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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 ofthe Professor's will. No guarantee is given regarding the completeness or correctness of thispaper. In particular, it is highly probable that many errors (likely conceptual!) will occurbecause of the unskillfulness of the curator. Use the information contained herein at yourown risk.

Tommaso R. Cesari

Contents

1 Commutative Banach Algebras 31.1 Basic denitions . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Basic properties of spectra . . . . . . . . . . . . . . . . . . . . . . 8

1.2.1 Abstract holomorphic functions . . . . . . . . . . . . . . . 101.3 Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.3.1 Multiplicative functionals . . . . . . . . . . . . . . . . . . 161.3.2 Quotient algebra . . . . . . . . . . . . . . . . . . . . . . . 17

1.4 Involutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271.5 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

1.5.1 Wiener algebra . . . . . . . . . . . . . . . . . . . . . . . . 321.5.2 Disk algebra . . . . . . . . . . . . . . . . . . . . . . . . . 341.5.3 Complex compact operators in Hilbert spaces . . . . . . . 36

2 Seminars 392.1 Gelfand and Fourier transform (Tommaso Cesari) . . . . . . . . . 392.2 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 (com-mutative) 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 xnn∈N ⊂ X, x, y ∈ X, there exists the limit

limn→+∞

xny = xy.

Problem 2. The last property states that in Banach algebras the product iscontinuous with respect to both components. Does this implies that the productis 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 anyBanach algebra.

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

∀ (x, λ) , (y, µ) ∈ X, (x, λ) + (y, µ) := (x+ y, λ+ µ) ,

∀ (x, λ) ∈ X, ∀µ ∈ C, µ (x, λ) := (µx, µλ) ,

∀ (x, λ) , (y, µ) ∈ X, (x, λ) (y, µ) := (xy + µx+ λy, λµ)

and the norm∀ (x, λ) ∈ X, ‖(x, λ)‖X = ‖x‖X + |λ| ;

X has unit (0, 1) and admits a subalgebra isometric to X. The isometric em-

bedding of X in X is given by

∀x ∈ X, x 7→ (x, 0) .

Proof. Trivial.

Remark 5 (Convention). Because of the previous proposition from now on wewill always assume that Banach algebras have units. We will always use theletter e to denote the unit element of a Banach algebra.

Proposition 6. Let X be Banach space and A,B : X → X two bounded linearoperators. Then, indicating with |||·||| the operator norm on BL (X;X),

|||AB||| ≤ |||A||| · |||B|||

Proof. For all x ∈ X, ‖x‖ = 1,

‖(AB) (x)‖ = ‖A (B (x))‖≤ |||A||| ‖B (x)‖≤ |||A||| · |||B||| · ‖x‖︸︷︷︸

=1

.

Theorem 7. Let (X, ‖·‖X) be a Banach algebra. Then there exists a norm ‖·‖equivalent to ‖·‖X such that, for all x, y ∈ X,

‖xy‖ ≤ ‖x‖ ‖y‖ (1.1.1)

and ‖e‖ = 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 componentwisecontinuity 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 thatBL (X;X) is a Banach space. We want to prove that V is a closed subspaceof BL (X;X). Take any V ∈ V ′. Than there exists xnn∈N ⊂ X such that

Vxn

n→+∞−→ V . This means that, for n→ +∞

supy∈X|xny − V (y)| → 0,

i.e. for all y ∈ X, if n→ +∞

xny → V (y) ∈ X.

This implies (by the componentwise continuity of the product and the complete-ness 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 checkthat V is a Banach algebra with respect to the pointwise product2. Considerthe mapping

T : V → X,

Vx 7→ x.

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

‖Vx‖V = sup‖y‖X=1

‖xy‖X ≥∥∥∥∥x e

‖e‖x

∥∥∥∥X

=1

‖e‖X‖x‖X .

This shows that for all x ∈ X,

‖T (Vx)‖X ≤ ‖e‖X ‖Vx‖V ,

hence T is continuous. By the open mapping theorem T is an homeomorphism,so∥∥T−1 (·)

∥∥V and ‖·‖X are equivalent norms. Clearly the unit of V is Ve and

‖Ve‖V = sup‖x‖X=1

‖ex‖X = 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, thatVxn → Vx implies xn → x.

1.1 Basic denitions 6

Remark 8 (Convention). Because of the previous result from now on we willalways assume that Banach algebras satisfy inequality (1.1.1) and that the unithas norm 1.

Example 9. Let's now see some examples of Banach agebras.

1. The Banach space (C [0, 1] , ‖·‖∞) is a Banach algebra with unit e ≡ 1 withrespect to the product (fg) (·) = f (·) g (·). The relations ‖e‖ = 1 and

∀f, g ∈ C [0, 1] , ‖fg‖∞ ≤ ‖f‖∞ ‖g‖∞also (clearly) hold.

2. Let D ⊂ C be the complex unit disk. The normed space

D :=

(f ∈ C

(D;C

)∣∣ f|D is holomorphic, supD

·

)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 Ck ([0, 1] ;C) ,

k∑n=0

∥∥∥(·)(n)∥∥∥∞

n!

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. Letxnn∈N , ynn∈N ⊂ A and x, y ∈ A. If xn → x and yn → y, then xnyn → xy.

Proof. For all n ∈ N

‖xnyn − xy‖ = ‖xnyn − xny + xny − xy‖≤ ‖xnyn − xny‖+ ‖xny − xy‖= ‖xn (yn − y)‖+ ‖(xn − x) y‖≤ ‖xn‖ ‖yn − y‖+ ‖xn − x‖ ‖y‖ .

Theorem 11. Let A be a Banach algebra and G :=x ∈ A

∣∣ ∃x−1 ∈ 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, ‖x‖ < 1, there exists (e− x)−1 ∈

G. I.e. there is a neighborhood of e consisting of invertible elements. Let'sprove that the series

y :=

+∞∑k=0

xk

converges. By induction it's easy to prove that, for all k ∈ N∥∥xk∥∥ ≤ ‖x‖k .Then, for all ε > 0 there exists n0 ∈ N such that, for all n,m ∈ N,n0 ≤ n < m, ∥∥∥∥∥

m∑k=n+1

xk

∥∥∥∥∥ ≤m∑

k=n+1

∥∥xk∥∥≤

m∑k=n+1

‖x‖k

< ε.

Being A a Banach space, the sum converges.

Claim. y = (e− x)−1

i.e. y (e− x) = e.

Indeed

y (e− x) = (e− x) limn→+∞

n∑k=0

xk

[The product is continuous]

= limn→+∞

(e− x)

n∑k=0

xk

= limn→+∞

(n∑k=0

xk −n+1∑k=1

xk

)= lim

n→+∞

(e− xn+1

)= e

Let's nally prove that G is open. Take x ∈ G and h ∈ A such that

‖h‖ < 1

2 ‖x−1‖.

Then(x+ h) = x

(e+ x−1h

).

The right hand side is invertible by hypothesis and the rst part of theproof, so x+ h is invertible.

1.2 Basic properties of spectra 8

2. Take x ∈ G and h ∈ A so that

‖h‖ ≤ 1

2 ‖x−1‖.

