Analyse Master Cours de Francis Clarke

Francis Clarke

Description

Niveau: Supérieur, Master
Chapter 7 Hilbert spaces Analyse Master 1 : Cours de Francis Clarke (2011) An inner product on X refers to a bilinear mapping · , · X : X?X ? R (that is, linear separately in each variable) such that x,y X = y,x X ?x, y ? X . A Banach space X is said to be a Hilbert space if it admits an inner product satis- fying x2 = x,x X ?x ? X . Canonical cases of Hilbert spaces include Rn, L2(?), and 2. We have, for exam- ple, u,v Rn = u • v, f ,g L2(?) = ? f (x)g(x)dx. Some rather remarkable consequences follow just from the existence of this scalar product. We suspect that the reader has seen the more immediate ones; we review them now, rather expeditiously. 7.1 Basic properties The first conclusion below is called the Cauchy-Schwarz inequality, and the sec- ond is known as the parallelogram identity. 121

• then


• distance between


• appropriate finite sums


• then any orthonormal


• ?y ?m


• hilbert space


• conclusion below


• finite-dimensional subspaces


• let u?

Sujets

• Clarke
• Hilbert space

Informations

• Publié par: cidur
• Nombre de lectures: 63
• Langue: English

Extrait

Chapter 7 Hilbert spaces

Analyse Master 1 : Cours de Francis Clarke (2011)

An inner product on X refers to a bilinear mapping ￿ ∙ , ∙  X : X × X → R (that is, linear separately in each variable) such that ￿ x , y  X = ￿ y , x  X ∀ x , y ∈ X . A Banach space X is said to be a Hilbert space if it admits an inner product satis- fying ￿ x ￿ 2 = ￿ x , x  X ∀ x ∈ X . Canonical cases of Hilbert spaces include R n , L 2 ( Ω ) , and ￿ 2 . We have, for exam- ple, ￿ u , v  R n = u • v , ￿ f , g  L 2 ( Ω ) =  Ω f ( x ) g ( x ) dx . Some rather remarkable consequences follow just from the existence of this scalar product. We suspect that the reader has seen the more immediate ones; we review them now, rather expeditiously.

7.1 Basic properties

The first conclusion below is called the Cauchy-Schwarz inequality , and the sec- ond is known as the parallelogram identity .

121

122 Cours de Francis Clarke : Hilbert spaces 7.1 Proposition. Let X be a Hilbert space, and let x , y be points in X . Then  ￿ x , y  X  ￿ ￿ x ￿ ￿ y ￿ and  x + 2 y  2 +  x − 2 y  2 = 1 2  ￿ x ￿ 2 + ￿ y ￿ 2  . Proof. We may suppose x , y  = 0. For any λ > 0, we have 0 ￿  x − λ y  2 = ￿ x − λ y , x − λ y  X = ￿ x ￿ 2 − 2 λ ￿ x , y  + λ 2 ￿ y ￿ 2 . This yields 2 ￿ x , y  ￿ ￿ x ￿ 2 / λ + λ ￿ y ￿ 2 . Putting λ = ￿ x ￿ / ￿ y ￿ gives the required inequality. The identify is proved by writing the norms in terms of the inner product and expanding.  It follows that the square of the norm is strictly convex, in the sense that  x + y 2 1 2   x  2 +  y  2  . x = y = ⇒  2  < 7.2 Theorem. Any Hilbert space is uniformly convex, and therefore reflexive. To every element ζ in X ∗ there corresponds a unique u ∈ X such that ζ ( x ) = ￿ u , x  X ∀ x ∈ X ,  ζ  ∗ = ￿ u ￿ . The mapping ζ → u is an isometry from X ∗ to X . It is natural, therefore, to identify the dual X ∗ of a Hilbert space X with the space itself, which we usually do. Proof. Let ε > 0 and x , y ∈ B satisfy ￿ x − y ￿ > ε . By the parallelogram identity we have x + 1 ε 2 x + y  2 y  2 < − 4 = ⇒  2  < 1 − δ , where δ : = 1 − [ 1 − ε 2 / 4 ] 1 / 2 . This confirms that X is uniformly convex. Let us now consider the linear operator T : X → X ∗ defined by ￿ T x , y  = ￿ x , y  X ∀ y ∈ X . We deduce  T x  ∗ = ￿ x ￿ , by the Cauchy-Schwarz inequality. Thus T is an isome- try, and T ( X ) is a closed subspace of X ∗ . To conclude the proof of the theorem, it suffices to prove that T ( X ) is dense in X ∗ . Assume the contrary; then there exists θ ∈ X ∗∗ different from 0 such that ￿ θ , T ( X )  = 0. Because X is reflexive, we may write θ = Jx ¯ for some point x ¯ ∈ X . Then 0 = ￿ θ , T x  = ￿ Jx ¯ , T x  = ￿ T x , x ¯  = ￿ x , x ¯  X ∀ x ∈ X , whence x = 0 and θ = 0, a contradiction.  ¯ 7.3 Proposition. Let C be a closed, convex, nonempty subset of a Hilbert space X , and let x ∈ X . Then there exists a unique u ∈ C satisfying d C ( x ) = ￿ x − u ￿ .

