Functional Analysis Zhejiang University

Transcription

1 Functional Analysis Zhejiang University

2 Contents 1 Elementary Spectral Theory Banach Algebras The Spectrum and the Spectral Radius The Gelfand Representation Compact and Fredholm Operators Spectral Theorem on the Hilbert Spaces C*-Algebras Positive Elements of C*-Algebras Operators and Sesquilinear Forms Compact Hilbert Space Operators The Spectral Theorem Unbounded Operator Basic Properties of Unbounded Operators Symmetric and Self-Adjoint Operators The Cayley Transform Unbounded Normal Operator Spectral Theory Stone Theorem GNS Construction and Fuglede-Putnam Theorem Gelfand-Naimark-Segal Construction Fuglede-Putnam Theorem and Applications of Spectral Theorem Tensor Product Theory 135 2

3 5.1 Tensor Product of Hilbert Spaces Tensor Product of Bounded Linear Operators

4 1 Elementary Spectral Theory In this chapter we cover the basic results of spectral theory. The most important of these are the non-emptiness of the spectrum, Beurling s spectral radius formula, and the Gelfand representation theory for commutative Banach algebras. 1.1 Banach Algebras We begin by setting up the basic vocabulary needed to discuss Banach algebras and by giving some examples. An algebra is a vector space A together with a bilinear map A 2 A, (A, B) AB, such that A(BC) = (AB)C (A, B, C A) A subalgebra of A is a vector subspace B of A such that B, B B BB B. Endowed with the multiplication got by restriction, B is itself an algebra. A norm. on A is said to be submultiplicative if AB A B (A, B A). In this case the pair (A, ) is called a normed algebra. If A admits a unit, that is, AI = IA = A for all A A, and I = 1, we say that A is a unital normed algebra. 4

5 If A is a normed algebra, then it is evident from the inequality AB A B A B B + A A B. that the multiplication operation (A, B) AB is jointly continuous. A complete normed algebra is called a Banach algebra. A complete unital normed algebra is called a unital Banach algebra. A subalgebra of a normed algebra is obviously itself a normed algebra with the norm got by restriction. The closure of a subalgebra in a normed algebra is a subalgebra. A closed subalgebra of a Banach algebra is a Banach algebra. Example If S is a set, l (S), the set of all bounded complex valued functions on S, is a unital Banach algebra where the operations are defined pointwise: and the norm is the sup-norm (f + g)(x) = f(x) + g(x), (fg)(x) = f(x)g(x), (λf)(x) = λf(x), f = sup f(x). x S Example If Ω is a topological space, the C b (Ω) of all bounded continuous complex-valued functions on Ω is a closed subalgebra of l (Ω). Thus, C b (Ω) is a unital Banach algebra. If Ω is compact, C(Ω), the set of continuous functions from Ω to C, is of course equal to C b (Ω). 5

6 Example If Ω is a locally compact Hausdorff space, we say that a continuous function f from Ω to C vanishes at infinity, if for each positive number ε the set {ω Ω f(ω) ε} is compact. We denote the set of such functions by C 0 (Ω). It is a closed subalgebra of C b (Ω), and therefore, a Banach algebra. It is unital if and only if Ω is compact, and in this case C 0 (Ω) = C(Ω). The algebra C 0 (Ω) is one of the most important examples for Banach algebras, and we shall see it used constantly in C -algebra theory. Example If (Ω, µ) is a measure space, the set L (Ω, µ) of classes of essentially bounded complex-valued measurable functions on Ω is a unital Banach algebra with the usual pointwise-defined operations and the essential supremum norm f f. Example If Ω is a measure space, let B (Ω) denote the set of all bounded complex-valued measurable functions on Ω. Then B (Ω) is a closed subalgebra of l (Ω), so it is a unital Banach algebra. This example will be used in connection with the spectral theorem in Chapter 2. Example The set A of all continuous functions on the closed unit disc D in the plane which are analytic on the interior of D is a closed subalgebra of C(D), so A is a unital Banach algebra, called the disc algebra. This is the motivating example in the theory of function algebras, where many aspects of the theory of analytic functions are extended to a Banach algebraic setting. All of the above examples are of course abelian that is, AB = BA for all elements A and B but the following examples are not, in general. Example If X is a normed vector space, denote by B(X ) the set of all bounded linear maps from X to itself. It is routine to show that B(X ) is 6

7 normed algebra with the pointwise-defined operations for addition and scalar multiplication, multiplication given by (U, V ) U V, and the operator norm: U(x) U = sup x 0 x = sup U(x). x 1 If X is a Banach space, B(X ) is complete and is therefore a Banach algebra. Example The algebra M n (C) of n n-matrices with entries in C is identified with B(C n ). It is therefore a unital Banach algebra. Recall that an upper triangular matrix is one of the form λ 11 λ λ 1n 0 λ λ 2n 0 0 λ λ 3n, λ nn that is, all entries below the main diagonal are zero. These matrices form a subalgebra of M n (C). We shall be seeing many more examples of Banach algebras as we proceed. Most often of these will be non-abelian, but in the first three sections of this chapter we shall be principally concerned with the abelian case. If (B λ ) λ Λ is a family subalgebras of an algebra A, then λ Λ B λ is a subalgebra, also. Hence, for any subset S of A, there is a smallest subalgebra B of A containing S, namely, the intersection of all the subalgebras containing S. This algebra is called the subalgebra of A generated by S. If S is the singleton set {A}, then B is the linear span of all powers A n of A, n = 1, 2,.... If A is 7

8 a normed algebra, the closed subalgebra C of A generated by a set S in A is the smallest closed subalgebra containing S. It is plain that C = B, where B is the subalgebra generated by S. If A = C(T), where T is the unit circle, and if z : T C is the inclusion function, then the closed algebra generated by z and its conjugate z is C(T) itself, this can be proved from the Stone-Weierstrass theorem. A left, respectively, right, ideal in an algebra A is a vector subspace I of A such that A A and B I AB I, respectively, BA I. An ideal in A is a vector subspace that is simultaneously a left and a right ideal in A. Obviously, 0 and A are ideals in A, called the trivial ideals. A maximal ideal in A is a proper ideal, that is, it is not A, that is not contained in any other proper ideal in A. Maximal left ideals are defined similarly. If A is unital, then it follows easily from Zorn s Lemma that every proper ideal is contained in a maximal ideal. If (I λ ) λ Λ is a family of ideals of an algebra A, then λ Λ I λ is an ideal of A. Hence, if S A, there is a smallest ideal I of A containing S, we call I the ideal generated by S. If A is a normed algebra, then the closure of an ideal is an ideal. The closed ideal J generated by a set S is the smallest closed ideal containing S. It is clear that J is the closure of the ideal generated by S. Theorem If I is a closed ideal in a normed algebra A, then A/I is a 8

