Commutative Banach Algebras

Tommaso R. Cesari

Unofficial notes1

(Seminar on Commutative Banach Algebras - Vladimir P. Fonf)

1Editor's note These notes were written during academic year 2014-2015. They are totally independent of the Professor's will. No guarantee is given regarding the completeness or correctness of this paper. In particular, it is highly probable that many errors (likely conceptual!) will occur because of the unskillfulness of the curator. Use the information contained herein at your own risk.

Tommaso R. Cesari Contents

1 Commutative Banach Algebras 3 1.1 Basic denitions ...... 3 1.2 Basic properties of spectra ...... 8 1.2.1 Abstract holomorphic functions ...... 10 1.3 Ideals ...... 13 1.3.1 Multiplicative functionals ...... 16 1.3.2 Quotient algebra ...... 17 1.4 Involutions ...... 27 1.5 Applications ...... 32 1.5.1 Wiener algebra ...... 32 1.5.2 Disk algebra ...... 34 1.5.3 Complex compact operators in Hilbert spaces ...... 36

2 Seminars 39 2.1 Gelfand and Fourier transform (Tommaso Cesari) ...... 39 2.2 Lomonosov's Invariant Subspace Theorem (Tommaso Russo) . . 41 Chapter 1

Commutative Banach Algebras

1.1 Basic denitions

Denition 1 (Banach algebra). A complex Banach space X is called a (commutative) Banach algebra if there exist a product

X × X → X, (x, y) 7→ xy such that

1. (commutativity) for all x, y ∈ X,

xy = yx;

2. (associativity) for all x, y, z ∈ X,

(xy) z = x (yz);

3. (distributivity) for all x, y, z ∈ X,

x (y + z) = xy + xz;

4. (continuity) for all {xn} ⊂ X, x, y ∈ X, there exists the limit n∈N

lim xny = xy. n→+∞

Problem 2. The last property states that in Banach algebras the product is continuous with respect to both components. Does this implies that the product is continuous? We will see later on that the answer is yes. 1.1 Basic denitions 4

Problem 3. Do Banach algebras always have a unit1? Sadly, the answer is no. Luckily enough, though, there is a standard way to add a unit element to any Banach algebra.

Proposition 4. Let X be a Banach algebra. Then, the set Xe := {(x, λ) ∈ X × C} is a Banach algebra with respect to the operations

∀ (x, λ) , (y, µ) ∈ X,e (x, λ) + (y, µ) := (x + y, λ + µ) , ∀ (x, λ) ∈ X,e ∀µ ∈ C, µ (x, λ) := (µx, µλ) , ∀ (x, λ) , (y, µ) ∈ X,e (x, λ)(y, µ) := (xy + µx + λy, λµ) and the norm ∀ (x, λ) ∈ X, k(x, λ)k = kxk + |λ| ; e Xe X Xe has unit (0, 1) and admits a subalgebra isometric to X. The isometric em- bedding of X in Xe is given by

∀x ∈ X, x 7→ (x, 0) .

Proof. Trivial.

Remark 5 (Convention). Because of the previous proposition from now on we will always assume that Banach algebras have units. We will always use the letter e to denote the unit element of a Banach algebra. Proposition 6. Let X be Banach space and A, B : X → X two bounded linear operators. Then, indicating with |||·||| the operator norm on BL (X; X),

|||AB||| ≤ |||A||| · |||B|||

Proof. For all x ∈ X, kxk = 1,

k(AB)(x)k = kA (B (x))k ≤ |||A||| kB (x)k ≤ |||A||| · |||B||| · kxk . |{z} =1

Theorem 7. Let be a Banach algebra. Then there exists a norm (X, k·kX ) k·k equivalent to such that, for all , k·kX x, y ∈ X kxyk ≤ kxk kyk (1.1.1) and kek = 1.

1I.e. an element e ∈ X such that, for all x ∈ X, ex = x. 1.1 Basic denitions 5

Proof. For all x ∈ X, the map

Vx : X → X, y 7→ xy. is linear and continuous (because of the distributivity and the componentwise continuity the product). Then

V := {Vx| x ∈ X} ⊂ BL (X; X) := {T : X → X| T is bounded and linear} . Clearly V is a complex linear space (distributivity again). It is well known that BL (X; X) is a Banach space. We want to prove that V is a closed subspace 0 of BL (X; X). Take any V ∈ V . Than there exists {xn} ⊂ X such that n∈N n→+∞ . This means that, for Vxn −→ V n → +∞

sup {|xny − V (y)|} → 0, y∈X i.e. for all y ∈ X, if n → +∞

xny → V (y) ∈ X. This implies (by the componentwise continuity of the product and the completeness of X) that there exists x ∈ X such that

xn → x. By the uniqueness of the limit and the componentwise continuity of the product,

V = Vx, thus V ∈ V and consequently V is a Banach space. It's easy to check that V is a Banach algebra with respect to the pointwise product2. Consider the mapping

T : V → X,

Vx 7→ x. Clearly T is linear. Note that, for all x ∈ X

e 1 kVxk = sup {kxyk } ≥ x = kxk . V X kek kek X kykX =1 x X X This shows that for all x ∈ X,

kT (Vx)kX ≤ kekX kVxkV , hence T is continuous. By the open mapping theorem T is an homeomorphism, so −1 and are equivalent norms. Clearly the unit of is and T (·) V k·kX V Ve

kVekV = sup {kexkX } = 1. kxkX =1 Finally, the inequality (1.1.1) follows by the previous proposition.

2For proving the componentwise continuity of the product use the fact, proven above, that implies . Vxn → Vx xn → x 1.1 Basic denitions 6

Remark 8 (Convention). Because of the previous result from now on we will always assume that Banach algebras satisfy inequality (1.1.1) and that the unit has norm 1. Example 9. Let's now see some examples of Banach agebras. 1. The Banach space is a Banach algebra with unit with (C [0, 1] , k·k∞) e ≡ 1 respect to the product (fg)(·) = f (·) g (·). The relations kek = 1 and

∀f, g ∈ C [0, 1] , kfgk∞ ≤ kfk∞ kgk∞ also (clearly) hold.

2. Let D ⊂ C be the complex unit disk. The normed space ! is holomorphic D := f ∈ C D; C f|D , sup {·} D is a Banach algebra with unit e ≡ 1 with respect to the pointwise product. D is called disk algebra. Cleary D satises (1.1.1). 3. The normed space

 (n)  k (·) X Ck ([0, 1] ; ) , ∞  C n!  n=0

is a Banach algebra with unit e ≡ 1 with respect to the pointwise product. It saties (1.1.1).

Corollary 10 (Continuity of the product). Let A be a Banach algebra. Let {xn} , {yn} ⊂ A and x, y ∈ A. If xn → x and yn → y, then xnyn → xy. n∈N n∈N Proof. For all n ∈ N

kxnyn − xyk = kxnyn − xny + xny − xyk

≤ kxnyn − xnyk + kxny − xyk

= kxn (yn − y)k + k(xn − x) yk

≤ kxnk kyn − yk + kxn − xk kyk .

−1 Theorem 11. Let A be a Banach algebra and G := x ∈ A ∃x ∈ A . Then 1. G is open in A; 2. the mapping

ψ : G → G, x 7→ ψ (x) := x−1 is continuous. 1.1 Basic denitions 7

Proof.

1. Let's start by proving that for all x ∈ A, kxk < 1, there exists (e − x)−1 ∈ G. I.e. there is a neighborhood of e consisting of invertible elements. Let's prove that the series +∞ X y := xk k=0 converges. By induction it's easy to prove that, for all k ∈ N k k x ≤ kxk . Then, for all there exists such that, for all ε > 0 n0 ∈ N n, m ∈ N, n0 ≤ n < m,

m m X k X k x ≤ x k=n+1 k=n+1 m X k ≤ kxk k=n+1 < ε. Being A a Banach space, the sum converges. Claim. y = (e − x)−1 i.e. y (e − x) = e. Indeed

n X y (e − x) = (e − x) lim xk n→+∞ k=0 [The product is continuous] n X = lim (e − x) xk n→+∞ k=0 n n+1 ! X X = lim xk − xk n→+∞ k=0 k=1 = lim e − xn+1 n→+∞ = e Let's nally prove that G is open. Take x ∈ G and h ∈ A such that 1 khk < . 2 kx−1k Then (x + h) = x e + x−1h . The right hand side is invertible by hypothesis and the rst part of the proof, so x + h is invertible. 1.2 Basic properties of spectra 8

2. Take x ∈ G and h ∈ A so that 1 khk ≤ . 2 kx−1k Then

−1 −1 −1 −1 −1 −1 (x + h) − x = x e + x h − x −1 −1 −1 = x e + x h − e

 −1

−1 −1 ≤ x e + x h − e | {z } =:y = (∗) . Being kyk < 1/2, by the previous part of the proof,

+∞ −1 X k (e + y) = (−1) yk. k=0 So

+∞ −1 X k k (∗) ≤ x (−1) y k=1

+∞ −1 X k+1 k = x y (−1) y k=0

+∞ −1 X k+1 k ≤ x kyk (−1) y k=0 +∞ −1 X k ≤ x kyk y k=0 +∞ −1 2 X 1 ≤ x khk 2k k=0 −1 2 = 2 x khk , and ψ is continuous.

1.2 Basic properties of spectra

Denition 12 (Spectrum and resolvent set). Let A be a Banch algebra and x ∈ A. The set

σ (x) := {λ ∈ C | x − λe is not invertible} 1.2 Basic properties of spectra 9 is called spectrum of x. The set n o −1 Ω(x) := λ ∈ ∃ (x − λe) is called resolvent set of x. For all λ ∈ Ω(x), we will also use the notation

−1 Rλ (x) := (x − λe) .

