Finite-dimensional spaces. C n is the space of n-tuples x = (x 1,..., x n ) of complex numbers. It is a Hilbert space with the inner product

Chapter 4 Hilbert Spaces 4.1 Inner Product Spaces Inner Product Space. A complex vector space E is called an inner product space (or a pre-hilbert space, or a unitary space) if there is a mapping (, ) : E E C, called an inner product, that satisfies the conditions: x, y, z E, α C: 1. (x, x) 0 2. (x, x) = 0 if and only if x = 0 3. (x, y + z) = (x, y) + (x, z) 4. (x, αy) = α(x, y) 5. (x, y) = (y, x) Finite-dimensional spaces. C n is the space of n-tuples x = (x 1,..., x n ) of complex numbers. It is a Hilbert space with the inner product (x, y) = n x j y j j=1 Orthogonal Vectors. Let E be an inner product space. Two vectors x, y E are orthogonal if (x, y) = 0. Notation. The orthogonality of the vectors x and y is denoted by x y 61

62 CHAPTER 4. HILBERT SPACES The relation is symmetric, that is, if x y then y x. Examples. Spaces of sequences The spaces of sequences l 0 (sequences with zero tails) and l 2 (sequences with finite 2 norm) are 0inner product spaces with the inner product Spaces of functions (x, y) = x j y j The space C([a, b]) is an inner product space with the inner product ( f, g) = The space C 0 (R) (continuous functions with compact support) is an inner product space with the inner product ( f, g) = f g Spaces of square integrable functions j=1 b The space L 2 ([a, b]) is inner product space with the inner product ( f, g) = a R b The space L 2 (R) is and inner product space with the inner product ( f, g) = f g L 2 ([a, b], µ) (space of square integrable functions with the measure µ) with the inner product ( f, g) = a b a R f g f g µ f g, where µ is an almost everywhere positive measurable function, µ > 0. mathphys15.tex; October 28, 2015; 15:35; p. 60

4.1. INNER PRODUCT SPACES 63 L 2 (R, µ) (space of square integrable functions with the measure µ) with the inner product ( f, g) = µ f g, where µ > 0 almost everywhere. Let Ω be an open set in R n (in particular, Ω can be the whole R n ). The space L 2 (Ω) is the set of complex valued functions such that f 2 <. It is an inner product space with the inner product ( f, g) = f g L 2 (R n ) with the inner product ( f, g) = Ω Let Ω be an open set in R n (in particular, Ω can be the whole R n ) and V be a finite-dimensional vector space. Let, be the inner product on the vector space V. The space L 2 (Ω, V, µ) is the set of vector valued functions f = ( f 1,..., f N ) on Ω such that R Ω R n f g µ f, f = µ Ω Ω N f i 2 <. i=1 It is an inner product space with the inner product ( f, g) = µ f, g = µ Ω Ω N i=1 f i g i Real Inner Product Spaces. The inner product in a real inner product space is symmetric. mathphys15.tex; October 28, 2015; 15:35; p. 61

64 CHAPTER 4. HILBERT SPACES Direct Sum of Inner Product Spaces. Let E 1 and E 2 be inner product spaces. The direct sum E = E 1 E 2 of E 1 and E 2 is an inner product space of ordered pairs z = (x, y) with x E 1 and y E 2 with the inner product defined by (z 1, z 2 ) E = (x 1, x 2 ) E1 + (y 1, y 2 ) E2. Tensor Products of Inner Product Spaces. Let E 1 and E 2 be inner product spaces. For each ϕ 1 E 1 and ϕ 2 E 2 let ϕ 1 ϕ 2 denote the conjugate bilinear form on E 1 E 2 defined by where ψ 1 E 1 and ψ 2 E 2. (ϕ 1 ϕ 2 )(ψ 1, ψ 2 ) = (ψ 1, ϕ 1 ) E1 (ψ 2, ϕ 2 ) E2 Let E = E 1 E 2 be the set of finite linear combinations of such bilinear forms. An inner product on E can be defined by (ϕ ψ, η µ) E = (ϕ, η) E1 (ψ, µ) E2 (with ϕ, η E 1 and ψ, µ E 2 ) and extending by linearity on E. Fock Space. Let E be an inner product space. The space is called the Fock space over E. F(E) = C E E } {{ } n 4.1.1 Norm in an Inner Product Space. Let E be an inner product space. The norm in E is a functional : E R defined by x = (x, x). Theorem 4.1.1 Every inner product space is a normed space with the norm x = (x, x) and a metric space with the metric d(x, y) = (x y, x y). Theorem 4.1.2 Schwarz s Inequality. Let E be an inner product space. Then for any x, y E we have (x, y) x y. mathphys15.tex; October 28, 2015; 15:35; p. 62

