A Brief Introduction to Functional Analysis

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1 A Brief Introduction to Functional Analysis Sungwook Lee Department of Mathematics University of Southern Mississippi July 5, 2007 Definition 1. An algebra A is a vector space over C with an additional operation : A A A; (a, b) a b, which is distributive over +: a(b + c) = ab + ac, (a + b)c = ac + bc, for all a, b, c A. Remark 1. In general, the product is not commutative, i.e., ab ba. Hereafter, we assume that A has a unit 1, i.e., 1 is the identity element for the product. Definition 2. The algebra A is called a -algebra if it admits an involution : A A with properties for all a, b A and α, β C. (a ) = a, (ab) = b a, (αa + βb) = ᾱa + βb, Definition 3. A normed algebra A is an algebra with a norm : A R + {0} which has the properties a = 0 a = 0, for all a, b A and α C. αa = α a, a + b a + b (triangle inequality), ab a b (product inequality) 1

2 If A is a normed algebra, the norm defines a metric d : A A R by d(x, y) = x y for x, y A. The topology induced by this metric is called the uniform topology. A sequence {a n } in a normed algebra A is said to be a Cauchy sequence if given ɛ > 0 there exists a positive integer N such that d(a m, a n ) = a m a n < ɛ for all m, n N. A Banach algebra is a normed algebra which is complete (i.e., every Cauchy sequence is a convergent sequence) in the uniform topology. A Banach -algebra A is a normed algebra -algebra which is complete and such that a = a (1) for a A. A C -algebra A is a Banach -algebra whose norm satisfies the additional identity a a = a 2 (2) for a A. Proposition 1. The identity (2) implies (1). Proof. For any a A, by the product inequality, This implies a 2 = a a a a. For any a A, a A. Then by the inequality (3), Therefore, a = a. a a. (3) a (a ) = a. Definition 4. A subspace I of the algebra A is a left ideal (resp. right ideal) if for any a A and b I, ab I (resp. ba I). If I is both left ideal and right ideal, it is called a two-sided ideal. Let A be a -algebra. An ideal I is called a -ideal if I I 1. Proposition 2. Let A be a -algebra and I a -ideal of A. Then I is a two-sided ideal. 1 In fact, this implies that I = I. 2

3 Proof. Assume that I is a left ideal. Let a A and b I. Then a A and b I. Since I is a left ideal, a b = (ba) I and so ba = ((ba) ) I. Let A be an algebra and I a two side ideal of A. Denote by A/I the set of all left (or right, since + is commutative) cosets of I in A, i.e., A/I = {a + I : a A}. Define +, scalar multiplication, and the product as follows: For any a + I, b + I A/I and α C, (a + I) + (b + I) := (a + b) + I, α(a + I) := αa + I, (a + I) (b + I) := ab + I. Then one can easily see that these operations are well-defined; hence A/I becomes an algebra over C with these operations. One will notice that in order for the product to be well-defined, I needs to be two-sided. Define : A/I R + {0} by a + I := inf a + b (4) b I for each a + I A/I. Clearly, a + I a. Then is a norm. Exercise 1. Show that defined as in (4) is a norm. If A is a Banach -algebra and I is a -ideal which is closed, then the quotient A/I can be made into a Banach -algebra. Furthermore, if A is a C*-algebra, then A/I is also a C*-algebra. Let A be an algebra with unit 1. The resolvant set r(a) of a A is r(a) = {λ C : a λ1is invertible}. For any λ r(a), the inverse (a λ1) 1 is the resolvant of a at λ. C \ r(a) is called the spectrum σ(a) of a. The spectral radius ρ(a) of a A is ρ(a) = sup{ λ : λ r(a)}. If A is a C*-algebra then ρ(a) = a for any a A. An element a A is called self-adjoint if a = a. Note that the spectrum of a self-adjoint element is real and that σ(a) [ a, a ], σ(a 2 ) [0, a 2 ]. An element a A is called positive if it is self-adjoint and its spectrum is a subset of [0, ). 3

