CHAPTER II HILBERT SPACES

CHAPTER II HILBERT SPACES 2.1 Geometry of Hilbert Spaces Definition 2.1.1. Let X be a complex linear space. An inner product on X is a function, : X X C which satisfies the following axioms : 1. y, x = x, y, the complex conjugate of x, y. 2. x + y, z = x, z + y, z. 3. αx, y = α x, y. 4. x, x 0 and x, x = 0 if and only if x = 0. Here x, y, z X and α is a complex number. An inner product space is a linear space with an inner product on it. Proposition 2.1.2. Let X be an inner product space. Then (i) x, y + z = x, y + x, z. (ii) x, αy = ᾱ x, y. (iii) x, z = y, z x = y.

42 (iv) x, y = 0 for all y X if and only if x = 0 (the zero-element). Proof. We have x, y + z = y + z, x = y, x + z, x = x, y + x, z, this proves (i). Since x, αy = αy, x = ᾱ y, x, then (ii) is true. Suppose that x, z = y, z for all z X. Then x y, z = x, z + y, z = x, z + ( 1) y, z = x, z y, z = 0. Taking z = x y, we have x y, x y = 0 and so x y = 0. Hence x = y. Therefore (iii) holds. Suppose that x, y = 0 for all y X. Putting y = x. Then x, x = 0, whence x = 0. Conversely, if x = 0, then for all y X, we have x, y = 0, y = 0 + 0, y = 0, y + 0, y.

43 Hence x, y = 0, y = 0. This proves (iv). Definition 2.1.3. Let X be an inner product space. For each x X, write x = x, x. Theorem 2.1.4. (Cauchy-Schwarz inequality). Let X be an inner product space. Then for any x, y X, x, y x y, and the equality holds if and only if x and y are linearly dependent, i.e., there exists α, β F, not both zero, such that αx + βy = 0. Proof. If y = 0, then the inequality reduces to an equality. Assume that y 0. For any α F, we have 0 x αy 2 (2.1) = x αy, x αy = x, x α y, x ᾱ x, y + αᾱ y, y = x 2 α y, x ᾱ [ x, y α y 2]. If we choose α so that α = x, y y 2, then the term in the parenthesis becomes zero and hence 0 x 2 x, y 2 y, x, y

44 which implies that x, y 2 x 2 y 2. It is easy to see that if x and y are linearly dependent, then the equality holds. Assume that the equality holds and y 0. Then the inequality in (2.1) must be equality. Hence, x αy = 0, that is, x αy = 0 for some α F. Theorem 2.1.5. 1. Every inner product space is a normed linear space. 2. Let X be an inner product space. Then the inner product space is a continuous function from X X into F. In particular, x n x and y n y x n, y n x, y. Proof. Let X be an inner product space. 1. It suffices to show that is a norm on X. It is easy to see that x 0 and x = 0 if and only if x = 0. For any x X and α F, αx 2 = αx, αx = αᾱ x, x = α 2 x 2. Hence, αx = α x.

45 To prove the triangle inequality, let x, y X. x + y 2 = x + y, x + y = x, x + x, y + y, x + y, y = x 2 + 2Re x, y + y 2 x 2 + 2 x y + y 2 = ( x + y ) 2. Hence, x + y x + y. Therfore, is a norm on X. This proves 1. 2. Let (x n ) and (y n ) be sequences in X such that x n x and y n y. Since (x n ) is convergent, it is bounded. Hence, there exists an M > 0 such that x n M for any n N. Then x n, y n x, y = x n, y n x n, y + x n, y x, y x n, y n y + x n x, y x n y n y + x n x y M y n y + x n x y. From this, it follows that x n, y n x, y. Hence 2 is true. Proposition 2.1.6. (Parallelogram Law). Let X be an inner product space. Then for any x, y X, x + y 2 + x y 2 = 2 x 2 + 2 y 2.