Then ∥∥∥(x+ h)−1 − x−1

∥∥∥ =∥∥∥x−1

(e+ x−1h

)−1 − x−1∥∥∥

=∥∥∥x−1

((e+ x−1h

)−1 − e)∥∥∥

≤∥∥x−1

∥∥∥∥∥∥∥∥∥e+ x−1h︸ ︷︷ ︸

=:y

−1

− e

∥∥∥∥∥∥∥= (∗) .

Being ‖y‖ < 1/2, by the previous part of the proof,

(e+ y)−1

=

+∞∑k=0

(−1)kyk.

So

(∗) ≤∥∥x−1

∥∥∥∥∥∥∥+∞∑k=1

(−1)kyk

∥∥∥∥∥=

∥∥x−1∥∥∥∥∥∥∥y

+∞∑k=0

(−1)k+1

yk

∥∥∥∥∥≤

∥∥x−1∥∥ ‖y‖ ∥∥∥∥∥

+∞∑k=0

(−1)k+1

yk

∥∥∥∥∥≤

∥∥x−1∥∥ ‖y‖ +∞∑

k=0

∥∥yk∥∥≤

∥∥x−1∥∥2 ‖h‖

+∞∑k=0

1

2k

= 2∥∥x−1

∥∥2 ‖h‖ ,

and ψ is continuous.

1.2 Basic properties of spectra

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

Ω (x) :=λ ∈

∣∣∣∃ (x− λe)−1

is called resolvent set of x. For all λ ∈ Ω (x), we will also use the notation

Rλ (x) := (x− λe)−1.

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

Ω (x) ⊃ C \B (0, ‖x‖) .

Proof. Let's x λ ∈ C, |λ| > ‖x‖. We need to prove the existance of Rλ (x).Being

x− λe = λ(xλ− e)

= −λ(e− x

λ

)and ∥∥∥x

λ

∥∥∥ =‖x‖|λ|

< 1,

by the proof of Theorem 11 on page 6 it follows that e− xλ is invertible, hence

there exists

(x− λe)−1= λ−1

(xλ− e)−1

.

Corollary 14. Let A be a Banch algebra and x ∈ A. If λ ∈ σ (x), then|λ| ≤ ‖x‖. In other words

σ (x) ⊂ B (0, ‖x‖).

Corollary 15. Let A be a Banch algebra and x ∈ A. Then the spectrum of xis a bounded set.

Lemma 16. Let A be a Banch algebra and x ∈ A. Then the set Ω (x) is openin C.

Proof. Suppose λ0 ∈ Ω (x). Let's nd a neigborhood of λ0 in Ω (x). By deni-

tion it exists (x− λ0e)−1. For all ε > 0, if λ ∈ C, |λ− λ0| < ε, then

‖e− λ0x− (e− λx)‖ = |λ− λ0| ‖x‖< ε ‖x‖ .

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 ⊂C open and f : D → A. Given z0 ∈ D, we say that f is abstractly holomorpicin z0 if the following limit exists:

f ′ (z0) := limh→0

f (z0 + h)− f (z0)

h.

If f is abstracly holomorphic in every point of D we say that f is an abstractholomorphic function. If f is an abstract holomorphic function and D = C wesay that f is an abstract entire function.

Remark 20. Almost every result which holds for holomorphic functions alsoholds 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 forall z0 ∈ D there exists r > 0 such that, for all z ∈ B (z0, r),

f (z) =

+∞∑k=0

f (k) (z0) (z − z0)k.

Notation 23 (Radius of convergence). What is the radius of convergence ofthe previous power series? Without loss of generality, suppose z0 = 0. We needto study the convergence of the series

z 7→+∞∑n=0

cn︸︷︷︸∈A

zn︸︷︷︸∈C

.

Dening

R :=1

lim supn→+∞n√‖cn‖

,

by the general theory of power series, it is possible to state that the seriesconverges 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

z 7→+∞∑n=0

cnzn.

Clearly, for all z ∈ C such that ‖cz‖ < 1, i.e. for all z ∈ B (0, 1/ ‖c‖), the seriesconverges. Since by the general theory if the series converges in z ∈ C, thenz ∈ B (0, R), the following inequality should hold:

1

‖c‖≤ 1

lim supn→+∞n√‖cn‖

. (1.2.1)

1.2 Basic properties of spectra 11

This is an easy exercise, in fact for all n ∈ N

n√‖cn‖ ≤ n

√‖c‖n = ‖c‖ .

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

Proposition 24. Let A a Banach algebra and x ∈ A. Then there exists thelimit

limn→+∞

n√‖xn‖.

Proof. For all n ∈ N, call αn := ‖xn‖. For all n ∈ N and for all k ∈ N thereexist m, ` ∈ N such that n = mk + ` and consequently

αn.= ‖xn‖=

∥∥xkm+`∥∥

=∥∥∥(xk)m x`∥∥∥

≤∥∥∥(xk)m∥∥∥∥∥x`∥∥

≤∥∥xk∥∥m ∥∥x`∥∥

.= αmk α`,

hence(αn)

1n ≤ αm/nk α

1/n` .

Note that if n, k,m, ` ∈ N satisfy n = mk + `, then mn = 1

k −`kn and

(αn)1n ≤ α

1k−

`kn

k α1/n` .

Since ` is bounded by 0 and k (it's the rest in the Euclidean division), takingboth sides to the lim sup, we derive, for all k ∈ N,

lim supn→+∞

(αn)1n ≤ a1/k

k .

Now, taking both member to the lim inf, we get

lim supn→+∞

(αn)1n ≤ lim inf

k→+∞a

1/kk

which proves the existence of the limit.

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

Lemma 26 (Hilbert). Let A be commutative Banach algebra and x ∈ A. Thenfor 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 thefunction λ 7→ Rλ is holomorphic on the whole Ω (x).

Proof. For all λ, µ ∈ Ω (x)

Rλ −Rµλ− µ

= RµRλ.

Since the inverse function (·) 7→ (·)−1is continuous where it's dened, for all

µ ∈ Ω (x) there exists the limit

limλ→µ

Rλ = Rµ.

Corollary 28. Let A be commutative Banach algebra and x ∈ A. Then. Wehave

∂

∂λ(λ 7→ Rλ) =

(λ 7→ R2

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 allx ∈ A, σ (x) 6= ∅.

Proof. Assume the contrary. Then there is x ∈ X s.t. σx = ∅, i.e. so Ω (x) = C.Hence λ 7→ Rλ

.= (x− λe)−1

is an entire function. Fix R > 0.

• For all λ ∈ C \B (0, R),

Rλ =1

λ

(xλ− e)−1

.

Note that if |λ| → +∞x

λ− e→ e.

So, if |λ| → +∞, (xλ− e)−1

→ (−e)−1= −e.

This imples that the function λ 7→(xλ − e

)−1is bounded. For |λ| → +∞

it follows

‖Rλ‖ =1

|λ|︸︷︷︸→0

∥∥∥∥(xλ − e)−1∥∥∥∥︸ ︷︷ ︸

bounded

,

hence λ 7→ ‖Rλ‖ is bounded on C \B (0, R).