7.1 Basic properties 123 Proof. The existence of a closest point u is known (Exer. 5.51). The uniqueness follows easily from the strict convexity of the square norm.  The point u is called the projection of x onto C , denoted proj C ( x ) . We proceed to characterize it geometrically. (Since we identify X ∗ with X , the normal cone N C ( u ) is viewed as lying in the space X itself.) 7.4 Proposition. Let C be a closed, convex, nonempty subset of a Hilbert space X , and let x ∈ X . Then u = proj C ( x ) ⇐⇒ x − u ∈ N C ( u ) ⇐⇒ ￿ x − u , y − u  X ￿ 0 ∀ y ∈ C . Proof. Let us first consider u = proj C ( x ) . Let y ∈ C and 0 < t < 1. Since C is convex, we have ￿ x − u ￿ 2 ￿  x − ( 1 − t ) u − ty  2 =  ( x − u ) + t ( u − y )  2 . We expand on the left in order to obtain t ￿ x − u , y − u  X ￿ t 2 ￿ y − u ￿ 2 . Now divide by t , and then let t decrease to 0; we arrive at x − u ∈ N C ( u ) . Conversely, let x − u ∈ N C ( u ) . Then for all y ∈ C , ￿ x − u ￿ 2 − ￿ y − x ￿ 2 = 2 ￿ x − u , y − u  X − ￿ u − y ￿ 2 ￿ 0 , whence u = proj C ( x ) . The last condition in the three-way equivalence is simply a restatement of the fact that x − u is a normal vector (Prop. 2.9).  7.5 Exercise. Let C be as in Prop. 7.4. Then  proj C ( x ) − proj C ( y )  ￿ ￿ x − y ￿ ∀ x , y ∈ X .  Another special feature of Hilbert spaces is that subspaces admit complements. For a subset A of X , we define A ⊥ =  ζ ∈ X : ￿ ζ , x  X = 0 ∀ x ∈ A  . The closed subspace A ⊥ (often pronounced “ A perp”) is the orthogonal to A . 7.6 Proposition. Let M be a closed subspace of a Hilbert space X . Then every point x ∈ X admits a unique representation x = m + µ where m ∈ M and µ ∈ M ⊥ ; the point m coincides with proj M ( x ) . Proof. Let x ∈ X , and set m = proj M ( x ) . By Prop. 7.4, we have ￿ x − m , y − m  X ￿ 0 ∀ y ∈ M .