4.1. INNER PRODUCT SPACES 65 The equality (x, y) = x y holds if and only if x and y are linearly dependent. Proof: 1. Corollary 4.1.1 Triangle Inequality. Let E be an inner product space. Then for any x, y E we have Proof: x + y x + y. 1. Theorem 4.1.3 Parallelogram Law. Let E be an inner product space. Then for any x, y E we have x + y 2 + x y 2 = 2 ( x 2 + y 2) Proof: 1. Theorem 4.1.4 Pythagorean Theorem. Let E be an inner product space. If two vectors x, y E are orthogonal then x + y 2 = x 2 + y 2. Proof: 1. Examples. mathphys15.tex; October 28, 2015; 15:35; p. 63

66 CHAPTER 4. HILBERT SPACES 4.2 Hilbert Spaces Hilbert Space. An inner product space is called a Hilbert space if it is complete as a normed space. Examples. C n is complete. Spaces of sequences The space l 2 of square summable sequences is complete. Proved before. The space l 0 of sequences with vanishing tails is not complete. Counterexample. Spaces of continuous functions C([a, b]) is not complete. Counterexample. C 0 (R) (space of continuous functions with compact support) is not complete. Counterexample. Spaces of square integrable functions. Remark. For any f, g L 2 (Ω) there holds ( f, g) f g 1 f 2 g 2 If Ω is compact then for any f L 2 (Ω) f 1 [µ(ω)] 1/2 f 2 Theorem 4.2.1 Proof: The space L 2 ([a, b]) is complete. 1. Take a Cauchy sequence f n in L 2 so that as n, m. f n f m 2 0 mathphys15.tex; October 28, 2015; 15:35; p. 64

4.2. HILBERT SPACES 67 2. By Schwarz (or Hoelder) inequality f n f m 1 (b a) f n f m 2 0 as n, m. 3. So, f n is Cauchy in L 1. 4. Since L 1 is complete it converges to some f L 1. 5. Then there is a subsequence f pn convergent to f (almost everywhere). 6. Clearly, the subsequence f pn is Cauchy in L 2, f pn f pm 2 0 as n, m. 7. Therefore, as m f pn f 2 0 as n. 8. Therefore, f L 2. 9. Finally, f f n 2 f f pn 2 + f pn f n 2 0. 10. Therefore, f n f. Theorem 4.2.2 The space L 2 (R) is complete. Proof. Idea: Construct a subsequence of a Cauchy sequence in L 2 (R) that converges almost everywhere by a diagonal argument. 1. Let f n be a Cauchy sequence in L 2 (R). 2. Then f n is Cauchy in L 2 ([ 1, 1]). 3. Then it has a subsequence f 1,n that converges in [ 1, 1] almost everywhere. 4. Since f 1,n is Cauchy in L 2 ([ 2, 2]) it has a subsequence f 2,n that converges in [ 2, 2] almost everywhere. 5. Then the subsequence f n,n of f n converges in R almost everywhere to some f. mathphys15.tex; October 28, 2015; 15:35; p. 65

68 CHAPTER 4. HILBERT SPACES 6. Then, one shows that f L 2 (R) and f n f in L 2 (R). Theorem 4.2.3 Let µ be a measurable function on [a, b] positive almost everywhere. The space L 2 ([a, b], µ) is complete. Proof. Rescale functions by µ. More generally, Theorem 4.2.4 The space L 2 (Ω, µ, V) is complete. Sobolev Spaces Let Ω be an open set in R n (in particular, Ω can be the whole R n ) and V a finite-dimensional complex vector space. Let C m (Ω, V) be the space of complex vector valued functions that have continuous partial derivatives of all orders less or equal to m. Let α = (α 1,..., α n ), α j Z +, be a multiindex of nonnegative integers, α i 0, and let Define α = α 1 + + α n. D α f = α x α 1 1 xα n n Then f C m (V, Ω) iff α, α m, i = 1,..., N, x Ω we have D α f i (x) <. The space H m (Ω, V) is the space of complex vector valued functions such that α, α m, D α f L 2 (Ω, V) i.e. such that α, α m, Ω dx D α f, D α f = Ω f. N D α f i (x) 2 dx <. i=1 mathphys15.tex; October 28, 2015; 15:35; p. 66