1.3 Ideals 13

• The function λ 7→ ‖Rλ‖ is continuous (composition of norm - continuous -with λ 7→ Rλ - holomorphic) on the compact set B (0, R), so it is boundedon B (0, R) also.

Putting together these two points, it follows thatλ 7→ ‖Rλ‖ is bounded on C.By Liouville's Theorem, it's constant. Since, for |λ| → +∞, Rλ → 0, necessarelyλ 7→ Rλ ≡ 0. This is a contraddiction because for all λ ∈ C Rλ is an inverse,and it can't happen that for some λ ∈ C

e = (x− λe) (x− λe)−1︸ ︷︷ ︸=0

= 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 asigleton 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 (nontrivial) 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. Westill prefer to add it to the denition of ideal to avoid trivial cases and make thedevelopment 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, 12 ]≡ 0

.

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 theris 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 equiv-alent:

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 iscalled 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 t0 ∈ [0, 1], Mt0 is maximal.

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)︸ ︷︷ ︸∈Mt0

.

This implies that the ideal is maximal.

Claim. For each maximal idealM ⊂ C [0, 1] there exists a unique t0 ∈ [0, 1] suchthat M = Mt0 .

Proof. Fix a maximal ideal M ⊂ C [0, 1]. Assume to the contrary that for allτ ∈ [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 suchthat, for all t ∈ Uτ ,

|fτ (t)| > δτ > 0.

Being Uττ∈[0,1] an open cover of the compact set [0, 1], there exists τk1 , . . . , τkn ∈[0, 1] such that

[0, 1] ⊂n⋃k=1

Uτk .

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

M ∩ R 3n∑k=1

fτkfτk︸ ︷︷ ︸=:f

=

n∑k=1

|fτk |2 ≥ min

k∈1,...,n|δτk | > 0.

This is a contraddiction because, being stricly positive, f ∈ M is invertible. Itfollows that there exists t0 ∈ [0, 1] such that all the functions in M vanish in t0,i.e. M ⊂Mt0 . Being M maximal, M = Mt0 . Noting that t0, t1 ∈ [0, 1], t0 6= t1imples Mt0 6= Mt1 , the uniqueness is trivial.

Proposition 41. Let A be a commutative Banach algebra. If I ⊂ A is an ideal,

the closure IAis also an ideal.

Proof. First of all, let's prove that IAis a linear subspace of A. For all x, y ∈ IA

there exist xnn∈N , ynn∈N ⊂ I such that, for n→ +∞,

xn → x, yn → y.

By the continuity of the sum,

x+ y = limn→+∞

(xn + yn)︸ ︷︷ ︸∈I

,

hence x+ y ∈ IA. Now take x ∈ IA and y ∈ X. There exists xnn∈N ⊂ I suchthat xn → x. Because of the continuity of the product, if n→ +∞, xny → xy.

Since I is an ideal, for all n ∈ N xny ∈ I, hence xy ∈ IA. Finally, let's prove that

IAis non trivial. Call G := x ∈ A| x is invertible. Being an ideal, I ⊂ A \G,

which is a closed set5, so

∅ ( I ⊂ IA ⊂ A \G ( A.

Corollary 42. Let A be a commutative Banach algebra. IfM ⊂ A is a maximalideal, 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 idealscontaining I, partially ordered by inclusion. ClearlyM is not empty (I ∈ M).Note that if any linearly ordered6 subfamily Iλλ∈Λ ⊂M has an upper bound,we are done (Zorn's Lemma). Denote I0 =

⋃α Iα. By denition ofM, I0 ⊃ I

and clearly I0 is an ideal upper bound to Iλλ∈Λ.

5Theorem 11 states that G is open.6Iλλ∈Λ ⊂M is linearly ordered if anf only if, for all λ1, λ2 ∈ Λ, Iλ1

⊂ Iλ2or Iλ2

⊂ Iλ1.

1.3 Ideals 16

Proposition 44. Let A be a commutative Banach algebra, G := x ∈ A| x is invertibleand 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 linearfunctional7 f ∈ A′ \ 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 ∈ A′ a multiplicative func-tional. Then

1. f (e) = 1;

2. for all x ∈ G, f (x) 6= 0 and

f(x−1

)=

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 ∈ A′ a multiplicative func-tional. If x ∈ A, ‖x‖ < 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− λe = λ(xλ− e).

Since ∥∥∥xλ

∥∥∥ < 1,

the right hand side is invertible and so is the left hend side. By the previousProposition f (x− λe) 6= 0 which leads to

f (x) 6= λ.

Absurd.

Remark 50. The previous result says that multiplicative functionals map theunit ball of the Banach algebra into the unit complex ball.

Corollary 51. Let A be a Banach algebra and f ∈ A′ a multiplicative func-tional. Then

‖f‖ = 1.

Proof. Because of the previous proposition

‖f‖ = sup

x∈BA

|f (x)| ≤ 1.

Proposition 48 on the preceding page clinches the result.

Remark 52. The previous result says that multiplicative functionals are alwayscontinuous!

Proposition 53. Let A be a Banach algebra and f ∈ A∗ a multiplicative func-tional. Then ker (f) is a maximal ideal.

Proof. It is well known that kernels of continuous linear functionals are closedsubspace of codim = 1. It's easy to check that ker (f) is an ideal. Havingcodimension 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. Thenthe 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, ‖x+ I‖ = inf

u∈x+I‖u‖A .

Furthermore, calling e the unit of A,

1. A/I has unit e+ I;

2. ‖e+ I‖ = 1;

3. for all (x+ I) , (y + I) ∈ A/I,

‖(x+ I) (y + I)‖ ≤ ‖x+ I‖ ‖y + I‖ .

Proof. Easy and boring.

Denition 55 (Quotient algebra). Let A be a Banach algebra and I ⊂ A aclosed ideal. The Banach algebra A/I dened in the previous proposition iscalled quotient algebra.

Remark 56. Proposition 54 on the previous page states that quotients of aclosed ideals are Banach algebras.

Problem 57. What can be said of the quotient of a Banach algebra with amaximal ideal?

Exercise 58. Let A be a Banach algebra and I ⊂ A a closed ideal. Prove thatthe 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 ′ ⊂ A/I aclosed ideal on the quotient algebra and q : A → A/I the quotient map. ThenJ := q−1 (J ′) is a closed ideal containing I.

I ⊂ A q→ A/I

I ⊂ J 7→q−1

J′

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 thatexists J ∈ A/M ideal. Without loss of generality, J is closed. By the previouslemma, 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. ThenA/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. ThenA/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. Thenq : 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 somemultiplicative functional. On the other hand, Proposition 53 on page 17 statedthat kernels of multiplicative functionals were maximal ideals. This leads to thefollowing, unexpected, result.

Corollary 67. Let A be a Banach algebra andM the set of maximal ideals inA. 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 topologyinduced on ∆ by the w∗ topology of A∗ is called Gelfand topology.

