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Notes on Operator Algebras Notes on operator algebras Jesse Peterson March 17, 2014 2 Chapter 1 Spectral theory If A is a complex unital algebra then we denote by G(A) the set of elements which have a two sided inverse. If x 2 A, the spectrum of x is σA(x) = fλ 2 C j x − λ 62 G(A)g: The complement of the spectrum is called the resolvent and denoted ρA(x). Proposition 1.0.1. Let A be a unital algebra over C, and consider x; y 2 A. Then σA(xy) [ f0g = σA(yx) [ f0g. Proof. If 1 − xy 2 G(A) then we have (1 − yx)(1 + y(1 − xy)−1x) = 1 − yx + y(1 − xy)−1x − yxy(1 − xy)−1x = 1 − yx + y(1 − xy)(1 − xy)−1x = 1: Similarly, we have (1 + y(1 − xy)−1x)(1 − yx) = 1; and hence 1 − yx 2 G(A). Knowing the formula for the inverse beforehand of course made the proof of the previous proposition quite a bit easier. But this formula is quite natural to consider. Indeed, if we just consider formal power series then we have 1 1 X X (1 − yx)−1 = (yx)k = 1 + y( (xy)k)x = 1 + y(1 − xy)−1x: k=0 k=0 1.1 Banach and C∗-algebras A Banach algebra is a Banach space A, which is also an algebra such that kxyk ≤ kxkkyk: 3 4 CHAPTER 1. SPECTRAL THEORY A Banach algebra A is involutive if it possesses an anti-linear involution ∗, such that kx∗k = kxk, for all x 2 A. If an involutive Banach algebra A additionally satisfies kx∗xk = kxk2; for all x 2 A, then we say that A is a C∗-algebra. If a Banach or C∗-algebra is unital, then we further require k1k = 1. Note that if A is a unital involutive Banach algebra, and x 2 G(A) then −1 ∗ ∗ −1 ∗ (x ) = (x ) , and hence σA(x ) = σA(x). Example 1.1.1. Let K be a locally compact Hausdorff space. Then the space ∗ C0(K) of complex valued continuous functions which vanish at infinity is a C - algebra when given the supremum norm kfk1 = supx2K jf(x)j. This is unital if and only if K is compact. Example 1.1.2. Let H be a complex Hilbert space. Then the space of all bounded operators B(H) is a C∗-algebra when endowed with the operator norm kxk = supξ2H;kξk≤1 kxξk. Lemma 1.1.3. Let A be a unital Banach algebra and suppose x 2 A such that k1 − xk < 1, then x 2 G(A). P1 k Proof. Since k1 − xk < 1, the element y = k=0(1 − x) is well defined, and it is easy to see that xy = yx = 1. Proposition 1.1.4. Let A be a unital Banach algebra, then G(A) is open, and the map x 7! x−1 is a continuous map on G(A). Proof. If y 2 G(A) and kx − yk < ky−1k then k1 − xy−1k < 1 and hence by the previous lemma xy−1 2 G(A) (hence also x = xy−1y 2 G(A)) and 1 X kxy−1k ≤ k(1 − xy−1)kn n=0 1 X 1 ≤ ky−1knky − xkn = : 1 − kyk−1ky − xk n=0 Hence, kx−1 − y−1k = kx−1(y − x)y−1k ky−1k2 ≤ ky−1(xy−1)−1kky−1kky − xk ≤ ky − xk: 1 − ky−1kky − xk ky−1k2 Thus continuity follows from continuity of the map t 7! 1−ky−1kt t, at t = 0. Proposition 1.1.5. Let A be a unital Banach algebra, and suppose x 2 A, then σA(x) is a non-empty compact set. 1.1. BANACH AND C∗-ALGEBRAS 5 x Proof. If kxk < jλj then λ − 1 2 G(A) by Lemma 1.1.3, also σA(x) is closed by Proposition 1.1.4, thus σA(x) is compact. ∗ To see that σA(x) is non-empty note that for any linear functional ' 2 A , −1 we have that f(z) = '((x − z) ) is analytic on ρA(x). Indeed, if z; z0 2 ρA(x) then we have −1 −1 −1 −1 (x − z) − (x − z0) = (x − z) (z − z0)(x − z0) : Since inversion is continuous it then follows that f(z) − f(z0) −2 lim = '((x − z0) ): z!z0 z − z0 We also have limz!1 f(z) = 0, and hence if σA(x) were empty then f would be a bounded entire function and we would then have f = 0. Since ' 2 A∗ were arbitrary this would then contradict the Hahn-Banach theorem. Theorem 1.1.6 (Gelfand-Mazur). Suppose A is a unital Banach algebra such that