From 777f9d3fd8caf56e6bc6999a4b05379307d0733f Mon Sep 17 00:00:00 2001 From: Justin Gassner Date: Tue, 12 Sep 2023 07:36:33 +0200 Subject: Initial commit --- pages/operator-algebras/banach-algebras/index.md | 259 +++++++++++++++++++++++ 1 file changed, 259 insertions(+) create mode 100644 pages/operator-algebras/banach-algebras/index.md (limited to 'pages/operator-algebras/banach-algebras') diff --git a/pages/operator-algebras/banach-algebras/index.md b/pages/operator-algebras/banach-algebras/index.md new file mode 100644 index 0000000..2fb8f03 --- /dev/null +++ b/pages/operator-algebras/banach-algebras/index.md @@ -0,0 +1,259 @@ +--- +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 %} + + -- cgit v1.2.3-54-g00ecf