Lemma 13. Let A be a Banch algebra and x ∈ A. If λ ∈ C, |λ| > kxk, then λ ∈ Ω(x). In other words

Ω(x) ⊃ C \ B (0, kxk) . Proof. Let's x , . We need to prove the existance of . λ ∈ C |λ| > kxk Rλ (x) Being x x − λe = λ − e λ x = −λ e − λ and x kxk = < 1, λ |λ| by the proof of Theorem 11 on page 6 it follows that x is invertible, hence e − λ there exists x −1 (x − λe)−1 = λ−1 − e . λ

Corollary 14. Let A be a Banch algebra and x ∈ A. If λ ∈ σ (x), then |λ| ≤ kxk. In other words σ (x) ⊂ B (0, kxk). Corollary 15. Let A be a Banch algebra and x ∈ A. Then the spectrum of x is a bounded set. Lemma 16. Let A be a Banch algebra and x ∈ A. Then the set Ω(x) is open in C.

Proof. Suppose λ0 ∈ Ω(x). Let's nd a neigborhood of λ0 in Ω(x). By denition it exists −1. For all , if , , then (x − λ0e) ε > 0 λ ∈ C |λ − λ0| < ε

ke − λ0x − (e − λx)k = |λ − λ0| kxk < ε kxk . Being G open3, (x − λe) is also invertible, hence λ ∈ Ω(x). Corollary 17. The set σ (x) is closed. Corollary 18. The set σ (x) is compact.

3G is the set of invertible elements in A. 1.2 Basic properties of spectra 10

1.2.1 Abstract holomorphic functions

Denition 19 (Abstract holomorphic function). Let A a Banach algebra, D ⊂ open and . Given , we say that is abstractly holomorpic C f : D → A z0 ∈ D f in z0 if the following limit exists:

0 f (z0 + h) − f (z0) f (z0) := lim . h→0 h If f is abstracly holomorphic in every point of D we say that f is an abstract holomorphic function. If f is an abstract holomorphic function and D = C we say that f is an abstract entire function. Remark 20. Almost every result which holds for holomorphic functions also holds for abstract holomorphic functions.

Theorem 21 (Dunford). Let A a Banach algebra, D ⊂ C open and f : D → A. Then f is an abstract holomorphic functions if and only if for all α ∈ A∗, α ◦ f : D → C is a (regular) holomorphic function. Theorem 22. Let A a Banach algebra, D ⊂ C open and f : D → A. Then for all z0 ∈ D there exists r > 0 such that, for all z ∈ B (z0, r),

+∞ X (k) k f (z) = f (z0)(z − z0) . k=0 Notation 23 (Radius of convergence). What is the radius of convergence of the previous power series? Without loss of generality, suppose z0 = 0. We need to study the convergence of the series

+∞ X n z 7→ cn z . n=0 |{z} |{z} ∈A ∈C Dening 1 R := , pn lim supn→+∞ kcnk by the general theory of power series, it is possible to state that the series converges in B (0,R). But let's look at this problem from a dierent prospective. Fix c ∈ A \{0} and consider the complex power series

+∞ X z 7→ cnzn. n=0

Clearly, for all z ∈ C such that kczk < 1, i.e. for all z ∈ B (0, 1/ kck), the series converges. Since by the general theory if the series converges in z ∈ C, then z ∈ B (0,R), the following inequality should hold: 1 1 ≤ . (1.2.1) pn n kck lim supn→+∞ kc k 1.2 Basic properties of spectra 11

This is an easy exercise, in fact for all n ∈ N q pn kcnk ≤ n kckn = kck .

The following proposition proves a useful fact. The lim sup in (1.2.1) always exists as a limit.

Proposition 24. Let A a Banach algebra and x ∈ A. Then there exists the limit lim pn kxnk. n→+∞ Proof. For all , call n . For all and for all there n ∈ N αn := kx k n ∈ N k ∈ N exist m, ` ∈ N such that n = mk + ` and consequently . n αn = kx k km+` = x

km ` = x x

km ` ≤ x x k m ` ≤ x x . m = αk α`, hence 1 n m/n 1/n (αn) ≤ αk α` . Note that if satisfy , then m 1 ` and n, k, m, ` ∈ N n = mk + ` n = k − kn

1 1 − ` n k kn 1/n (αn) ≤ αk α` . Since ` is bounded by 0 and k (it's the rest in the Euclidean division), taking both sides to the lim sup, we derive, for all k ∈ N,

1 n 1/k lim sup (αn) ≤ ak . n→+∞ Now, taking both member to the lim inf, we get

1 n 1/k lim sup (αn) ≤ lim inf ak n→+∞ k→+∞ which proves the existence of the limit.

Notation 25 (Rλ). Let A be a commutative Banach algebra. Whenever x ∈ A is xed, for all λ ∈ Ω(x) we will write Rλ instead of Rλ (x) in order to shorten the notation.

Lemma 26 (Hilbert). Let A be commutative Banach algebra and x ∈ A. Then for all λ, µ ∈ Ω(x) Rλ − Rµ = (λ − µ) RµRλ. 1.2 Basic properties of spectra 12

Proof. For all λ, µ ∈ Ω(x),

Rλ [(x − µe) − (x − λe)] Rµ = Rλ [(λ − µ) e] Rµ.

Corollary 27. Let A be commutative Banach algebra and x ∈ A. Then the function λ 7→ Rλ is holomorphic on the whole Ω(x). Proof. For all λ, µ ∈ Ω(x) R − R λ µ = R R . λ − µ µ λ Since the inverse function (·) 7→ (·)−1 is continuous where it's dened, for all µ ∈ Ω(x) there exists the limit

lim Rλ = Rµ. λ→µ

Corollary 28. Let A be commutative Banach algebra and x ∈ A. Then. We have ∂ (λ 7→ R ) = λ 7→ R2 . ∂λ λ λ Theorem 29 (Liouville). Let A be a commutative Banach algebra and f : C → A an abstract entire function. If f is is bounded, the f is constant. Exercise 30. Prove the previous result. Theorem 31. Let A 6= {0} be commutative Banach algebra. Then For all x ∈ A, σ (x) 6= ∅. Proof. Assume the contrary. Then there is s.t. , i.e. so . x ∈ X σx = ∅ Ω(x) = C . −1 Hence λ 7→ Rλ = (x − λe) is an entire function. Fix R > 0. • For all λ ∈ C \ B (0,R), 1 x −1 R = − e . λ λ λ Note that if |λ| → +∞ x − e → e. λ So, if |λ| → +∞, x −1 − e → (−e)−1 = −e. λ This imples that the function x −1 is bounded. For λ 7→ λ − e |λ| → +∞ it follows

1 x −1 kRλk = − e , |λ| λ |{z} | {z } →0 bounded hence is bounded on . λ 7→ kRλk C \ B (0,R) 1.3 Ideals 13

• The function λ 7→ kRλk is continuous (composition of norm - continuous - with λ 7→ Rλ - holomorphic) on the compact set B (0,R), so it is bounded on B (0,R) also. Putting together these two points, it follows that is bounded on . λ 7→ kRλk C By Liouville's Theorem, it's constant. Since, for |λ| → +∞, Rλ → 0, necessarely . This is a contraddiction because for all is an inverse, λ 7→ Rλ ≡ 0 λ ∈ C Rλ and it can't happen that for some λ ∈ C e = (x − λe)(x − λe)−1 = 0. | {z } =0

Corollary 32 (Gelfand-Mazur Theorem). Let A 6= {0} be a Banach algebra. If each x ∈ A \{0} is invertible, then A is isometric to C. Proof. For all x ∈ A, by hypothesis and the denition of spectrum, σ (x) is a sigleton containing the only λ ∈ C satisfying x − λe = 0.

So for all x ∈ A\{0} there exist a unique λ ∈ C such that x = λe. The mapping

A → C, x 7→ λ is an isometry.

1.3 Ideals

Denition 33 (Ideal). Let A be a Banach algebra and I ⊂ A. I is called (non trivial) ideal if

1. I is a subspace4 of A; 2. for all x ∈ A and for all y ∈ I, xy ∈ I; 3. I 6= {0} and I 6= A. Remark 34. The last axiom is not necessary for the denition of ideal. We still prefer to add it to the denition of ideal to avoid trivial cases and make the development of the theory cleaner.

Example 35. In C [0, 1] is easy to check that the following set is an ideal

I := f ∈ C [0, 1] f| ≡ 0 . 0, 1 [ 2 ] 4Not necessarely closed. 1.3 Ideals 14

Lemma 36. Let A be a Banach algebra. If I ⊂ A is an ideal, then for all x ∈ I, x is not invertible. Proof. Trivial.

Lemma 37. Let A be a Banach algebra. For all non invertible x ∈ A \{0} ther is an ideal I ⊂ A such that x ∈ I. Proof. For all x ∈ A not invertible, consider I = {xy | y ∈ A}. Corollary 38. Let A be a Banach algebra. The following statements are equivalent:

1. all x ∈ A \{0} are invertible 2. no ideals exist in A. Proof. Follows directly from the previous two lemmas.

Denition 39 (Maximal). Let A be a Banach algebra. An ideal M ⊂ A is called maximal if it is maximal with respect to the inclusion.

Example 40. Let A = C [0, 1]. For all t0 ∈ [0, 1], dene the ideal

Mt0 := {f ∈ C [0, 1]| f (t0) = 0} . Claim. For all , is maximal. t0 ∈ [0, 1] Mt0

Proof. Fix t0 ∈ [0, 1]. Note that,

C [0, 1] = C + Mt0 , in fact, for all f ∈ C [0, 1],

f = f (t0) + f − f (t0) . | {z } ∈Mt0 This implies that the ideal is maximal.