4.2. HILBERT SPACES 69 It is an inner product space with the inner product ( f, g) = α, α m Ω D α f, D α g = α, α m Ω N i=1 D α f i D α g i The Sobolev space H m (Ω, V) is the completion of the space H m (Ω, V) defined above. 4.2.1 Strong and Weak Convergence Strong Convergence. A sequence (x n ) of vectors in an inner product space E is strongly convergent to a vector x E if Notation. x n x. lim n x n x = 0 Theorem 4.2.5 Let E be an inner product space. Then for every y E the linear functional ϕ y : E C defined by ϕ y (x) = (y, x) x E is continuous (and, therefore, bounded). Proof: Use Schwarz inequality. Weak Convergence. A sequence (x n ) of vectors in an inner product space E is weakly convergent to a vector x E if for any y E lim n (y, x n x) = 0 Notation. x n w x Theorem 4.2.6 A strongly convergent sequence is weakly convergent to the same limit. Proof: Use Schwarz inequality. Converse is not true. Counterexample later. mathphys15.tex; October 28, 2015; 15:35; p. 67

70 CHAPTER 4. HILBERT SPACES Theorem 4.2.7 Let (x n ) be a sequence in an inner product space E and x E. Suppose that 1. x n w x and 2. x n x. Then x n x. Proof: Easy. Theorem 4.2.8 Let S be a subset of an inner product space E such that span S is dense in E, (x n ) be a bounded sequence in E and x E. Suppose that for any y S, lim n (y, x n x) = 0. Then x n w x. Proof: Let z E. Then there is y m span S such that y m z. Then converges to zero as n, m. (x n x, z) = (x n x, y m ) + (x n x, z y m ) w It is possible that x n x but xn does not converge to x. But at least the sequence x n has to be bounded. Theorem 4.2.9 Weakly convergent sequences in a Hilbert space are bounded. Proof: Read in functional analysis books. 4.3 Orthogonal and Orthonormal Systems Orthogonal and Orthonormal Systems. Let E be an inner product space. A set S of vectors in E is called an orthogonal system if any pair of distinct vectors in S is orthogonal to each other. An orthogonal system of unit vectors is an orthonormal system. Every orthogonal system can be made orthonormal. mathphys15.tex; October 28, 2015; 15:35; p. 68

4.3. ORTHOGONAL AND ORTHONORMAL SYSTEMS 71 Let S be a set of vectors in E. We say that x S if x y for any y S. If x S then x span S. The orthogonal complement of S is the space S = {x E x S } Theorem 4.3.1 Orthogonal systems are linearly independent. Proof: Easy. Orthonormal Sequence. A sequence of vectors which is an orthonormal system is an orthonormal sequence; then (x n, x m ) = δ nm Examples. 1. Canonical basis in l 2. (e n ) i = (δ i n) 2. Fourier basis. The functions f n (x) = 1 2π e inx are orthonormal in L 2 ([ π, π]). 3. Legendre polynomials. P n (x) = 1 2 n n! n x(x 2 1) n Show that the functions f n = n + 1 2 P n are orthonormal in L 2 ([ 1, 1]). mathphys15.tex; October 28, 2015; 15:35; p. 69

72 CHAPTER 4. HILBERT SPACES 4. Hermite polynomials. H n = ( 1) n e x2 n xe x2 Show that the functions f n (x) = ( ) 1 2n n!π exp x2 H n (x) 1/4 2 are orthonormal in L 2 (R). Any orthogonal sequence can be always made orthonormal. Gram-Schmidt orthonormalization process. Any sequence of linearly independent vectors can be made orthonormal. 1. Let (y n ) be a linearly independent sequence. 2. Let z n = y n y n 3. Recall that for any orthonormal sequence e n the operator P n defined by n 1 P n = I e k e k or k=1 n 1 P n x = x (e k, x)e k is the projections to the orthogonal complement of span {e 1,..., e n 1 }. 4. Then the sequence (e n ) defined by k=1 e 1 = z 1, n 1 e n = P n z n = z n (e k, z n )e k k=1 is orthonormal. mathphys15.tex; October 28, 2015; 15:35; p. 70

