kvadmits the weak derivativecDu+kDv. Let 1￿p￿∞. TheSobolev spaceW 1,p (Ω)is by definition the vector space of all functionsu∈L p (Ω,R)which admit weak derivatives inL p , equipped with the following norm:

 u  W 1,p (Ω) =  u  L p (Ω) + ∑ i n =1  D i u  L p (Ω) . The spaceW 1,2 (Ω)is usually denotedH 1 (Ω).

5.9 Exercise.Letu j be a sequence inW 1,p (Ω)(1￿p<∞) such thatu j converges weakly inL p (Ω)to a limitu, and such that, for eachi=1,2, . . . ,n, the sequence

1 See for exampleAnalyse fonctionnelle, B  ´ i rez s.

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D i u j converges weakly inL p (Ω)to a limitv i . Prove thatu∈W 1,p (Ω), and that Du=  v 1 ,v 2 , . . . ,v n  .

The following exercise shows that, whenn=1, withΩ= (a,b), the functionsuin W 1,1 (Ω)are essentially the elements of AC[a,b]. 5.10 Exercise.Prove thatulies inW 1,1 (a,b)if and only if there is a functionf∈ AC[a,b]such thatu(t) =f(t),t∈[a,b]a.e., in which case we haveDu=f  .

In a context such as this, bearing in mind that the elements ofL p (a,b)are really equivalence classes, it is common to say thatuadmits a continuousrepresentative. Whenn>no longer that case that an element of1, it is W 1,p (Ω)necessarily admits a continuous representative, or even a locally bounded one. 5.11 Exercise.Prove thatW 1,p (Ω)is a Banach space.

The Uniform Boundedness principle.The following highly useful Banach tool, also known as the Banach-Steinhaus theorem, plays a central role in the the- ory.

5.12 Theorem.LetXbe a Banach space andYa normed space. LetΓbe a col- lection of operators inL c (X,Y). IfΓis simply bounded(meaning that, for every x∈X, the image set  Λx:Λ∈Γ  is bounded inY), thenΓis uniformly bounded: there existsM0such that  Λ  L(X,Y) ￿M∀Λ∈Γ.

The theorem affirms that if the setΓinL c (X,Y)is simply bounded, then it is, quite simply, bounded.

Proof.For each integern, we define a subset ofXby F n =  x∈X:  Λx  Y ￿n∀Λ∈Γ  . Note thatF n is closed, since, for eachΛ∈Γ, the functionx→  Λx  Y is continuous. By hypothesis, we haveX=∪ n1 F n . SinceXis complete, Baire’s theorem 2 tells us that, for someN, there existsz∈Xandr>0 such thatB  z,r  ⊂F N . Thus we have  Λ(z+ru)  Y ￿N∀u∈B,Λ∈Γ. We deduce r  Λu  Y ￿N+  Λz  Y ∀u∈B,Λ∈Γ.

2 Theorem.Let  E,d  be a complete metric space, andF n a sequence of closed subsets ofEsuch thatint   n F n  =0/.Then there existsNsuch thatintF N =/0.

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ButK:=sup Λ∈Γ  Λz  Y is finite by hypothesis, so the preceding inequality leads to  Λ  =s ∈ up B  Λu  Y ￿(N+K)/r=:M∀Λ∈Γ.  u The next three exercises are rather immediate corollaries of the Uniform Bounded- ness principle. In each case, one may find counterexamples to show the necessity of the completeness hypothesis.

5.13 Exercise.LetΛ n be a sequence of linear operators mapping a Banach space Xto a normed spaceY. Suppose that for eachx∈X, the limitΛx:=lim n→∞ Λ n x exists. Prove that the (linear) mappingΛis continuous.

5.14 Exercise.Prove that a weakly compact subset of a normed space is bounded. Deduce that a weakly convergent subsequence in a normed space is bounded. 5.15 Exercise.LetXbe a Banach space and letζ n be a sequence inX ∗ converging weak ∗  that is, such that for every ;to 0x∈X, we have lim n→∞ ￿ ζ n ,x  =0. Prove thatζ n is bounded inX ∗ .

Support functions and boundedness.As we now see, in a normed space, and in the dual of a Banach space, the boundedness of a (nonempty) set is equivalent to its support function being finite-valued.

5.16 Proposition.LetXbe a normed space. A subsetSofXis bounded if and only if H S (ζ) =sup ￿ ζ,x  <∞∀ζ∈X ∗ . x∈S IfXis a Banach space, then a subsetΣofX ∗ is bounded if and only if H Σ (x) =sup ￿ σ,x  <∞∀x∈X. σ∈Σ

Proof.to prove is that the given propertyIt is clear that, in each case, the only thing of the support function implies the boundedness of the set. We may assume that the sets involved are nonempty; thus, the support function is finite-valued. Everyx∈Sengenders a  n Λ e x le  x m ∈ e S is ntΛ s x of imp L c (X ∗ ,R)viaΛ x (ζ) = ￿ ζ,x  , with  Λ x  =  x  . The familyΓ=ly bounded since, for a givenζ∈X ∗ , we have x i ∈ nf S ￿ ζ,x  =−H S (−ζ)>−∞, x s ∈ up S ￿ ζ,x  =H S (ζ)<+∞. Then, sinceX ∗ is a Banach space (Theorem 5.1), the familyΓis bounded by Theo- rem 5.12, whence the boundedness ofS. The argument forΣis similar, using the family  Λ ζ  , whereΛ ζ (x) = ￿ ζ,x  . Note that now we need to posit thatXis complete.

