Chapter 1 Banach algebras Whilst we are primarily concerned with C∗-algebras, we shall begin with a study of a more general class of algebras, namely, Banach algebras. These are of interest in their own right and, in any case, many of the concepts introduced in their analysis are needed for that of C∗-algebras. Furthermore, some feeling for the kind of behaviour that can occur in various Banach algebras helps one to appreciate how well-behaved C∗-algebras are. Denition 1.1. A Banach algebra is a complex Banach space A together with an associative and distributive multiplication such that λ(ab) = (λa)b = a(λb) and kabk ≤ kak kbk for all a; b 2 A, λ 2 C. For any x; x0; y; y0 2 A, we have kxy − x0y0k = kx(y − y0) + (x − x0)yk ≤ kxk ky − y0k + kx − x0k ky0k and so we see that multiplication is jointly continuous. The algebra A is said to be commutative (or abelian) if ab = ba for all a,b in A, and A is said to be unital if it possesses a (multiplicative) unit (this is also called an identity). Note that if A has an identity, then it is unique: since if 1l and 1l0 are units, then 1l = 1l1l0 = 1l0. Example 1.2. If E is a complex Banach space, then B(E), the set of bounded linear operators on E is a unital Banach algebra when equipped with the usual linear structure and operator norm. If 1l denotes the unit in the unital Banach algebra A, then 1l = 1l2 and so we have k1lk ≤ k1lk k1lk, which implies that k1lk ≥ 1. 1 2 Chapter 1 Lemma 1.3. Let A be a Banach algebra with identity 1l. Then there is a norm jjj · jjj on A, equivalent to the original norm, such that (A; jjj · jjj) is a unital Banach algebra with jjj1ljjj = 1. Proof. For each x 2 A, let Lx denote the linear operator Lx : y 7! xy 2 A, 0 y 2 A. Then if Lx = Lx0 , it follows that Lx1l = Lx0 1l and so x = x . Hence x 7! Lx is an injective map from A into the set of linear operators on A. Now, kLxyk = kxyk ≤ kxk kyk ; for y 2 A which implies that Lx is bounded, and kLxk ≤ kxk. Put jjjxjjj = kLxk. Then we have just shown that jjjxjjj ≤ kxk, for any x 2 A. On the other hand, jjjxjjj = kLxk = supfkLxyk : kyk ≤ 1g = supfkxyk : kyk ≤ 1g 1l ≥ kxy0k; where y0 = k1lk kxk = : k1lk Hence, kxk=k1lk ≤ jjjxjjj ≤ kxk, for all x 2 A, which shows that the two norms jjj· jjj and k · k are equivalent. Moreover, for any x; y 2 A, jjjxyjjj = kLxyk = kLxLyk ≤ kLxk kLyk = jjjxjjjjjjyjjj and so A with norm jjj · jjj is a Banach algebra. To complete the proof, we have jjj1ljjj = kL1l k = 1. This lemma allows us to assume that the unit of a unital Banach algebra has norm 1. In fact, this is often taken as part of the denition of a unital Banach algebra. If A does not have a unit, then we can adjoin one as follows. Lemma 1.4. A Banach algebra A without a unit can be embedded into a unital Banach algebra AI as an ideal of codimension one. Proof. Let AI = A ⊕ C as a linear space, and dene a multiplication in AI by (x; λ)(y; µ) = (xy + µx + λy; λµ): Department of Mathematics Banach algebras 3 It is easily checked that this is associative and distributive. Moreover, the element (0; 1) is a unit for this multiplication: (x; λ)(0; 1) = (x0 + x + λ0; λ1) = (x; λ) = (0; 1)(x; λ): Put k(x; λ)k = kxk + jλj . Then AI is a Banach space when equipped with this norm. Furthermore, k(x; λ)(y; µ)k = k(xy + µx + λy; λµ)k = kxy + µx + λyk + jλµj ≤ kxkkyk + jµjkxk + jλjk + kjλjjµj = (kxk + jλj)(kyk + jµj) = k(x; λ)k k(y; µ)k : Hence AI is a Banach algebra with unit. We may identify A with the ideal f(x; 0) : x 2 Ag in AI via the isometric isomorphism x 7! (x; 0). We write (x; λ) as (x; λ) = x + λ1l 2 AI . (Compare this with complex numbers a + ib $ (a; b).) Note that k1lk = 1. We will see, later, that an analogous result holds for C∗-algebras, but more care has to be taken regarding the norm. Examples 1.5. 