4 Proposition 3. a A is positive if and only if a = b b for some b A. Let H be a complex vector space, i.e., a vector space over the complex field C. H is called an inner product space or unitary space if there is a mapping : H H C such that Some immediate consequences are: y x = x y (5) x + y z = x z + y z (6) αx y = α x y, x, y H, α C (7) x x 0, x H (8) x x = 0 iff x = 0. (9) 1. (7) implies that 0 y = 0 for all y H. 2. By (6) and (7), the mapping x x y for each y H is a linear functional on H. 3. By (5) and (7), 4. By (5) and (6), x αy = αy x = α y x = ᾱ y x = ᾱ x y. z x + y = x + y z = x z + y z = x z + y z = z x + z y. 5. By (8), we may define x, the norm of the vector x H: x := x x. Proposition 4. For any x, y H, the Cauchy-Schwarz inequality holds. x y x y (10) 4

5 Proposition 5. For any x, y H, the triangle inequality holds. x + y x + y Denote by H(1) the closed unit ball centered at 0 in H: H(1) := {x H : x 1}. For any subset S of H the closed linear subspace is the orthogonal space of S. S := {y H : x y = 0 for all x S} Proposition 6. Let S be a subset of H. Then 1. S (S ). 2. If S T then S T, for all S, T H. 3. If S is a linear subspace of H, then (S ) = S, the closure of S. For each x H, define ϕ x : H C by ϕ x (y) = x y for any y H. Then ϕ x is a continuous 2 linear functional. ϕ x is called the dual vector of x H. Exercise 2. Show that S = x S ϕ 1 x (0). Hence it is closed in H. If H is complete as a metric space, it is called a (complex) Hilbert space. Example 1. Let C n = {(ξ 1,, ξ n ) : ξ i C, i = 1,, n}, i.e., C n = C } C {{ C }. n copies For any x = (ξ 1,, ξ n ), y = (η 1,, η n ) C n, define x y = n ξ j η j. The inner product is called the hermitian product. C n with the hermitian product is a Hilbert space. 2 See Proposition 7. j=1 5

6 Example 2. Let (X, M, µ) be a measure space. L 2 (X, µ) = {ξ : X C : ξ(x) 2 dµ(x) < } is a Hilbert space with the inner product ξ η = ξηdµ(x). l 2 = {ξ = (ξ n ) n 0 : ξ n C and is a Hilbert space with the inner product ξ η = X X ξ i η i. n=0 ξ n 2 < 0} Example 3. Let C([0, 1]) := {f : [0, 1] C : f is continous.}. For each f, g C([0, 1]), define f g = 1 0 n=0 f(t)g(t)dt. C([0, 1]) with is not a Hilbert space. It is a prehilbert space, i.e., an inner product space which is not complete. Exercise 3. In Example 3, find a sequence {f n } in C([0, 1]) of which limit is a step function which is not continuous on [0, 1]. Proposition 7. Let H be a (pre)hilbert space. For any y H, the mappings are uniformly continuous. Proof. x x y, x y x, x x x 1 y x 2 y = x 1 x 2 y x 1 x 2 y by Cauchy-Schwarz inequality (10). Given ɛ > 0, choose any positive number if y = 0, δ = ɛ if y 0. y 6

7 Then x x y is uniformly continuous. It follows from the inequality x 1 x 2 x 1 x 2, that the mapping x x is uniformly continuous. Hereafter, H denotes a complex Hilbert space. Theorem 1 (Riesz Representation Theorem). If φ : H C is a continuous linear functional, then there exists uniquely ξ H such that for all η H. Moreover, φ(η) = ξ η ξ = sup{ φ(η) : η H(1)}. Remark 2. In quantum mechanics, states of a system form a complex Hilbert space H. State vectors in H are represented as kets ψ and dual vectors are represented as bras ψ. Using ket and bra notations, the Riesz Representation Theorem can be stated as: If φ : H C is a continuous linear functional, then there exists uniquely a bra ξ such that for all kets η H. Moreover, φ η = ξ η ξ = sup{ φ η : η H(1)}. Theorem 2. Let a : H H be a linear mapping. Then the following are equivalent: (i) a = sup{ aξ : ξ H(1)} (11) is bounded. (ii) a is continuous. (iii) a is continuous at one point of H. Definition 5. If a : H H satisfies (i)-(iii), a is said to be a bounded linear operator from H to H and a is said to be its norm. 7