Remark 70. From now on we will always consider ∆ as a topological spacesendowed with the Gelfand topology. As the follow result shows, this choicemakes ∆ 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 byBanach-Alaoglu theorem. Then we just need to prove that ∆ is w∗-closed. Wedo that using nets. Fix a net αλλ∈Λ ⊂ ∆ and a functional α ∈ A′ such that

αλw∗−→ α.

Remember that the w∗-convergence of a net is just its pointwise convergence:

αλw∗−→ 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,

where

x : ∆ → C,α 7→ xα := α (x)

is called Gelfand transform.

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

Λ (x+ y) = x+ y,

Λ (xy) = xy.

Proof. Trivial.

Proposition 74. Let A be a Banach algebra. Then

|||Λ||| = 1.

Proof. We want to prove that

supx∈BA

‖x‖C(∆;C)

= 1.

This follows from noting that for all x ∈ BA

‖x‖C(∆;C) = supα∈∆|α (x)| ≤ ‖x‖A ≤ 1

and‖e‖C(∆;C) = 1.

Proposition 75. Let A be a Banach algebra. The Gelfand topology on ∆ isthe weakest topology that makes every x ∈ A continuous.

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 byn⋂k=1

α ∈ A∗| γxk(α) > ε

ε>0

.

This means that a base for the Gelfand topolofy of ∆ is simplyn⋂k=1

α ∈ A∗| γxk(α) > ε ∩∆

ε>0

=

n⋂k=1

α ∈ ∆| γxk(α) > ε

ε>0

.=

n⋂k=1

α ∈ ∆| α (xk) > ε

ε>0

.=

n⋂k=1

α ∈ ∆| xk (α) > ε

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 σ (x) ⊂ α (x)α∈∆. Take λ ∈ σ (x). Thenx − λe is not invertible. So it lies in some maximal ideal, kernel of somemultiplicative functional α. Since α (x− λe) = 0 it immediately follows thatα (x) = λ. Viceversa, take an arbitrary α ∈ ∆ and consider λ := α (x) ∈ C. Bydenition λ satises

0 = α (x)− λ= α (x− λe) ,

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

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

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 t0 ∈ [0, 1], δt0 is a multiplicative functional. With the samenotation 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 func-tionals, it follows

∆ = δt0t0∈[0,1] .

By Theorem 76 on the previous page, for all f ∈ C [0, 1],

σ (f) = δt0 (f)t0∈[0,1] = f (t0)t0∈[0,1] = f ([0, 1]) .

This makes perfect sense! For all f ∈ C [0, 1], λ ∈ σ (f) if and only if f − λe isnot invertible, which happens if and only if f − λe is null at some point, i.e. ifand only if there exists t0 ∈ [0, 1] such that

f (t0)− λ e (t0)︸ ︷︷ ︸e≡1

= 0 ⇐⇒ λ = f (t0) .

Example 78. L1 [0, 1] is a Banach algebra with respect to the convolutionproduct

∀f, g ∈ L1 [0, 1] , x 7→ (f ∗ g) (x) =

ˆ x

0

f (t) g (x− t) dt

and the natural ‖·‖1 norm. Sadly this Banach algebra has no unit. It is possibleto plug in an idendity e the standard way. Dening a new Banach algebraL1 [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 kindof weird!

Denition 79 (Radical). Let A be a Banach algebra. The intersection of allmaximal ideals of A is called radical of A and indicated with Rad (A).

Proposition 80. Let A be a Banach algebra. Then

1. 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 ∈ ∆, xα = 0= ker (Λ) .

11Denition 72 on page 20.

1.3 Ideals 23

Theorem 81. Let A be a Banach algebra and x ∈ A. Then

‖x‖ = limn→+∞

n

√‖xn‖A

Proof. By Theorem 76 on page 21

‖x‖ .= max

α∈∆|xα|

.= max

α∈∆|α (x)|

= maxλ∈σ(x)

|λ|︸ ︷︷ ︸=:a

;

a is the so called spectral radius. We want to prove that

a = limn→+∞

n

√‖xn‖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

λx− e = λ

(x− 1

λe

)(1.3.1)

and 1/ |λ| > |a|, so 1/λ ∈ Ω (x). Note that for all λ ∈ C such that |λ| a < 1and ‖λx‖ < 1,

+∞∑k=0

(λx)k

=1

e− λx. (1.3.2)

The biggest set on which the function

λ 7→+∞∑k=0

(λx)k

(1.3.3)

is holomorphic is B (0, R), with

R :=1

limn→+∞n√‖xn‖

(dened the obvious way if the limit is null or innite), as discussed inNotation 23 on page 10 and Proposition 24 on page 11. On the otherhand, by Corollary 27 on page 12, the mapping µ 7→ Rµ (x) := (x− µe)−1

is holomorphic on its domain. Hence, by equation (1.3.1), the function

λ 7→ 1

e− λx(1.3.4)

1.3 Ideals 24

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

1

a≤ R,

i.e. thata ≥ lim

n→+∞n√‖xn‖.

Assume to the contrary that 1/a > R. It is well known that any holo-morphic function f dened on a disk B (x0, r) concides in the whole diskwith 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 isdened 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 Taylorseries, 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 lowerthan 1/a. Absurd.

Corollary 82. Let A be a Banach algebra. Then x ∈ Rad (A) if and only if,for n→ +∞,

n

√(‖x‖A)

n → 0.

Remark 83. Clearly nilpotent elements of A belong to Rad (A). For thisreason elements of Rad (A) are sometimes called generalized nilpotents. Notethat, belonging to Rad (A), every nilpotent is annihilated by all α ∈ ∆.

Denition 84 (Semisimple algebra). Let A be a Banach algebra. A is calledsemisimple 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 usedthroughout the paper:

A := Λ (A) .

Corollary 87. Let A be a semisimple Banach algebra. Then Λ is a linearisomorphism onto A ⊂ C (∆;C).

Proof. Trivial.

Proposition 88. Let A be a Banach algebra. Then A is an algebra with respectto the pointwise sum and product.

1.3 Ideals 25

Proof. Trivial.

Corollary 89. Let A be a semisimple Banach algebra. Then A is a Banachalgebra 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 thatD ⊂ 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 onlyif ∆ is total.

Proof. Suppose A semisimple, then ker (Λ) = 0, hence the only x ∈ A such thatx ≡ 0 is x = 0. If x ∈ A and for all α ∈ ∆

0 = α (x) = xα,

then x ≡ 0, hence x = 0; if x = 0, for all α ∈ ∆

α (0) = 0.

Viceversa, suppose that ∆ is total. Assume to the contrary that A is notsemisimple, i.e. that there exists x ∈ A \ 0 such that x ≡ 0. By denition, itfollows that for all α ∈ ∆,

0 = xα = 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 allx ∈ A, ∥∥x2

∥∥ = ‖x‖2 . (1.3.5)

Then Λ: A→ C (∆;C) is an isometry.

Proof. Since the rst equation holds and the rst limit exists

‖x‖ = limn→+∞

n√‖xn‖

= limn→+∞

2n√‖x2n‖

= limn→+∞

2n

√∥∥∥(x2n−1)2∥∥∥

= limn→+∞

2n√∥∥x2n−1

∥∥2

= . . .