46 Proof. For any x, y X, we have x + y 2 = x + y, x + y = x 2 + x, y + y, x + y 2 and x y 2 = x y, x y = x 2 x, y y, x + y 2. We immediately see that x + y 2 + x y 2 = 2 x 2 + 2 y 2. The Parallelogram Law is a fundamental property of inner product spaces. It characterizes the subclass of the Hilbert spaces in the class of the Banach spaces. Indeed, a Banach space is a Hilbert space if and only if its norm satisfies the Parallelogram Law. The next theorem shows that we can write an inner product in terms of the norm. It is called Polarization Identity. Proposition 2.1.7. (Polarization Identity). Let X be an inner product space. Then if F = C, for all x, y X, we get 4 x, y = x + y 2 x y 2 + i x + iy 2 i x iy 2. Proof. x + y 2 = x + y, x + y = x 2 + x, y + y, x + y 2 and x y 2 = x y, x y = x 2 x, y y, x + y 2.

47 Therefore, x + y 2 x y 2 = 2 x, y + 2 y, x. (2.2) Replacing y by iy in (2.2), we obtain x + iy 2 x iy 2 = 2 x, iy + 2 iy, x = 2ī x, y + 2i y, x = 2i x, y + 2i y, x. (2.3) Multiplying (2.3) by i throughout, we get i x + iy 2 i x iy 2 = 2 x, y 2 y, x. (2.4) Adding (2.2) and (2.4), we get x + y 2 x y 2 + i x + iy 2 i x iy 2 = 4 x, y. Remark 2.1.8. If F = R, then x + y 2 x y 2 = 4 x, y. Definition 2.1.9. A complete inner product space is called a Hilbert space. In other words, a Hilbert space is a Banach space whose norm is defined by an inner product. Proposition 2.1.10. A closed subspace of a Hilbert space is a Hilbert space. A

48 finite-dimentional inner product space is a Hilbert space. Example 2.1.11. F n is an inner product space with respect to the following inner product x, y = n x i ȳ i = x 1 ȳ 1 + x 2 ȳ 2 +... + x n ȳ n, i=1 where x = (x 1, x 2,...x n ) and y = (y 1, y 2,..., y n ) F. Since F n is finite-dimentional, it is a Hilbert space. Note that when F = R, then the inner product is simply x, y = n x i y i. i=1 Example 2.1.12. In the space l 2, for any two sequence x = (x k ) and y = (y k ), let x, y = x k ȳ k. k=1 Then l 2 is a Hilbert space since it is a Banach space with respect to the 2 norm. Example 2.1.13. Let (X, µ) be a measure space. Then L 2 (X, µ) is a Hilbert space with respect to the inner product f, g = fḡdµ (f, g L 2 (X, µ)). X Example 2.1.14. The space C[0, 1] of all complex-valued continuous function defined on [0, 1] is an inner product space under the inner product f, g = 1 0 f(t)g(t)dt. Here f = ( 1 0 f(t) 2 dt) 1/2. Since it is not complete under this norm (we leave

49 these as an exercise), C[0, 1] is not a Hilbert space. Definition 2.1.15. Let X be an inner product space. If x, y = 0, then x is said to be orthogonal to y. We express this fact by writing x y. Let A X. We say that x A if x, y = 0 for all y A. A denotes the set of all those element in X which are orthogonal to A, that is A = {x X x A}. A is called the orthogonal complement of A. We remark that inner product spaces are distinguished from general normed spaces by orthogonality. Thus orthogonality is an important concept. Example 2.1.16. Let C[ π, π] be the inner product space of all real valued functions defined on [ π, π] with inner product f, g = π π f(x)g(x)dx. Since cos x, sin x = π cos x sin xdx = 0, cos x and sin x are orthogonal in C[ π, π]. π Note that sin x = π. Theorem 2.1.17. If M is a subset of an inner product space X, then M is a closed subspace of X. Proof. It is easy to see that M is a subspace of X. To show that it is closed, let (x n ) be a sequence in M which converges to x X.