Claim. For each maximal ideal M ⊂ C [0, 1] there exists a unique t0 ∈ [0, 1] such that . M = Mt0 Proof. Fix a maximal ideal M ⊂ C [0, 1]. Assume to the contrary that for all τ ∈ [0, 1] there is fτ ∈ M such that fτ (τ) 6= 0. This implies that for all τ ∈ [0, 1] there is fτ ∈ M, a neighborhood Uτ of τ and a positive constant δτ > 0 such that, for all t ∈ Uτ , |fτ (t)| > δτ > 0. Being an open cover of the compact set , there exists {Uτ }τ∈[0,1] [0, 1] τk1 , . . . , τkn ∈ [0, 1] such that n [ [0, 1] ⊂ Uτk . k=1 1.3 Ideals 15

Note that if f ∈ C [0, 1], also the complex conjucate f ∈ C [0, 1]. Being M an ideal, this means that if f ∈ M, also the real function ff = |f|2 ∈ M. Hence

n n X X 2 M ∩ R 3 fτk fτk = |fτk | ≥ min {|δτk |} > 0. k∈{1,...,n} k=1 k=1 | {z } =:f This is a contraddiction because, being stricly positive, f ∈ M is invertible. It follows that there exists t0 ∈ [0, 1] such that all the functions in M vanish in t0, i.e. . Being maximal, . Noting that , M ⊂ Mt0 M M = Mt0 t0, t1 ∈ [0, 1] t0 6= t1 imples , the uniqueness is trivial. Mt0 6= Mt1 Proposition 41. Let A be a commutative Banach algebra. If I ⊂ A is an ideal, A the closure I is also an ideal. A A Proof. First of all, let's prove that I is a linear subspace of A. For all x, y ∈ I there exist {xn} , {yn} ⊂ I such that, for n → +∞, n∈N n∈N

xn → x, yn → y. By the continuity of the sum,

x + y = lim (xn + yn), n→+∞ | {z } ∈I

A A hence x + y ∈ I . Now take x ∈ I and y ∈ X. There exists {xn} ⊂ I such n∈N that xn → x. Because of the continuity of the product, if n → +∞, xny → xy. Since is an ideal, for all , hence A. Finally, let's prove that I n ∈ N xny ∈ I xy ∈ I A I is non trivial. Call G := {x ∈ A| x is invertible}. Being an ideal, I ⊂ A \ G, which is a closed set5, so

A ∅ ( I ⊂ I ⊂ A \ G ( A.

Corollary 42. Let A be a commutative Banach algebra. If M ⊂ A is a maximal ideal, M is closed. Theorem 43. Let A be a commutative Banach algebra and I ⊂ A an ideal. Then there exists a maximal ideal M ⊂ A containing I. Proof. Assume that I is not maximal. Consider the family M of all ideals containing I, partially ordered by inclusion. Clearly M is not empty (I ∈ M). Note that if any linearly ordered6 subfamily has an upper bound, {Iλ}λ∈Λ ⊂ M we are done (Zorn's Lemma). Denote S . By denition of , I0 = α Iα M I0 ⊃ I and clearly is an ideal upper bound to . I0 {Iλ}λ∈Λ 5Theorem 11 states that G is open. 6 is linearly ordered if anf only if, for all , or . {Iλ}λ∈Λ ⊂ M λ1, λ2 ∈ Λ Iλ1 ⊂ Iλ2 Iλ2 ⊂ Iλ1 1.3 Ideals 16

Proposition 44. Let A be a commutative Banach algebra, G := {x ∈ A| x is invertible} and x ∈ A. The following statements are equivalent: 1. x ∈ G; 2. for all M ⊂ A maximal ideals, x∈ / M. Proof. Both implications are direct consequences of previously proven results.

Example 45. In Example 40 on page 14 we showed all maximal ideals in C [0, 1].

1.3.1 Multiplicative functionals

Denition 46 (Multiplicative functional). Let A be a Banach algebra. A linear functional7 f ∈ A0 \{0} is called multiplicative if, for all x, y ∈ A,

f (xy) = f (x) f (y) .

Example 47. Let A = C [0, 1] and t0 ∈ [0, 1]. The evaluation functional

Vt0 : C [0, 1] → C,

f 7→ Vt0 (f) := f (t0) is multiplicative. Indeed for all f, g ∈ A

Vt0 (fg) = (fg)(t0) = f (t0) g (t0) = Vt0 (f) Vt0 (g) .

Proposition 48. Let A be a Banach algebra and f ∈ A0 a multiplicative functional. Then

1. f (e) = 1; 2. for all x ∈ G, f (x) 6= 0 and 1 f x−1 = . f (x)

Proof.

1. Trivial: f (e) = f (e · e). 2. Trivial: for all x ∈ G, f (e) = f x · x−1.

Proposition 49. Let A be a Banach algebra and f ∈ A0 a multiplicative functional. If x ∈ A, kxk < 1, then |f (x)| < 1.

7Not necessarely bounded. 1.3 Ideals 17

Proof. Call λ := f (x). Assume to the contrary that |λ| ≥ 1 and consider x x − λe = λ − e . λ Since x < 1, λ the right hand side is invertible and so is the left hend side. By the previous Proposition f (x − λe) 6= 0 which leads to

f (x) 6= λ. Absurd. Remark 50. The previous result says that multiplicative functionals map the unit ball of the Banach algebra into the unit complex ball.

Corollary 51. Let A be a Banach algebra and f ∈ A0 a multiplicative functional. Then kfk = 1. Proof. Because of the previous proposition

kfk = sup |f (x)| ≤ 1. ◦ x∈BA Proposition 48 on the preceding page clinches the result. Remark 52. The previous result says that multiplicative functionals are always continuous!

Proposition 53. Let A be a Banach algebra and f ∈ A∗ a multiplicative functional. Then ker (f) is a maximal ideal. Proof. It is well known that kernels of continuous linear functionals are closed subspace of codim = 1. It's easy to check that ker (f) is an ideal. Having codimension 1, ker (f) is clearly maximal8.

1.3.2 Quotient algebra

Proposition 54. Let A be a Banach algebra and I ⊂ A a closed ideal. Then the set A/I := {x + I| x ∈ A} is a Banach algebra with respect to the operations

∀ (x + I) , (y + I) ∈ A/I, (x + I) + (y + I) := (x + y) + I, ∀ (x + I) ∈ A/I, ∀µ ∈ C, µ (x + I) := µx + I, ∀ (x + I)(y + I) ∈ A/I (x + I)(y + I) := xy + I

8It can't be enlarged without getting the whole A 1.3 Ideals 18 and the norm ∀ (x + I) ∈ A/I, kx + Ik = inf {kuk } . u∈x+I A Furthermore, calling e the unit of A, 1. A/I has unit e + I; 2. ke + Ik = 1; 3. for all (x + I) , (y + I) ∈ A/I,

k(x + I)(y + I)k ≤ kx + Ik ky + Ik .

Proof. Easy and boring.

Denition 55 (Quotient algebra). Let A be a Banach algebra and I ⊂ A a closed ideal. The Banach algebra A/I dened in the previous proposition is called quotient algebra. Remark 56. Proposition 54 on the previous page states that quotients of a closed ideals are Banach algebras. Problem 57. What can be said of the quotient of a Banach algebra with a maximal ideal?

Exercise 58. Let A be a Banach algebra and I ⊂ A a closed ideal. Prove that the quotient map q : A → A/I is multiplicative, i.e. for all x, y ∈ A,

q (xy) = q (x) q (y) .

Exercise 59. Let A be a Banach algebra , I ⊂ A a closed ideal and q : A → A/I. Prove that for all ideal J ⊂ A/I, q−1 (J) ⊂ A is an ideal. Lemma 60. Let A be a Banach algebra, I ⊂ A a closed ideal, J 0 ⊂ A/I a closed ideal on the quotient algebra and q : A → A/I the quotient map. Then J := q−1 (J 0) is a closed ideal containing I.

q I ⊂ A → A/I 0 I ⊂ J 7→ J q−1

Proof. Clearly J ⊃ q−1 (0) = I. By the continuity of q and the previous excer- cise J is a closed ideal. Proposition 61. Let A be a Banach algebra and M ⊂ A a maximal ideal. Then A/M has no ideals. Proof. Let q : A → A/M be the quotient map. Assume to the contrary that exists J ∈ A/M ideal. Without loss of generality, J is closed. By the previous lemma, q−1 (J) is an ideal containing M. Absurd. 1.3 Ideals 19

Corollary 62. Let A be a Banach algebra and M ⊂ A a maximal ideal. Then A/M is a eld. Proof. It follows directly from Proposition 44 on page 16.

Corollary 63. Let A be a Banach algebra and M ⊂ A a maximal ideal. Then A/M is isometric to C. Proof. It follows directly from Gelfand-Mazur Theorem (Corollary 32 on page 13).

Corollary 64. Let A be a Banach algebra and M ⊂ A a maximal ideal. Then q : A → A/M = C is a multiplicative functional with ker (q) = M. Denition 65 (∆). Let A be a Banach algebra. We dene the set

∆ := {α: A → C| α is multiplicative} . Remark 66. Corollary 64 states that any maximal ideal is the kernel of some multiplicative functional. On the other hand, Proposition 53 on page 17 stated that kernels of multiplicative functionals were maximal ideals. This leads to the following, unexpected, result.

Corollary 67. Let A be a Banach algebra and M the set of maximal ideals in A. There is a one-to-one correspondence

Φ: M ↔ ∆.

Example 68. In C [0, 1], Card (∆) = Card [0, 1]. Denition 69 (Gelfand topology). Let A be a Banach algebra. The topology induced on ∆ by the w∗ topology of A∗ is called Gelfand topology. Remark 70. From now on we will always consider ∆ as a topological spaces endowed with the Gelfand topology. As the follow result shows, this choice makes ∆ a compact (Hausdor) topological space. Theorem 71. Let A be a Banach algebra. Then ∆ is w∗-compact.

∗ Proof. Because of Corollary 51 on page 17, ∆ ⊂ BA∗ which is w -compact by Banach-Alaoglu theorem. Then we just need to prove that ∆ is w∗-closed. We do that using nets. Fix a net and a functional 0 such that {αλ}λ∈Λ ⊂ ∆ α ∈ A

w∗ αλ −→ α. Remember that the w∗-convergence of a net is just its pointwise convergence:

w∗ αλ −→ α ⇐⇒ ∀x ∈ A, αλ (x) → α (x) . We want to prove that α ∈ ∆. Clearly α 6= 0, indeed

1 = αλ (e) → α (e) 6= 0. 1.3 Ideals 20

By the continuity of the product, it's also easy to show that for all x, y ∈ A, α (xy) = α (x) α (y), ideed

αλ (xy) = αλ (x) αλ (y) ↓ ↓ α (xy) α (x) α (y) .