= limn→+∞

2n√‖x‖2

n

= ‖x‖ .

1.3 Ideals 26

Corollary 93. Let A be a Banach algebra. If (1.3.5) holds, then A 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 allBanach algebras. C [0, 1] satises it, but operators algebras, where the productis the composition, usually do not.

Denition 95 (Symmetric space). Let A be a Banach algebra. A is calledsymmetric 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 existsf ∈ 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 A is symmetric and forall x ∈ A,

∥∥x2∥∥ = ‖x‖2. Then A = C (∆;C), i.e. A and C (∆;C) are isometric

via Λ.

Proof. Let's apply Stone-Weierstrass theorem for X = A ⊂ C (∆;C). Since

(1.3.5) holds, A is complete, hence closed. By Proposition on page 24, A is alinear space. Let's check all other conditions:

1. the constant function 1 belongs to A, indeed

Λe.= e

.= [α 7→ α (e) = 1] ≡ 1;

2. A separates points of ∆, indeed if α, β ∈ ∆ and for all x ∈ A

α (x) = x (α) = x (β) = β (x) ,

then α = β;

3. by Proposition on page 24, A is an algebra with respect to the pointwiseproduct;

1.4 Involutions 27

4. x x ∈ A; since A is symmetric there is y ∈ X such that, for all α ∈ ∆,

x (α).= α (x) = α (y)

.= y (α) ,

hence x = y ∈ A.

Example 99. We stress again that the hypothesis aren't trivial. The Banachalgebra of functions continue on the closed disk and holomorphic in the interiordoesn't satisfy them. Same goes for the Banach algebra of functions which aresum 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 algebrawith involution.

Remark 101. From now on we will always assume that the Banach algebraswe 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 wrotethe third property of an involution in such a way that makes sense even for noncommutative ones.

1.4 Involutions 28

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

BL (H;H) → BL (H;H) ,

D 7→ D∗,

where D∗ is the unique operator such that, for all x, y ∈ H

〈D∗x, y〉 = 〈x,Dy〉 .

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

u =1

2(x+ x∗) and v = − i

2(x− x∗)

do the trick. For the uniqueness take another u′, v′ ∈ A self-adjoint andsuch that x = u′ + iv′. Then

u′ − iv′ = x∗ = u− iv.

Hence2w = x+ x∗ = 2u

and this easily brings us the uniqueness.

1.4 Involutions 29

3. Note thate = (e∗)

∗= (e∗e)

∗= e∗ e∗∗︸︷︷︸

=e

= 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 theprevious point also (x− λe)∗ is invertible and(

(x− λe)∗)−1

=(

(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 realnumbers 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, thefunctional

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, thefunctional

ϕ : A → C,x 7→ h (x∗)

is continuous.

Proposition 112. Let A be a semisimple Banach algebra. Then the involution

is continuous, i.e. if xnn∈N ⊂ A, x ∈ A and xnn→+∞−→ x, then

x∗nn→+∞−→ x∗.

Proof. We use the Closed Graph Theorem12. Take the convolution as an anti-linear operator on A. The theorem states that, if[

xnn→+∞−→ x, x∗n

n→+∞−→ y =⇒ y = x∗],

then the involution operator is continuous. Let's check the premises. Fix any

h ∈ ∆. If xnn→+∞−→ x and x∗n

n→+∞−→ y, by the previous corollary

h (x∗) = ϕ (x) = limn→+∞

ϕ (xn) = limn→+∞

h (x∗n) = h (y),

henceh (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

‖x · x∗‖ = ‖x‖2 .

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 forall f ∈ C [0, 1] by the complex conjugate

f∗ = f.

Lemma 115. Let A be a B∗-algebra, u ∈ A self-adjoint and h ∈ ∆. Thenh (u) ∈ R.

12The Closed Graph Theorem is usually stated for linear operators. Check that the sametheorem 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 ≤ ‖z‖2

= ‖zz∗‖≤

∥∥u2∥∥+ t2

but also

|h (z)|2 = |h (u) + it|2

= |α+ i (β + t)|2

= α2 + β2 + 2βt+ t2.

Since, for all t ∈ R

α2 + β2 + 2βt+ t2 ≤∥∥u2∥∥+ t2

mα2 + β2 + 2βt ≤

∥∥u2∥∥ .

This clearly implies that β = 0 (the right hand side is independent of t).

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

Theorem 117 (Gelfand-Naimark). Let A be a B∗-algebra. Then A is isometricto 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 A = C (∆;C) via the Stone-Weierstrass theorem:

1. the constant function 1 = e belongs to A;

2. A separates the points of ∆, i.e. for all α, β ∈ ∆, α 6= β, there existsx ∈ A such that

α (x) = x (α) 6= x (β) = β (x) ;

assume to the contrary that there are α, β ∈ ∆, α 6= β such that, for allx ∈ A,

α (x) = x (α) = x (β) = β (x) ,

i.e. α = β, absurd;

1.5 Applications 32

3. for all x, y ∈ A, the pointwise product xy ∈ A;for all α ∈ ∆,

(xy) (α).= x (α) y (α)

.= α (x)α (y) = α (xy) = [Λ (xy)]α;

4. for all x ∈ A, x ∈ A;x any x ∈ A and u, v ∈ A self-adjoint such that x = u + iv; thenx∗ = u− iv and x = u+ iv; by previous corollary

Λ (x∗) = u− iv = x

which also proves formulae (1.4.1) and (1.4.2).

We now want to prove that Λ: A → A is an isometry. If we do that, we provethat A is closed in C (∆;C), by Stone-Weierstrass theorem we can conclude that

A = C (∆;C) and consequently that A and C (∆;C) are isometric via Λ.

• Take any y ∈ A self-adjoint. Then∥∥y2∥∥ = ‖yy∗‖ = ‖y‖2 .

Cleary, for all n ∈ N, also y2n

is self-adjoint and∥∥∥y2n∥∥∥ = ‖y‖2

n

.

By Theorem 81 on page 23,

‖y‖A = limn→+∞

n√‖yn‖

= limn→+∞

2n√‖y2n‖

= limn→+∞

2n√‖y‖2

n

= ‖y‖ .

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

‖x‖2 = ‖xx∗‖ = ‖y‖ = ‖y‖A = ‖xΛ (x∗)‖A =∥∥∥xx∥∥∥

A=∥∥∥|x|2∥∥∥

A= ‖x‖2A .

1.5 Applications

1.5.1 Wiener algebra

Example 118. The so called Wiener algebra W is the set of 2π-periodic com-plex functions wich are absolute convergent sum of their Fourier seriesf : [0, 2π)→ C| ∃ cnn∈Z ⊂ C,

∑k∈Z|cn| < +∞,∀t ∈ [0, 2π) , f (t) =

∑k∈Z

cneikt

1.5 Applications 33

endowed with the pointwise sum, pointwise product and the norm dened forall f ∈W such that f =

∑k∈Z cne

ik(·) by

‖f‖ =∑k∈Z|cn| .

The function f ≡ 1 ∈W is the unit of W . It's easy to check that W is Banachalgebra13.

