9. Banach algebras and C -algebras

Transcription

1 Institut for Matematiske Fag Det Naturvidenskabelige Fakultet Aarhus Universitet September 2005

2 We read in W. Rudin: Functional Analysis Based on parts of Chapter 10 and parts of Chapter 11.

3 Banach algebras - the definition A complex algebra A is a vector space over C in which a multiplication is defined such that and x(yz) = (xy)z, (x + y)z = xz + yz, x(y + z) = xy + xz, α(xy) = (αx)y = x(αy) for all x,y,z A and all α C. Examples to have in mind: i) The complex n n matrices is an algebra M n (C) with the usual multiplication of matrices: n (AB) ij = A ik B kj. k=1 ii) The complex functions f : X C on any set X: (fg)(x) = f (x)g(x).

4 Banach algebras - the definition If a complex algebra A in addition is a Banach space with respect to a norm which satisfies that xy x y for all x,y A, then A is a Banach algebra. A is a unital Banach algebra when it is a Banach algebra and there is an element e A such that ex = x = xe for all x A. e is then called the unit of A. In contrast to Rudin, I usually denote the unit by 1. (Note that we slightly deviate from Rudin s terminology since he only allow Banach algebras that are unital.) An important class of examples: Let X be a Banach space. A linear map L : X X is often called an operator and it is bounded when there is a constant K > 0 such that Lx K x for all x X. The set of such bounded operators is denoted by B(X) and it is a complex vector space in an obvious way.

5 B(X) B(X) is a complex algebra when the product is given by composition: (AB)(x) = A(Bx), and a Banach algebra when B(X) is given the operator norm: L = sup { Lx : x 1}. It follows, for example, that M n (C) is a Banach algebra in the operator norm A = sup { Av : v C n, v 1}. Then norm v on C n can be any norm - showing that an algebra can be a Banach algebra in more than one way! We shall later see the signifigance of choosing the euclidian norm on C n in this case.

6 Basic theory of Banach algebras Let A be a complex algebra. A non-zero linear map φ : A C which satisfies that φ(ab) = φ(a)φ(b), a,b A, is called a complex homomorphism or a character. Example to think of: Let X be a compact Hausdorff space, and C(X) the algebra of continuous complex-valued functions on X. Let x X, and define φ x : C(X) C by φ x (f ) = f (x). Then φ x is a character. An element a of a unital algebra A (with unit 1) is invertible when there is an element b A such that ab = ba = 1. The element b, which is then unique, is called the inverse of a, and denoted by a 1. Thus in M n (C) a matrix a is inverible if and only if the determinant of a is not zero. And in the algebra of all functions on a set X, a function f : X C is invertible if and only if f (x) 0 for all x X.

7 A lemma Lemma (Proposition 10.6) Let A be a complex algebra with unit 1, and φ : A C a character. Then φ(1) = 1 and φ(x) 0 when x A is invertible. Proof. Since φ is not identical zero there is a y A such that φ(y) 0. Since φ(y) = φ(1y) = φ(1)φ(y), we conclude that φ(1) = 1, When x A is invertible we find that 1 = φ(1) = φ ( xx 1) = φ(x)φ ( x 1). This is only possible when φ(x) 0.

8 A fundamental theorem on Banach algebras Theorem Let A be a Banach algebra with unit 1, and x A an element with x < 1. Then (a) 1 x is invertible, (b) (1 x) 1 1 x x 2 1 x, (c) φ(x) < 1 for every character φ on A. Proof. (a) Set s n = 1 + x + x x n, and note that m s n s m = x j m x j m x j when m > n. j=n+1 j=n+1 j=n+1

9 A fundamental theorem on Banach algebras This yields (b) when we let n tend to infinity. Proof. Since j=0 x j <, we conclude that {s n } is a Cauchy sequence in A. Set s = lim n s n A. Since s n x = x + x x n+1 = s n+1 1, we find be letting n tend to, that sx = s 1 1 = s(1 x). Similarly, the equality xs n = x + x x n+1 = s n+1 1 leads to the conclusion that 1 = (1 x)s. It follows that 1 x is invertible and that (1 x) 1 = s. (b) Note that s n 1 x = x 2 + x x n x j = j=2 x j 1 x = j=0 n x j j=2 1 x 2 1 x = 1 x 1 x.

10 A fundamental theorem on Banach algebras Proof. (c) Let λ C, λ 1. Then λ 1 x < 1 and hence 1 λ 1 x is invertible by (a). By Lemma 1 this implies that 1 λ 1 φ(x) = φ ( 1 λ 1 x ) 0. Thus φ(x) λ. We conclude that φ(x) < 1.