8 Proposition 8. a : H H is a bounded linear operator if and only if there is a constant C > 0 such that aξ H C ξ H for all ξ H. Proof. Suppose that a : H H is a bounded linear operator. Since a is bounded, there exists a constant C > 0 such that a C. This implies that aξ H C for all ξ H(1). Let ψ( 0) H. Then ψ ψ H H(1). So, 1 ψ H aψ H = a ψ ψ H C. H This implies that The converse is clear. aψ H C ψ H. Denote by B(H, H ) the set of all bounded linear mappings from H to H. Then B(H, H ) is a Banach space with the norm in (i) of Theorem 2. B(H, H) is simply denoted by B(H), i.e., B(H) is the set of all bounded operators on H. Exercise 4. Show that B(H, H ) is a Banach space. Proposition 9. Let a B(H, H ) and b B(H, H ). Then ba B(H, H ). aξ Proof. For any ξ H, aξ H ( ) (1) and aξ b aξ = 1 aξ baξ. On the ( other hand, b aξ ) b. So for all ξ H, baξ b a. This implies that for all ξ H(1), Therefore, This implies that ba B(H, H ). baξ b aξ b a. ba b a. (12) Remark 3. Proposition 9 tells that B(H) is closed under function composition. Function composition can be regarded as a product operation on B(H). 8

9 Exercise 5. Show that B(H) is a Banach algebra. Proposition 10. For any bounded linear operator a : H H, there exists uniquely a bounded linear operator a : H H such that a η ξ = η aξ for all ξ H, η H. a is called the adjoint of a. Proof. For any η H, there is a continuous linear functional H C; ξ η aξ. By the Riesz Representation Theorem (Theorem 1), there exists uniquely ζ H such that ζ ξ = η aξ for any ξ H. Define a : H H by Then a is linear: a η = ζ. By the uniqueness of a (αη 1 + βη 2 ), αη 1 + βη 2 aξ = α η 1 aξ + β η 2 aξ = α ζ 1 ξ + β ζ 2 ξ = αζ 1 + βζ 2 ξ. a (αη 1 + βη 2 ) = αζ 1 + βζ 2 = αa η 1 + βa η 2. The boundedness of a follows from the following proposition. Proposition 11. a = sup{ η aξ : ξ H(1), η H (1)} = a. Proof. By Cauchy-Schwarz inequality (10), η aξ η aξ η a ξ for all ξ H, η H. So, for all ξ H(1), η H (1), η aξ a. 9

10 This implies that a sup{ η aξ : ξ H(1), η H (1)}. Assume that a 0 and choose 0 < ɛ < a 2. Then from the definition of a in (11). there exists ξ H(1) such that aξ a ɛ. Set η = aξ aξ H (1). Then η aξ = aξ a ɛ and so, Hence, Since a = a. sup{ η aξ : ξ H(1), η H (1)} a. a = sup{ η aξ : ξ H(1), η H (1)}. η aξ = a η ξ = ξ a η, Proposition 12. For a, b B(H, H ), λ, µ C, (a ) = a and (λa + µb) = λa + µb. Proof. Let ξ H. The operator (a ) : H H satisfies for all η H. So, (a ) ξ η = ξ a η = aξ η (a ) ξ aξ η = 0 for all η H. This implies that (a ) ξ aξ = 0. Since (a ) ξ = aξ for all ξ H, (a ) = a. Let a, b B(H, H ) and λ, µ C. Then (λa + µb) η ξ = η (λa + µb)ξ = η λaξ + η µbξ = λ η aξ + µ η bξ = λ a η ξ + µ b η ξ = λa η + µb η ξ = ( λa η + µb )η ξ for all ξ H, η H. Therefore, (λa + µb) = λa + µb. 10

