# Banach Algebras

Definition (Banach Algebra)

A

$∥xy∥≤∥x∥∥y∥∀x,y∈A.$Banach algebra$A$ is a complex Banach space endowed with a binary operation $(x,y)↦xy$, calledproduct, that makes the underlying vector space into an associative algebra, and that satisfies

The algebraic properties required of the product are explicitly:

$x(y+y_{′})(x+x_{′})y =xy+xy_{′}=xy+x_{′}y (λx)yx(λy) =λ(xy)=λ(xy) (xy)z=x(yz)$The topological property is sometimes described by saying that the norm is *submultiplicative*.

Definition (Commutative Banach Algebra)

A Banach algebra $A$ is said to be

commutative(orabelian) if $xy=yx$ holds for all $x,y∈A$.

Definition (Unital Banach Algebra)

An element $e$ of a Banach algebra $A$ is called a

unit(or anidentity), if $∥e∥=1$ and $ex=x=xe$ for all $x∈A$. We say that $A$ is anunitalBanach algebra, if $A$ contains a unit.

It is easy to see that a Banach algebra has at most one unit.

Proposition (Neumann Series)

Let $A$ be a unital Banach algebra and let $x∈A$ satisfy $∥x∥<1$. Then $1−x$ is invertible, and the inverse is given by the series

$(1−x)_{−1}=n=0∑∞ x_{n},$which converges absolutely in norm. Moreover, we have the estimate

$∥(1−x)_{−1}∥≤1−∥x∥1 .$

Proof Since the Banach algebra norm is submultiplicative, we have $∥x_{n}∥≤∥x∥_{n}$ for all $n∈N$. This implies that the series $∑∥x_{n}∥$ is majorized by the geometric series $∑∥x∥_{n}$, which is known to be convergent for $∥x∥<1$. It follows that the series $∑x_{n}$ is absolutely convergent. Denote its limit by $s=lim_{n→∞}s_{n}=∑_{n=0}x$, where $s_{n}=1+x+⋯+x_{n}$ is the $n$th partial sum. Clearly,

$(1−x)s_{n}=s_{n}(1−x)=1−x_{n+1}.$In the limit $n→∞$ we obtain $(1−x)s=s(1−x)=1$, because multiplication in a Banach algebra is continuous, and because $y_{n}→0$ when $∥y∥<1$. This proves that $s$ is the inverse of $1−x$.

The estimate follows from $∥s∥≤∑∥x∥_{n}=1/(1−∥x∥)$. $□$

## The Spectrum

Definition (Spectrum, Resolvent Set)

Suppose $x$ is an element of a unital Banach algebra $A$.

- The
spectrumof $x$ is the set $σ(x)={λ∈C:x−λ$ is not invertible in $A}$.

The elements of $σ(x)$ are calledspectral valuesof $x$.- The
resolvent setof $x$ is the set $ρ(x)=C∖σ(x)$.

For $λ∈ρ(x)$ theresolventof $x$ is the algebra element $R_{λ}=(λ−x)_{−1}$.

The mapping $R:ρ(x)→A$, $λ↦R_{λ}$, is calledresolvent map.

Suppose $x$ is an element of a unital Banach algebra $A$. If $λ$ lies in the resolvent set of $x$, then so do all complex numbers $μ$ with the property that

$∣λ−μ∣<∥(λ−x)_{−1}∥1 .(∗)$For such $μ$ the resolvent of $x$ is represented by the absolutely convergent power series

$(μ−x)_{−1}=n=0∑∞ (μ−λ)_{n}(λ−x)_{−(n+1)}.$

Proof Let $λ$ be in the resolvent set of $x$. Then $λ−x$ is invertible, and we have for all $μ∈C$

$μ−x=(1−(λ−μ)(λ−x)_{−1})(λ−x).$If $μ$ satisfies condition ($∗$), the first factor is invertible and the inverse is given by a Neumann series:

$(1−(λ−μ)(λ−x)_{−1})_{−1}=n=0∑∞ (λ−μ)_{n}(λ−x)_{−n}.$As a product of invertible algebra elements, $μ−x$ must itself be invertible; the claimed formula for its inverse follows by an application of the rule $(ab)_{−1}=b_{−1}a_{−1}$ for invertible $a,b∈A$. $□$

The resolvent set $ρ(x)$ is open and the spectrum $σ(x)$ is closed.

Suppose $x$ is an element of a unital Banach algebra $A$. The resolvent map

$R:ρ(x)⟶A,λ⟼R_{λ}=(λ−x)_{−1},$is (strongly) analytic.

Suppose $x$ is an element of a unital Banach algebra. Then its spectrum $σ(x)$ is not empty.

Proof We assume that $σ(x)$ is empty and derive a contradiction. Observe that the resolvent map $R$ is defined on the whole complex plane. By this corollary, $R$ is analytic, hence entire. Analytic functions are continuous; therefore $R$ is bounded on the compact disk $∣λ∣≤2∥x∥$. For $∣λ∣>2∥x∥$ we may expand $R_{λ}$ into a Neumann series,

$R_{λ}=(λ−x)_{−1}=λ_{−1}(1−λ_{−1}x)_{−1}=λ_{−1}n=0∑∞ (λ_{−1}x)_{n},$and make the estimate

$∥R_{λ}∥≤∣λ∣_{−1}(1−∥λ_{−1}x∥)_{−1}=(∣λ∣−∥x∥)_{−1}<∥x∥_{−1}.$This shows that $R$ is a bounded entire function. Now Liouville’s Theorem (for vector-valued functions) implies that $R$ is constant. This is contradictory because XXX $□$

Gelfand–Mazur Theorem

Every Banach algebra in which all nonzero elements are invertible is isometrically isomorphic to $C$.

Proof For any Banach algebra $A$, the mapping $φ:C→A$, $λ↦λ1$, is linear, multiplicative and isometric, hence injective. Let $x$ be any element of $A$. Since its spectrum is not empty, there must exist a complex number $λ$ such that $x−λ1$ is not invertible. Now suppose that all nonzero elements of $A$ are invertible. Then necessarily $x−λ1=0$, or $x=λ1$. This proves that the mapping $φ$ is also surjective and thus an isometric isomorphism. $□$

Other ways of stating that all nonzero elements of a Banach algebra $A$ are invertible include:

- $A$ is a division algebra.
- The underlying ring of $A$ is a field.

Spectral Radius Formula

For every Banach algebra element $x$ the spectral radius is given by

$r(x)=n→∞lim ∥x_{n}∥_{1/n}.$

## Gelfand’s Theory

Let $A$ be a unital commutative Banach algebra. If $ϕ$ is a nonzero multiplicative linear functional on $A$, then its kernel $kerϕ$ is a maximal ideal in $A$. Every maximal ideal $I$ in $A$ is of the form $I=kerϕ$ for some nonzero multiplicative linear functional $ϕ$ on $A$.

In other words, the mapping $ϕ↦kerϕ$ gives a bijection between the sets of nonzero multiplicative linear functionals and maximal ideals.

The

Gelfand space$Γ_{A}$ of a unital commutative Banach algebra $A$ is the set of maximal ideals of $A$; its topology is inherited from the weak* topology on the dual of $A$ via the correspondence described above.

The

maximal ideal space$M_{A}$ of a unital commutative Banach algebra $A$ is the set of maximal ideals of $A$; its topology is inherited from the weak* topology on the dual of $A$ via the correspondence described above.

The

Gelfand spaceof a unital commutative Banach algebra is a compact Hausdorff space.