Lemma 119. Let W be the Wiener algebra. Then

∆ = αt0 := [f 7→ f (t0)]t0∈[0,2π) .

Proof. Clearly∆ ⊃ αt0t0∈[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

α(e−i(·)

)=(α(ei(·)

))−1

=1

a

and that multplicative functional has norm 1, hence

|a| .=

∣∣∣α(ei(·))∣∣∣ ≤ ∥∥∥ei(·)∥∥∥ = 1,

1/ |a| =∣∣∣α(e−i(·))∣∣∣ ≤ ∥∥∥e−i(·)∥∥∥ = 1,

and consequently |a| = 1. This means that there exists t0 ∈ [0, 2π) such thata = eit0 . We have proved that

α(ei(·)

)= eit0 .

Being α multiplicative, for all n ∈ Z,

α(ein(·)

)= eint0 .

This means that for all cnn∈Z ⊂ C such that∑n∈Z |cn| < +∞,

α

(∑n∈Z

cnein(·)

)=∑n∈Z

cneint0 .

13For the completeness, note that W is isomorphic to `1 (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 allt ∈ [0, 2π), f 6= 0, then

1

f∈W.

Proof. By previous exercise and Corollary 64 on page 19 it follows that for anymaximal 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 continuousfunctions dened on the closed complex disk whose restriction to the interior ofthe disk is holomorphic

A :=f ∈ C

(D;C

)∣∣ f|D is holomorphic

endowed with the pointwise sum, pointwise product and the norm dened forall f ∈ A by

‖f‖ = maxz∈D|f (z)|

= maxz∈∂D

|f (z)|

(last idendity holds because of the maximum modulus principle). The functionf ≡ 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

∆ = αz0 := [f 7→ f (z0)]z0∈D .

Proof. Clearly∆ ⊃ αz0z0∈D .

Viceversa, x any α ∈ ∆. We want to prove that there exists z0 ∈ D such thatα = αz0 . Call

id : D → D,

z 7→ z

andα (id) =: z0 ∈ C.

Note that z0 ∈ D, indeed|α (g)| ≤ ‖id‖ = 1.

Claim. For all p ∈ A polynomial, α (p) = αz0 (p).

Proof. Fix any p ∈ A polynomial. Then there exist aknk=0 ⊂ C such that

p = anidn + an−1idn−1 + . . .+ 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 psuch that ‖f − p‖ < ε. Being f continuous on a compact set, by Heine-Cantortheorem its uniformly continuous. Then there is δ ∈ (0, 1) such that, for allz1, z2 ∈ D with |z1 − z2| < δ,

|f (z1)− f (z2)| < ε

2.

Fix that δ. Note that the function

f : (1 + δ)D → C,

z 7→ f (z) := f

(z

1 + δ

)is holomorphic on

(1 + δ)D ⊃ D.

Hence f Taylor series centered at 0 uniformly converges to f on compact subsetsincluded in (1 + δ)D (and including 0). So there exists a polynomial p such that

supz∈D

∣∣∣∣f ( z

1 + δ

)− p (z)

∣∣∣∣ < ε

2.

Note that for all z ∈ D ∣∣∣∣z − z

1 + δ

∣∣∣∣ =δ

1 + δ|z|

≤ δ

1 + δ︸ ︷︷ ︸>1

< δ.

Consequently, for all z ∈ D,

|f (z)− p (z)| =

∣∣∣∣f (z)− f(

z

1 + η

)+ f

(z

1 + η

)− p (z)

∣∣∣∣≤

∣∣∣∣f (z)− f(

z

1 + η

)∣∣∣∣+

∣∣∣∣f ( z

1 + η

)− p (z)

∣∣∣∣≤ ε.

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

1.5 Applications 36

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

α (f) = α

(lim

n→+∞pn

)= lim

n→+∞α (pn)

= limn→+∞

αz0 (pn)

= αz0

(lim

n→+∞pn

)= αz0 (f) .

Theorem 123. Let A be the disk algebra and f1, . . . , fm ∈ A such tath for allz ∈ D there is j ∈ 1, . . . ,m such that fj (z) 6= 0. Then there is h1, . . . , hm ∈ Asuch that

m∑k=1

hkfk ≡ 1.

Proof. Consider the set

I =g =

∑gkfk

∣∣∣ g1, . . . gm ∈ A.

Clearly I is a subspace of A, is dierent from 0 and is closed under the productwith elements in A. We want to prove that I = A. If this is true, then 1 ∈ Iand the theorem is proved. Assume to the contrary that I ( A. Then I is anideal (by denition). By Theorem 43 on page 15 there is M maximal ideal suchthat M ⊃ I. By previous proposition and Corollary 64 on page 19 there existsz0 ∈ 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 andg1, . . . , 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 operatorT : 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

〈Tx, y〉 = 〈x, Ty〉 .

1.5 Applications 37

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

PT :=anT

n + an−1Tn−1 + . . .+ a0

∣∣ n ∈ N0, a0, . . . , an ∈ C.

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

‖P‖ = sup‖x‖≤1

‖Px‖ .

PT is an incomplete normed algebra (check it). Denote with A the completionof 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 forp = T . Note that (check it), for all B : H → H self-adjoint,

‖B‖ = sup‖x‖≤1

|〈Bx, x〉| .

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

‖x‖≤1

∣∣⟨T 2x, x⟩∣∣

[self − adjoint]

= sup‖x‖≤1

|〈Tx, Tx〉|

= sup‖x‖≤1

‖Tx‖2

= ‖T‖2 .

The same equalities holds for all X ∈ A (check it).

Remark 127. By Gelfand-Naimark theorem17, since A is a B∗-algebra, A isisomorphic to C (∆;C) via the Gelfand transform. We want to say explicitlywho is C (∆,C). In order to to that, we need the spectral theorem for compactself-adjoint operators.

Denition 128 (Eigenvalues, eigenvectors, spectrum and eigenspaces). Let Hbe a Hilbert space and T : H → H a linear operator. We say that λ ∈ C isan eigenvalue for T if there exists x ∈ H \ 0, called eigenvector, such thatTx = λx. The set of all eigenvalues is called spectrum. Let λ be an eigenvalue forT , then the linear space Eλ := x ∈ H| Tx = λx is called eigenspace (relativeto λ).

Proposition 129. Let H be a Hilbert space and T a compact self-adjoint op-erator. 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 bea Hilbert space and T a compact self-adjoint operator. Assume that the spectrumis countable18 and ker (T ) = 019. If σ := λii∈N ⊂ C is the spectrum, forall λ ∈ σ we denote with Eλ the eigenspace relative to λ and with Pλ : H → Eλthe orthogonal projector on the eigenspace. Assuming without loss of generalitythat |λ1| > |λ2| > . . ., it follows

T =

+∞∑i=1

λiPλi.