11 The spectrum Let A be a unital Banach algebra, and x A. The spectrum of x is by definition the set of complex numbers λ such that λ1 x is not invertible in A. We denote the spectrum of x by σ(x), i.e. σ(x) = {λ C : λ1 x is not invertible}. The spectral radius of x is ρ(x) = sup { λ : λ σ(x)}. At this point it is not at all clear if the spectrum can be empty - the next goal is to show that this can never happen. I.e. we will show that σ(x).

12 A lemma Lemma (Theorem ) (a) Let A be a unital Banach algebra, x A an invertible element, and h A an element such that h < 1 2 x 1 1. Then x + h is invertible and (x + h) 1 x 1 + x 1 hx 1 2 x 1 3 h 2. (b) The set A 1 of invertible elements of A is an open subset of A and the map x x 1 is a homeomorphism of A 1 onto A 1.

13 A lemma - the proof Proof. (b) follows immediately from (a) so we concentrate on (a): Since x + h = x ( 1 + x 1 h ) = x ( 1 ( x 1 h )) and x 1 h h x 1 x x 1 1 = 1 2, it follows from (a) of Theorem 2 that x + h is invertible and from (b) of the same theorem that (x + h) 1 x 1 + x 1 hx 1 = ( 1 ( x 1 h )) 1 x 1 x 1 + x 1 hx 1 = (1 ( [ x 1 h )) ] x 1 h x 1 x 1 x 1 h 2 1 x 1 h x 1 x 1 2 h 2 1/2 = 2 x 1 3 h 2. In order to compensate for the lack of background in complex function theory we now deviate briefly from Rudins presentation.

14 Another lemma Lemma Let A be a unital Banach algebra and let x A be an invertible element. Then every element y A with x y < 1 is also x 1 invertible and x 1 y 1 x 1 2 x y 1 x 1 x y. Proof. Note that 1 x 1 y = x 1 (x y) x 1 x y < 1 so that x 1 y is invertible. y must therefore also be invertible with y 1 = (x 1 y) 1 x 1 = n=0 (1 x 1 y) n x 1, so that x 1 y 1 x 1 n=1 1 x 1 y n = x 1 1 x 1 y 1 1 x 1 y x 1 2 x y 1 x 1 x y.

15 On the last lemmas + another one Note that the qualitative statements of the last lemmas is that the invertible elements form an open subset of A and that the map x x 1 is continuous on this subset. In particular we see that for any sequence {x n } in A we have the equivalence lim x n = 0 lim (1 x n) 1 = 1. (1) n n Lemma Let {α n } be a sequence of real numbers such that 1 α n+m α n + α m for all n,m N. Then the limit lim n n α n 1 exists and equals inf n N n α n.

16 Proof of the last lemma Fix ǫ > 0 and m N. Choose N N so large that 2 m N max { α 1, 1 2 α 2,..., 1 m α m } < ǫ. If n N we write n = lm + r where l N and r {1,2,...,m}. Then 1 n α n 1 n α lm + 1 n α r l n α m + 1 n α r = lm 1 n m α m + r 1 n r α r = 1 m α m + r 1 n r α r r 1 n m α m 1 m α m + 2 m N max { α 1, 1 2 α 2,..., 1 m α m } < 1 m α m + ǫ.

17 Proof of the last lemma Thus lim sup n 1 n α 1 n sup n N n α n 1 m α m + ǫ. If we let ǫ 0 and take the infimum over all m N we see that lim sup n 1 n α 1 n inf n N n α n. 1 Since obviously inf n N n α 1 n lim inf n n α n we conclude that 1 lim n n α 1 n = inf n N n α n.

18 The hard lemma Lemma Let A be a unital Banach algebra and a A. There is an element λ σ(a) such that λ lim n a n 1 n = inf n N a n 1 n. To prove this we prove first that lim n a n 1 n = inf n N a n 1 n. If a k = 0 for some k N, a n = 0 n k and the conclusion is trivial. So we assume that a n > 0 n N and set α n = log a n. Note that α n+k α n + α k, n,k N. α By one of the preceding lemmas lim n n n = inf n N It follows that α n n. lim n an 1 1 n = exp( lim n n α 1 n) = exp( inf n N n α n) = inf n N an n. 1

19 The proof of the hard lemma Set r = lim n a n 1 n. We will prove that σ(a) contains an element λ with λ r. If r = 0 it follows that 0 σ(a). Indeed, if a is invertible we have that 1 = a n a n so that 1 a n a n for all n N and this implies that 1 r lim n a n n, 1 which is impossible because r = 0. We can therefore assume that r > 0. Assume to get a contradiction that λ1 a is invertible whenever λ r. Fix first n N and let ω = e 2πi n. 1 Set M k = {1,2,...,n}\{k}, k = 1,2,...,n. Then n (1 ω k z) = 1 z n, z C,and k=1 n k=1 i M k (1 ω i z) = n, z C.

