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Hilbert-Spaces.Pdf 222 BRUCE K. DRIVER† 12. Hilbert Spaces 12.1. Hilbert Spaces Basics. Definition 12.1. Let H be a complex vector space. An inner product on H is a function, , : H H C, such that h· ·i × → (1) ax + by, z = a x, z + b y, z i.e. x x, z is linear. h i h i h i → h i (2) x, y = y, x . (3) hx 2i hx, x i 0 with equality x 2 =0iff x =0. k k ≡ h i ≥ k k Notice that combining properties (1) and (2) that x z,x is anti-linear for fixed z H, i.e. → h i ∈ z,ax + by =¯a z,x + ¯b z,y . h i h i h i We will often find the following formula useful: x + y 2 = x + y,x + y = x 2 + y 2 + x, y + y, x k k h i k k k k h i h i (12.1) = x 2 + y 2 +2Re x, y k k k k h i Theorem 12.2 (Schwarz Inequality). Let (H, , ) be an inner product space, then for all x, y H h· ·i ∈ x, y x y |h i| ≤ k kk k and equality holds iff x and y are linearly dependent. Proof. If y =0, the result holds trivially. So assume that y =0. First off notice 2 6 that if x = αy for some α C, then x, y = α y and hence ∈ h i k k x, y = α y 2 = x y . |h i| | |k k k kk k x,y h i Moreover, in this case α := y 2 . k k 2 Now suppose that x H is arbitrary, let z x y − x, y y. (So z is the “orthogonal projection”∈ of x onto y, seeFigure28.)Then≡ − k k h i Figure 28. The picture behind the proof. x, y 2 x, y 2 x, y 0 z 2 = x h iy = x 2 + |h i| y 2 2Re x, h iy ≤ k k − y 2 k k y 4 k k − h y 2 i ° k k ° k k k k ° x, y° 2 = °x 2 |h °i| k° k − y °2 k k from which it follows that 0 y 2 x 2 x, y 2 with equality iff z =0or 2 ≤ k k k k − |h i| equivalently iff x = y − x, y y. k k h i ANALYSIS TOOLS WITH APPLICATIONS 223 Corollary 12.3. Let (H, , ) be an inner product space and x := x, x . Then is a norm on H. Moreoverh· ·i , is continuous on H H,kwherek H his viewedi as thek·k normed space (H, ). h· ·i × p k·k Proof. The only non-trivial thing to verify that is a norm is the triangle inequality: k·k x + y 2 = x 2 + y 2 +2Re x, y x 2 + y 2 +2 x y k k k k k k h i ≤ k k k k k kk k =( x + y )2 k k k k where we have made use of Schwarz’s inequality. Taking the square root of this inequality shows x + y x + y . For the continuity assertion: k k ≤ k k k k x, y x0,y0 = x x0,y + x0,y y0 |h i − h i| |h − i h − i| y x x0 + x0 y y0 ≤ k kk − k k kk − k y x x0 +( x + x x0 ) y y0 ≤ k kk − k k k k − k k − k = y x x0 + x y y0 + x x0 y y0 k kk − k k kk − k k − kk − k from which it follows that , is continuous. h· ·i Definition 12.4. Let (H, , ) be an inner product space, we say x, y H are orthogonal and write x h· y·i iff x, y =0. More generally if A H ∈is a set, x H is orthogonal to⊥A andh writei x A iff x, y =0for all⊂ y A. Let ∈ ⊥ h i ∈ A⊥ = x H : x A be the set of vectors orthogonal to A. We also say that a set S { H∈ is orthogonal⊥ } if x y for all x, y S such that x = y. If S further satisfies,⊂ x =1for all x S, then⊥ S is said to∈ be orthonormal.6 k k ∈ Proposition 12.5. Let (H, , ) be an inner product space then h· ·i (1) (Parallelogram Law) (12.2) x + y 2 + x y 2 =2 x 2 +2 y 2 k k k − k k k k k for all x, y H. (2) (Pythagorean∈ Theorem) If S H is a finite orthonormal set, then ⊂ (12.3) x 2 = x 2. k k k k x S x S X∈ X∈ (3) If A H isaset,thenA⊥ is a closed linear subspace of H. ⊂ Remark 12.6. See Proposition 12.37 in the appendix below for the “converse” of the parallelogram law. Proof. IwillassumethatH is a complex Hilbert