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5.17 Exercise.In the second assertion of the preceding theorem, it is assumed that XBanach space. Show that this is essential: give an example of an unboundedis a setΣin the dual of a normed spaceXwhose support functionH Σ is everywhere finite-valued.

Continuity of convex functions.In a Banach space, the convexity of a lower semi- continuous function automatically implies its Lipschitz continuity (compare with Theorem 2.34):

5.18 Theorem.LetXbe a Banach space, and letf:X→R ∞ be convex and lsc. Thenfis locally Lipschitz in the setint domf.

Proof.[Exercise] Letx 0 ∈int domf. In view of Theorem 2.34, it suffices to prove thatfis continuous atx 0 . a) T  o x d 0 o  s + o, ε i  ibhgienasfodoohrohy?)0.(W ufficestotsof,tynarvorpahteε>0, the set  y∈X:f  x 0 +y  ￿ f b) It suffices therefore to show that the following setCis a neighborhood of 0: C=  y∈X:f  x 0 +y  ￿f  x 0  +ε,f  x 0 −y  ￿f  x 0  +ε  . Show thatCis convex, closed, and symmetric (−C=C).

c) Prove that, for every pointz∈X, there existst>0 such thattz∈C. (Cor. 2.35 is relevant here.)

d) Invoke Baire’s theorem to deduce that intC=0/. e) Letα∈intC. Then 0∈int  C−α  and 0∈int  C/2−α/2  ⊂int  C/2−C/2  =int  C/2+C/2  =intC.

We deduce thatCis a neighborhood of 0.



It follows from Theorem 5.18 that a discontinuous linear functional on a Banach space cannot be lower semicontinuous. Things are otherwise on general normed spaces: 5.19 Exercise.We setX=￿ c ∞ and, forx=  x 1 ,x 2 , . . .  ∈X, we definef(x) = ∑ i  x i  . Show that the functionf:X→Ris lsc, but not continuous.

We deduce from this example that the completeness hypothesis in Theorem essential (as is the lower semicontinuity).

5.18 is

5.2 Perturbed minimization

5.2 Perturbed minimization

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It is clear that the completeness of a normed space is an important factor in the quest to identify a set of ingredients allowing us to affirm that minimization problems ad- mit a solution. But completeness is not enough: we generally require a pinch of compactness. Later, we shall find a way to obtain it by exploiting the weak topol- ogy. Here, however, we examine a more modern consideration that has been very fruitful, an approach in which the original problem (that may not admit a minimum) is slightly perturbed to obtain a new problem thatdoeshave one. We refer to a the- orem making this type of assertion as aminimization principle(for short; a more descriptive term would be “perturbed minimization principle”). We shall prove two such results. R  trfofauhtsepuopcallthate tyastahancfuontiϕ tcoinn saib ϕ u : m X p f → un R ct i i s o t n losu hwne hce ϕonti re i o s f c these nu t o  u x: sa ϕ n ( d x) h  a = s 0 . We nonempty bounded support.

The first minimization principle we examine is the following.

5.20 Theorem. (Deville, Godefroy, and Zizler)LetXbe a Banach space, and let Gbe a Banach space of bounded continuous functionsg:X→Rwith the following properties:

a)For everyt>0andg∈G, the functionh:x→h(x):=g(tx)belongs toG. b)We have  g  G sup x∈X  g(x)  ∀g∈G. c)The norm onGis invariant under translation:for everyu∈Xandg∈G, the functionh:x→h(x):=g(x+u)belongs toGand satisfies  h  G =  g  G . d)Gcontains a bump functionϕ. Then, for any proper lsc functionf:X→R ∞ that is bounded below, the set of functionsg∈Gfor whichf+gattains a minimum overXis dense inG.

Proof.Consider the set 3 U n =  g∈G:∃x 0 ∈Xs.t.(f+g)(x 0 )<inf  (f+g)(x):x∈X\B(x 0 ,1/n)  . It follows from hypothesis (b) thatU n is an open set inG. We proceed to show that it is also dense. Letg∈Gandε>0. We wish to exhibith∈Gwith  h  G <ε, together withx 0 ∈X such thatg+h∈U n ; that is, such that, for somex 0 ,  f+g+h  (x 0 )<inf   f+g+h  (x):x∈X\B  x 0 ,1/n   .

3 We reproduce the proof given in [?, p.11].