path: root/pages/operator-algebras/banach-algebras/index.md

---
title: Banach Algebras
parent: Operator Algebras
nav_order: 1
has_children: true
has_toc: false
---

# {{ page.title }}

{% definition Banach Algebra %}
A *Banach algebra* $\mathcal{A}$ is a complex Banach space
endowed with a binary operation $(x,y) \mapsto xy$, called *product*,
that makes the underlying vector space into an associative algebra,
and that satisfies

$$
\norm{xy} \le \norm{x} \norm{y} \quad \forall x,y \in \mathcal{A}.
$$
{% enddefinition %}

The algebraic properties required of the product are explicitly:

$$
\begin{align*}
x(y+y') &= xy + xy' &\quad
(\lambda x)y &= \lambda (xy) &\quad
(xy)z &= x(yz) \\
(x+x')y &= xy + x'y &
x(\lambda y) &= \lambda (xy)
\end{align*}
$$

The topological property is sometimes described by saying
that the norm is *submultiplicative*.

{% definition Commutative Banach Algebra %}
A Banach algebra $\mathcal{A}$ is said to be *commutative* (or *abelian*) if
$xy = yx$ holds for all $x,y \in \mathcal{A}$.
{% enddefinition %}

{% definition Unital Banach Algebra %}
An element $e$ of a Banach algebra $\mathcal{A}$ is called a *unit* (or an *identity*),
if $\norm{e} = 1$ and $ex=x=xe$ for all $x \in \mathcal{A}$.
We say that $\mathcal{A}$ is an *unital* Banach algebra, if $\mathcal{A}$ contains a unit.
{% enddefinition %}

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

{: .proposition-title #neumann-series }
> Proposition (Neumann Series)
>
> Let $\mathcal{A}$ be a unital Banach algebra
> and let $x \in \mathcal{A}$ satisfy $\norm{x} < 1$.
> Then $\mathbf{1}-x$ is invertible
> and the inverse is given by the series
> 
> $$
> (\mathbf{1}-x)^{-1} = \sum_{n=0}^{\infty} x^n,
> $$
>
> which converges absolutely in norm.
> Moreover, we have the estimate
> 
> $$
> \norm{(\mathbf{1}-x)^{-1}} \le \frac{1}{1 - \norm{x}}.
> $$
> {: .katex-display .mb-0 }

{% proof %}
Since the Banach algebra norm is submultiplicative,
we have $\norm{x^n} \le \norm{x}^n$ for all $n \in \NN$.
This implies that the series $\sum \norm{x^n}$
is majorized by the geometric series $\sum \norm{x}^n$,
which is known to be convergent for $\norm{x} < 1$.
It follows that the series $\sum x^n$ is absolutely convergent.
Denote its limit by $s = \lim_{n \to \infty} s_n = \sum_{n=0}^{\infty} x$,
where $s_n = \mathbf{1} + x + \cdots + x^n$ is the $n$th partial sum.
Clearly,

$$
(\mathbf{1}-x) s_n = s_n (\mathbf{1}-x) = \mathbf{1} - x^{n+1}.
$$

In the limit $n \to \infty$ we obtain $(\mathbf{1}-x) s = s (\mathbf{1}-x) = \mathbf{1}$, 
because multiplication in a Banach algebra is continuous, and because $y^n \to 0$ when $\norm{y} < 1$.
This proves that $s$ is the inverse of $\mathbf{1}-x$.

The estimate follows from $\norm{s} \le \sum \norm{x}^n = 1 / (1 - \norm{x})$.
{% endproof %}

## The Spectrum

{: .definition-title }
> Definition (Spectrum, Resolvent Set)
>
> Suppose $x$ is an element of a unital Banach algebra $\mathcal{A}$.
> {: .mb-0 }
> 
> {: .my-0 }
> - The *spectrum* of $x$ is the set $\sigma(x) = \lbrace\lambda \in \CC : x - \lambda$ is not invertible in $\mathcal{A}\rbrace$. \
>   The elements of $\sigma(x)$ are called *spectral values* of $x$.
> - The *resolvent set* of $x$ is the set $\rho (x) = \CC \setminus \sigma(x)$. \
>   For $\lambda \in \rho(x)$ the *resolvent* of $x$ is the algebra element $R_{\lambda} = (\lambda - x)^{-1}$. \
>   The mapping $R : \rho(x) \to \mathcal{A}$, $\lambda \mapsto R_{\lambda}$, is called *resolvent map*.

{% theorem %}
Suppose $x$ is an element of a unital Banach algebra $\mathcal{A}$.
If $\lambda$ lies in the resolvent set of $x$,
then so do all complex numbers $\mu$ with the property that

$$
\abs{\lambda - \mu} < \frac{1}{\norm{(\lambda - x)^{-1}}}. \tag{$*$}
$$

For such $\mu$ the resolvent of $x$ is represented by the absolutely convergent power series

$$
(\mu - x)^{-1} = \sum_{n=0}^{\infty} (\mu - \lambda)^n (\lambda - x)^{-(n+1)}.
$$
{% endtheorem %}

{% proof %}
Let $\lambda$ be in the resolvent set of $x$.
Then $\lambda - x$ is invertible and we have for all $\mu \in \CC$ 

$$
\mu - x = \bigl(\mathbf{1} - (\lambda - \mu) (\lambda - x)^{-1}\bigr) (\lambda - x).
$$

If $\mu$ satisfies condition ($*$), the first factor is invertible
and the inverse is given by a [Neumann series](#neumann-series):

$$
\bigl(\mathbf{1} - (\lambda - \mu) (\lambda - x)^{-1}\bigr)^{-1}
= \sum_{n=0}^{\infty} (\lambda - \mu)^n (\lambda - x)^{-n}.
$$

As a product of invertible algebra elements, $\mu - x$ 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 $a,b \in \mathcal{A}$.
{% endproof %}

{: .corollary #resolvent-set-is-open #spectrum-is-closed }
> The resolvent set $\rho(x)$ is open and the spectrum $\sigma(x)$ is closed.