50 Claim that x M. Then x, z = lim x n, z n = lim n x n, z = 0, for any z M by the continuity of the inner product. Thus x M. Proposition 2.1.18. Let X be an inner product space. 1. {0} = X and X = {0}. 2. If A is a subset of X, then A A = or A A = {0} ; if A is a subspace of X, then A A = {0}. 3. For any subsets A, B of X, if A B, then B A. 4. For any subset A of X, A A. Proof. 1. is trivial. 2. Assume that A A. Let x A A. Since x A, we have x, y = 0 for each y A. In particular, x, x = 0. Hence x = 0. This shows that A A {0}. Now, assume that A is a subspace of A. Since both A and A are subspaces of X, 0 A A. Hence A A = {0}. 3. Assume that A B. Let x B. If y A, then y B and hence x, y = 0. This shows that x A. Thus, B A. 4. Let x A. Then x, y = 0 for any y A. Hence, x A.

51 Proposition 2.1.19. If a vector x is orthogonal to A, then x is also orthogonal to Ā. Proof. Exercise. Solution. Let x A. Then x, y = 0 for all y A.. If z Ā, then there is a sequence (x n ) in A converging to z. But then x, x n = 0 for all n. Hence lim n x, x n = 0, whence x, z = 0. Thus, x is orthogonal to Ā. Remark 2.1.20. If x x, then x, x = 0. That is, x 2 = 0. Hence x = 0. Thus, 0 is the only vector which is orthogonal to itselt. Proposition 2.1.21. Let A be a nonempty subset of a Hilbert space H. Then A is a closed linear subspace of H. Proof. Let a A be arbitrary. Then a = {x H x, a = 0}. If x, y a, then x, a = 0 and y, a = 0. But then, αx + βy, a = αx, a + βy, a = α x, a + β y, a = 0, and so αx + βy a. Hence, a is a linear subspace. Let p be an element of the closure of a. Then there is a sequence (x n ) in a converging to p. We have p, a = lim x n, a n = lim n x n, a = 0. Hence, p a. Accordingly the closure of a is a subset of a. Thus, the closure of a = a.

52 In order words, a is a closed in H. Since A = a A a, it follows that A is a closed linear subspace of H. Theorem 2.1.22. Let A be a closed subspace of a Hilbert space H and let x 0 H. Then there exists a unique element x in A such that x 0 x = d, where d = inf a A { x 0 a }. Also, x 0 x A and x is the only element in A such that x 0 x is orthogonal to A. Proof. By definition of d, there is an element x n in A such that d x 0 x n < d + 1 n n = 1, 2,... (2.5) We claim that (x n ) is a Cauchy sequence. By the Parallelogram law, we get 2 x 0 x n 2 + 2 x 0 x m 2 = x n x m 2 + 2x 0 x n x m 2. Hence, ( ) x n x m 2 = 2 x 0 x n 2 + 2 x 0 x m 2 4 x xn + x m 2 0. (2.6) 2 But x n + x m 2 belongs to A, thus ( ) d x xn + x m 0. (2.7) 2

53 Using (2.5) and (2.7) in (2.6), we obtain that ( x n x m 2 2 d + n) 1 2 ( + 2 d + 1 ) 2 4d 2 m = 2 n + 2 ( 1 2 m + 4d 2 n + 1 ) 0 as n, m. m Thus (x n ) is a Cauchy sequence. Since A is a closed subspace of the complete space H, A is complete. Consequently, x n x A as n. (2.8) Letting n in (2.5), we infer that lim n x n x 0 = d. Therefore, by using (2.8), we have lim n x n x 0 = d = x x0. It remains to prove the uniqueness of x. If there were y in A with y x 0 = d, then we would obtain the following impossibility : d 2 x 0 x + y 2 2 = 2 x 0 2 x 2 + 2 x 0 2 2 y 2 x 2 2 y 2 < x 0 x 2 + x 0 y 2 = d 2. 2 2 2 Hence, x is unique. Now let y A. If y = 0, then x 0 x, y = 0, so (x 0 x) A.