9 normed algebra when endowed with the quotient norm A + I = inf A + B. B I Proof. Let ε > 0 and suppose that A, B belong to A. Then ε + A + I > A + A and ε + B + I > B + B for some A, B I. Hence, (ε + A + I )(ε + B + I ) > A + A B + B AB + C, where C = A B+AB +A B I. Thus, (ε+ A+I )(ε+ B+I ) AB+I. Letting ε 0, we get A+I B +I AB +I ; that is, the quotient norm is submultiplicative. A homomorphism from an algebra A to an algebra B is a linear map ϕ : A B such that ϕ(ab) = ϕ(a)ϕ(b) for all A, B A. Its kernel ker(ϕ) is an ideal in A and its image ϕ(a) is a subalgebra of B. We say ϕ is unital if A and B are unital and ϕ(i) = I. If I is an ideal in A, the quotient map π : A A/I is a homomorphism. If ϕ, ψ are continuous homomorphisms from a normed algebra A to a norm algebra B, then ϕ = ψ if ϕ and ψ are equal on a set S that generates A as a normed algebra, that is, A is the closed algebra generated by S. This follows from the observation that the set {A A ϕ(a) = ψ(a)} is a closed subalgebra of A and contains S. 9

10 1.2 The Spectrum and the Spectral Radius Let C[z] denote the algebra of all polynomials in an indeterminate z with complex coefficients. If A is an element of a unital algebra A and p C[z] is the polynomial p = λ 0 + λ 1 z λ n z n, we set p(a) = λ 0 I + λ 1 A λ n A n. The map C[z] A, p p(a), is a unital homomorphism. We say that A A is invertible if there is an element B in A such that AB = BA = I. In this case B is unique and written by A 1. The set Inv(A) = {A A A is invertible} is a group under multiplication. We define the spectrum of an element A in A to be the set σ(a) = σ A (A) = {λ C λi A / Inv(A)}. We shall henceforth find it convenient to write λi simply as λ. Example Let A = C(Ω), where Ω is a compact Hausdorff space. Then σ(f) = f(ω) for all f A. 10

11 Example Let A = l (S), where S is a non-empty set, Then σ(f) = f(s), the closure in C, for all f A. Example Let A be the algebra of upper triangular n n-matrices. If A A, say λ 11 λ λ 1n 0 λ A = λ 2n λ nn it is elementary that σ(a) = {λ 11, λ 22,..., λ nn }. Similarly, if A = M n (C) and A A, then σ(a) is the set of eigenvalues of A. Thus, one thinks of the spectrum as simultaneously a generalisation of the range of a function and the set of eigenvalues of a finite square matrix. Remark If A, B are elements of a unital algebra A, then I AB is invertible if and only if I BA is invertible. This follows from the observation that if I AB has inverse C, then I BA has inverse I + BCA. A consequence of this equivalence is that σ(ab) \ {0} = σ(ba) \ {0} for all A, B A. Theorem Let A be an element of a unital algebra A. If σ(a) is nonempty and p C[z], then σ(p(a)) = p(σ(a)). 11

12 Proof. We may suppose that p is not constant. If µ C, there are elements λ 0,..., λ n in C, where λ 0 0, such that p µ = λ 0 (z λ 1 )... (z λ n ), and therefore, p(a) µ = λ 0 (A λ 1 )... (A λ n ). It is clear that p(a) µ is invertible if and only if A λ 1,..., A λ n are. It follows that µ σ(p(a)) if and only if µ = p(λ) for some λ σ(a), and therefore, σ(p(a)) = p(σ(a)). The spectral mapping property for polynomials is generalized to continuous functions in Chapter 2, but only for certain elements in certain algebras. Theorem Let A be a unital Banach algebra and A an element of A such that A < 1. Then I A Inv(A) and (I A) 1 = A n. n=0 Proof. Since n=0 An n=0 A n = (1 A ) 1 <, the series n=0 An is convergent, to B say, in A, and since (I A)(I A n ) = I A n+1 converges to (I A)B = B(I A) and to I as n, the element B is the inverse of I A. The series in Theorem is called the Neumann series for (I A) 1. Theorem If A is a unital Banach algebra, then Inv(A) is open in A, and the map Inv(A) A, A A 1, 12

13 is differentiable. Proof. Suppose that A Inv(A) and B A < A 1 1. Then BA 1 I B A A 1 < 1, so BA 1 Inv(A) is open in A. If B A and B < 1, then I + B Inv(A) and Inv(A), and therefore, B Inv(A), thus, (I + B) 1 I + B = ( 1) n B n I + B = ( 1) n B n n=0 B n = B 2 /(1 B ). n=2 Let A Inv(A) and suppose that C < 1 2 A 1 1. Then A 1 C < 1/2 < 1, with B = A 1 C, so, (I + A 1 C) 1 I + A 1 C A 1 C 2 /(1 A 1 C ) 2 A 1 C 2, since 1 A 1 C > 1/2. Now define U to be the linear operator on A given by U(B) = A 1 BA 1. Then, (A+C) 1 A 1 U(C) = (I + A 1 C) 1 A 1 A 1 + A 1 CA 1 Consequently, n=2 (I + A 1 C) 1 I + A 1 C A 1 2( A 1 3 C 2 ). (A + C) 1 A 1 U(C) lim C 0 C = 0, and therefore, the map τ : B B 1 is differentiable at B = A with derivative τ (A) = U. The algebra C[z] is a normed algebra where the norm is defined by setting p = sup p(λ). λ 1 13

14 Observe that Inv(C[z]) = C \ {0}, so the polynomials p n = 1 + z/n are not invertible. But lim n p n = 1, which shows that Inv(C[z]) is not open in C[z]. Thus, the norm on C[z] is not complete. Lemma Let A be a unital Banach algebra and let A A. The spectrum σ(a) of A is a closed subset of the disc in the plane of centre the origin and radius A, and the map is differentiable. C \ σ(a) A, λ (A λ) 1, Proof. If λ > A, then λ 1 A < 1, so I λ 1 A is invertible, and therefore, so is λ A. Hence, λ / σ(a). Thus, λ σ(a) λ A. The set σ(a) is closed, that is, C \ σ(a) is open, because Inv(A) is open in A. Differentiability of the map λ (A λ) 1 follows from Theorem The following result can be thought of as the fundamental theorem of Banach algebras. Theorem (Gelfand) If A is an element of a unital Banach algebra A, then the spectrum σ(a) of A is non-empty. Proof. Suppose that σ(a) = and we shall obtain a contradiction. If λ > 2 A, then λ 1 A < 1 2, A 0 since σ(a) = and therefore, 1 λ 1 A > 1 2. Hence, (I λ 1 A) 1 I = (λ 1 A) n n=1 λ 1 A 1 λ 1 A 2 λ 1 A < 1, 14