{: .corollary #resolvent-map-is-analytic }
> Suppose $x$ is an element of a unital Banach algebra $\mathcal{A}$.
> The resolvent map
>
> $$
> R : \rho(x) \longrightarrow \mathcal{A}, \quad \lambda \longmapsto R_{\lambda} = (\lambda - x)^{-1},
> $$
>
> is (strongly) analytic.

 ---

{: .proposition #spectrum-is-not-empty }
> Suppose $x$ is an element of a unital Banach algebra.
> Then its spectrum $\sigma(x)$ is not empty.

{% proof %}
We assume that $\sigma(x)$ is empty
and derive a contradiction.
Observe that the resolvent map $R$ is defined on the whole complex plane.
By [this corollary](#resolvent-map-is-analytic), $R$ is analytic, hence entire.
Analytic functions are countinuous;
therefore $R$ is bounded on the compact disk $\abs{\lambda} \le 2 \norm{x}$.
For $\abs{\lambda} > 2 \norm{x}$ we may expand $R_{\lambda}$ into a [Neumann series](#neumann-series),

$$
R_{\lambda}
= (\lambda - x)^{-1}
= \lambda^{-1} (\mathbf{1} - \lambda^{-1} x)^{-1} 
= \lambda^{-1} \sum_{n=0}^{\infty} (\lambda^{-1} x)^n,
$$

and make the estimate

$$
\norm{R_{\lambda}}
\le \abs{\lambda}^{-1} (1 - \norm{\lambda^{-1} x})^{-1} 
= (\abs{\lambda} - \norm{x})^{-1}
< \norm{x}^{-1}.
$$

This shows that $R$ is a bounded entire function. Now
[Liouville's Theorem](/pages/complex-analysis/one-complex-variable/cauchys-integral-formula.html#liouvilles-theorem)
(for vector-valued functions) implies that $R$ is constant.
This is contradictiory because XXX
{% endproof %}

{: .theorem-title }
> Gelfand–Mazur Theorem
>
> Every Banach algebra in which all nonzero elements are invertible is isometrically isomorphic to $\CC$.

{% proof %}
For any Banach algebra $A$,
the mapping $\varphi : \CC \to A$, $\lambda \mapsto \lambda \mathbf{1}$,
is linear, multiplicative and isometric, hence injective.
Let $x$ be any element of $A$.
Since its 
[spectrum is not empty](/pages/operator-algebras/banach-algebras/index.html#spectrum-is-not-empty),
there must exist a complex number $\lambda$
such that $x - \lambda \mathbf{1}$ is not invertible.
Now suppose that all nonzero elements of $A$ are invertible.
Then necessarily $x - \lambda \mathbf{1} = 0$, or $x = \lambda \mathbf{1}$.
This proves that the mapping $\varphi$ is also surjective
and thus an isometric isomorphism.
{% endproof %}

Other ways of stating that
all nonzero elements of a Banach algebra $\mathcal{A}$ are invertible
include:
{: .mb-0 }

{: .mt-0 }
- $\mathcal{A}$ is a division algebra.
- The underlying ring of $\mathcal{A}$ is a field.

{: .theorem-title }
> Spectral Radius Formula
>
> For every Banach algebra element $x$ the spectral radius is given by
> 
> $$
> r(x) = \lim_{n \to \infty} \norm{x^n}^{1/n}.
> $$
> {: .katex-display .mb-0 }

## Gelfand’s Theory

Proposition
Let $\mathcal{A}$ be a unital commutative Banach algebra.
If $\phi$ is a nonzero multiplicative linear functional on $\mathcal{A}$,
then its kernel $\ker \phi$ is a maximal ideal in $\mathcal{A}$.
Every maximal ideal $\mathcal{I}$ in $\mathcal{A}$ is of the form
$I = \ker \phi$ for some nonzero multiplicative linear functional $\phi$ on $\mathcal{A}$.

In other words, the mapping $\phi \mapsto \ker \phi$ is gives a bijection
between the sets of nonzero multiplicative linear functionals and maximal ideals.


Definition
The *maximal ideal space* $\mathcal{M}_{\mathcal{A}}$ of a unital commutative Banach algebra $\mathcal{A}$
is the set of maximal ideals of $\mathcal{A}$; its topology is inherited from
the weak* topology on the dual of $\mathcal{A}$ via the correspondece described above.

Proposition
The *maximal ideal space* of a unital commutative Banach algebra is a compact Hausdorff space.

{% definition bla, blubb %}
a
b
{% enddefinition %}