space, the real case being easier. Items 1. and 2. are proved by the following elementary computations: x + y 2 + x y 2 = x 2 + y 2 +2Re x, y + x 2 + y 2 2Re x, y k k k − k k k k k h i k k k k − h i =2 x 2 +2 y 2, k k k k and x 2 = x, y = x, y k k h i h i x S x S y S x,y S X∈ X∈ X∈ X∈ = x, x = x 2. h i k k x S x S X∈ X∈ 224 BRUCE K. DRIVER† Item 3. is a consequence of the continuity of , and the fact that h· ·i A⊥ = x A ker( ,x ) ∩ ∈ h· i where ker( ,x )= y H : y,x =0 — a closed subspace of H. h· i { ∈ h i } Definition 12.7. A Hilbert space is an inner product space (H, , ) such that the induced Hilbertian norm is complete. h· ·i Example 12.8. Let (X, ,µ) be a measure space then H := L2(X, ,µ) with inner product M M (f,g)= f gdµ¯ · ZX is a Hilbert space. In Exercise 12.6 you will show every Hilbert space H is “equiv- alent” to a Hilbert space of this form. Definition 12.9. A subset C of a vector space X is said to be convex if for all x, y C the line segment [x, y]:= tx +(1 t)y :0 t 1 joining x to y is contained∈ in C as well. (Notice that any{ vector− subspace≤ of≤X }is convex.) Theorem 12.10. Suppose that H is a Hilbert space and M H beaclosedconvex subset of H. Then for any x H there exists a unique y M⊂ such that ∈ ∈ x y = d(x, M)= inf x z . z M k − k ∈ k − k Moreover, if M is a vector subspace of H, then the point y may also be characterized astheuniquepointinM such that (x y) M. − ⊥ Proof. By replacing M by M x := m x : m M we may assume x =0. − { − ∈ } Let δ := d(0,M)=infm M m and y, z M, seeFigure29. ∈ k k ∈ Figure 29. Thegeometryofconvexsets. By the parallelogram law and the convexity of M, y + z (12.4) 2 y 2 +2 z 2 = y+z 2 + y z 2 =4 2 + y z 2 4δ2 + y z 2. k k k k k k k − k k 2 || k − k ≥ k − k Hence if y = z = δ, then 2δ2 +2δ2 4δ2 + y z 2, so that y z 2 =0. k k k k ≥ k − k k − k Therefore, if a minimizer for d(0, ) M exists, it is unique. · | ANALYSIS TOOLS WITH APPLICATIONS 225 Existence. Let yn M be chosen such that yn = δn δ d(0,M). Taking ∈ 2 k2 k 2 → ≡ 2 y = ym and z = yn in Eq. (12.4) shows 2δm +2δn 4δ + yn ym . Passing to the limit m, n in this equation implies, ≥ k − k →∞ 2 2 2 2 2δ +2δ 4δ + lim sup yn ym . ≥ m,n k − k →∞ Therefore yn ∞ is Cauchy and hence convergent. Because M is closed, y := { }n=1 lim yn M and because is continuous, n →∞ ∈ k·k y = lim yn = δ = d(0,M). n k k →∞ k k So y is the desired point in M which is closest to 0. Now for the second assertion we further assume that M is a closed subspace of H and x H. Let y M be the closest point in M to x. Then for w M, the function ∈ ∈ ∈ g(t) x (y + tw) 2 = x y 2 2tRe x y, w + t2 w 2 ≡ k − k k − k − h − i k k has a minimum at t =0. Therefore 0=g0(0) = 2Re x y,w . Since w M is arbitrary, this implies that (x y) M. Finally suppose− h −y Miis any point∈ such that (x y) M. Then for z −M,⊥by Pythagorean’s theorem,∈ − ⊥ ∈ x z 2 = x y + y z 2 = x y 2 + y z 2 x y 2 k − k k − − k k − k k − k ≥ k − k which shows d(x, M)2 x y 2. That is to say y is the point in M closest to x. ≥ k − k Definition 12.11. Suppose that A : H H is a bounded operator. The adjoint → of A, denote A∗, is the unique operator A∗ : H H such that Ax, y = x, A∗y . → h i h i (The proof that A∗ exists and is unique will be given in Proposition 12.16 below.) A bounded operator A : H H is self - adjoint or Hermitian if A = A∗.
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