15 Consequently, (I λ 1 A) 1 < 2, since σ(a) =, so A 0, therefore, (A λ) 1 = λ 1 (I λ 1 A) 1 < 2/ λ < A 1. Moreover, since the map λ (A λ) 1 is continuous, it is bounded on the compact disc 2 A D. Thus, we have shown that this map is bounded on all of C; that is, there is a positive number M such that (A λ) 1 M, λ C. If τ A, the function λ τ((a λ) 1 ) is entire, and bounded by M τ, so by Liouville s theorem in complex analysis, it is constant. In particular, τ(a 1 ) = τ((a I) 1 ). Because this is true for all τ A, we have A 1 = (A I) 1, so A = A I, which is a contradiction. Theorem (Gelfand-Mazur) If A is a unital Banach algebra in which every non-zero element is invertible, then A = CI. Proof. This is immediate from Theorem to be If A is an element of a unital Banach algebra A, its spectral radius is defined r(a) = sup λ. λ σ(a) By Remark 1.2.4, r(ab) = r(ba) for all A, B A. Example If A = C(Ω), where Ω is a compact Hausdorff space, then r(f) = f (f A). Example Let A = M 2 (C) and A = Then A = 1, but r(a) = 0, since A 2 =

16 Theorem (Beurling) If A is an element of a unital Banach algebra A, then r(a) = inf n 1 An 1/n = lim n A n 1/n. Proof. If λ σ(a), then λ n σ(a n ), so λ n A n, and therefore, r(a) inf n 1 A n 1/n lim inf n A n 1/n. Let be the open disc in C centered at 0 and of radius 1/r(A), we use the usual convention that 1/0 = +. If λ, then I λa Inv(A). If τ A, then the map f : C, λ τ((i λa) 1 ) is analytic, so there are unique complex numbers λ n such that f(λ) = λ n λ n, λ. n=0 However, if λ < 1/ A, since 1/ A 1/r(A), then λa < 1, so (I λa) 1 = λ n A n, n=0 and therefore, f(λ) = λ n τ(a n ). n=0 It follows that λ n = τ(a n ) for all n 0. Hence, the sequence (τ(a n )λ n ) converges to 0 for each λ, and therefore a fortiori, it is bounded. Since this is true for each τ A, it follows from the principle of uniform boundedness that (λ n A n ) is a bounded sequence. Hence, there is a positive number M such that λ n A n M for all n 0, and therefore, if λ 0, A n 1/n M 1/n / λ. 16

17 Consequently, lim sup n A n 1/n 1/ λ. We have thus shown that if r(a) < λ 1, then lim sup n A n 1/n λ 1. It follows that lim sup n A n 1/n r(a), and since r(a) lim inf n A n 1/n, therefore r(a) = inf n 1 A n 1/n = lim n A n 1/n. Example Let A be the set of C 1 -functions on the interval [0, 1]. This is an algebra when endowed with the pointwise-defined operations, and a submultiplicative norm on A is given by f = f + f (f A). It is elementary that A is complete under this norm, and therefore, A is a unital Banach algebra. Let x : [0, 1] C be the inclusion, so x A. Clearly, x n = 1 + n for all n, so r(x) = lim(1 + n) 1/n = 1 < 2 = x. Recall that if K is a non-empty compact set in C, its complement C \ K admits exactly one unbounded component, and that the bounded components of C \ K are called the holes of K. Theorem Let B be a closed subalgebra of a unital Banach algebra A containing the unit of A. (1) The set Inv(B) is a clopen subset of B Inv(A). (2) For each B B, σ A (B) σ B (B) and σ B (B) σ A (B). (3) If B B and σ A (B) has no holes, then σ A (B) = σ B (B). Proof. Clearly Inv(B) is an open set in B Inv(A). To see that it is also closed, Let (B n ) be a sequence in Inv(B) converging to a point B B Inv(A). Then 17

18 (Bn 1 ) converges to B 1 in A, so B 1 B, which implies that B Inv(B). Hence, Inv(B) is clopen in B Inv(A). If B B, the inclusion σ A (B) σ B (B) is immediate from the inclusion Inv(B) Inv(A). If λ σ B (B), then there is sequence (λ n ) in C \ σ B (B) converging to λ. Hence, λ n B Inv(B), and λ B / Inv(B), so λ B / Inv(A), by Condition (1). Also, λ n B Inv(A), so λ n C \ σ A (B). Therefore, λ σ A (B). This proves Condition (2). If B B and σ A (B) has no holes, then C \ σ A (B) is connected. Since C \ σ B (B) is a clopen subset of C \ σ A (B) by Condition (1) and (2), it is follows that C \ σ A (B) = C \ σ B (B), and therefore, σ A (B) = σ B (B). Example Let C = C(T) and let A be the disc algebra. If f A, let ϕ(f) be its restriction to T. One easily checks that the map ϕ : A C, f ϕ(f), is an isometric homomorphism onto the closed subalgebra B of C generated by the unit and the inclusion z : T C, the equation ϕ(f) = f is given by the maximum modulus principle. Clearly, σ B (z) = σ A (z) = D, and σ C (z) = T. Let A be an element of a unital Banach algebra A. Since A n /n! A n /n! <, n=0 the series n=0 An /n! is convergent in A. We denote its sum by e A. n=0 In proving the next theorem, we shall use some elementary results concerning differentiation. Suppose that f, g are differentiable maps from R to A with 18

19 derivatives f, g, respectively. Then fg is differentiable and (fg) = fg + f g. To prove this, just mimic the proof of the scalar-valued case. If f = 0, then f is constant. We prove this: If τ A, then the function R C, t τ(f(t)), is differentiable with zero derivative, and therefore, τ(f(t)) = τ(f(0)) for all t. Since τ was arbitrary, this implies that f(t) = f(0). Theorem Let A be a unital Banach algebra. (1) If A A and f : R A is differentiable, f(0) = 1, and f (t) = Af(t) for all t R, then f(t) = e ta for all t R. (2) If A A, then e A is invertible with inverse e A, and if A, B are commuting elements of A, then e A+B = e A e B. Proof. First we observe that if f : R A is defined by f(t) = e ta, then f(t) = n=0 tn A n /n!, so differentiating term by term we get f (t) = Af(t). Now suppose f, g are any pair of differentiable maps from R to A such that f (t) = Af(t) and g (t) = Ag(t) and f(0) = g(0) = 1. Then the map h : R A, t f(t)g( t), is differentiable with zero derivative. Hence, h(t) = 1 for all t R. Applying this to the map t e ta we get e ta e ta = 1; in particular, e A e A = 1. It follows that if f : R A is differentiable, f(0) = 1, and f (t) = Af(t) for all t, then f(t) = e ta as set g(t) = e ta and get f(t)e ta = 1, so f(t) = e ta. Now suppose that A and B are commuting elements of A and set f(t) = e ta e tb. Then f(0) = 1 and f (t) = e ta Be tb + Ae ta e tb = (A + B)f(t). Hence, f(t) = e t(a+b) for all t R, so, in particular, e A+B = f(1) = e A e B. 19