11 Proposition 12 tells that defines an involution on B(H), i.e., B(H) is a -algebra. The following is a corollary to Proposition 11. Corollary 1. For any a B(H, H ), a a = a 2. Proof. By the product inequality (12), On the other hand, a a a a = a 2. a 2 = sup{ aξ 2 : ξ H(1)} = sup{ aξ aξ : ξ H(1)} = sup{ ξ a aξ : ξ H(1)} a a, since ξ a aξ a aξ for all ξ H(1). This completes the proof. Finally we can conclude that B(H) is a C*-algebra. Example 4 (Examples of C*-algebra). 1. The commutative algebra of C(M) of continuous functions on a Hausdorff space M (vanishing at infinity 3 if M is not compact) with denoting complex conjugation and the norm given by f = sup x M f(x). 2. Given an integer n 1 and the Hilbert space C n, the algebra B(C n ) can be identified with the matrix algebra The involution is given by M n (C) = {(a ij ) : a ij C, i, j = 1,, n}. (a ) ij = a ji, i, j = 1,, n, for every a M n (C). The norm is given by a a = sup aξ = max µ j, ξ C n, ξ 1 1 j n where µ 1,, µ n denote the eigenvalues of a a. 3 A real or complex valued function f on a locally compact Hausdorff space M is said to vanish at infinity if for every ɛ > 0 there exists a compact set K S such that f(x) < ɛ for x / K. 11

12 Exercise 6. Show that a a = a 2 for every a M n (C). Exercise 7. Show that for every a M n (C), a = max 1 j n µ j, where µ 1,, µ n denote the eigenvalues of a a. It is interesting to see that: Theorem 3. On the -algebra M n (C), the only norm ν such that ν(a a) = ν(a) 2 for every a M n (C) is the operator norm a a. Let A and B be C*-algebras. A -morphism π : A B is a C-linear map such that for all a, b A. π(ab) = π(a)π(b), π(a ) = π(a), Proposition 13. π is positive, namely if a A is positive then so is π(a). Proof. If a A is positive then by Proposition 3, a = b b for some b A. Now, π(a) = π(b b) = π(b )π(b) = π(b) π(b). Note that if π : A B is a -morphism then for any a A, π(a) B a A. This inequality implies that every -morphism is continuous. If π : A B (A and B are C*-algebras) is -morphism, then the image π(a) is C*-subalgebra of B. If -morphism is bijective then it is called a -isomorphism. Theorem 4 (Fundamental Morphism Theorem). Let A, B be C*- algebras and π : A B an onto -morphism. Then ker π = {a A : π(a) = 0} = π 1 (0) is a -ideal of A and A/ ker π is -isomorphic to B. Exercise 8. Prove Theorem 4. A representation of a C*-algebra A is a pair (H, π) where H is a Hilbert space and π is a -morphism π : A B(H), with B(H) the C*-algebra of bounded operators on H. The representation (H, π) is called faithful if ker π = 0, so that π is a -isomorphism between A and π(a). If a A is positive, we denote it by a > 0. 12

13 Proposition 14. A representation is faithful if and only if π(a) = a for any a A or π(a) > 0 for all a > 0. Exercise 9. Prove Proposition 14. The representation (H, π) is irreducible if the only closed subspace of H which are invariant under the action of π(a) are {0} and H. Two representations (H 1, π 1 ) and (H 2, π 2 ) are said to be (unitary) equivalent if there exists a unitary operator U : H 1 H 2 such that π 1 (a) = U π 2 (a)u for each a A. 13