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+---
+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 %}
+
+