20 1.3 The Gelfand Representation The idea of this section is to represent a unital abelian Banach algebra as an algebra of continuous functions on a compact Hausdorff space. This is an extremely useful way of looking at these algebras. We begin by proving some results on ideals and multiplicative linear functions. Theorem Let I be an ideal of a unital Banach algebra A. proper, so is its closure I. If I is maximal, then it is closed. If I is Proof. If I I, there exists an element B I such that I B < 1 and this implies that B is invertible in A. Since I is an ideal, I I and so I is not proper. This contradiction shows that I is proper. If I is maximal, then I = I, as I is a proper ideal containing I. Lemma If I is a maximal ideal of a unital abelian algebra A, then A/I is a field. Proof. The algebra A/I is unital and abelian, with unit I + I say. If J is an ideal of A/I and π is the quotient map from A to A/I, then π 1 (J ) is an ideal of A containing I. Hence, π 1 (J ) = A or I, by maximality of I. Therefore, J = A/I or 0. Thus, A/I and 0 are the only ideals of A/I. Now suppose that π(a) is a non-zero element of A/I. Then J = π(a)(a/i) is a non-zero ideal of A/I, and therefore, J = A/I. Hence, there is an element B of A such that (A + I)(B + I) = I + I, so A + I is invertible. This shows that A/I is a field. 20

21 If ϕ : A B is a unital homomorphism between unital algebras, then ϕ(inv(a)) Inv(B), so σ(ϕ(a)) σ(a), A A. A character on an abelian algebra A is a non-zero homomorphism τ : A C. We denote by Ω(A) the set of characters on A. Theorem Let A be a unital abelian Banach algebra. (1) If τ Ω(A), then τ = 1. (2) The set Ω(A) is non-empty, and the map τ ker(τ) defines a bijection from Ω(A) onto the set of all maximal ideals of A. Proof. If τ Ω(A) and A A, then τ(a) σ(a), so τ(a) r(a) A. Hence, τ 1. Also, τ(i) = 1, since τ(i) = τ(i) 2 and τ(i) 0. Hence, τ = 1. Let I denote the closed ideal ker(τ). This is proper, since τ 0, and I + CI = A, since A τ(a) I for all A A. It follows that I is a maximal ideal of A. If τ 1, τ 2 Ω(A) and ker(τ 1 ) = ker(τ 2 ), then for each A A we have τ 1 (A τ 2 (A)) = 0, so τ 1 (A) = τ 2 (A). Thus, τ 1 = τ 2. If I is an arbitrary maximal ideal of A, then I is closed by Theorem and A/I is a unital Banach algebra in which every non-zero element is invertible, by Lemma Hence, by Theorem A/I = C(I + I). It follows that A = I CI. Define τ : A C by τ(a + λ) = λ, A I, λ C. Then τ is a character and ker(τ) = I. Thus, we have shown that the map τ ker(τ) is a bijection from the characters onto the maximal ideals of A. 21

22 We have seen already that A admits maximal ideals since it is unital. Therefore, Ω(A). Theorem Let A be a unital abelian Banach algebra. Then σ(a) = {τ(a) τ Ω(A)} (A A). Proof. If A is an element of A whose spectrum contains λ, then the ideal I = (A λ)a is proper, so I is contained in a maximal ideal ker(τ), where τ Ω(A). Hence, τ(a) = λ. This shows that the inclusion σ(a) {τ(a) τ Ω(A)} holds, and the reverse inclusion is clear. If A is a unital abelian Banach algebra, it follows from Theorem that Ω(A) is contained in the closed unit ball of A. We endow Ω(A) with the relative weak topology, and call the topological space Ω(A) the character space, or spectrum, of A. Theorem If A is a unital abelian Banach algebra, then Ω(A) is a compact Hausdorff space. Proof. It is easily checked that Ω(A) is weak closed in the closed unit ball S of A. Since S is weak compact by the Banach-Alaoglu theorem, Ω(A) is weak compact. Suppose that A is a unital abelian Banach algebra, thus the space Ω(A) is non-empty. If A A, we define the function  by  : Ω(A) C, τ τ(a). 22

23 Clearly the topology on Ω(A) is the smallest one making all of the functions  continuous. Hence,  C(Ω(A)). We call  the Gelfand transform of A. The following result is very important: Theorem (Gelfand Representation) Suppose that A is a unital abelian Banach algebra. Then the map A C(Ω(A)), A Â, is a norm-decreasing homomorphism, r(a) =  (A A), and σ(a) = Â(Ω(A)). Proof. By Theorem the spectrum σ(a) is the range of Â. Hence, r(a) = Â, which implies that the map A  is norm-decreasing. That this map is a homomorphism is easily checked. The kernel of the Gelfand representation is called the radical of the algebra A. It consists of the elements A such that r(a) = 0. It therefore contains the nilpotent elements. If the radical is zero, A is said to be semisimple. In a general algebra an element whose spectrum consists of the set {0} is said to be quasinilpotent. Let A, B be commuting elements of an arbitrary unital Banach algebra A. Then r(a + B) r(a) + r(b), and r(ab) r(a)r(b). To see this, we may suppose A is abelian, if necessary, restrict to the closed subalgebra generated 23

24 by I, A, and B. Then r(a+b) = (A+B)ˆ  + ˆB = r(a)+r(b) by Theorem Similarly, r(ab) = (AB)ˆ  ˆB = r(a)r(b). The spectral radius is neither subadditive nor submultiplicative in general: Let A = M 2 (C) and suppose A = 0 1 and B = Then r(a) = r(b) = 0, since A and B have square zero, but r(a + B) = r(ab) = 1. The interpretation of the character space as a sort of generalised spectrum is motivated by the following result. Theorem Let A be a unital Banach algebra generated by I and an element A. Then A is abelian and the map  : Ω(A) σ(a), τ τ(a), is a homeomorphism. Proof. It is clear that A is abelian and that  is a continuous bijection, and because Ω(A) and σ(a) are compact Hausdorff spaces,  is therefore a homeomorphism. 1.4 Compact and Fredholm Operators This section is concerned with the elementary spectral theory of compact operators, a class that plays an important and fundamental role in operator theory. These operators behave much like operators on finite-dimensional vector spaces, and for this reason they are relatively easy to analysis. 24