Denition 72 (Gelfand transform). Let A be a Banach algebra. The mapping

Λ: A → C (∆; C) , x 7→ Λx := x,b where

xb: ∆ → C, α 7→ xαb := α (x) is called Gelfand transform.

Proposition 73. Let A be a Banach algebra. Then the Gelfand transform Λ is a linear multiplicative mapping, i.e. for all x, y ∈ A

Λ(x + y) = xb + y,b Λ(xy) = xby.b Proof. Trivial.

Proposition 74. Let A be a Banach algebra. Then

|||Λ||| = 1. Proof. We want to prove that n o sup kxk = 1. b C(∆;C) x∈BA

This follows from noting that for all x ∈ BA

kxbkC(∆; ) = sup {|α (x)|} ≤ kxkA ≤ 1 C α∈∆ and kek = 1. b C(∆;C)

Proposition 75. Let A be a Banach algebra. The Gelfand topology on ∆ is the weakest topology that makes every continuous. xb ∈ Ab 1.3 Ideals 21

Proof. If for all x ∈ A,

∗ γx : A → A,

α 7→ γx (α) := α (x) , a base9 for the w∗ topology of A∗ is given by

( n ) \ ∗ {α ∈ A | γxk (α) > ε} . k=1 ε>0 This means that a base for the Gelfand topolofy of ∆ is simply

( n ) ( n ) \ ∗ \ {α ∈ A | γxk (α) > ε} ∩ ∆ = {α ∈ ∆| γxk (α) > ε} k=1 ε>0 k=1 ε>0 ( n ) . \ = {α ∈ ∆| α (xk) > ε} k=1 ε>0 ( n ) . \ = {α ∈ ∆| xck (α) > ε} . k=1 ε>0

Theorem 76. Let A be a Banach algebra and x ∈ A. Then10

σ (x) = {α (x)}α∈∆ . Proof. Let's start by proving that . Take . Then σ (x) ⊂ {α (x)}α∈∆ λ ∈ σ (x) x − λe is not invertible. So it lies in some maximal ideal, kernel of some multiplicative functional α. Since α (x − λe) = 0 it immediately follows that α (x) = λ. Viceversa, take an arbitrary α ∈ ∆ and consider λ := α (x) ∈ C. By denition λ satises

0 = α (x) − λ = α (x − λe) , hence x − λe ∈ ker (α), which is a maximal ideal. By Proposition 44 on page 16 it follows that λ ∈ σ (x).

Example 77. Consider A = C [0, 1]. For all t0 ∈ [0, 1] consider the Dirac δ distribution

δt0 : C [0, 1] → C, f 7→ f (t0) . 9For the open neighborhoods of the origin. 10Pure mathemagics! 1.3 Ideals 22

Clearly, for all , is a multiplicative functional. With the same t0 ∈ [0, 1] δt0 notation as Example 40 on page 14, for all t0 ∈ [0, 1],

Mt0 = ker (δt0 ) . Being maximal ideals in a one-to-one correspondence with multiplicative functionals, it follows ∆ = {δ } . t0 t0∈[0,1] By Theorem 76 on the previous page, for all f ∈ C [0, 1],

σ (f) = {δ (f)} = {f (t )} = f ([0, 1]) . t0 t0∈[0,1] 0 t0∈[0,1] This makes perfect sense! For all f ∈ C [0, 1], λ ∈ σ (f) if and only if f − λe is not invertible, which happens if and only if f − λe is null at some point, i.e. if and only if there exists t0 ∈ [0, 1] such that

f (t0) − λ e (t0) = 0 ⇐⇒ λ = f (t0) . | {z } e≡1 Example 78. L1 [0, 1] is a Banach algebra with respect to the convolution product

x ∀f, g ∈ L1 [0, 1] , x 7→ (f ∗ g)(x) = f (t) g (x − t) dt ˆ0 and the natural norm. Sadly this Banach algebra has no unit. It is possible k·k1 to plug in an idendity e the standard way. Dening a new Banach algebra L1 [0, 1] + e one can see that there is just one maximal ideal, namely L1 [0, 1]. Consequently L1 [0, 1] + e has just one multiplicative functional, which is kind of weird!

Denition 79 (Radical). Let A be a Banach algebra. The intersection of all maximal ideals of A is called radical of A and indicated with Rad (A). Proposition 80. Let A be a Banach algebra. Then 1. T ; Rad (A) = α∈∆ ker (α) 2. Rad (A) = ker (Λ). Proof. 1. Trivial. 2. Remembering the denition of Gelfand transform11, it's clear that

Rad (A) =1. {x ∈ A| ∀α ∈ ∆, α (x) = 0} = {x ∈ A| ∀α ∈ ∆, xαb = 0} = ker (Λ) . 11Denition 72 on page 20. 1.3 Ideals 23

Theorem 81. Let A be a Banach algebra and x ∈ A. Then q kxk = lim n kxnk b n→+∞ A Proof. By Theorem 76 on page 21 . kxk = max {|xα|} b α∈∆ b . = max {|α (x)|} α∈∆ = max {|λ|}; λ∈σ(x) | {z } =:a a is the so called spectral radius. We want to prove that q a = lim n kxnk . n→+∞ A Let's prove the two inequalities separately.

≥) If λ ∈ C, |λ| a < 1, then λx − e is invertible. Indeed that is tue if λ = 0; if λ 6= 0 1 λx − e = λ x − e (1.3.1) λ and 1/ |λ| > |a|, so 1/λ ∈ Ω(x). Note that for all λ ∈ C such that |λ| a < 1 and kλxk < 1, +∞ X k 1 (λx) = . (1.3.2) e − λx k=0 The biggest set on which the function

+∞ X k λ 7→ (λx) (1.3.3) k=0 is holomorphic is B (0,R), with 1 R := pn n limn→+∞ kx k (dened the obvious way if the limit is null or innite), as discussed in Notation 23 on page 10 and Proposition 24 on page 11. On the other −1 hand, by Corollary 27 on page 12, the mapping µ 7→ Rµ (x) := (x − µe) is holomorphic on its domain. Hence, by equation (1.3.1), the function

1 λ 7→ (1.3.4) e − λx 1.3 Ideals 24

is holomorphic on B (0, 1/a) (dened the obvious way if a is null or innite). We want to prove that

1 ≤ R, a i.e. that a ≥ lim pn kxnk. n→+∞ Assume to the contrary that 1/a > R. It is well known that any holomorphic function f dened on a disk B (x0, r) concides in the whole disk with its Taylor series T centered at the center of the disk. In other words, if f is holomorphic in B (x0, r), then its Taylor series T centered at x0 is dened in the whole B (x0, r) and for all x ∈ B (x0, r),

f (x) = T (x) .

In our case, the function (1.3.4) is holomorphic on B (0, 1/a), so its Taylor series, which by (1.3.2) is (1.3.3), should converge on the whole B (0, 1/a). Sadly we supposed that it's radius of convergence R were strictly lower than 1/a. Absurd.

Corollary 82. Let A be a Banach algebra. Then x ∈ Rad (A) if and only if, for n → +∞, q n n (kxkA) → 0. Remark 83. Clearly nilpotent elements of A belong to Rad (A). For this reason elements of Rad (A) are sometimes called generalized nilpotents. Note that, belonging to Rad (A), every nilpotent is annihilated by all α ∈ ∆. Denition 84 (Semisimple algebra). Let A be a Banach algebra. A is called semisimple if Rad (A) = {0}. Proposition 85. Let A be a semisimple Banach algebra. Then Λ is injective. Proof. Trivial.

Notation 86. Let A be a Banach algebra. The following notation will be used throughout the paper: Ab := Λ (A) . Corollary 87. Let A be a semisimple Banach algebra. Then Λ is a linear isomorphism onto Ab ⊂ C (∆; C). Proof. Trivial.

Proposition 88. Let A be a Banach algebra. Then Ab is an algebra with respect to the pointwise sum and product. 1.3 Ideals 25

Proof. Trivial.

Corollary 89. Let A be a semisimple Banach algebra. Then Ab is a Banach algebra with respect to the pointwise sum and product. Proof. Trivial.

Denition 90 (Total set). Let V be a linear space and x ∈ V . We say that D ⊂ V ∗ is total if [∀α ∈ D, α (x) = 0] ⇐⇒ x = 0. Proposition 91. Let A be a Banach algebra. Then A is semisimple if and only if ∆ is total. Proof. Suppose A semisimple, then ker (Λ) = 0, hence the only x ∈ A such that is . If and for all xb ≡ 0 x = 0 x ∈ A α ∈ ∆

0 = α (x) = xα,b then , hence ; if , for all xb ≡ 0 x = 0 x = 0 α ∈ ∆ α (0) = 0.

Viceversa, suppose that ∆ is total. Assume to the contrary that A is not semisimple, i.e. that there exists such that . By denition, it x ∈ A \{0} xb ≡ 0 follows that for all α ∈ ∆, 0 = xαb = α (x) . Being x 6= 0, this implies that ∆ is not total. Absurd. Corollary 92 (Of Theorem 81). Let A be a Banach algebra such that, for all x ∈ A, 2 2 x = kxk . (1.3.5) Then Λ: A → C (∆; C) is an isometry. Proof. Since the rst equation holds and the rst limit exists

kxk = lim pn kxnk b n→+∞ = lim 2pn kx2n k n→+∞ r n 2 = lim 2 x2n−1 n→+∞

qn 2 n−1 2 = lim x2 n→+∞ = ... qn n = lim 2 kxk2 n→+∞ = kxk . 1.3 Ideals 26

Corollary 93. Let A be a Banach algebra. If (1.3.5) holds, then Ab is complete. Proof. It follows directly from the previous result. Remark 94. The (1.3.5) is a strong request, which is not trivially satied by all Banach algebras. C [0, 1] satises it, but operators algebras, where the product is the composition, usually do not.