124 Cours de Francis Clarke : Hilbert spaces Since M is a subspace, we obtain ￿ x − m , y  X = 0 ∀ y ∈ M ; that is, x − m ∈ M ⊥ . We have written x in the desired fashion. If x = m  + µ  is another such decomposition, then m − m  = µ  − µ ∈ M ∩ M ⊥ = { 0 } , whence m  = m and µ  = µ . The represen- tation is thereby unique.  7.7 Exercise. In the context of Prop. 7.6, show that the mapping proj M : X → X belongs to L C ( X , X ) .  Orthonormal sets. Two points u and v in a Hilbert space X are orthogonal if ￿ u , v  X = 0. A set  u α : α ∈ A  in X is said to be orthonormal if ￿ u α ￿ = 1 ∀ α , ￿ u α , u β  X = 0 when α  = β . The following result characterizes projection onto finite-dimensional subspaces. 7.8 Proposition. Let X be a Hilbert space, and let M be a subspace of X generated by a finite orthonormal set  u i : i = 1 , . . . , n  . Then proj M ( x ) = ∑ i n = 1 ￿ x , u i  X u i , x − proj M ( x ) ∈ M ⊥ , and 2 d 2 M ( x ) = ￿ x ￿ 2 − ∑ i n = 1  ￿ x , u i  X  . Proof. By expressing the norm in terms of the inner product and expanding, we find  x − ∑ i n = 1 λ i u i  2 = ￿ x ￿ 2 − 2 ∑ i n = 1 λ i ￿ x , u i  X + ∑ i n = 1 λ i 2 . attains a Th g e lo r b ig al ht m s i i n d i e m d u e m fin w es he a re co it n s ve g x ra f d u ie n n c t tio v n an o is f he λ s; = th  at λ 1 i , s λ , 2 f , . o . r . , λ λ i n  = w ￿ h x i , c u h i  X ( i = 1 , . . . , n ) . The corresponding linear combination of the u i is therefore proj M ( x ) . Explicit cal- culation yields ￿ x − proj M ( x ) , u i  X = 0 ∀ i , so that x − proj M ( x ) ∈ M ⊥ . Finally, the expression for d 2 M ( x ) follows from expanding  x − proj M ( x )  2 .  a Hilbert x ∈ X , w Gi e v d e e n fi a n n e o ˆ x r α th = on ￿ o x r , m u α al  X se . t T  h u es α e : ar α e t ∈ he A F  ou in rier coeffici s e p n a t c s e of X , x a w n i d th a re p s o p i e nt cttothe orthonormal set. 7.9 Theorem. (Bessel) Let  u α : α ∈ A  be an orthonormal set. Then the set A ( x ) =  α ∈ A : x ˆ α  = 0  is finite or countable, and we have 2 ∑  x ˆ α  2 = ∑  ˆ x α  ￿ ￿ x ￿ 2 ∀ x ∈ X . α ∈ A α ∈ A ( x )

125

7.1 Basic properties Proof. The sum on the left is defined to be sup  ∑ α ˆ  2 : F ⊂ A , F finite  , ∈ F  x α so the inequality follows from Prop. 7.8. We also deduce that for each i , the set of indices α for which  ˆ x α  > 1 / i is finite, which implies that A ( x ) is countable.  A maximal orthonormal set  u α : α ∈ A  is called a Hilbert basis for X . One may use Zorn’s lemma to prove that a Hilbert basis exists. Theorem 7.9 implies that the expression ∑ α ∈ A x ˆ α u α is absolutely convergent; it defines therefore an element of X , by completeness. The next result shows that when  u α : α ∈ A  is a Hilbert basis, we can recover x from its Fourier coefficients. A Hilbert space isomorphism between X and Y refers to an isometry T that also preserves the inner product: ￿ T x , Tu  Y = ￿ x , u  X ∀ x , u ∈ X . 7.10 Theorem. (Parseval) Let  u α : α ∈ A  be a Hilbert basis for the Hilbert space X . The set of (finite) linear combinations of the u α is a dense subset of X . We have x = ∑ ˆ x α u α , ￿ x ￿ 2 = ∑  x ˆ α  2 , ￿ x , y  X = ∑ ˆ x α y ˆ α ∀ x , y ∈ X . α ∈ A α ∈ A α ∈ A X is finite dimensional if and only if X admits a finite Hilbert basis, in which case X is isomorphic as a Hilbert space to R n ( for some n ) . When X is infinite dimensional, then X is separable if and only if X admits a countable Hilbert basis; in this case, X is isomorphic as a Hilbert space to ￿ 2 . Proof. We merely sketch the proof of the theorem. The density assertion is that the closed subspace generated by the u α coincides with X . This is easily proved by contradiction, using Prop. 7.6. As mentioned above, the sum ∑ α ∈ A ˆ x α u α defines an element u ∈ X ; it satisfies ￿ u − x , u β  X = 0 ∀ β ∈ A , since this holds when u is replaced by any of the finite sums ∑ α ∈ F ˆ x α u α that converge to it. It follows that u = x . Similarly, the expression for ￿ x , y  X is obtained by passing to the limit in the appropriate finite sums that converge respectively to x and y . The assertion in the finite dimensional case follows from Theorem 1.21. If X ad- mits a countable Hilbert basis, then the set of (finite) linear combinations of the u α with rational coefficients defines a countable dense subset of X , which is therefore separable. Finally, suppose that X is separable. Then any orthonormal set is count- able, since the distance between any two distinct elements of the set is √ 2. Let u i be a sequence constituting a Hilbert basis. The preceding conclusions then show that x → ￿ x , u 1  X , ￿ x , u 2  X , . . .  defines an isomorphism from X to ￿ 2 .  7.11 Exercise. The goal is to prove the following result, the Lax-Milgram theorem . (It is a tool designed to solve certain linear equations in Hilbert space.) Theorem. Let b ( u , v ) be a bilinear form on a Hilbert space X . We suppose that b is continuous, and coercive in the following sense: there exist c > 0 , C > 0 such that

