Banach Algebras

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

$$\Vert xy\Vert \le \Vert x\Vert \Vert y\Vert \forall x,y\in \mathcal{A}.$$

The algebraic properties required of the product are explicitly:

$$\begin{array}{rlrlrl}x(y+y^{\prime })& =xy+x{y}^{\prime }& (\lambda x)y& =\lambda (xy)& (xy)z& =x(yz)\\ (x+x^{\prime })y& =xy+x^{\prime }y& x(\lambda y)& =\lambda (xy)& & \end{array}$$

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 $x,y\in \mathcal{A}$.

Definition (Unital Banach Algebra)

An element $e$ of a Banach algebra $\mathcal{A}$ is called a unit (or an identity), if $\Vert e\Vert =1$ 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.

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

Proposition (Neumann Series)

Let $\mathcal{A}$ be a unital Banach algebra and let $x\in \mathcal{A}$ satisfy $\Vert x\Vert <1$. Then $1-x$ is invertible, and the inverse is given by the series

$$(1-x{)}^{-1}=\sum _{n=0}^{\infty }{x}^n,$$

which converges absolutely in norm. Moreover, we have the estimate

$$\Vert (1-x{)}^{-1}\Vert \le \frac{1}{1-\Vert x\Vert }.$$

Proof   Since the Banach algebra norm is submultiplicative, we have $\Vert x^n\Vert \le \Vert x{\Vert }^n$ for all $n\in \mathbb{N}$. This implies that the series $\sum \Vert x^n\Vert$ is majorized by the geometric series $\sum \Vert x{\Vert }^n$, which is known to be convergent for $\Vert x\Vert <1$. 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 $s_n=1+x+\cdots +x^n$ is the $n$th partial sum. Clearly,

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

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

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

The Spectrum

Definition (Spectrum, Resolvent Set)

Suppose $x$ is an element of a unital Banach algebra $\mathcal{A}$.

• The spectrum of $x$ is the set $\sigma (x)=\{\lambda \in \mathbb{C}:x-\lambda$ is not invertible in $\mathcal{A}\}$.
  The elements of $\sigma (x)$ are called spectral values of $x$.
• The resolvent set of $x$ is the set $\rho (x)=\mathbb{C}\setminus \sigma (x)$.
  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.

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

$$|\lambda -\mu |<\frac{1}{\Vert (\lambda -x{)}^{-1}\Vert }.\text{\hspace{1em}(∗)}$$

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

$$(\mu -x{)}^{-1}=\sum _{n=0}^{\infty }(\mu -\lambda {)}^n(\lambda -x{)}^{-(n+1)}.$$

Proof   Let $\lambda$ be in the resolvent set of $x$. Then $\lambda -x$ is invertible, and we have for all $\mu \in \mathbb{C}$

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

If $\mu$ satisfies condition ($\ast$), the first factor is invertible and the inverse is given by a Neumann series:

$$(1-(\lambda -\mu )(\lambda -x{)}^{-1}{)}^{-1}=\sum _{n=0}^{\infty }(\lambda -\mu {)}^n(\lambda -x{)}^{-n}.$$

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 $a,b\in \mathcal{A}$. $\square$

The resolvent set $\rho (x)$ is open and the spectrum $\sigma (x)$ is closed.

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

$$R:\rho (x)⟶\mathcal{A},\lambda ⟼R_{\lambda }=(\lambda -x{)}^{-1},$$

is (strongly) analytic.

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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, $R$ is analytic, hence entire. Analytic functions are continuous; therefore $R$ is bounded on the compact disk $|\lambda |\le 2\Vert x\Vert$. For $|\lambda |>2\Vert x\Vert$ we may expand $R_{\lambda }$ into a Neumann series,

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

and make the estimate

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

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 $\square$

Gelfand–Mazur Theorem

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

Proof   For any Banach algebra $A$, the mapping $\phi :\mathbb{C}\to A$, $\lambda \mapsto \lambda 1$, 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 $\lambda$ such that $x-\lambda 1$ is not invertible. Now suppose that all nonzero elements of $A$ are invertible. Then necessarily $x-\lambda 1=0$, or $x=\lambda 1$. This proves that the mapping $\phi$ is also surjective and thus an isometric isomorphism. $\square$

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

• $\mathcal{A}$ is a division algebra.
• The underlying ring of $\mathcal{A}$ is a field.

Spectral Radius Formula

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

$$r(x)=\underset{n\to \infty }{\lim }\Vert x^n{\Vert }^{1/n}.$$

Gelfand’s Theory

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

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

The Gelfand space ${\Gamma }_{\mathcal{A}}$ 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 correspondence described above.

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 correspondence described above.

The Gelfand space of a unital commutative Banach algebra is a compact Hausdorff space.