Denition 95 (Symmetric space). Let A be a Banach algebra. Ab is called symmetric if for all x ∈ A there is y ∈ A such that for all α ∈ ∆

α (x) = α (y)

Example 96. C\[0, 1] is symmetric. Indeed multiplicative functionals of C [0, 1] are Dirac deltas, so for all x ∈ C [0, 1], y = x satises the property.

Theorem 97 (Stone-Weierstrass). Let K be a compact T2 topological space, X ⊂ C (K; C) a closed subspace and assume that: 1. the constant function 1 belongs to X;

2. X separates the points of K, i.e. for all t1, t2 ∈ K, t1 6= t2, there exists f ∈ X such that f (t1) 6= f (t2); 3. for all f, g ∈ X, the pointwise product fg ∈ X; 4. for all g ∈ X, g ∈ X.

Then X = C (K; C).

Theorem 98. Let A be a Banach algebra. Assume that Ab is symmetric and for 2 all x ∈ A, x2 = kxk . Then Ab = C (∆; C), i.e. A and C (∆; C) are isometric via Λ.

Proof. Let's apply Stone-Weierstrass theorem for X = Ab ⊂ C (∆; C). Since (1.3.5) holds, Ab is complete, hence closed. By Proposition on page 24, Ab is a linear space. Let's check all other conditions:

1. the constant function 1 belongs to Ab, indeed . . Λe = eb = [α 7→ α (e) = 1] ≡ 1;

2. separates points of , indeed if and for all Ab ∆ α, β ∈ ∆ xb ∈ Ab

α (x) = xb (α) = xb (β) = β (x) , then α = β;

3. by Proposition on page 24, Ab is an algebra with respect to the pointwise product; 1.4 Involutions 27

4. x ; since is symmetric there is such that, for all , xb ∈ Ab Ab y ∈ X α ∈ ∆ . . xb (α) = α (x) = α (y) = yb(α) , hence . xb = yb ∈ Ab

Example 99. We stress again that the hypothesis aren't trivial. The Banach algebra of functions continue on the closed disk and holomorphic in the interior doesn't satisfy them. Same goes for the Banach algebra of functions which are sum of Fourier series.

1.4 Involutions

Denition 100 (Involution). Let A be a Banach algebra. A mapping

A → A, x 7→ x∗, such that

1. for all x, y ∈ A, (x + y)∗ = x∗ + y∗;

2. for all x ∈ A, for all λ ∈ C, (λx)∗ = λx∗;

3. for all x, y ∈ A, (xy)∗ = y∗x∗;

4. for all x ∈ A, (x∗)∗ = x is called involution. If such a mapping is dened on A, A is called an algebra with involution. Remark 101. From now on we will always assume that the Banach algebras we consider are algebras with involutions.

Example 102. In C [0, 1] we dene the involution as the conjugate map:

∀g ∈ C [0, 1] , g∗ = g.

Remark 103. Even if we work with commutative Banach algebras, we wrote the third property of an involution in such a way that makes sense even for non commutative ones. 1.4 Involutions 28

Exercise 104. Let H be a Hilbert space and consider the Banach algebra of bounded linear operator A = BL (H; H). This is an important example of a non commutative Banach algebra. Here one can get an involution via the dual operator

BL (H; H) → BL (H; H) , D 7→ D∗, where D∗ is the unique operator such that, for all x, y ∈ H

hD∗x, yi = hx, Dyi .

We will study some commutative subalgebras of B (H; H). Denition 105 (Self-adjoint element). Let A be a Banach algebra and x ∈ A. If x∗ = x we say that x is self-adjoint. Exercise 106. In C [0, 1] all real functions are self-adjoint. Proposition 107. Let A be a Banach algebra. Then 1. for all x ∈ A, (x + x∗), i (x − x∗) and xx∗ are self-adjoint; 2. for all x ∈ A, exist two unique self-adjoint u, v ∈ A such that x = u + iv; 3. e∗ = e; 4. for all x ∈ A, x is invertible if and only if x∗ is invertible and ∗ (x∗)−1 = x−1 ;

5. for all x ∈ A, σ (x) = σ (x∗), where σ (x∗) denotes the complex conjugate of σ (x∗). Proof. 1. Trivial.

2. Fix x ∈ A. Clearly 1 i u = (x + x∗) and v = − (x − x∗) 2 2 do the trick. For the uniqueness take another u0, v0 ∈ A self-adjoint and such that x = u0 + iv0. Then

u0 − iv0 = x∗ = u − iv. Hence 2w = x + x∗ = 2u and this easily brings us the uniqueness. 1.4 Involutions 29

3. Note that e = (e∗)∗ = (e∗e)∗ = e∗ e∗∗ = e∗. |{z} =e 4. For all x ∈ A, ∗ ∗ e = xx−1 = x∗ x−1 . 5. Fix any x ∈ A. Take any λ∈ / σ (x). Then x − λe is invertible. By the previous point also (x − λe)∗ is invertible and

−1 ∗ (x − λe)∗ = (x − λe)−1 . Since (x − λe)∗ = x∗ − λe, λ∈ / σ (x∗) or equivalently λ∈ / σ (x∗). Clearly the converse holds too, i.e. λ∈ / σ (x) if and only if λ∈ / σ (x∗).

Remark 108. Note that self-adjoints elements play the same role that real numbers play in the eld of complex number. Remark 109. Remember that a Banach algebra A is semisimple if and only if ∆ is total, i.e. for all x ∈ A \{0} there exists h ∈ ∆ such that h (x) 6= 0. Lemma 110. Let A be a simisimple Banach algebra and h ∈ ∆. Then, the functional

ϕ: A → C, x 7→ h (x∗) is multiplicative (i.e. ϕ ∈ ∆). Proof. For all x, y ∈ A,

ϕ (x + y) = h (x + y)∗ = h (x∗ + y∗) = h (x∗) + h (y∗) = ϕ (x) + ϕ (y) . Clearly replacing the sum with the multiplication in A, the same equalities hold. For all λ ∈ C, for all x ∈ A ϕ (λx) = h (λx)∗ = h λx∗ = λh (x∗) = λϕ (x) . Finally, ϕ 6= 0, indeed ϕ (e) ≡ 1. 1.4 Involutions 30

Corollary 111. Let A be a simisimple Banach algebra and h ∈ ∆. Then, the functional

ϕ: A → C, x 7→ h (x∗) is continuous.

Proposition 112. Let A be a semisimple Banach algebra. Then the involution n→+∞ is continuous, i.e. if {xn} ⊂ A, x ∈ A and xn −→ x, then n∈N

∗ n→+∞ ∗ xn −→ x . Proof. We use the Closed Graph Theorem12. Take the convolution as an anti- linear operator on A. The theorem states that, if

h n→+∞ ∗ n→+∞ ∗i xn −→ x, xn −→ y =⇒ y = x , then the involution operator is continuous. Let's check the premises. Fix any . If n→+∞ and ∗ n→+∞ , by the previous corollary h ∈ ∆ xn −→ x xn −→ y

∗ ∗ h (x ) = ϕ (x) = lim ϕ (xn) = lim h (xn) = h (y), n→+∞ n→+∞ hence h (x∗) = h (y) ⇐⇒ h (x∗ − y) = 0. Being A semisimple, ∆ is total. Being h arbitrary, it follows x∗ = y. Denition 113 (B∗-algebra). Lets A a be Banach algebra. A is called a B∗- algebra if A is with involution and for all x ∈ A

kx · x∗k = kxk2 .

In non commutative cases those algebras are also known as C∗-algebras. Exercise 114. C [0, 1] is a B∗-algebra with respect to the involution dened for all f ∈ C [0, 1] by the complex conjugate

f ∗ = f.

Lemma 115. Let A be a B∗-algebra, u ∈ A self-adjoint and h ∈ ∆. Then h (u) ∈ R. 12The Closed Graph Theorem is usually stated for linear operators. Check that the same theorem holds for anti-linear operators, i.e. operators that are additive and anti-homogeneous, like the involution. 1.4 Involutions 31

Proof. Fix α, β ∈ R such that h (u) = α + iβ. For all t ∈ R, take z := u + ite.

Then, for all t ∈ R z∗ = u − ite, zz∗ = u2 + t2e, |h (z)|2 ≤ kzk2 = kzz∗k 2 2 ≤ u + t but also

|h (z)|2 = |h (u) + it|2 = |α + i (β + t)|2 = α2 + β2 + 2βt + t2.

Since, for all t ∈ R 2 2 2 2 2 α + β + 2βt + t ≤ u + t m 2 2 2 α + β + 2βt ≤ u . This clearly implies that β = 0 (the right hand side is independent of t).

Corollary 116. Let be a ∗-algebra and self-adjoint. Then is A B u ∈ A ub ∈ Ab a real functional.

Theorem 117 (Gelfand-Naimark). Let A be a B∗-algebra. Then A is isometric to C (∆; C) via the Gelfand transform. Furthermore, for all x ∈ A Λ(x∗) = Λ(x) (1.4.1) i.e. for all h ∈ ∆, h (x∗) = h (x). (1.4.2)

Proof. Let's prove that Ab = C (∆; C) via the Stone-Weierstrass theorem: 1. the constant function belongs to ; 1 = eb Ab 2. Ab separates the points of ∆, i.e. for all α, β ∈ ∆, α 6= β, there exists such that xb ∈ Ab α (x) = xb (α) 6= xb (β) = β (x); assume to the contrary that there are α, β ∈ ∆, α 6= β such that, for all , xb ∈ Ab α (x) = xb (α) = xb (β) = β (x) , i.e. α = β, absurd; 1.5 Applications 32

3. for all , the pointwise product ; x,b yb ∈ Ab xbyb ∈ Ab for all α ∈ ∆, . . (xbyb)(α) = xb (α) yb(α) = α (x) α (y) = α (xy) = [Λ (xy)] α; 4. for all , ; xb ∈ Ab xb ∈ Ab x any x ∈ A and u, v ∈ A self-adjoint such that x = u + iv; then ∗ and ; by previous corollary x = u − iv xb = ub + ivb ∗ Λ(x ) = ub − ivb = xb which also proves formulae (1.4.1) and (1.4.2).