25 A linear map U : X Y between Banach spaces X and Y is compact if U(S) is relatively compact in Y, where S is the closed unit ball of X. Equivalently, U(S) is totally bounded. therefore, U is bounded. In this case U(S) is bounded, and Remark Note that the range of a compact operator is separable. This is immediate from the fact that a compact metric space is separable, and that the closure of the image of the ball under a compact operator is compact. The theory of compact operators arose out of the analysis of linear integral equations. The following example illustrates the connection. Example Let I = [0, 1] and let X be the Banach space C(I), where the norm is the supremum norm. If k C(I 2 ), define U B(X ) by setting U(f)(s) = 1 0 k(s, t)f(t)dt We show that U(f) X. Observe first that U(f)(s) U(f)(s ) = (f X, s I). (k(s, t) k(s, t))f(t)dt k(s, t) k(s, t) f(t) dt sup k(s, t) k(s, t) f. t I Now k is uniformly continuous because I 2 is compact, so if ε > 0, there exists δ > 0 such that max{ s s, t t } < δ k(s, t) k(s, t ) < ε. Hence, s s < δ U(f)(s) U(f)(s ) ε f. (1) Thus, U(f) is continuous, that is, U(f) X, but more is true, for it is immediate from Inequality (1) that U(S) is equicontinuous, where S is the closed 25

26 unit ball of X. Also, U(S) is bounded, since U(f) 1 0 k(s, t)f(t) dt k f. By the Arzelà-Ascoli theorem the set U(S) is totally bounded. Therefore, U is a compact operator on X. The function k is called the kernel of the operator U, and U is called an integral operator. If X, Y are Banach spaces, we denote by B(X, Y) the vector space of all bounded linear maps from X to Y. This is a Banach space when endowed with the operator norm. The set of all compact operators from X to Y is denoted by K(X, Y). The proof of the following is a routine exercise. Theorem Let X and Y be Banach spaces and U B(X, Y). Then the following conditions are equivalent: (1) U is compact; (2) For each bounded set S in X, the set U(S) is relatively compact in Y; (3) For each bounded sequence (x n ) in X, the sequence (U(x n )) admits a subsequence that converges in Y. It follows easily from (3) in Theorem that K(X, Y) is a vector subspace of B(X, Y). Also, if X V X U Y W Y are bounded linear maps between Banach spaces and U is compact, then W UV are compact. Hence K(X ) = K(X, X ) is an ideal in B(X ). 26

27 Theorem If X is a Banach space, then K(X ) = B(X ) if and only if X is finite-dimensional. Proof. If S denotes the closed unit ball of X, then K(X ) = B(X ) id X compact S is compact X is finite-dimensional. is Theorem If X, Y are Banach spaces, then K(X, Y) is a closed vector space of B(X, Y). Proof. We show that if a sequence (U n ) in K(X, Y) converges to an operator U in B(X, Y), then U is compact. Let S denote the closed unit ball of X and let ε > 0. Choose an integer N such that U N U < ε/3. Since U N (S) is totally bounded, there are elements x 1,..., x n S, such that for each x in S, the inequality U N (x) U N (x j ) < ε/3 holds for some index j. Hence, U(x) U(x j ) U(x) U N (x) + U N (x) U N (x j ) + U N (x j ) U(x j ) < ε/3 + ε/3 + ε/3 = ε. Thus, U(S) is totally bounded, and therefore, U K(X, Y). Recall that a linear map U : X Y is of finite rank if U(X ) is finitedimensional and that rank(u) = dim(u(x )). If X and Y are Banach spaces and U B(X, Y) is of finite rank, then U K(X, Y). This is immediate from the fact that the closed unit ball of the finite-dimensional space U(X ) is compact. It follows from this remark and Theorem that norm-limits of finiterank operators are compact, and it is natural to ask whether the converse is true. This is the case for Hilbert spaces, as we shall see in the next chapter, but it is not true for arbitrary Banach spaces. P.Enflo has given an example of 27

28 a Banach space for which there are compact operators that are not norm-limits of finite-rank operators. If U : X Y is a bounded linear map between Banach spaces, we define its transpose U B(Y, X ) by U (τ) = τ U. Theorem Let X, Y be Banach spaces and let U K(X, Y). U K(Y, X ). Then Proof. Let S be the closed unit ball of X and let ε > 0. Since U(S) is totally bounded, there exist elements x 1,..., x n in S, such that if x S, then U(x) U(x i ) < ε/3 for some index i. Define V B(Y, C n ) by setting V (τ) = (τ U(x 1 ),..., τ U(x n )). Since the rank of V is finite, V is compact, and therefore V (Y1) is totally bounded, where Y1 is the closed unit ball of Y. Hence, there exist functionals τ 1,..., τ m in Y1, such that if τ Y1, then V (τ) V (τ j ) < ε/3 for some index j. Observe that V (τ) V (τ j ) = [ i U (τ)(x i ) U (τ j )(x i ) ] 1/2. Now suppose that x S. Then U(x) U(x i ) < ε/3 for some index i, and U (τ)(x i ) U (τ j )(x i ) < ε/3. Hence, U (τ)(x) U (τ j )(x) U (τ)(x) U (τ)(x i ) + U (τ)(x i ) U (τ j )(x i ) + U (τ j )(x i ) U (τ j )(x) ε/3 + ε/3 + ε/3 = ε. It follows that U (τ) U (τ j ) ε, so U (Y1) is totally bounded and therefore U is compact. 28

29 A linear bounded map U : X Y between Banach spaces is bounded below if there is a positive number δ such that U(x) δ x (x X ). Note that in this case U(X ) is necessarily closed, for if (U(x n )) is a Cauchy sequence in U(X ), then (x n ) is a Cauchy sequence in X and therefore converges to some element x X, because X is complete. Hence, the sequence (U(x n )) converges to U(x) by continuity of U. Thus, U(X ) is complete and therefore closed in Y. Observe that every invertible linear bounded map is bounded below, as is every isometric linear map. It is easily checked that U : X Y is not bounded below if and only if there is a sequence of unit vectors (x n ) in X such that lim n U(x n ) = 0. These remarks will be used in the following theorem. Theorem Let U be a compact operator on a Banach space X and suppose that λ C \ {0}. (1) The space ker(u λ) is finite-dimensional. (2) The space (U λ)(x ) is closed and finite-codimensional in X. In fact, the codimension of (U λ)(x ) in X is the dimension of ker(u λ). Proof. Let Z = ker(u λ). Then U(Z) Z, and the restriction U Z of U to Z is in K(Z). Since U Z = λid Z and λ 0, the map id Z is compact. Hence, Z is finite-dimensional by Theorem Because Z is finite-dimensional, there is a closed vector space Y in X such that Z Y = X. Observe that (U λ)x = (U λ)y, so to show that (U λ)x is closed in X it suffices to show that the restriction (U λ) Y : Y X is bounded below. Suppose otherwise, and we shall obtain a contradiction. There is a 29