4.3. ORTHOGONAL AND ORTHONORMAL SYSTEMS 73 4.3.1 Properties of Orthonormal Systems Theorem 4.3.2 Pythagorean Formula. Let E be an inner product space and {x n } N be an orthogonal set in E. Then 2 N N x n = x n 2 Proof: By induction. Theorem 4.3.3 Bessel s Equality and Inequality. Let E be an inner product space and {e n } N be an orthonormal set in E. Then x E N 2 x 2 = x n 2 + N x x n e n and where x n = (e n, x). Proof: N x n 2 x 2 Remarks. Let (e n ) be an orthonormal sequence in an inner product space E. The complex numbers x n = (e n, x) are called the generalized Fourier coefficients of x with respect to the orthonormal sequence (e n ). An orthonormal sequence (e n ) in E induces a mapping ϕ : E l 2 defined by ϕ(x) = x = (x n ) The sequence of Fourier coefficients x = (x n ) is square summable, that is x l 2 since for any x E x n 2 x 2 and, therefore, this series converges. mathphys15.tex; October 28, 2015; 15:35; p. 71

74 CHAPTER 4. HILBERT SPACES The expansion x x n e n is called the generalized Fourier series of x. The question is whether the mapping ϕ is bijective and whether the Fourier series converges. Theorem 4.3.4 Let (e n ) be an orthonormal sequence in a Hilbert space H and (x n ) be a sequence of complex numbers. Then the series x n e n converges if and only if (x n ) l 2, that is, the series x n 2 converges. In this case 2 x n e n = x n 2 Proof: By Pythagorean theorem 2 m x k e k = k=n m x k 2 k=n Fourier series of any x H in a Hilbert space H converges. Fourier series of x may converge to a vector different from x! Example. Let (e n ) be an orthonormal sequence. Then f n = e 2n is also orthonormal. Let x = e 1. Then x n = ( f n, x) = 0 and 0 = x n f n x. Let (e n ) be an orthonormal sequence in an inner product space E. The sequence of Fourier coefficients x n = (e n, x) is square summable, and, therefore, lim (e n, x) = 0 x E n Thus, every orthonormal sequence weakly converges to zero. Orthonormal sequences are not strongly convergent since e n = 1 n Z +. mathphys15.tex; October 28, 2015; 15:35; p. 72

4.3. ORTHOGONAL AND ORTHONORMAL SYSTEMS 75 Complete Orthonormal Sequence. Let E be an inner product space. An orthonormal sequence (e n ) in E is complete if x E the Fourier series of x converges to x, that is, x = x n e n, more explicitly, lim n n x x k e k = 0 Example. L 2 ([ π, π]) does not imply pointwise convergence. k=1 Orthonormal Basis. Let E be an inner product space. An orthonormal system B in E is an orthonormal basis if for any x E there exists a unique orthonormal sequence (e n ) in B and a unique sequence (x n ) of nonzero complex numbers such that x = x n e n. Remarks. A complete orthonormal sequence in an inner product space is an orthonormal basis. Let E be an inner product space and (e n ) be a complete orthonormal sequence. Then the set S = span {e n n Z + } is dense in E. Theorem 4.3.5 Let H be a Hilbert space. An orthonormal sequence (e n ) in H is complete if and only if the only vector orthogonal to this sequence is the zero vector, that is, span {e n n Z + } = {0}. Proof: Easy. To prove converse, let y = n(e n, x)e n. Then (x y, e n ) = 0. So, x = y. mathphys15.tex; October 28, 2015; 15:35; p. 73

76 CHAPTER 4. HILBERT SPACES Theorem 4.3.6 Parseval s Formula. Let H be a Hilbert space. An orthonormal sequence (e n ) in H is complete if and only if x H x 2 = x n 2 where x n = (e n, x). Proof: Use Bessel equality. Theorem 4.3.7 Let H 1 and H 2 be Hilbert spaces. If {ϕ k } and {ψ l } are orthonormal bases for H 1 and H 2 respectively, then {ϕ k ψ l } is an orthonormal basis for the tensor product H 1 H 2. Examples. The sequence f n (x) = 1 2π e inx, n Z is orthonormal and complete in L 2 ([ π, π]). Proof later. The sequence g 0 = 1 2π, g n (x) = 1 2π cos(nx), f n (x) = 1 2π sin(nx), n Z + is orthonormal and complete in L 2 ([ π, π]). Proof later. The sequence h 0 = 1 π, h n (x) = 2 π cos(nx), n Z + is orthonormal and complete in L 2 ([0, π]). Proof later. The sequence ψ n (x) = 2 sin(nx), n = 0, 1, 2,..., π is orthonormal and complete in L 2 ([0, π]). Proof later. mathphys15.tex; October 28, 2015; 15:35; p. 74