We now want to prove that Λ: A → Ab is an isometry. If we do that, we prove that Ab is closed in C (∆; C), by Stone-Weierstrass theorem we can conclude that Ab = C (∆; C) and consequently that A and C (∆; C) are isometric via Λ. • Take any y ∈ A self-adjoint. Then

2 ∗ 2 y = kyy k = kyk .

n Cleary, for all n ∈ N, also y2 is self-adjoint and

n n 2 2 y = kyk . By Theorem 81 on page 23,

kyk = lim pn kynk b Ab n→+∞ = lim 2pn ky2n k n→+∞

qn n = lim 2 kyk2 n→+∞ = kyk .

• Take any x ∈ A and call y := xx∗. Note that y is self-adjoint. Then

kxk2 = kxx∗k = kyk = kyk = kxΛ(x∗)k = xx = |x|2 = kxk2 . Ab Ab Ab b b bb Ab b Ab b

1.5 Applications 1.5.1 Wiener algebra

Example 118. The so called Wiener algebra W is the set of 2π-periodic complex functions wich are absolute convergent sum of their Fourier series ( ) X X ikt f : [0, 2π) → | ∃ {cn} ⊂ , |cn| < +∞, ∀t ∈ [0, 2π) , f (t) = cne C n∈Z C k∈Z k∈Z 1.5 Applications 33 endowed with the pointwise sum, pointwise product and the norm dened for P ik(·) all f ∈ W such that f = cne by k∈Z X kfk = |cn| . k∈Z The function f ≡ 1 ∈ W is the unit of W . It's easy to check that W is Banach algebra13.

Lemma 119. Let W be the Wiener algebra. Then

∆ = {α := [f 7→ f (t )]} . t0 0 t0∈[0,2π) Proof. Clearly ∆ ⊃ {α } , t0 t0∈[0,2π) indeed for all t0 ∈ [0, 2π) and for all f, g ∈ W . . . αt0 (fg) = (fg)(t0) = f (t0) g (t0) = αt0 (f) αt0 (g) . Let's prove the converse. Fix anyα ∈ ∆. Note that ei(·) ∈ W and call a := α ei(·) .

Remember14 that −1 1 α e−i(·) = α ei(·) = a and that multplicative functional has norm 1, hence . i(·) i(·) |a| = α e ≤ e = 1, −i(·) −i(·) 1/ |a| = α e ≤ e = 1, and consequently |a| = 1. This means that there exists t0 ∈ [0, 2π) such that a = eit0 . We have proved that α ei(·) = eit0 .

Being α multiplicative, for all n ∈ Z, α ein(·) = eint0 . P This means that for all {cn} ⊂ such that |cn| < +∞, n∈Z C n∈Z ! X in(·) X int0 α cne = cne . n∈Z n∈Z

13 1 For the completeness, note that W is isomorphic to ` (C). 14See Proposition 48 on page 16. 1.5 Applications 34

Theorem 120 (Wiener). Let W be the Wiener algebra. If f ∈ W and for all t ∈ [0, 2π), f 6= 0, then 1 ∈ W. f Proof. By previous exercise and Corollary 64 on page 19 it follows that for any maximal ideal M ⊂ W , f∈ / M. By Proposition 44 on page 16, 1/f ∈ W .

1.5.2 Disk algebra

Exercise 121. The so called disk algebra A is the set of complex continuous functions dened on the closed complex disk whose restriction to the interior of the disk is holomorphic

is holomorphic A := f ∈ C D; C f|D endowed with the pointwise sum, pointwise product and the norm dened for all f ∈ A by

kfk = max |f (z)| z∈D = max |f (z)| z∈∂D (last idendity holds because of the maximum modulus principle). The function f ≡ 1 ∈ A is the unit of A. It's easy to check that A is Banach algebra. Proposition 122. Let A be the disk algebra. Then

∆ = {α := [f 7→ f (z )]} . z0 0 z0∈D Proof. Clearly ∆ ⊃ {α } . z0 z0∈D

Viceversa, x any α ∈ ∆. We want to prove that there exists z0 ∈ D such that . Call α = αz0 id: D → D, z 7→ z and α (id) =: z0 ∈ C.

Note that z0 ∈ D, indeed |α (g)| ≤ kidk = 1. Claim. For all polynomial, . p ∈ A α (p) = αz0 (p) Proof. Fix any polynomial. Then there exist n such that p ∈ A {ak}k=0 ⊂ C n n−1 p = anid + an−1id + ... + a1id + a0 and α is linear and moltiplicative. 1.5 Applications 35

Claim. Polynomials are dense in A15. Proof. Fix f ∈ A and ε > 0. We want to prove that there exist a polynomial p such that kf − pk < ε. Being f continuous on a compact set, by Heine-Cantor theorem its uniformly continuous. Then there is δ ∈ (0, 1) such that, for all z1, z2 ∈ D with |z1 − z2| < δ, ε |f (z ) − f (z )| < . 1 2 2 Fix that δ. Note that the function

fe: (1 + δ) D → C, z z 7→ fe(z) := f 1 + δ is holomorphic on (1 + δ) D ⊃ D.

Hence feTaylor series centered at 0 uniformly converges to feon compact subsets included in (1 + δ) D (and including 0). So there exists a polynomial p such that z ε sup f − p (z) < . z∈D 1 + δ 2

Note that for all z ∈ D

z δ z − = |z| 1 + δ 1 + δ δ ≤ 1 + δ | {z } >1 < δ.

Consequently, for all z ∈ D, z z |f (z) − p (z)| = f (z) − f + f − p (z) 1 + η 1 + η z z ≤ f (z) − f + f − p (z) 1 + η 1 + η ≤ ε.

15Note that this doesn't follow from Stone-Weierstrass theorem (as in the real case) because the algebra of complex polynomials is not closed under complex conjugation. 1.5 Applications 36

Being both and continuous the result easily follows from density. Indeed, α αz0 if {pn} is a sequence of polynomials such that, for n → +∞, pn → f, then n∈N α (f) = α lim pn n→+∞

= lim α (pn) n→+∞

= lim αz (pn) n→+∞ 0 = αz lim pn 0 n→+∞

= αz0 (f) .

Theorem 123. Let A be the disk algebra and f1, . . . , fm ∈ A such tath for all z ∈ D there is j ∈ {1, . . . , m} such that fj (z) 6= 0. Then there is h1, . . . , hm ∈ A such that m X hkfk ≡ 1. k=1 Proof. Consider the set n o X I = g = gkfk g1, . . . gm ∈ A .

Clearly I is a subspace of A, is dierent from {0} and is closed under the product with elements in A. We want to prove that I = A. If this is true, then 1 ∈ I and the theorem is proved. Assume to the contrary that I ( A. Then I is an ideal (by denition). By Theorem 43 on page 15 there is M maximal ideal such that M ⊃ I. By previous proposition and Corollary 64 on page 19 there exists z0 ∈ D such that M = {f ∈ A| f (z0) = 0} .

Hence all functions in I vanishes in z0. For all j ∈ {1, . . . , m} take gj = 1 and g1, . . . , gj−1, gj+1, . . . , gm = 0. It follows that for all j ∈ {1, . . . , m}, fj (z0) = 0. Absurd.

1.5.3 Complex compact operators in Hilbert spaces

Denition 124 (Compact operator). Let H be a Hilbert space. An operator T : H → H is called compact if T (BH ) is relatively compact. Denition 125 (Self-adjoint operator). Let H be a Hilbert space and T : H → H a linear operator. T is called self-adjoint if for all x, y ∈ H

hT x, yi = hx, T yi . 1.5 Applications 37

Exercise 126. Let H be a Hilbert space and T : H → H a compact self-adjoint operator. Consider the set of polynomials in T

n n−1 PT := anT + an−1T + ... + a0 n ∈ N0, a0, . . . , an ∈ C .

The setPT , endowed with the obvious sum and scalar multiplication is a normed algebra (check it) with respect to the norm dened for all P ∈ PT by

kP k = sup kP xk . kxk≤1

PT is an incomplete normed algebra (check it). Denote with A the completion of PT . By denition polynomials in T are dense in A. Clearly A has a unit. A also has an involution16. Furthermore, A is a B∗-algebra. Let's check it for p = T . Note that (check it), for all B : H → H self-adjoint,

kBk = sup |hBx, xi| . kxk≤1

Nothe that (check it) if T is self-adjoint, also T 2 is self-adjoint. Hence

2 2 T = sup T x, x kxk≤1 [self − adjoint] = sup |hT x, T xi| kxk≤1 = sup kT xk2 kxk≤1 = kT k2 .

The same equalities holds for all X ∈ A (check it). Remark 127. By Gelfand-Naimark theorem17, since A is a B∗-algebra, A is isomorphic to C (∆; C) via the Gelfand transform. We want to say explicitly who is C (∆, C). In order to to that, we need the spectral theorem for compact self-adjoint operators.

Denition 128 (Eigenvalues, eigenvectors, spectrum and eigenspaces). Let H be a Hilbert space and T : H → H a linear operator. We say that λ ∈ C is an eigenvalue for T if there exists x ∈ H \{0}, called eigenvector, such that T x = λx. The set of all eigenvalues is called spectrum. Let λ be an eigenvalue for T , then the linear space Eλ := {x ∈ H| T x = λx} is called eigenspace (relative to λ). Proposition 129. Let H be a Hilbert space and T a compact self-adjoint operator. Then

16Which doesn't do much since T ∗ = T . 17Theorem 117 on page 31. 1.5 Applications 38

1. T has has most contably many eigenvalues; 2. all eigenvalues are reals; 3. all eigenspaces are nite-dimensional.

Theorem 130 (Spectral theorem for compact self-adjoint operators). Let H be a Hilbert space and T a compact self-adjoint operator. Assume that the spectrum 18 19 is countable and ker (T ) = {0} . If σ := {λi} ⊂ is the spectrum, for i∈N C all λ ∈ σ we denote with Eλ the eigenspace relative to λ and with Pλ : H → Eλ the orthogonal projector on the eigenspace. Assuming without loss of generality that |λ1| > |λ2| > . . ., it follows

+∞ X T = λiPλi . i=1 Remark 131. To describe the set ∆ of multiplicative functionals of A we should think about the previous result in a weak sense. Fix any x ∈ H. Then

+∞ X T x = λiPλi x i=1 and this series is convergent in the norm of the Hilbert space20. By Theorem 76 on page 21 we have

σ (x) = {α (x)}α∈∆ . Since, by previous theorem, the spectrum is countable, also ∆ is countable, say

∆ =: {α0, α1, α2,...} .