every non-zero element is invertible, then A =∼ C. Proof. Fix x 2 A, and take λ 2 σ(x). Since x − λ is not invertible we have that x − λ = 0, and the result then follows. Pn k If f(z) = k=0 akz is a polynomial, and x 2 A, a unital Banach algebra, Pn k then we define f(x) = k=0 akx 2 A. Proposition 1.1.7. Let A be a unital Banach algebra, x 2 A and f a polyno- mial. then σA(f(x)) = f(σA(x)). Pn k Proof. If λ 2 σA(x), and f(z) = k=0 akz then n X k k f(x) − f(λ) = ak(x − λ ) k=1 k−1 X X j k−j−1 = (x − λ) ak x λ ; k=1 j=0 hence f(λ) 2 σA(x). conversely if µ 62 f(σA(x)) and we factor f − µ as f − µ = αn(x − λ1) ··· (x − λn); then since f(λ) − µ 6= 0, for all λ 2 σA(x) it follows that λi 62 σA(x), for 1 ≤ i ≤ n, hence f(x) − µ 2 G(A). If A is a unital Banach algebra and x 2 A, the spectral radius of x is r(x) = sup jλj: λ2σA(x) Note that by Proposition 1.1.5 the spectral radius is finite, and the supremum is attained. Also note that by Proposition 1.0.1 we have the very useful equality r(xy) = r(yx) for all x and y in a unital Banach algebra A. A priori the spectral radius depends on the Banach algebra in which x lives, but we will show now that this is not the case. 6 CHAPTER 1. SPECTRAL THEORY Proposition 1.1.8. Let A be a unital Banach algebra, and suppose x 2 A. n 1=n Then limn!1 kx k exists and we have r(x) = lim kxnk1=n: n!1 Proof. By Proposition 1.1.7 we have r(xn) = r(x)n, and hence r(x)n ≤ kxnk; n 1=n showing that r(x) ≤ lim infn!1 kx k . n 1=n To show that r(x) ≥ lim supn!1 kx k , consider the domain Ω = fz 2 C j jzj > r(x)g, and fix a linear functional ' 2 A∗. We showed in Proposition 1.1.5 that z 7! '((x−z)−1) is analytic in Ω and as such we have a Laurent expansion 1 X an '((z − x)−1) = ; zn n=0 for jzj > r(x). However, we also know that for jzj > kxk we have 1 X '(xn−1) '((z − x)−1) = : zn n=1 By uniqueness of the Laurent expansion we then have that 1 X '(xn−1) '((z − x)−1) = ; zn n=1 for jzj > r(x). '(xn−1) Hence for jzj > r(x) we have that limn!1 jzjn = 0, for all linear functionals ' 2 A∗. By the uniform boundedness principle we then have n−1 kx k n 1=n limn!1 jzjn = 0, hence jzj > lim supn!1 kx k , and thus n 1=n r(x) ≥ lim sup kx k : n!1 Exercise 1.1.9. Suppose A is a unital Banach algebra, and I ⊂ A is a closed two sided ideal, then A=I is again a unital Banach algebra, when given the norm ka + Ik = infy2I ka + yk, and (a + I)(b + I) = (ab + I). Exercise 1.1.10. Let A be a unital Banach algebra and suppose x; y 2 A such that xy = yx. Show that r(xy) ≤ r(x)r(y). 1.2 The Gelfand transform Let A be a abelian Banach algebra, the spectrum of A, denoted by σ(A), is the set of continuous ∗-homomorphsims ' : A ! C such that k'k = 1, which we endow with the weak*-topology as a subset of A∗. 1.3. CONTINUOUS FUNCTIONAL CALCULUS 7 Note that if A is unital, and ' : A ! C is a ∗-homomorphism, then it follows easily that ker(') \ G(A) = ;. In particular, this shows that '(x) 2 σ(x), since x − '(x) 2 ker('). Hence, for all x 2 A we have j'(x)j ≤ r(x) ≤ kxk. Since, '(1) = 1 this shows that the condition k'k = 1 is automatic in the unital case. It is also easy to see that when A is unital σ(A) is closed and bounded, by the Banach-Alaoglu theorem it is then a compact Hausdorff space. Proposition 1.2.1. Let A be a unital Banach algebra. Then the association ' 7! ker(') gives a bijection between the spectrum of A and the space of maximal ideals. Proof. If ' 2 σ(A) then ker(') is clearly an ideal, and if we have a larger ideal I, then there exists x 2 I such that '(x) 6= 0, hence 1 − x='(x) 2 ker(') ⊂ I and so 1 = (1 − x='(x)) + x='(x) 2 I which implies I = A.
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