Remark 131. To describe the set ∆ of multiplicative functionals of A we shouldthink about the previous result in a weak sense. Fix any x ∈ H. Then

Tx =

+∞∑i=1

λiPλix

and this series is convergent in the norm of the Hilbert space20. By Theorem 76on 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 itbehaves on polynomals in T . Without loss of generality, for all P ∈ PT and forall i ∈ N

αi (P ) = P (λi)

and, if P := anTn + . . .+ a0I,

α0 (P ) = α0 (anTn + . . .+ 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 makesit diagonal. This is a generalization of the same concept. Indeed the previous identity denesa diagonal matrix.

Chapter 2

Seminars

2.1 Gelfand and Fourier transform (Tommaso Ce-sari)

Proposition 132.(L1 (Rn) , ‖·‖1

)is a Banach algebra without unit with respect

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 beenlarged enough to admit a unit element. Note that in the particular case ofL1 (Rn), the complex component added to L1 (Rn) in order to get the unit canbe thought as a coecient of a function δ, dened for all x ∈ Rn as the δx Diracmeasure on Rn. Indeed the set

L :=f + αδ

∣∣ f ∈ L1 (Rn) , 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 , ‖f + αδ‖ = ‖f‖1 + |α|

and L1 (Rn) is isometrically embedded in L via f 7→ f + 0 · δ. Note that, forall x ∈ Rn,

(f ∗ δ) (x)︸ ︷︷ ︸=f(x)

=

ˆRn

f (t) dδx (t)︸ ︷︷ ︸=f(x)

,

so the convolution f ∗ δ actually coincide with the integration of f with respectto the Dirac measure.

2.1 Gelfand and Fourier transform (Tommaso Cesari) 40

Notation 134. We x the notation L to indicate the smallest Banach algebrawith unit containing

(L1 (Rn) ; ∗

), as described in the previous remark.

Notation 135 (Fourier transform). For all f ∈ L1 (Rn), f and F (f) will denotethe Fourier transform of f .

Proposition 136. The function

h : Rn → ∆,

t 7→ ht :=

[f + αδ 7→ f (t) + α

], if t ∈ Rn,

[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 theusual Gelfand topology. I.e., as topological spaces

∆ = Rn.

Proof. For all f ∈ L1 (Rn), dene

f (∞) := limt→∞

f (t) = 0.

Let's start by proving that h is well dened. For all (f + αδ) , (g + βδ) ∈ Land 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

= f (t) g (t) + βf (t) + αg (t) + αβ

=(f (t) + α

)(g (t) + β)

= ht (f + αδ)ht (f + βδ) .

Now we prove that h is injective. If t1, t2 ∈ Rn, t1 6= t2, assume (by contradic-tion) [

f + αδ 7→ f (t1) + α]

= [g + βδ 7→ g (t2) + β] ,

this means in particular that for all f ∈ L1 (Rn)

f (t1) = f (t2) .

Of course this cannot happen for all f ∈ L1 (Rn) (take for instance x 7→ e−πx2

and y 7→ e−π(y−1)2). The proof that h is surjective can be easily derived fromthe result in Section 9.22 of W. Rudin - Real and Complex Analysis. Nowlet's prove that h is an homeomorphism. Call τ the topology of the one-pointcompactication 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 L , that is, the

weakest topology that makes every Λ (f + βδ) ∈ L continuous. Since, for all

Λ (f + βδ) ∈ L

Λ (f + βδ) = [α 7→ α (f + βδ)]

=[ht 7→ ht (f + βδ) = f (t) + β

],

then h (τ) makes every Λ (f + βδ) ∈ L continuous, hence, unsurprisingly, theweak topology γ is weaker then h (τ). Since γ ⊂ h (τ) and both topologies arecompact 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) = f (t) + α.

This shows how the Fourier transform of an L1 function coincides with itsGelfand transform1, making the Gelfand transform a proper generalization ofthe 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 inthe course will not be applied to X, but to B(X), which has a clear structureof 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 mentionthat the following result holds also with no assumptions of commutativity andwill have a very main role in the sequel.

Theorem 138 (Spectral Radius Formula). Let A be a Banach Algebra withunit and let x ∈ A. Then the spectral radius of x, dened by

ρ(x) := maxλ∈σ(x)

|λ|

satises the following Spectral Radius Formula

ρ(x) = limn→+∞

‖xn‖1/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 subspacefor T , if M is a closed subspace of X and T (M) ⊆M . It's said to be nontrivialif M 6= 0 ,M 6= X.

The main topic of this section will be to discuss the existence of nontrivial in-variant subspaces for elements of B(X). These are quite interesting in OperatorTheory according to the following trivial remark: if M is an invariant subspacefor T , then T|M ∈ B(M) and M is a Banach Space too, so one can study theoperator T on a smaller space and then try to put together the behaviour of theoperator on the various invariant subspaces in order to deduce the behaviouron the whole space. This is a very familiar machinery since is exactly what onedoes in nite dimensional spaces when tries to diagonalize a matrix or to nd itsJordan Canonical Form. However in innite dimensional spaces, it may happenthat 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 bepresent. As a simple (but userful for what follows) example consider:

Example 140. Let X = `2 and SR ∈ B(`2) the right shift, dened by:

(SRx)k

=

0 k = 1

xk−1 k ≥ 2∀x ∈ `2

if x ∈ `2 is the sequence x = (xk)+∞k=1. Explicitly the action is the following:(

x1, x2, x3, · · ·)7→(0, x1, x2, x3, · · ·

).

If one tries to solve SRx = λx, it comes out λx1 = 0, λxk+1 = xk,∀k ≥ 1. Nowthere are two cases: λ = 0 immediately implies 0 = xk, ∀k ≥ 1, while λ 6= 0gives 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). Someevidence of this is given by the fact that SNR has norm 1, ∀N ≥ 1 and so the

Spectral Radius Formula gives σ(SR) ⊆ D(0, 1).

Then it's natural to ask the following question: x X Banach Space andT ∈ B(X). Is it true that T admits a non trivial invariant subspace? Underwhich conditions on T and or on X there exists an invariant subspace? Let'sbegin by some cases in which the answer is easy and we can answer with nocondition 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 BanachSpace, 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 givenby the following matrix (wrt the canonical basis of course)(

0 −11 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 wouldbe an eigenspace, but there are no real eigenvalues and so no nontrivialinvariant subspaces.

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

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

• The rst innite dimensional case that can be treated easily is when X isnon separable and the answer is yes also here3. To see this, take x0 ∈ Xdierent from zero and consider the closed linear span of Tnx0+∞n=0 whichis 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 acomplex 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, thenT has a non trivial invariant subspace. Their proof used nonstandardanalysis, a proof of the same result using only classical concepts was givenby Halmos in the same year.

Now we turn to the key result, from which the last two statements follow im-mediately . The original proof, given by Lomonosov in 1973, made use of theSchauder 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). LetX be a innite-dimensional complex Banach Space and let T ∈ B(X) be a nonzero compactoperator. Then there exists a closed proper subspace M ⊆ X which is invariantfor 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 nonzerocompact 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.

4i.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 ev-ery compact operator admits a nontrivial invariant subspace, so we haveAronszajn and Smith result.