30 sequence (x n ) of unit vectors in Y such that lim n U(x n ) λx n = 0. Using the compactness of U and going to a subsequence if necessary, we may suppose that (U(x n )) is convergent. It follows from the equation x n = λ 1 (U(x n ) (U λ)(x n )) that the sequence (x n ) is convergent, to x say, and, since Y is closed in X, it contains x. Obviously, U(x) = λx, so x Y ker(u λ) and therefore x = 0. However, x is the limit of unit vectors and is therefore itself a unit vector, a contradiction. This shows that (U λ) Y is bounded below. Now let W = X /(U λ)(x ). To show that (U λ)(x ) is finite-codimensional in X, we have to show W is finite-dimensional, and we do this by showing W is finite-dimensional. Let π : X W be the quotient map. It is clear that the image of π is contained in the kernel of U λ. In fact these spaces are equal. For suppose that σ ker(u λ). Then σ annihilates (U λ)(x ) and therefore induces a bounded linear functional τ : W C such that σ = τ π = π (τ). Since U is compact by Theorem 1.4.5, ker(u λ) is finitedimensional by the first part of this proof. Thus, π has finite-dimensional range, and clearly π is injective, so W is finite-dimensional, and therefore dim(w) = dim(w ) = dim(π (W )) = dim(ker(u λ)). If U : X X is a linear map on a vector space X, then the sequence of spaces (ker(u n )) is clearly increasing. If ker(u n ) ker(u n+1 ) for all n N, we say that U has infinite ascent and set ascent(u) = +. Otherwise we say U has finite ascent and we define ascent(u) to be the least p such that ker(u p ) = ker(u p+1 ). In this case, ker(u p ) = ker(u n ) for all n p. The sequence of spaces (U n (X )) is decreasing. We say that U has infinite descent, and we set descent(u) = +, if U n (X ) U n+1 (X ) for all n N. 30

31 Otherwise, we say that U has finite descent and we define descent(u) to be the least p N such that U p (X ) = U p+1 (X ). In this case U p (X ) = U n (X ) for all n p. We recall now a theorem of F. Riesz from elementary functional analysis: If Y is a proper closed vector subspace of a normed vector space X and ε > 0, then there exists a unit vector x X such that x + Y > 1 ε. This simple result plays a key role in the theory of compact operators. Theorem Let U be a compact operator on a Banach space X and suppose that λ C \ {0}. Then U λ has finite ascent and descent. Proof. Suppose the ascent is infinite, and we deduce a contradiction. If N n = ker(u λ) n, then N n 1 is a proper subspace of N n, and therefore, by the theorem of Riesz discussed earlier, there is a unit vector x n N n such that x n + N n 1 1/2. If m < n, then U(x n ) U(x m ) = λx n + (U λ)(x n ) (U λ)(x m ) λx m = λx n z, where z N n 1. Hence, U(x n ) U(x m ) = λx n z = λ x n λ 1 z λ /2 > 0. It follows that (U(x n )) has no convergent subsequence, contradicting the compactness of U. Consequently, ascent(u) < +. The proof that U λ has finite descent is completely analogous and is left as an exercise. We shall have more to say about compact operators presently. One can give direct proofs of these later results, but the details are tedious and a little 31

32 messy, whereas when one use the homomorphism property of the Fredholm index, which we are now going to introduce, they drop out very nicely. The index which we shall also introduce, is the indispensible item in the operator theory. Nevertheless, many of the proofs in Fredholm theory are elementary, although often neither trivial or obvious. Let X, Y be Banach spaces and U B(X, Y). We say U is Fredholm if ker(u) is finite-dimensional and U(X ) is finite-codimensional in Y. We define the nullity of U to be dim(ker(u)) and denote it by nul(u). The defect of U is the codimension of U(X ) in Y, and is denoted by def(u). The index of U is defined to be ind(u) = nul(u) def(u). Note that because there is a finite-dimensional vector subspace Z of Y, such that U(X ) Z = Y, it is a consequence of the following theorem that U(X ) is closed in Y. Theorem Let X, Y be Banach spaces and U B(X, Y). Suppose that there is a closed vector subspace Z such that U(X ) Z = Y. Then U(X ) is closed in Y. Proof. The bounded linear map X / ker(u) Y, x + ker(u) U(x), has the same range as U and is injective, so we may suppose without loss of generality that U is injective. The map V : X Z Y, (x, z) U(x) + z, 32

33 is a continuous linear isomorphism between Banach spaces, so by the open mapping theorem, V 1 is also continuous.. If x X, then x = V 1 U(x) V 1 U(x), so U(x) V 1 1 x. Thus, U is bounded below, and therefore U(X ) is closed in Y. The following theorem is a fundamental result of Fredholm theory. Theorem Let X U Y V Z be Fredholm linear maps between Banach spaces X, Y, Z. Then V U is Fredholm and ind(v U) = ind(v ) + ind(u). Proof. Set Y 2 = ker(v ) U(X ) and choose suitable closed vector subspaces Y 1, Y 3, Y 4 of Y, such that U(X ) = Y 2 Y 3, ker(v ) = Y 1 Y 2, and Y = Y 1 U(X ) Y 4. Note that Y 1, Y 2, Y 4 are finite-dimensional. The map ker(v U) Y 2, x U(x), is surjective and it has the same kernel as U, so the kernel of V U is finitedimensional and nul(v U) = nul(u) + dim(y 2 ). Since V (Y) = V (Y 3 ) V (Y 4 ) and V (Y 3 ) = V U(X ), therefore V (Y) = V U(X ) V (Y 4 ). Choose a finite-dimensional vector subspace Z of Z such that V (Y) Z = Z, so Z = V U(X ) V (Y 4 ) Z. Because V (Y 4 ) Z is finite-dimensional, V U(X ) is finite-codimensional in Z. therefore, V U is a Fredholm operator. The map Y 4 V (Y 4 ), y V (y), 33

34 is a linear isomorphism, so dim(y 4 ) = dim(v (Y 4 )). Hence, def(v U) = dim(y 4 )+ dim(z ) = dim(y 4 ) + def(v ). Consequently, nul(v U) + def(u) + def(v ) = nul(u) + dim(y 4 ) + nul(v ) + def(v ) = nul(u) + nul(v ) + def(v U), and therefore, ind(v U) = nul(v U) def(v U) = nul(u) + nul(v ) def(u) def(v ) = ind(u) + ind(v ). We give an immediate easy application of the index: Theorem Let U be a compact operator on a Banach space X, and let λ C \ {0}. (1) The operator U λ is Fredholm of index zero. (2) If p denotes the (finite) ascent of U λ, then X = ker(u λ) p (U λ) p (X ). Proof. That U λ is Fredholm follows from Theorem 1.4.7, and the ascent and descent of U λ are finite by Theorem If we suppose that m, n are integers greater than max{ascent(u λ), descent(u λ)}, then we have nul(u λ) m = nul(u λ) n and def(u λ) m = def(u λ) n, so ind((u λ) m ) = ind((u λ) n ), and therefore m ind(u λ) = n ind(u λ) by Theorem It follows that ind(u λ) = 0. Thus, Condition (1) is proved. If x ker(u λ) p (U λ) p (X ), then there is an element y X such that x = (U λ) p (y) and (U λ) 2p (y) = 0. Since ker(u λ) p = ker(u λ) 2p, it follows that (U λ) p (y) = 0; that is, x = 0. Moreover, since nul(u λ) p = def(u λ) p, because ind(u λ) p = 0, it follows that X = ker(u λ) p (U λ) p (X ). 34

