Lecture 15 & 16 : Examples of Hilbert Spaces. Projection Theorem. Riesz Represen- tation Theorem. Adjoint Operators. Example 1. 1. The space Rn is a Hilbert space over R with the standard inner product defined by n n x, y := xkyk for x, y R . È Í œ kÿ=1 2. The space Cn is a Hilbert space over C with inner product defined by n n x, y := xkyk for x, y C . È Í œ kÿ=1 b 2 3. The space L2[a, b]= f :[a, b] K s.t f(t) dt < is a Hilbert space over K with inner I æ a | | ŒJ b s product defined by f,g := f(t)g(t) dt. È Í a s Œ 2 4. The space ¸2 = x =(x1,x2,...) s.t xj < is a Hilbert space with inner product defined I j=1 | | ŒJ by q Œ x, y := x y for x =(x ) ¸ ,y =(y ) ¸ . È Í k k k œ 2 k œ 2 kÿ=1 5. The space (¸p, ) with p =2is not an inner product space. ηÎp ” Proof. We will show that the norm does not satisfy the parallelogram equality. Take x =(1, 1, 0, 0, ) ¸ and y =(1, 1, 0, 0, ) ¸ .Then ··· œ p ≠ ··· œ p x = y =21/p, x + y = x y =2 Î Î Î Î Î Î Î ≠ Î So the parallelogram equality is not satisfied. 6. The space (C[a, b], ) is not an inner product space, hence not a Hilbert space. ηΌ t a Proof. Take f(t) = 1 and g(t)= ≠ .Wehave f = g = 1 and f + g =2, f g = 1. So b a Î Î Î Î Î Î Î ≠ Î the parallelogram equality is not satisfied.≠ Proposition 22. If in an inner product space, x x and y y then x ,y x, y . n æ n æ È n nÍæÈ Í Proof. We have x ,y x, y = x ,y x ,y + x ,y x, y = x ,y y + x x, y |È n nÍ≠È Í| |È n nÍ≠È n Í È n Í≠È Í| |È n n ≠ Í È n ≠ Í|Æ x ,y y + x x, y x y y + x x y Æ|È n n ≠ Í| |È n ≠ Í|ÆÎ nÎÎ n ≠ Î Î n ≠ ÎÎ Î Let n , we get æŒ 0 lim xn,yn x, y x 0+0 y =0. n Æ æŒ |È Í≠È Í|ÆÎ Î Î Î Hence, lim xn,yn x, y =0, lim xn,yn = x, y , which completes the proof. n n æŒ |È Í≠È Í| æŒÈ Í È Í 65 3.2 Orthogonal Projection Definition 1. Let A be a subset of an inner product space X.Theorthogonal complement of A is defined as A‹ := x X : x, a =0 for all a A . { œ È Í œ } Proposition 23. Let A be a subset of an inner product space X. Then A‹ is a closed linear subspace of X and A A 0 . fl ‹ µ{ } Proof. Exercise. Theorem 1. Let Y be a closed linear subspace of the real or complex Hilbert space X and x X be given. œ Then the following holds (i) There exists a unique y Y such that œ x y =min x z Î ≠ Î z Y Î ≠ Î œ (ii) The point y in part (i) is the unique vector in Y such that x y Y . ≠ œ ‹ The point y is called the orthogonal projection of x onto the subspace Y . Proof. (i) Existence. Denote d =inf x z . By the definition of the infimum, there exists a sequence z Y Î ≠ Î y Y such that x y d asœ n . We will prove that y is a Cauchy sequence. { n}µ Î ≠ nÎæ æŒ { n} Using parallelogram law, we have 2 2 1 2 2 yn ym +4 x (yn + ym) = 2( x yn + x ym ) for n, m 1. Î ≠ Î . ≠ 2 . Î ≠ Î Î ≠ Î Ø . 1 1 Since Y is a linear subspace of X and yn,ym Y ,wehave (yn +ym) Y . Therefore, x (yn + ym) œ 2 œ . ≠ 2 . Ø d. Hence . 2 . 2 2 2 1 2 2 2 yn ym = 2( x yn + x ym ) 4 x (yn + ym) 2( x yn + x ym ) 4d . Î ≠ Î Î ≠ Î Î ≠ Î ≠ . ≠ 2 . Æ Î ≠ Î Î ≠ Î ≠ . Let n, m ,wehave x yn d, x ym . d and . æŒ Î ≠ Îæ Î ≠ Îæ 2 2 2 0 lim yn ym 4d 4d =0. n,m Æ æŒ Î ≠ Î Æ ≠ Therefore, lim yn ym = 0 and yn is a Cauchy sequence. Since X is a Hilbert space, there exists n,m æŒ Î ≠ Î { } y X such that lim yn = y.Sinceyn Y and Y is closed, y Y . In conclusion, we have n œ æŒ œ œ x y =min x z . Î ≠ Î z Y Î ≠ Î œ Uniqueness. Suppose there is alsoy ˆ Y such that œ x yˆ =min x z . Î ≠ Î z Y Î ≠ Î œ 66 1 1 Applying the parallelogram law and using x (y +ˆy) d (since (y +ˆy) Y ), we have . ≠ 2 . Ø 2 œ . 