4.4. SEPARABLE HILBERT SPACES 77 4.4 Separable Hilbert Spaces Separable Spaces. An infinite-dimensional Hilbert space is separable if it contains a complete orthonormal sequence. Examples. L 2 ([ π, π]) l 2 Example (Non-separable Hilbert Space). Let H be the space of all complex valued functions f : R C on R such that they vanish everywhere except a countable number of points in R and f (x) 2 <. f (x) 0 Define the inner product by (g, f ) = f (x)g(x) 0 ḡ(x) f (x). Let f n be an orthonormal sequence in H. Then there are non-zero functions f such that ( f, f n ) = 0 for all n Z +. Therefore, f n cannot be complete and H is not separable. Theorem 4.4.1 Let H be a separable Hilbert space. Then H contains a countable dense subset. Proof: 1. Let (e n ) be a complete orthonormal sequence in H. 2. Define the set n S = (α k + iβ k )e k α k, β k Q, n Z + 3. Then S is countable. 4. Also, x H, k=1 lim n n (e k, x)e k x = 0. k=1 mathphys15.tex; October 28, 2015; 15:35; p. 75

78 CHAPTER 4. HILBERT SPACES 5. Therefore, S is dense in H. Theorem 4.4.2 Let H be a separable Hilbert space. Then every orthogonal set S in H is countable. Proof: 1. Let S be an orthogonal set in H. 2. Let 3. Then x, y S 1, x y, S 1 = { } x x x S. x y 2 = 2. 4. Consider the collection of balls B ε (x) with ε = 2 1/2 /3 for every x S 1. 5. Then, for any x, y S 1, if x y, then the corresponding balls are disjoint, B ε (x) B ε (y) =. 6. Since H is separable, it has a countable dense subset A. 7. Since A is dense in H it must have at least one point in every ball B ε (x). 8. Therefore, S 1 must be countable. 9. Thus S is countable. Unitary Linear Transformations. Let H 1 and H 2 be Hilbert spaces. A linear map T : H 1 H 2 is unitary if x, y H 1 (T(x), T(y)) H2 = (x, y) H1. Hilbert Space Isomorphism. Let H 1 and H 2 be Hilbert spaces. Then H 1 is isomorphic to H 2 if there exists a linear unitary bijection T : H 1 H 2 (called a Hilbert space isomorphism). Remark. Every Hilbert space isomorphism has unit norm T = 1. mathphys15.tex; October 28, 2015; 15:35; p. 76

4.5. TRIGONOMETRIC FOURIER SERIES 79 Theorem 4.4.3 Every infinite-dimensional separable Hilbert space is isomorphic to l 2. Proof: 1. Let (e n ) be a complete orthonormal sequence in H. 2. Let x H. 3. Let x n = (e n, x). 4. This defines a linear bijection T : H l 2 by 5. Let x, y H. 6. Then 7. On another hand (x, y) H = ( T(x) = (x n ). (T(x), T(y)) l 2 = x n e n, y) = x n y n x n (e n, y) = x n y n 8. Therefore T is unitary, and is, therefore, an isomorphism from H onto l 2. Remarks. Isomorphism of Hilbert spaces is an equivalence relation. All separable infinite-dimensional Hilbert spaces are isomorphic. 4.5 Trigonometric Fourier Series Consider the Hilbert space L 2 ([ π, π]). The sequence ϕ n (x) = 1 2π e inx, n Z is an orthonormal sequence in L 2 ([ π, π]). mathphys15.tex; October 28, 2015; 15:35; p. 77

80 CHAPTER 4. HILBERT SPACES Note that L 2 L 1. Consider the space L 1 ([ π, π]). Identify the elements of L 1 ([ π, π]) with 2π periodic functions on R. Then for any f L 1 ([ π, π]), π π dt f (t) = π For any k Z + let P n be the projections such that where is the integral kernel of P k. π P k = ϕ k ϕ k (P k f )(x) = (ϕ k, f )ϕ k (x) = dt f (t x) π π P k (x) = 1 2π eikx Let G k be the projections (not orthogonal) dt P k (x t) f (t) with the integral kernel G j = j k= j G j (x) = 1 2π P k j k= j e ikx Also define the sequence of operators K n = 1 n G j = 1 n + 1 n + 1 with the integral kernel K n (x) = called the Fejer s kernel. j=0 1 2π(n + 1) n (n + 1 k )P k k= n n (n + 1 k ) e ikx k= n mathphys15.tex; October 28, 2015; 15:35; p. 78