54 We suppose that y 0. Put x 0 x = z. Then for any complex number α, x + αy A. Accordingly, x 0 (x + αy) 2 x 0 x 2, that is, z αy 2 z 2. Or equivalently, z 2 z, αy αy, z + α 2 y 2 z 2. z, y z, y 2 Taking α = 2. Then y y 2 0. Or, z, y 2 0 which gives z, y = 0. In words, (x 0 x 0 ) A. Now, let (x 0 x) A. Assume that there is another element a A such that x 0 a A. Then x 0 x, y = x 0 a, y, and so, x, y = a, y y A. This implies that x = a. Theorem 2.1.23. (Projection Theorem). If A is a closed subspace of a Hilbert space H, then H = A A. Proof. Let x 0 H. If x 0 A, then x 0 A + A. Hence, H = A + A. If x 0 H A, then by the previous theorem, there is a vector x in A such that x 0 x A.

55 Hence, x 0 = x + (x 0 x) A + A and so H = A + A. Furthermore, A A = {0}. Hence we have the result. Theorem 2.1.24. If A is a closed linear subspace of a Hilbert space H, then A = A. Proof. Exercise. Solution. Since A is a closed linear subspace of H, we have H = A A. We show that A A. To this end, let z A. Then z = x + y, where x A and y A. Now z, x = x + y, x = x, x + y, x = x 2. Similarly, z, y = y 2. Since z A and y A, we have z, y = 0. Therefore, y 2 = 0 and so y = 0. Hence, z = x A. Thus A A. (2.9) If x A, then x A and so x A. Hence, A A. (2.10) From (2.9) and (2.10), we get A = A. Corollary 2.1.25. If B is a nonempty subset of a Hilbert space H, then B = B. Proof. Taking A = B in the previous theorem, then A = A.

56 Theorem 2.1.26. (Pythagorean Theorem). If x and y are two vectors in a Hilbert space H, such that x y, then x + y 2 = x y 2 = x 2 + y 2. Proof. We have x + y 2 = x + y, x + y = x, x + y, x + x, y + y, y = x 2 + y 2 because y, x = x, y = 0. Also, x y 2 = x y, x y = x, x x, y y, x + y, y = x 2 + y 2 because y, x = x, y = 0. This completes the proof. Theorem 2.1.27. (Riesz s Theorem). If ϕ is a bounded linear functional on H, then there exists a unique y H such that ϕ(x) = x, y x H. Moreover, ϕ = y. Proof. If ϕ = 0, then choose y = 0. Assume that ϕ 0. Let M = ker ϕ.

57 Then M is a closed proper subspace of H. Hence, M {0}. Choose a M with a = 1. Note that ϕ(a)x ϕ(x)a M for each x H. This implies that ϕ(a)x ϕ(x)a, a = 0 for each x H. Let y = ϕ(a)a. Then ϕ(x) = ϕ(a) x, a = x, ϕ(a)a = x, y for any x H. If y 1, y 2 H are such that x, y 1 = x, y 2 for any x H, then y 1 = y 2. Next, we will show that ϕ = y. Let x H. Then ϕ(x) = x, y y x. This implies that ϕ y. On the other hand, y 2 = y, y = ϕ(y) ϕ y. Hence, ϕ = y.

58 2.2 Exercises 1. Show that the following Banach spaces are not Hilbert spaces (a) C[a, b] under the supremum norm; (b) l p, for p 2. 2. Show that a normed linear space is an inner product space if and only if the Parallelogram law holds. 3. Prove Proposition 2.1.19 4. Prove Theorem 2.1.24. 5. Show that in a complex inner product space, x y if and only if x + αy = x αy for all scalars α. 6. If M and N are closed subspaces of a Hilbert space and M N, show that M +N is a closed subspace. 7. In an inner product space, prove that if x n y and x n, y y 2, then x n y.