126 Cours de Francis Clarke : Hilbert spaces  b ( u , v )  ￿ C ￿ u ￿ ￿ v ￿ , b ( u , u )  c ￿ u ￿ 2 ∀ u , v ∈ X . Then for any ϕ ∈ X , there exists a unique u ϕ ∈ X such that b ( u ϕ , v ) = ￿ ϕ , v  ∀ v ∈ X . If b is symmetric ( that is, if b ( u , v ) = b ( v , u ) ∀ u , v ∈ X ) , then u ϕ may be character- ized as the unique point in X minimizing the function u → ( 1 / 2 ) b ( u , u ) − ￿ ϕ , u  . a) Show that the map u → Tu : = b ( u , ∙ ) defines an element of L C ( X , X ∗ ) satisfying  Tu  ∗  c ￿ u ￿ . b) Prove that T X is closed. c) Prove that T is onto, and then deduce the existence and uniqueness of u ϕ . d) Now let b be symmetric, and define f ( u ) = b ( u , u ) / 2. Show that f is strictly convex, and that f  ( u ; v ) = b ( u , v ) ∀ u , v . e) Prove that the function u → ( 1 / 2 ) b ( u , u ) − ￿ ϕ , u  attains a unique minimum over X at a point u ϕ . Write Fermat’s Rule to conclude.  A crucial property of Hilbert spaces is that their norms are smooth. 7.12 Exercise. Let X be a Hilbert space. Prove that the squared norm function θ : X → R , x → θ ( x ) = ￿ x ￿ 2 is continuously differentiable, with θ  ( x ) = 2 x . Deduce that X admits a bump func- tion belonging to C b 1 , 1 ( X ) (see Exer. 5.23).  It follows from the above that, in a Hilbert space, the norm is differentiable . (One must hear, sotto voce, the words “except at the origin” in such a sentence; a norm can never be differentiable at 0, because of positive homogeneity.) In a general Ba- nach space, however, this is not necessarily the case. In fact, a space may admit no equivalent norm which is differentiable.

7.2 The proximal subdifferential

We proceed to extend in a local fashion, to functions that are not necessarily convex, the notion of subgradient. Let X be a normed space. 7.13 Definition. Let f : X → R ∞ be given, with x ∈ dom f . We say that ζ ∈ X ∗ is a proximal subgradient of f at x if, for some σ = σ ( x , ζ )  0 , for some neighbor- hood V = V ( x , ζ ) of x , we have f ( y ) − f ( x ) + σ ￿ y − x ￿ 2  ￿ ζ , y − x  ∀ y ∈ V .

7.2 The proximal subdifferential 127 The proximal subdifferential of f at x , denoted ∂ P f ( x ) , is the set of all such ζ . The above is very reminiscent of the subdifferential ∂ f ( x ) of convex analysis: when ζ ∈ ∂ f ( x ) , the inequality above holds globally, and for σ = 0. It is decidedly not the case that f is assumed convex here, but let us note that if it happens to be, we obtain the same construct as before: 7.14 Proposition. Let f be convex. Then ∂ P f ( x ) = ∂ f ( x ) . Proof. Clearly it suffices to show that an element ζ ∈ ∂ P f ( x ) belongs to ∂ f ( x ) . To do so, note that for such a ζ , the convex function y → g ( y ) = f ( y ) + σ ￿ y − x ￿ 2 − ￿ ζ , y  attains a local minimum at y = x (by definition of proximal subgradient). Thus 0 ∈ ∂ g ( x ) , which, by the Sum Rule (Theorem 4.11) and Exer. 7.12, yields ζ ∈ ∂ f ( x ) , as required.  Let us turn now to the relation between proximal subgradients and derivatives. Sup- pose that f is Gaˆteaux differentiable at x . We claim that the only possible element of ∂ P f ( x ) is f G  ( x ) . To see this, fix any v ∈ R n . Observe that we may set y = x + tv in the proximal subgradient inequality to obtain  f ( x + tv ) − f ( x )  / t  ￿ ζ , v  − σ t 2 ￿ v ￿ 2 for all t sufficiently small . Passing to the limit as t ↓ 0, this yields ￿ f G  ( x ) , v    ζ , v  . Since v is arbitrary, we must have ζ = f G  ( x ) . We have proved: 7.15 Proposition. If f is Gaˆteaux differentiable at x , then ∂ P f ( x ) ⊂  f G  ( x )  . 7.16 Example. The last proposition may fail to hold with equality; in general, ∂ P f ( x ) may be empty even when f G  ( x ) exists. To develop more insight into this question, bear in mind that the proximal subdifferential is philosophically linked to (local) convexity, as mentioned above. At points where f has a “concave cor- ner”, there will be no proximal subgradients. A simple example is provided by the function f ( x ) = −￿ x ￿ . If ζ ∈ ∂ P f ( 0 ) , then, by definition, −￿ y ￿ − 0 + σ ￿ y ￿ 2  ￿ ζ , y  for all y near 0 . Fix any point v ∈ X , and substitute y = tv in the inequality above, for t > 0 sufficiently small. Dividing across by t and then letting t ↓ 0 leads to ￿ ζ , v  ￿ −￿ v ￿ ∀ v ∈ X , a condition that no ζ can satisfy. Thus, we have ∂ P f ( 0 ) = /0. The proximal subdifferential ∂ P f ( x ) can be empty even when f is continuously differentiable. Consider the function f ( x ) = −￿ x ￿ 3 / 2 , which is C 1 , with derivative 0 at 0. Yet, we claim, ∂ P f ( 0 ) = /0.Toseethis,let ζ ∈ ∂ P f ( 0 ) . By Prop. 7.15, ζ must be 0. But then the proximal subgradient inequality becomes