35 Corollary (Fredholm Alternative) The operator U λ is injective if and only if it is surjective. Proof. Since the index of U λ is zero, the nullity is zero if and only if the defect is zero; that is, U λ is injective if and only if it is surjective. Remark If Y, Z are complementary vector subspaces of a vector space X, and V, U are linear maps on Y, Z, respectively, we denote by U V the linear map on X given by (U V )(y + z) = U(y) + V (z) (y Y, z Z). Clearly, U V is invertible if and only if U and V are invertible. If X is a Banach space and W B(X ), we write σ(w ) for σ B(X ) (W ). If Y, Z are closed complementary vector subspaces of X, and if U B(Y), V B(Z), W B(X ), and W = U V, then σ(w ) = σ(u) σ(v ), by the preceding observation. Theorem Let U be a compact operator on a Banach space X. Then σ(u) is countable, and each non-zero point of σ(u) is an eigenvalue of U and an isolated point of σ(u). Proof. If λ is a non-zero point of σ(u), the by the Fredholm alternative, Corollary , U λ is not injective, and therefore λ is an eigenvalue of U. The operator U λ has finite ascent, p say, and by Theorem we can write X = Y Z, where Y = ker(u λ) p and Z = (U λ) p (X ). The spaces Y, Z are closed and invariant for U (that is, U(Y) Y and U(Z) Z). Hence, U λ = (U Y λid Y ) (U Z λid Z ), where U Y, U Z are the restrictions of 35

36 U to Y, Z, respectively. Since (U Y λid Y ) p = 0, the spectrum σ(u Y ) is the singleton set {λ}. Also, the operator U Z is compact and ker(u Z λid Z ) p = 0, so (U Z λid Z ) p is invertible (as it is injective and Fredholm of index zero), and therefore U Z λid Z is invertible. Hence, λ / σ(u Z ). This implies that σ(u) \ {λ} = σ(u Z ), so λ is an isolated point of σ(u) because σ(u Z ) is closed in σ(u). Countability of σ(u) follows by elementary topology. Example Let us interpret our results now in terms of integral equations. Let I = [0, 1] and suppose k C(I 2 ). Consider the integral equation 1 0 k(s, t)f(t)dt λf(s) = g(s). Here λ is a non-zero scalar, g C(I) is a known function, and f C(I) is the unknown. If U is the compact integral operator corresponding to the kernel k, as in Example 1.4.2, then we can rewrite our equation as (U λ)(f) = g. The non-zero spectrum of U is of the form {λ n 1 n N}, where N is an integer or. If 0 λ λ n for all n, then the integral equation has a unique solution: f = (U λ) 1 (g). If on the other hand λ = λ n say, then 1 0 k(s, t)f(t)dt λf(s) = 0. has a non-zero solution by the Fredholm alternative (Corollary ), and by Theorem the solution set is finite-dimensional. Observe that if N =, then lim n λ n = 0 by Theorem

37 Example One should not be misled by Theorem the spectral behaviour of compact operators is not typical of all operators. To illustrate this, let H be a separable Hilbert space with an orthonormal basis (e n ) n=1. If (λ n ) is a bounded sequence of scalars, define U B(H) by setting U(x) = n=1 λ na n e n when x = n=1 a ne n. We call U the diagonal operator with diagonal (λ n ) with respect to the basis (e n ). It is readily verified that U = sup n λ n, and that U is invertible if and only if inf n λ n > 0, and in this case U 1 is the diagonal operator with respect to (e n ) with diagonal (λ 1 n ). These observations imply that σ(u) is the closure of the set {λ n n = 1, 2,...}. Suppose that a non-empty compact set K in C is given and choose a dense sequence λ n in K. If U is the corresponding diagonal operator, then σ(u) = K. Thus, the spectrum is an arbitrary non-empty compact set in general. 37

38 2 Spectral Theorem on the Hilbert Spaces In this chapter we commence our study of C*-Algebras and of operators on Hilbert Spaces. Hilbert Spaces are very well-behaved compared with general Banach spaces, and the same is even more true of C*-Algebras as compared with general Banach algebras. The main results of this chapter are a theorem of Gelfand, which asserts that, up to isomorphism, all unital abelian C*-Algebras are of the form C(Ω), where Ω is a compact Hausdorff space, and the spectral theorem. This spectral theorem enables us to synthesize a normal operator from linear combinations of projections where the coefficients lie in the spectrum. It is a very powerful result. 2.1 C*-Algebras We begin by defining a number of concepts that make sense in any algebra with an involution. An involution on an algebra A is a conjugate-linear map A A on A, such that A = A and (AB) = B A for all A, B A. The pair (A, ) is called an involutive algebra, or a -algebra. If S is a subset of A, we set S = {A A S}, and if S = S we say S is self-adjoint. A self-adjoint subalgebra B of A is a -subalgebra of A and is a -algebra when endowed with the involution got by restriction. Because the intersection of a family of -subalgebras of A is itself one, there is for every subset S of A a smallest -algebra B of A containing S, called the -algebra generated by S. If I is self-adjoint ideal of A, then the quotient algebra A/I is a -algebra with the involution given by (A + I) = A + I (A A). 38

39 An element A in A is self-adjoint or hermitian if A = A. For each A A there exist unique hermitian elements B, C A such that A = B + ic, where B = 1 2 (A+A ) and C = 1 2i (A A ). The elements A A and AA are hermitian. The set of hermitian elements of A is denoted by A sa. We say A is normal if A A = AA. In this case the -algebra it generates is abelian and is in fact the linear span of all A m A n, where m, n N and n + m 0. An element P is a projection if P = P = P 2. If A is unital, then I = I (I = (II ) = I). If A Inv(A), then (A ) 1 = (A 1 ). Hence, for any A A, σ(a ) = σ(a) = {λ C λ σ(a)}. An element U in A is a unitary if U U = UU = I. If U U = I, then U is an isometry; and if UU = I, then U is a co-isometry. If ϕ : A B is a homomorphism of -algebras A and B and ϕ preserves adjoints, that is, ϕ(a ) = (ϕ(a)) (A A), then ϕ is a -homomorphism. If in addition ϕ is a bijection, it is a -isomorphism. If ϕ : A B is a -homomorphism, then ker(ϕ) is a self-adjoint ideal in A and ϕ(a) is a - subalgebra of B. An automorphism of a -algebra A is a -isomorphism ϕ : A A. If A is unital and U is a unitary in A, then Ad U : A A, A UAU, is an automorphism of A. Such automorphisms are called inner. We say elements A, B of A are unitarily equivalent if there exists a unitary U of A 39