4.5. TRIGONOMETRIC FOURIER SERIES 81 Lemma 4.5.1 We have K n (x) = Proof: Direct calculation. 1 sin [ ] 2 (n + 1) x 2 2π(n + 1) sin ( ) 2 x 2 A sequence K n of 2π-periodic continuous functions is a summability kernel if it satisfies the conditions: 1. π dt K n (t) = 1, n Z + π 2. There is M R such that n Z + 3. For any δ (0, π) π π lim n dt K n (t) M, 2π δ A summability kernel converges formally to δ lim K n(x) = n dt K n (t) = 0 δ(x 2πk) k= Theorem 4.5.1 Let (K n ) be a summability kernel. Then for any f L 1 ([ π, π]) the sequence F n = K n f strongly converges to f in L 1 norm, that is, lim F n f = lim (K n I) f = 0 n n Proof: Use the properties of the summability kernel. Basically because K n I (delta function). Lemma 4.5.2 The Fejer s kernel is a summability kernel. Proof: Direct calculation. mathphys15.tex; October 28, 2015; 15:35; p. 79

82 CHAPTER 4. HILBERT SPACES Theorem 4.5.2 Let f L 1 ([ π, π]). If all Fourier coefficients f n = (ϕ n, f ) vanish, then f = 0 almost everywhere. Proof: Let F n = K n f. Suppose f n = 0. Then F n = 0. Since F n f, then f a.e. = 0. Theorem 4.5.3 The sequence ϕ n (x) = 1 2π e inx, n Z is a complete orthonormal sequence, (an orthonormal basis), in L 2 ([ π, π]). Proof: Let f L 2. Then f L 1. Suppose (ϕ n, f ) = 0. Then f a.e. = 0. That is, f = 0 in L 2. Let f L 2 ([ π, π]). The series where f (x) = f n = π π n= f n ϕ n (x), dt ϕ n (t) f (t), is the Fourier series. The scalars f n are the Fourier coefficients. Fourier series does not converge pointwise! Fourier series of a function f L 2 ([ π, π]) converges almost everywhere. 4.6 Orthonormal Complements and Projection Theorem A subspace of a Hilbert space is an inner product space. A closed subspace of a Hilbert space is a Hilbert space. Orthogonal Complement. Let H be a Hilbert space and S H be a nonempty subset of H. We say that x H is orthogonal to S, denoted by x S, if y S, (x, y) = 0. mathphys15.tex; October 28, 2015; 15:35; p. 80

4.6. ORTHONORMAL COMPLEMENTS AND PROJECTION THEOREM 83 The set S = {x H x S } of all vectors orthogonal to S is called the orthogonal complement of S. Two subsets A and B of H are orthogonal, denoted by A B, if every vector of A is orthogonal to every vector of B. If x H, then x = 0, that is If A B, then A B = {0} or. H = {0}, {0} = H. Orthogonal decomposition. If every element of H can be uniquely represented as the sum of an element of S and an element of S, then H is the direct sum of S and S, which is denoted by H = S S The union of a basis of S and a basis of S gives a basis of H. Orthogonal projection. An orthogonal decomposition H = S S induces a projection map P : H S defined by where y S and z S. Examples. P(y + z) = y, Theorem 4.6.1 The orthogonal complement of any subset of a Hilbert space is a Hilbert subspace. Proof: 1. Let H be a Hilbert space and S H. 2. Check directly that S is a vector subspace. 3. Claim: S is closed. 4. Let (x n ) be a sequence in S such that x n x H. mathphys15.tex; October 28, 2015; 15:35; p. 81

84 CHAPTER 4. HILBERT SPACES 5. Then y S 6. Thus x S. Remarks. (x, y) = lim n (x n, y) = 0. S does not have to be a vector subspace. Theorem 4.6.2 Orthogonal Projection. Let H be a Hilbert space and S be a closed subspace of H. Then H = S S. Proof: We need to show that x H there exist unique y S and z S such that x = y + z. 1. Let (e i, f j ) be an orthonormal basis in H such that e i S and f j S. 2. Let x H. Then 3. Let 4. Then x = y i e i + i=1 y = z = z j f j. j=1 y i e i i=1 z j f j j=1 x = y + z Theorem 4.6.3 Let H be a Hilbert space and S be a closed subspace of H. Then (S ) = S. Proof: mathphys15.tex; October 28, 2015; 15:35; p. 82