128 Cours de Francis Clarke : Hilbert spaces −￿ y ￿ 3 / 2 − 0 + σ ￿ y ￿ 2  0 for all y near 0 . We let the reader verify that this cannot hold; thus, ∂ P f ( 0 ) = 0/ once again.  It might be thought that a subdifferential which can be empty is not going to be of much use; for example, its calculus might be very poor. In fact, the possible empti- ness of ∂ P f is a positive feature in some contexts, as in characterizing certain proper- ties (we shall see this in connection with viscosity solutions later). And the calculus of ∂ P f is complete and rich (but fuzzy, in a way that will be made clear). It is evident that if f has a (finite) local minimum at x , then ∂ P f ( x ) is nonempty, since we have 0 ∈ ∂ P f ( x ) (Fermat’s Rule). This simple observation will be the key to proving the existence (for certain Banach spaces) of a dense set of points at which ∂ P f is nonempty. First, we require a simple rule in proximal calculus: 7.17 Proposition. Let x ∈ dom f , and let g : X → R be differentiable in a neighbor- hood of x , with g  Lipschitz near x . Then ∂ P  f + g  ( x ) = ∂ P f ( x ) +  g  ( x )  . Proof. We begin with Lemma. There exists δ > 0 and M such that y , z ∈ B ( x , δ ) = ⇒  g ( u ) − g ( x ) − ￿ g  ( x ) , u − x   ￿ M ￿ u − x ￿ 2 . To see this, we invoke the Lipschitz hypothesis on g  to find δ > 0 and M such that y , z ∈ B ( x , δ ) = ⇒  g  ( y ) − g  ( z )  ∗ ￿ M ￿ y − z ￿ . For any u ∈ B ( x , δ ) , by the Mean Value theorem, there exists z ∈ B ( x , δ ) such that g ( u ) = g ( x ) + ￿ g  ( z ) , u − x  . Then, by the Lipschitz condition for g  ,  g ( u ) − g ( x ) − ￿ g  ( x ) , u − x   =  ￿ g  ( z ) − g  ( x ) , u − x   ￿ M ￿ z − x ￿ ￿ u − x ￿ ￿ M ￿ u − x ￿ 2 , which proves the lemma. Now let ζ ∈ ∂ P  f + g  ( x ) . Then, for some σ  0 and neighborhood V of x , we have f ( y ) + g ( y ) − f ( x ) − g ( x ) + σ ￿ y − x ￿ 2  ￿ ζ , y − x  ∀ y ∈ V . It follows from the lemma that f ( y ) − f ( x ) + ( σ + M ) ￿ y − x ￿ 2  ￿ ζ − g  ( x ) , y − x  ∀ y ∈ V δ : = V ∩ B ( x , δ ) , whence ζ − g  ( x ) ∈ ∂ P f ( x ) . Conversely, if ψ ∈ ∂ P f ( x ) , then, for some σ  0 and neighborhood V of x ,