1. Consider C([0; 1]), the Banach space of continuous complex-valued func- tions dened on the interval [0; 1] equipped with the sup-norm, namely, , and with multiplication dened pointwise: kfk = sups2[0;1] jf(s)j (fg)(s) = f(s) g(s) ; for s 2 [0; 1]: Then C([0; 1]) is a commutative unital Banach algebra; the constant function 1 is the unit element. 2. As above, but replace [0; 1] by any compact topological space. 3. Let D denote the closed unit disc in C, and let A denote the set of continuous complex-valued functions on D which are analytic in the interior of D. Equip A with pointwise addition and multiplication and the norm kfk = supfjf(z)j : z 2 @Dg where @D is the boundary of D, that is, the unit circle. (That this is, indeed, a norm follows from the maximum modulus principle.) Then A is complete, and so is a (commutative) unital Banach algebra. A is called the disc algebra. King's College London 4 Chapter 1 4. Let be the Banach space 1 and dene by P A ` (Z) xy (xy)n = m xmyn−m for x = (xn) and y = (yn) in A. Then X X X j(xy)nj ≤ jxmjjyn−mj n n m X X = jxmj jyn−mj m n X = jxmj kyk m = kxk kyk : 1 Thus xy 2 ` (Z) and so A is a Banach algebra. Furthermore, A has a unit given by (xn) = (δ0n) = (:::; 0; 0; 1; 0; 0;::: ) where the 1 appears in the 0th position. Denition 1.6. An element x in a unital Banach algebra A is said to be invertible (or non-singular) in A if there is some z 2 A such that xz = zx = 1l. Note that if such a z exists, then it is unique; if z0x = xz0 = 1l, then z = z1l = zxz0 = 1lz0 = z0 . z is called the inverse of x, and is written x−1, as usual. Evidently, the set of invertible elements forms a group. Non-invertible elements are also called singular. Proposition 1.7. The set G(A) of invertible elements in a unital Banach al- gebra A is open in A, and the inverse operation x 7! x−1 is a continuous map from G(A) to G(A). Proof. First let with , and put Pn k, . Then y 2 A kyk < 1 sn = k=0 y n 2 N (sn) is a Cauchy sequence in A and so converges, since A is complete. Let denote its limit; P1 k. We claim that is the inverse of 1l . w w = k=0 y w − y Indeed, we have (1l − y)w = lim(1l − y)sn n = lim(1l − yn+1) = 1l n and w(1l − y) = lim sn(1l − y) n = lim(1l − yn+1) = 1l n which establishes the claim. Hence, if x 2 A with k1l − xk < 1, then writing x = 1l − (1l − x) and arguing as above (with y = 1l − x), we see that x is invertible and its inverse −1 is given by the convergent series P1 1l k. x k=0( −x) Let . Then for any , we have −1 . Now, x0 2 G(A) x 2 A x = x0x0 x 1l −1 −1 −1 k − x0 xk = kx0 (x0 − x)k ≤ kx0 k kx0 − xk; Department of Mathematics Banach algebras 5 and so we conclude that if −1 −1 then −1 is invertible with kx − x0k < kx0 k x0 x inverse given by 1 −1 −1 X k (x0 x) = y k=0 with 1l −1 . Hence is invertible and y = ( − x0 x) x 1 −1 X −1 k −1 x = [x0 (x0 − x)] x0 : k=0 −1 To see that x 7! x is continuous on G(A), suppose that x0 2 G(A) and that (xn) is a sequence in G(A) such that xn ! x0 as n ! 1. Then for all suciently large , −1 −1 and so n kxn − x0k < kx0 k 1 −1 −1 X −1 k −1 kx − x k = x (x0 − xn) x n 0 0 0 k=1 1 X −1 k −1 ≤ kx0 kkx0 − xnk kx0 k k=1 ! 0 as n ! 1. Denition 1.8. Let A be a unital algebra, and let x 2 A. The spectrum of x is the subset σA(x) of C given by σA(x) = fλ 2 C : x − λ1l 2= G(A)g: The resolvent set ρA(x) of x is the complement of the spectrum of x; ρA(x) = C n σA(x): The spectral radius rA(x) of an element x 2 A is dened as rA(x) = supfjλj : λ 2 σA(x)g provided σA(x) is not empty. Example 1.9. Let A be the unital Banach algebra Mn(C) of n × n complex matrices. For a 2 A and λ 2 C, a − λ1l is invertible in A if and only if λ is not an eigenvalue of a. In other words, σA(a) is just the set of eigenvalues of the matrix a. Example 1.10.