4.7. LINEAR FUNCTIONALS AND THE RIESZ REPRESENTATION THEOREM85 1. Let x S. 2. Then x S, or x S. 3. So, S S. 4. Let x S. 5. Since S is closed, there exist y S and z S such that x = y + z. 6. Then y S. 7. Since S is a vector space, z = x y S. 8. Since z S and z S. 9. Thus, z = 0, and x = y S. 10. Therefore, S S. 4.7 Linear Functionals and the Riesz Representation Theorem A linear functional on a Hilbert space H is a linear map F : H C. Examples. C 0 (R) F( f ) = f (x 0 ) Note that there is no g C 0 such that F( f ) = (g, f ) L 2 ([a, b]) l 2 0. F(x) = There is no y = (y n ) such that F(x) = (y, x). x n n mathphys15.tex; October 28, 2015; 15:35; p. 83

86 CHAPTER 4. HILBERT SPACES g is a linear bounded functional defined by ϕ g ( f ) = (g, f ) with the norm ϕ g = g Let x 0 (a, b). Then ϕ( f ) = f (x 0 ) is a linear but unbounded functional. Remarks. The set H of all bounded linear functionals on a Hilbert space is a Banach space, called the dual space. The dual space H of a Hilbert space H is isomorphic to H. If f = 0, then the null space N( f ) = E, and, therefore, (N( f )) = {0} and dim(n( f )) = 0. Lemma 4.7.1 Let E be an inner product space and f : E C be a nonzero bounded linear functional on E. Then the orthogonal complement of the null space dim(n( f )) = 1. Proof: 1. Since f is bounded and linear it is continuous. 2. Therefore, N( f ) is a closed subspace of E. 3. Thus, (N( f )) is not empty. 4. Let x, y (N( f )) be two nonzero vectors. 5. Then f (x) 0 and f (y) 0. 6. Therefore, there exists α 0 C such that f (x + αy) = 0. 7. Hence, x + αy N( f ). 8. Since x, y (N( f )), we also have x + αy (N( f )). 9. Thus x + αy = 0. 10. Therefore, x and y are linearly dependent, and, therefore, dim(n( f )) = 1. mathphys15.tex; October 28, 2015; 15:35; p. 84

4.7. LINEAR FUNCTIONALS AND THE RIESZ REPRESENTATION THEOREM87 There holds E = N( f ) (N( f )) Theorem 4.7.1 Riesz Representation Theorem. Let H be a Hilbert space and f : H C be a bounded linear functional on H. There exists a unique x 0 H such that f (x) = (x 0, x) for all x H. Moreover, Proof: 1. (I). Existence. If f = 0, then x 0 = 0. 2. Suppose f 0. 3. Then dim(n( f )) = 1. 4. Let u (N( f )). 5. Then x H, f = x 0. x = y + z where y = x (x, u)u N( f ) and z = (x, u)u (N( f )). 6. Therefore, f (y) = 0. 7. Further, where f (x) = f (z) = (x, u) f (u) = (x, x 0 ), x 0 = ( f (u)) u. 8. (II). Uniqueness. Suppose there exists x 0 and x 1 such that x H f (x) = (x, x 0 ) = (x, x 1 ). 9. Then x H (x, x 0 x 1 ) = 0. 10. Thus, (x 0 x 1 ) H = {0}. 11. So, x 0 = x 1. mathphys15.tex; October 28, 2015; 15:35; p. 85

88 CHAPTER 4. HILBERT SPACES 12. Finally, we have f = sup x 0 f (x) x = sup (x, x 0 ) x 0 x x 0. 13. On another hand f (x 0 ) x 0 = (x 0, x 0 ) x 0 = x 0. 14. Thus, f = x 0. Theorem 4.7.2 Let H be a Hilbert space and S be a subspace of H. Let f : S C be a bounded linear functional on S. There exists a unique linear functional g : H C such that g(x) = f (x) for all x S and g = f. Proof: Use Riesz theorem. mathphys15.tex; October 28, 2015; 15:35; p. 86