Since PT is dense in A, to describe each element of ∆ it's enough to say how it behaves on polynomals in T . Without loss of generality, for all P ∈ PT and for all i ∈ N αi (P ) = P (λi) n and, if P := anT + ... + a0I,

n α0 (P ) = α0 (anT + ... + a0I) = a0. Checking that everything makes sense it is possible to conclude that

∆ = c.

18The nite version should be well-kwown from previous courses. 19This hypothesis is unnecessary and only assumed for the sake of simplicity. It should be clear how the results would change if this hypothesis was dropped. 20If you think about nite-dimentional case, the well-known spectral theorem says that if a matrix is symmetric (which is the equivalent of being self-adjoint) there is a basis that makes it diagonal. This is a generalization of the same concept. Indeed the previous identity denes a diagonal matrix. Chapter 2

Seminars

2.1 Gelfand and Fourier transform (Tommaso Ce- sari)

Proposition 132. 1 n is a Banach algebra without unit with respect L (R ) , k·k1 to the convolution. Proof. It should be well known from previous courses. Remark 133. Proposition 4 on page 4 states that any Banach algebra can be enlarged enough to admit a unit element. Note that in the particular case of L1 (Rn), the complex component added to L1 (Rn) in order to get the unit can be thought as a coecient of a function , dened for all n as the Dirac δ x ∈ R δx measure on Rn. Indeed the set 1 n L := f + αδ f ∈ L (R ) , α ∈ C is a Banach algebra with unit δ with respect to the operations

∀f + αδ, g + βδ ∈ L , (f + αδ) + (g + βδ) := (f + g) + (α + β) δ, ∀f + αδ ∈ L , ∀β ∈ C, β (f + αδ) := βf + (αβ) δ, ∀f + αδ, g + βδ ∈ L , (f + αδ) ∗ (g + βδ) := (f ∗ g + βf + αg) + (αβ) δ, the norm

∀f + αδ ∈ L , kf + αδk = kfk1 + |α| and L1 (Rn) is isometrically embedded in L via f 7→ f + 0 · δ. Note that, for all x ∈ Rn,

(f ∗ δ)(x) = f (t) dδx (t), ˆ n | {z } R =f(x) | {z } =f(x) so the convolution f ∗ δ actually coincide with the integration of f with respect to the Dirac measure. 2.1 Gelfand and Fourier transform (Tommaso Cesari) 40

Notation 134. We x the notation L to indicate the smallest Banach algebra 1 n with unit containing L (R ); ∗ , as described in the previous remark.

Notation 135 (Fourier transform). For all f ∈ L1 (Rn), fband F (f) will denote the Fourier transform of f. Proposition 136. The function

h: Rn → ∆, (h i f + αδ 7→ fb(t) + α , if t ∈ Rn, t 7→ ht := [f + αδ 7→ α] , if t=∞ is a homeomorphism between Rn = Rn ∪{∞} with the topology of the one-point- compactication of Rn and the set ∆ of multiplicative functional on L with the usual Gelfand topology. I.e., as topological spaces

∆ = Rn.

Proof. For all f ∈ L1 (Rn), dene

fb(∞) := lim fb(t) = 0. t→∞

Let's start by proving that h is well dened. For all (f + αδ) , (g + βδ) ∈ L and for all t ∈ Rn

ht ((f + αδ) ∗ (g + βδ)) = ht ((f ∗ g + βf + αg) + (αβ) δ) = F (f ∗ g + βf + αg) + αβ = F (f ∗ g) + F (βf) + F (αg) + αβ

= fb(t) gb(t) + βfb(t) + αgb(t) + αβ = fb(t) + α (gb(t) + β)

= ht (f + αδ) ht (f + βδ) .

Now we prove that is injective. If n, , assume (by contradic- h t1, t2 ∈ R t1 6= t2 tion) h i f + αδ 7→ fb(t1) + α = [g + βδ 7→ gb(t2) + β] , this means in particular that for all f ∈ L1 (Rn)

fb(t1) = fb(t2) .

2 Of course this cannot happen for all f ∈ L1 (Rn) (take for instance x 7→ e−πx 2 and y 7→ e−π(y−1) ). The proof that h is surjective can be easily derived from the result in Section 9.22 of W. Rudin - Real and Complex Analysis. Now let's prove that h is an homeomorphism. Call τ the topology of the one-point compactication of Rn and γ the usual Gelfand topology on ∆. To prove that 2.2 Lomonosov's Invariant Subspace Theorem (Tommaso Russo) 41

Rn, τ is topologically equivalent to (∆, γ) we just need to note that h (τ) is a compact Hausdor topology and γ ⊂ h (τ). Since h is one-to-one, h (τ) is a compact Hausdor topology. To prove the inclusion remember that the Gelfand topology of ∆ is just the weak topology induced by Lc, that is, the weakest topology that makes every Λ(f + βδ) ∈ Lccontinuous. Since, for all Λ(f + βδ) ∈ Lc Λ(f + βδ) = [α 7→ α (f + βδ)] h i = ht 7→ ht (f + βδ) = fb(t) + β , then h (τ) makes every Λ(f + βδ) ∈ Lccontinuous, hence, unsurprisingly, the weak topology γ is weaker then h (τ). Since γ ⊂ h (τ) and both topologies are compact and T2, it follows γ = h (τ).

Remark 137. Proposition on the preceding page proves that in L , ∆ = Rn. Hence for all f + αδ ∈ L and for all t ∈ R,

Λ(f + αδ)(t) = Λ (f + αδ)(ht) = fb(t) + α. This shows how the Fourier transform of an L1 function coincides with its Gelfand transform1, making the Gelfand transform a proper generalization of the Fourier transform.

2.2 Lomonosov's Invariant Subspace Theorem (Tom- maso Russo)

In what follows X is a Banach Space and we denote

B(X) = {T : X → X linear and bounded} . Note that X is not assumed to be a Banach Algebra, indeed the theory seen in the course will not be applied to X, but to B(X), which has a clear structure of Banach Algebra with unit. What is missing in this Algebra is commutativity, however some of the results that we discussed still hold. In particular we mention that the following result holds also with no assumptions of commutativity and will have a very main role in the sequel. Theorem 138 (Spectral Radius Formula). Let A be a Banach Algebra with unit and let x ∈ A. Then the spectral radius of x, dened by

ρ(x) := max |λ| λ∈σ(x) satises the following Spectral Radius Formula

ρ(x) = lim kxnk1/n . n→+∞

1If α = 0, the equation above says that Λf = Ff. 2.2 Lomonosov's Invariant Subspace Theorem (Tommaso Russo) 42

Denition 139. Let T ∈ B(X). M ⊆ X is said to be an invariant subspace for T , if M is a closed subspace of X and T (M) ⊆ M. It's said to be nontrivial if M 6= {0} ,M 6= X. The main topic of this section will be to discuss the existence of nontrivial invariant subspaces for elements of B(X). These are quite interesting in Operator Theory according to the following trivial remark: if M is an invariant subspace for T , then T|M ∈ B(M) and M is a Banach Space too, so one can study the operator T on a smaller space and then try to put together the behaviour of the operator on the various invariant subspaces in order to deduce the behaviour on the whole space. This is a very familiar machinery since is exactly what one does in nite dimensional spaces when tries to diagonalize a matrix or to nd its Jordan Canonical Form. However in innite dimensional spaces, it may happen that an operator has no eigenvalues (even if X is a Complex Banach Space) and so one can not use eigenspaces, while generic invariant subspaces could be present. As a simple (but userful for what follows) example consider:

Example 140. Let X = `2 and SR ∈ B(`2) the right shift, dened by: ( k 0 k = 1 (SRx) = ∀x ∈ `2 xk−1 k ≥ 2 if is the sequence k +∞ . Explicitly the action is the following: x ∈ `2 x = (x )k=1 x1, x2, x3, ··· 7→ 0, x1, x2, x3, ··· .

1 k+1 k If one tries to solve SRx = λx, it comes out λx = 0, λx = x , ∀k ≥ 1. Now there are two cases: λ = 0 immediately implies 0 = xk, ∀k ≥ 1, while λ 6= 0 gives x1 = 0 and then inductively xk = 0, ∀k. Hence there are no eigenvalues. For the sake of completeness, let us mention that σ(SR) = D(0, 1). Some evidence of this is given by the fact that N has norm and so the SR 1, ∀N ≥ 1 Spectral Radius Formula gives σ(SR) ⊆ D(0, 1). Then it's natural to ask the following question: x X Banach Space and T ∈ B(X). Is it true that T admits a non trivial invariant subspace? Under which conditions on T and or on X there exists an invariant subspace? Let's begin by some cases in which the answer is easy and we can answer with no condition on T , in particular the answer is the same for all elements of B(X). So in these cases we actually solve a more dicult problem2: given X Banach Space, is it true that every T ∈ B(X) admits an invariant subspace?

• In R2 the answer is no. Indeeed it suces to consider the operator given by the following matrix (wrt the canonical basis of course)

0 −1 . 1 0

2This is sometimes referred to as the Invariant Subspace Problem. 2.2 Lomonosov's Invariant Subspace Theorem (Tommaso Russo) 43

Every nontrivial invariant subspace would have dimension 1 and so would be an eigenspace, but there are no real eigenvalues and so no nontrivial invariant subspaces.