1 . 2 . y yˆ 2 +4 x (y +ˆy) = 2( x y 2 + x yˆ 2)=4d2, Î ≠ Î . ≠ 2 . Î ≠ Î Î ≠ Î . 1 2 0 y yˆ 2 =4d2 4 x (y +ˆy) 0. ÆÎ ≠ Î ≠ . ≠ 2 . Æ . So y yˆ 2 =0,y=ˆy. Î ≠ Î . (ii). Orthogonality. Clearly, x y,0 = 0. Take z Y,z = 0. We will prove that x y, z = 0. By the È ≠ Í œ ” È ≠ Í construction of y in part (i), we have x y 2 x (y + ⁄z) 2 = x y 2 + ⁄ 2 z 2 ⁄ z,x y ⁄ x y, z , Î ≠ Î ÆÎ ≠ Î Î ≠ Î | | Î Î ≠ È ≠ Í≠ È ≠ Í 0 ⁄ 2 z 2 ⁄ x y, z ⁄ x y, z . Æ| | Î Î ≠ È ≠ Í≠ È ≠ Í x y, z Plugging ⁄ = È ≠ Í into the above inequality, we conclude z 2 Î Î x y, z 2 |È ≠ Í| 0, z 2 Æ Î Î which only happens when x y, z = 0. È ≠ Í Uniqueness. Assume there is also y Y such that x y Y .Theny y =(x y ) (x y) Y . ú œ ≠ ú œ ‹ ≠ ú ≠ ú ≠ ≠ œ ‹ On the other hand, y y Y since y, y Y .Soy y Y Y 0 . Therefore, y y = 0 and ≠ ú œ ú œ ≠ ú œ fl ‹ µ{ } ≠ ú y = yú. Note in part (i), we only need the condition that if y ,y Y then its average is also in Y . Therefore, n m œ we have a more general result for part (i). Theorem 2 (Hilbert’s Projection Theorem). Given a closed convex set Y in a Hilbert space X and x X. œ There exists a unique y Y such that œ x y =min x z . Î ≠ Î z Y Î ≠ Î œ Corollary 5 (Orthogonal Decomposition). Let Y be a closed linear subspace of the real or complex Hilbert space X. Then every vector x X can be uniquely represented as œ x = y + w, y Y, w Y ‹. œ œ The orthogonal decomposition is usually written as X = Y Y . ü ‹ Note: We can prove that X/Y =≥ Y ‹ (linear isomorphism). Definition 2 (Orthogonal Projection). Let Y be a closed linear subspace of the real or complex Hilbert space X.ThemapP : X X, P (x)=y, where x = y +w and (y, w) Y Y , is called the orthogonal Y æ Y œ ◊ ‹ projection in X onto Y . 67 3.3 Riesz Representation Theorem Lemma 7. Let (X, , ) be an inner product space. Then È Í 1. x, 0 = 0,x =0, x X È Í È Í ’ œ 2. If there are y ,y X such that x, y = x, y for all x X, then y = y . 1 2 œ È 1Í È 2Í œ 1 2 Proof. Exercise. Theorem 1 (Riesz Representation Theorem). Let X be a Hilbert space over K, where K = R or K = C. 1. For every y X, the functional f : X K, f(x)= x, y is an element in Xú and f = y . œ æ È Í Î Î Î Î 2. Conversely, for every f X , there exists a unique y X such that f(x)= x, y for every x X. œ ú œ È Í œ Moreover, f = y . Î Î Î Î Proof. (1). If y = 0, then the function f(x)= x, 0 = 0, for every x X, is an element in X and È Í œ ú f =0= y . Î Î Î Î If y = 0, we first verify that f(x)= x, y is linear. (prove this). ” È Í Using Cauchy-Schwarz inequality, we have f(x) = x, y x y . | | |È Í|ÆÎ ÎÎ Î So f is bounded and f y . Î ÎÆÎ Îf(x) f(y) On the other hand, f =sup| | | | = y . Therefore, f = y . Î Î x=0 x Ø y Î Î Î Î Î Î ” Î Î Î Î (2). Existence. Consider f X .Iff = 0, we can take y = 0 and f(x)= x, 0 for every x X. œ ú È Í œ Consider f X ,f = 0. œ ú ” Step 1: Prove that dim(X/ ker f) = 1. • Since f = 0, Imf = 0 . Moreover, since Imf is a subspace of K and dim K = 1, we have dim Imf ” ” { } K Æ 1.