Analyse Master Cours de Francis Clarke
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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
Voir
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
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- distance between
- appropriate finite sums
- then any orthonormal
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- hilbert space
- conclusion below
- finite-dimensional subspaces
- let u?
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 = 12 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 12 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 ) = ∑ in = 1 x , u i X u i , x − proj M ( x ) ∈ M ⊥ , and 2 d 2 M ( x ) = x 2 − ∑ in = 1 x , u i X . Proof. By expressing the norm in terms of the inner product and expanding, we find x − ∑ in = 1 λ i u i 2 = x 2 − 2 ∑ in = 1 λ i x , u i X + ∑ in = 1 λ i 2 . attains aThgelorbigalhtmsiindiemduemfinweshearecoitnsvegxrafduienncttiovnanoisfhe λ s; = th at λ 1 i , s λ , 2 f , . o . r . , λλ in = 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 , wGievdeenfianneoˆ x r α th = on o x r , m u α al X se.tT h u es α e:ar α e t ∈ he A F ou in rier coeffici s e p n a t c s eof X , x awnidtharepsopientcttothe 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 )
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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 ,