• Also the other statement we mentioned is easy to deduce: indeed we havetwo cases. If p(T ) 6= 0, then it is a nontrivial compact operator whichof course commutes with T and so by Lomonosov Theorem T admits aproper invariant subspace. If p(T ) = 0, let p(z) =

∑Nn=0 αnz

n with αN 6=0, N ≥ 1 so that we have 0 =

∑Nn=0 αnT

n, hence TN = −∑N−1n=0

αn

αNTn.

By this it's clear that the linear span of Tn(x0) : n = 0, · · · , N − 1 forsome 0 6= x0 ∈ X is invariant under T and is closed and proper since isnite dimensional.

Proof. Denote by Γ := S ∈ B(X) : ST = TS, which is easily seen to be a sub-algebra of B(X) and for y ∈ X denote Γy := Sy : S ∈ Γ. These are subspacesof X (since Γ is a subspace of B(X)) and are invariant under all S ∈ Γ (sinceΓ is closed under composition). They may miss being closed, but S (Γy) ⊆ Γyimplies5 S

(Γy)⊆ Γy, so that ∀y ∈ X, Γy is closed, invariant under all S ∈ Γ

and Γy 6= 0 if y 6= 0. If for at least one y 6= 0, Γy $ X the proof is concluded,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 ‖Tx0‖ > 1

2 ‖Tx0‖ > 0, ‖x0‖ > 12 ‖x0‖ > 0 and by continuity of T there exists

B open ball centered at x0 such that ∀x ∈ B, we have ‖Tx‖ ≥ 12 ‖Tx0‖ , ‖x‖ ≥

12 ‖x0‖. In particular 0 /∈ K := T (B) which is compact, since T is a compactoperator. By our previous assumption ∀y ∈ K, Γy∩B 6= ∅, so there exists Sy ∈ Γsuch that Syy ∈ B and by continuity of Sy there exists an open neighborhoodWy of y s.t. Sy (Wy) ⊆ B. Of course Wyy∈K is an open cover of K, so itadmits a nite subcover: there exist open sets W1, · · · ,Wn whose union coversK and S1, · · · , Sn ∈ Γ with Si(Wi) ⊆ B, ∀i = 1, · · · , n.

Start with x0 ∈ B, so that Tx0 ∈ K, hence in some Wi1 and so x1 :=Si1Tx0 ∈ B; then Tx1 ∈ K, so in some Wi2 and x2 := Si2Tx1 ∈ B. Goingon in this way, we nd a sequence xN = SiNTSiN−1

T · · ·Si1Tx0 ∈ B, so that∀N ‖xN‖ ≥ 1

2 ‖x0‖. But of course by commutativity xN = SiNSiN−1· · ·Si1TNx0

and this implies

1

2‖x0‖ ≤ ‖xN‖

≤ ‖SiN ‖∥∥SiN−1

∥∥ · · · ‖Si1‖ ∥∥TN∥∥ ‖x0‖≤ µN

∥∥TN∥∥ ‖x0‖

if µ := max ‖S1‖ , · · · , ‖Sn‖. Thus we have

1

21N

≤ µ∥∥TN∥∥ 1

N , ∀N ≥ 1

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

ρ(T ) ≥ 1

µ> 0.

The proof is now almost concluded, it suces to recall some spectral prop-erties 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 eigen-value; moreover it has nite multiplicity. So here we deduce that T admits aneigenvalue λ 6= 0 and the corresponding eigenspace Mλ := x ∈ X : Tx = λxis nite dimensional hence is closed and proper. Finally Mλ is invariant underall S ∈ Γ, since if S ∈ Γ and x ∈ Mλ we have TSx = STx = Sλx = λSx, soSx ∈Mλ. This concludes the proof.

As we have previously seen, this is quite a striking result. However it doesn'tsettle the invariant subspace problem: indeed there exist operators which com-mute with no nontrivial compact operator, for example the right shift alreadyintroduced.

Example 142. Let as before SR ∈ B(`2) and let T ∈ B(`2) be compact, suchthat SRT = TSR. Then T = 0.

Indeed let N > M ≥ 1 and compute, for x ∈ X,∥∥TSMR x− TSNR x∥∥ =∥∥SMR Tx− SNR Tx∥∥. Set y := Tx and note6

SMR y − SNR y = (0, 0, · · · , 0︸ ︷︷ ︸M terms

, y1, y2, y3, · · · )− (0, 0, · · · , 0︸ ︷︷ ︸N terms

, y1, y2, y3, · · · )

= (0, 0, · · · , 0︸ ︷︷ ︸M terms

, y1, y2, y3, · · · , yN−M︸ ︷︷ ︸N−M terms,

, yN−M+1 − y1, yN−M+2 − y2, · · · )

Hence,

∥∥SMR Tx− SNR Tx∥∥2=

N−M∑j=1

∣∣yj∣∣2 +

+∞∑j=1

∣∣yN−M+j − yj∣∣2

≥N−M∑j=1

∣∣yj∣∣2and now we can let N −M → +∞ and deduce

lim infN−M→+∞

∥∥TSMR x− TSNR x∥∥2 ≥+∞∑j=1

∣∣yj∣∣2 = ‖Tx‖2 .

6probably the notation here is not the most beautiful, but is hopefully clear.

2.2 Lomonosov's Invariant Subspace Theorem (Tommaso Russo) 46

But SnRxn∈N is a bounded sequence and so TSnRxn∈N is precompact andhence TSnk

R x→ x as k → +∞. Passing to another subsequence if necessary wecan assume nk+1 − nk → +∞ as k → +∞, so that N = nk+1, M = nk in theprevious inequality gives

0 = ‖x− x‖ = lim infk→+∞

∥∥TSnk

R x− TSnk+1

R x∥∥2 ≥ ‖Tx‖2 .

Thus we have Tx = 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 exis-tence of invariant subspaces, it's clear that SR admits many invariant subspaces,for example Mk :=

x = (xi)+∞

i=1 ∈ `2 : x1 = · · ·xk = 0. This is an interesting

case also because all invariant subpsaces of SR are known, this is a theorem byBeurling.

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 itwithout nontrivial invariant subspaces and Read in 1989 produced examples inthe 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 thequestion is still open, in particular even in Hilbert Spaces.

Index

Algebra,B∗-, 30C∗-, 30Banach, 3disk, 6, 34quotient, 18semisimple, 24Wiener, 32

Eigenspace, 37Eigenvalue, 37Eigenvector, 37

Function,Abstract entire, 10Abstract holomorphic, 10

Functional,multiplicative, 16

Gelfandtopology, 19transform, 20

Ideal, 13maximal, 14

Invariant subspace, 42Involution, 27

Lemma,Hilbert, 11

Nilpotents,generalized, 24

Operator,compact, 36self-adjoint, 36

Radical, 22

Self-adjoint element, 28Set,

Resolvent, 8Total, 25

Spectral radius, 23formula, 41

Spectrum, 8, 37Symmetric space, 26

Theorem,Dunford, 10Gelfand-Mazur, 13Gelfand-Naimark, 31Lomonosov's invariant subspace, 43spectral, for compact self-adjoint

operators, 38Stone-Weierstrass, 26Wiener, 34

Transform,Fourier, 40Gelfand, 20