• Instead in Rn n ≥ 3 the answer is yes; this is proved by standard (and quite boring) linear algebra arguments.

• In Cn the answer is trivial, since by the Fundamental Theorem of Algebra the characteristic polynomial has roots, so there are eigenspaces.

• The rst innite dimensional case that can be treated easily is when X is 3 non separable and the answer is yes also here . To see this, take x0 ∈ X dierent from zero and consider the closed linear span of n +∞ which {T x0}n=0 is closed, invariant and separable, hence a proper subspace. Now we have some example in which we ask something also on the operator:

• If we start with a Hilbert Space H, then it follows form the Spectral The- orem that every normal operator4 admits nontrivial invariant subspaces.

• In 1954 Aronszajn and Smith proved that every compact operator on a complex Banach Space admits a non trivial invariant subspace.

• In 1966 Bernstein and Robinson gave a generalization of the above result: if there exists a non zero polynomial p such that p(T ) is compact, then T has a non trivial invariant subspace. Their proof used nonstandard analysis, a proof of the same result using only classical concepts was given by Halmos in the same year. Now we turn to the key result, from which the last two statements follow immediately . The original proof, given by Lomonosov in 1973, made use of the Schauder Fixed Point Theorem in order to show the existence of an eigenvalue; the proof to be presented here instead deduces the same from the Spectral Ra- dius Formula, this argument was proposed by Hilden in 1977.

Theorem 141 (Lomonosov's Invariant Subspace Theorem). Let X be a innite- dimensional complex Banach Space and let T ∈ B(X) be a nonzero compact operator. Then there exists a closed proper subspace M ⊆ X which is invariant for all S ∈ B(X) that commute with T . In formula:

S ∈ B(X),ST = TS ⇒ S(M) ⊆ M

Before proving the statement, let us make some comment:

• In particular we have that every operator which commutes with a nonzero compact one admits a proper invariant subspace.

3Do not get surprised that non separability makes things easier: if the space is bigger, there are more subspaces, so is easier some to be invariant. 4 i.e. T ∈ B(H) such that TT ∗ = T ∗T , where T ∗ is the adjoint of T . 2.2 Lomonosov's Invariant Subspace Theorem (Tommaso Russo) 44

• Of course we can choose S to be equal to T and so we deduce that every compact operator admits a nontrivial invariant subspace, so we have Aronszajn and Smith result.

• Also the other statement we mentioned is easy to deduce: indeed we have two cases. If p(T ) 6= 0, then it is a nontrivial compact operator which of course commutes with T and so by Lomonosov Theorem T admits a proper invariant subspace. If , let PN n with p(T ) = 0 p(z) = n=0 αnz αN 6= N N−1 so that we have P n, hence N P αn n. 0,N ≥ 1 0 = n=0 αnT T = − n=0 α T n N By this it's clear that the linear span of {T (x0): n = 0, ··· ,N − 1} for some 0 6= x0 ∈ X is invariant under T and is closed and proper since is nite dimensional.

Proof. Denote by Γ := {S ∈ B(X): ST = TS}, which is easily seen to be a subalgebra of B(X) and for y ∈ X denote Γy := {Sy : S ∈ Γ}. These are subspaces of X (since Γ is a subspace of B(X)) and are invariant under all S ∈ Γ (since Γ is closed under composition). They may miss being closed, but S (Γy) ⊆ Γy 5 implies S Γy ⊆ Γy, so that ∀y ∈ X, Γy is closed, invariant under all S ∈ Γ and if . If for at least one , the proof is concluded, Γy 6= {0} y 6= 0 y 6= 0 Γy $ X otherwise we have that ∀y ∈ X, y 6= 0, Γy is dense in X. Let's assume this.

Now, we use that T is nonzero: there exists x0 6= 0 in X such that T (x0) 6= 0, so 1 1 and by continuity of there exists kT x0k > 2 kT x0k > 0, kx0k > 2 kx0k > 0 T open ball centered at such that , we have 1 B x0 ∀x ∈ B kT xk ≥ 2 kT x0k , kxk ≥ 1 . In particular which is compact, since is a compact 2 kx0k 0 ∈/ K := T (B) T operator. By our previous assumption ∀y ∈ K, Γy∩B 6= ∅, so there exists Sy ∈ Γ such that Syy ∈ B and by continuity of Sy there exists an open neighborhood of s.t. . Of course is an open cover of , so it Wy y Sy (Wy) ⊆ B {Wy}y∈K K admits a nite subcover: there exist open sets W1, ··· ,Wn whose union covers K and S1, ··· ,Sn ∈ Γ with Si(Wi) ⊆ B, ∀i = 1, ··· , n. Start with , so that , hence in some and so x0 ∈ B T x0 ∈ K Wi1 x1 := ; then , so in some and . Going Si1 T x0 ∈ B T x1 ∈ K Wi2 x2 := Si2 T x1 ∈ B on in this way, we nd a sequence , so that xN = SiN TSiN−1 T ··· Si1 T x0 ∈ B 1 . But of course by commutativity N ∀N kxN k ≥ 2 kx0k xN = SiN SiN−1 ··· Si1 T x0 and this implies

1 kx k ≤ kx k 2 0 N N ≤ kSiN k SiN−1 · · · kSi1 k T kx0k N N ≤ µ T kx0k if µ := max {kS1k , ··· , kSnk}. Thus we have

1 1 N N 1 ≤ µ T , ∀N ≥ 1 2 N

5S is continuous, so S A ⊆ S(A) 2.2 Lomonosov's Invariant Subspace Theorem (Tommaso Russo) 45 and we can pass to limit and use the Spectral Radius Formula to deduce

1 ≤ µρ(T ) which gives 1 ρ(T ) ≥ > 0. µ

The proof is now almost concluded, it suces to recall some spectral properties for compact operators. What we need is the following: σ(T ) \{0} = σp(T ) \{0}, i.e. every element of the spectrum dierent from 0 is an eigenvalue; moreover it has nite multiplicity. So here we deduce that T admits an eigenvalue λ 6= 0 and the corresponding eigenspace Mλ := {x ∈ X : T x = λx} is nite dimensional hence is closed and proper. Finally Mλ is invariant under all S ∈ Γ, since if S ∈ Γ and x ∈ Mλ we have T Sx = ST x = Sλx = λSx, so Sx ∈ Mλ. This concludes the proof. As we have previously seen, this is quite a striking result. However it doesn't settle the invariant subspace problem: indeed there exist operators which commute with no nontrivial compact operator, for example the right shift already introduced.

Example 142. Let as before SR ∈ B(`2) and let T ∈ B(`2) be compact, such that SRT = TSR. Then T = 0. Indeed let and compute, for , M N N > M ≥ 1 x ∈ X TSR x − TSR x = M N . Set and note6 SR T x − SR T x y := T x

M N 1 2 3 1 2 3 SR y − SR y = (0, 0, ··· , 0, y , y , y , ··· ) − (0, 0, ··· , 0, y , y , y , ··· ) | {z } | {z } M terms N terms = (0, 0, ··· , 0, y1, y2, y3, ··· , yN−M , yN−M+1 − y1, yN−M+2 − y2, ··· ) | {z } | {z } M terms N−M terms, Hence,

N−M +∞ M N 2 X j 2 X N−M+j j 2 SR T x − SR T x = y + y − y j=1 j=1 N−M X j 2 ≥ y j=1 and now we can let N − M → +∞ and deduce

+∞ M N 2 X j 2 2 lim inf TSR x − TSR x ≥ y = kT xk . N−M→+∞ j=1

6probably the notation here is not the most beautiful, but is hopefully clear. 2.2 Lomonosov's Invariant Subspace Theorem (Tommaso Russo) 46

But {Sn x} is a bounded sequence and so {TSn x} is precompact and R n∈N R n∈N hence nk as . Passing to another subsequence if necessary we TSR x → xe k → +∞ can assume nk+1 − nk → +∞ as k → +∞, so that N = nk+1,M = nk in the previous inequality gives

nk nk+1 2 2 0 = kx − xk = lim inf TSR x − TSR x ≥ kT xk . e e k→+∞

Thus we have T x = 0, ∀x ∈ X so T = 0, which is what we wanted to prove. Note that, altough we can not use Lomonosov Theorem to deduce the existence of invariant subspaces, it's clear that SR admits many invariant subspaces, for example i +∞ 1 k . This is an interesting Mk := x = (x )i=1 ∈ `2 : x = ··· x = 0 case also because all invariant subpsaces of SR are known, this is a theorem by Beurling. In fact operators with no nontrivial invariant subspaces have been found: for example Eno in 1987 built a non reexive space and an operator on it without nontrivial invariant subspaces and Read in 1989 produced examples in the Classical Spaces c0 and `1. So in all cases the space X was non reexive, under the assumptions of reexivity, no counterexample has been found and the question is still open, in particular even in Hilbert Spaces. Index

Algebra, Self-adjoint element, 28 B∗-, 30 Set, C∗-, 30 Resolvent, 8 Banach, 3 Total, 25 disk, 6, 34 Spectral radius, 23 quotient, 18 formula, 41 semisimple, 24 Spectrum, 8, 37 Wiener, 32 Symmetric space, 26

Eigenspace, 37 Theorem, Eigenvalue, 37 Dunford, 10 Eigenvector, 37 Gelfand-Mazur, 13 Gelfand-Naimark, 31 Function, Lomonosov's invariant subspace, 43 Abstract entire, 10 spectral, for compact self-adjoint Abstract holomorphic, 10 operators, 38 Functional, Stone-Weierstrass, 26 multiplicative, 16 Wiener, 34 Transform, Gelfand Fourier, 40 topology, 19 Gelfand, 20 transform, 20

Ideal, 13 maximal, 14 Invariant subspace, 42 Involution, 27

Lemma, Hilbert, 11

Nilpotents, generalized, 24

Operator, compact, 36 self-adjoint, 36

Radical, 22