40 such that B = UAU. Since the unitaries form a group, this is an equivalence relation on A. Note that σ(a) = σ(b) if A and B are unitarily equivalent. A Banach -algebra is a -algebra A together with a complete submultiplicative norm such that A = A (A A). If, in addition, A has a unit such that I = 1, we call A a unital Banach -algebra. A C*-algebra is a Banach -algebra such that A A = A 2 (A A). (1) A closed -subalgebra of a C*-algebra is obviously also a C*-algebra. We shall therefore call a closed -subalgebra of a C*-algebra a C*-subalgebra. If a C*-algebra has a unit I, then automatically I = 1, because I = I I = I 2. Similarly, if P is a non-zero projection, then P = 1. If U is a unitary of A, then U = 1, since U 2 = U U = I = 1. Hence, σ(u) T, for if λ σ(u), then λ 1 σ(u 1 ) = σ(u ), so λ and λ 1 1; that is, λ = 1. The seemingly mild requirement on a C*-algebra in Eq. (1) is in fact very strong far more is known about the nature and structure of these algebras than perhaps of any other non-trivial class of algebras. Because of the existence of the involution, C*-algebra theory can be thought of as infinite-dimensional real analysis. For instance, the study of linear functionals on C*-algebras is non-commutative measure theory. Example The scalar field C is a unital C*-algebra with involution given by complex conjugation λ λ. Example If Ω is a locally compact Hausdorff space, then C 0 (Ω) is a C*-algebra with involution f f. Similarly, all of the following algebras are 40

41 C*-algebras with involution given by f f: (a) l (S), where S is a set; (b) L (Ω, µ), where (Ω, µ) is a measure space; (c) C b (Ω), where Ω is a topological space; (d) B (Ω), where Ω is a measurable space. Example If H is a Hilbert space, then B(H) is a C*-algebra. Example If Ω is a non-empty set and A is a C*-algebra, then l (Ω, A) is a C*-algebra with the pointwise-defined involution. This of course generalises Example (a). If Ω is a locally compact Hausdorff space, we say a continuous function f : Ω A vanishes at infinity if, for each ε > 0, the set {ω Ω f(ω) ε} is compact. Denote by C 0 (Ω, A) the set of all such functions. This is a C*-subalgebra of l (Ω, A). The following easy result has a surprising and important corollary: Theorem If A is a self-adjoint element of a C*-algebra A, then r(a) = A. Proof. Clearly, A 2 = A 2, and therefore by induction A 2n = A 2n, so r(a) = lim n A n 1/n = lim n A 2n 1/2n = A. Corollary There is at most one norm on a -algebra making it a C*- algebra. Proof. If. 1 and. 2 are norms on a -algebra A making it a C*-algebra, the so A 1 = A 2. A 2 j = A A j = r(a A) = 41 sup λ (j = 1, 2), λ σ(a A)

42 Lemma Let A be a Banach algebra endowed with an involution such that A 2 A A (A A).Then A is a C*-algebra. Proof. The inequalities A 2 A A A A imply that A A for all A. Hence, A = A, and therefore A 2 = A A. Theorem A unital -homomorphism ϕ : A B from a unital Banach -algebra A to a unital C*-algebra B is necessarily norm-decreasing. Proof. If A A, then σ(ϕ(a)) σ(a), so ϕ(a) 2 = ϕ(a) ϕ(a) = ϕ(a A) = r(ϕ(a A)) r(a A) = A A A 2. Hence, ϕ(a) A. Theorem If A is a hermitian element of a unital C*-algebra A, then σ(a) R. Proof. Since e ia is unitary, σ(e ia ) T. If λ σ(a) and B = n=1 in (A λ) n 1 /n!, then e ia e iλ = (e i(a λ) 1)e iλ = (A λ)be iλ. Since B commutes with A, and since A λ is non-invertible, e ia e iλ is non-invertible. Hence, e iλ T, and therefore λ R. Thus, σ(a) R. Theorem If τ is a character on a unital C*-algebra A, then it preserves adjoints. Proof. If A A, then A = B + ic, where B, C are hermitian elements of A. The numbers τ(b) and τ(c) are real because they are in σ(b) and σ(c) respectively, so τ(a ) = τ(b ic) = τ(b) iτ(c) = (τ(b)+iτ(c)) = τ(a). 42

43 We shall now completely determine the abelian C*-algebras. This result can be thought of as a preliminary form of the spectral theorem. It allows us to construct the functional calculus, a very useful tool in the analysis of non-abelian C*-algebras. Theorem (Gelfand) If A is a non-zero unital abelian C*-algebra, then the Gelfand representation ϕ : A C(Ω(A)), A Â, is an isometric -isomorphism. Proof. That ϕ is a norm-decreasing homomorphism, such that ϕ(a) = r(a), is given by Theorem If τ Ω(A), then ϕ(a )(τ) = τ(a ) = τ(a) = ϕ(a) (τ), so ϕ is a -homomorphism. Moreover, ϕ is isometric, since ϕ(a) 2 = ϕ(a )ϕ(a) = ϕ(a A) = r(a A) = A A = A 2. Clearly, then, ϕ(a) is a closed -subalgebra of C(Ω(A)) separating the points of Ω(A), and having the property that for any τ Ω(A) there is an element A A such that ϕ(a)(τ) 0. The Stone-Weierstrass theorem implies, therefore, that ϕ(a) = C(Ω(A)). The following result is important. Theorem Let B be a C*-subalgebra of a unital C*-algebra A containing the unit of A. Then σ B (B) = σ A (B) (B B). 43

44 Proof. First suppose that B is a hermitian element of B. Since in this case σ A (B) is contained in R, it has no holes, and therefore by Theorem , σ A (B) = σ B (B). Therefore, B is invertible in B if and only if it is invertible in A. Now suppose that B is an arbitrary element of B, that is invertible in A, so there is an element A A such that BA = AB = I. Then A B = B A = I, so BB A A = I BB is invertible in A and therefore in B. Hence, there is an element C B such that BB C = I. Consequently, B C = A, so A B, which implies that B is invertible in B. Thus, for any element of B, its invertibility in A is equivalent to its invertibility in B. The theorem follows. Let S be a subset of a C*-algebra A. The C*-algebra generated by S is the smallest C*-subalgebra of A containing S. If S = {A}, we denote by C (A) the C*-subalgebra generated by S. If A is a normal, then C (A) is abelian. Similarly, if A is unital and A normal, then the C*-subalgebra generated by I and A is abelian. Observe that r(a) = A if A is a normal element of a unital C*-algebra, this can be proved by applying Theorem to C (A, I). If A is a unital C*-algebra and A A sa, then e ia is a unitary, but not all unitaries are of this form. Using Theorem , we can give some useful conditions that ensure a unitary does have a logarithm. Theorem Let U be a unitary in a unital C*-algebra A. If σ(u) T, then there exists A A sa such that U = e ia. If I U < 2, then σ(u) T. Proof. By replacing U by λu for some λ T if necessary, we may suppose that 1 / σ(u). Since U is normal, we may also suppose that A is abelian 44