20 The proof of the hard lemma If the reader find it difficult to prove these algebraic identities, we offer the following arguments : The identity n k=1 (1 ωk z) = 1 z n follows from the fact that the two polynomials have the same roots, namely {ω k : k = 1,2,,n}. The second identity follows by observing that the left hand side is a polynomial of degree n 1, which takes the same value at n distinct points, namely at z = ω k, k = 1,2,,n. It must therefore be the constant polynomial, and the constant value, n, is easily identified by checking for z = 0. By assumption 1 a λ is invertible when λ r and the above polynomial identities show that n (1 ω k a λ ) = 1 an λ n (2) and k=1 n (1 ω i a ) = n1. (3) λ i M k k=1

21 The proof of the hard lemma Substituting (2) into (3) yields that 1 n n (1 an λ n)(1 a ωk λ ) 1 = 1 k=1 or, alternatively, that (1 an λ n ) is invertible with (1 an λ n ) 1 = 1 n n (1 ω k a λ ) 1 (4) k=1 when λ r. Set K = sup { (λ1 a) 1 : λ r}. This is a finite number for the following reason. When λ > a + 1 we have that (λ1 a) 1 = λ 1 (1 a λ ) 1 = λ 1 n=0 ( a λ )n λ 1 n=0 ( a λ )n = 1 λ a 1.

22 The proof of the hard lemma Since M = {λ C : r λ a + 1} is a compact set and λ (λ1 a) 1 is continuous by any of the two previous lemmas, we see that K max {1, sup (λ1 a) 1 } <. λ M For each k N and λ r we have that (1 ω k a r ) 1 (1 ω k a λ ) 1 = ω k a( 1 r 1 λ )(1 ωk a r ) 1 (1 ω k a λ ) 1 = ω k a(λ r)(r1 ω k a) 1 (λ1 ω k a) 1 which gives the estimate (1 ω k a r ) 1 (1 ω k a λ ) 1 λ r a K 2 when λ r. By substituting this estimate into (4) we get that (1 an r n ) 1 (1 an λ n ) 1 λ r a K 2 (5)

23 The proof of the hard lemma We prove next that an lim 1 (1 n r n ) 1 = 0. (6) Let ǫ > 0. Choose λ C such that λ > r and a K 2 λ r ǫ 2. a Since λ > r, it follows that lim n ] n λ = 0. To see this, let n t r λ [.,1 Since lim n a n 1 n λ = λ 1 lim n n a n 1 n = r λ < t, there is an N so large that a n 1 n < t n for all n N. Hence lim n a n λ n = 0 because lim n t n = 0.

24 The proof of the hard lemma Now an application of (1) gives N N such that 1 (1 an λ n ) 1 ǫ 2, n N. Combined with (5) this shows that 1 (1 an r n ) 1 1 (1 an λ n ) 1 + (1 an λ n ) 1 (1 an r n ) 1 ǫ, n N proving (6). Combined with (1), (6) implies that lim n a n r n = 0. Thus ( an r n ) 1 n = r 1 ( a n ) 1 n < 1 for all sufficiently large n. But this is impossible since r = inf n a n 1 n.

25 The fundamental theorem on spectra Theorem Let A be a unital Banach algebra and a A. Then σ(a) is a non-empty compact subset of C and ρ(a) = inf n N an 1 n = lim n an 1 n a. Proof: If λ n > a n for some n N, λ n 1 a n = λ n (1 an λ ) is n invertible by Theorem 2. Since λ n 1 a n = (λ1 a) n 1 k=0 λk a n k 1, it follows that λ1 a is invertible, i.e. λ / σ(a). This shows that ρ(a) inf n N an 1 n = lim n an 1 n. But Lemma 6 implies that lim n a n 1 n ρ(a), i.e. we have that ρ(a) = inf n N an 1 n = lim n an 1 n.

26 Proof of the fundamental theorem on spectra and a corollary Since a n 1 n a, this identity implies that ρ(a) a. In particular, ρ(a) is a bounded subset of C. The invertible elements of A form an open subset of A and hence C\σ(a) = {λ C : λ1 a is invertible} is open in C; i.e. σ(a) is closed and therefore compact. σ(a) is non-empty by Lemma 6. Corollary Let A be a unital Banach algebra. If all non-zero a A are invertible we have that A = C1. Proof. Let a A. By Theorem 7 σ(a), so there is a λ C such that λ1 a is not invertible. By assumption this implies that λ1 a = 0, i.e. a = λ1.