From 28407333ffceca9b99fae721c30e8ae146a863da Mon Sep 17 00:00:00 2001 From: Justin Gassner Date: Wed, 14 Feb 2024 07:24:38 +0100 Subject: Update --- pages/operator-algebras/banach-algebras/index.md | 113 +++++++++++------------ 1 file changed, 54 insertions(+), 59 deletions(-) (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 index 2fb8f03..3335d78 100644 --- a/pages/operator-algebras/banach-algebras/index.md +++ b/pages/operator-algebras/banach-algebras/index.md @@ -47,25 +47,23 @@ We say that $\mathcal{A}$ is an *unital* Banach algebra, if $\mathcal{A}$ contai 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 } +{% 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 + +$$ +\norm{(\mathbf{1}-x)^{-1}} \le \frac{1}{1 - \norm{x}}. +$$ +{% endproposition %} {% proof %} Since the Banach algebra norm is submultiplicative, @@ -91,18 +89,17 @@ The estimate follows from $\norm{s} \le \sum \norm{x}^n = 1 / (1 - \norm{x})$. ## 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*. +{% 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*. +{% enddefinition %} {% theorem %} Suppose $x$ is an element of a unital Banach algebra $\mathcal{A}$. @@ -122,7 +119,7 @@ $$ {% proof %} Let $\lambda$ be in the resolvent set of $x$. -Then $\lambda - x$ is invertible and we have for all $\mu \in \CC$ +Then $\lambda - x$ is invertible, and we have for all $\mu \in \CC$ $$ \mu - x = \bigl(\mathbf{1} - (\lambda - \mu) (\lambda - x)^{-1}\bigr) (\lambda - x). @@ -165,7 +162,7 @@ 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; +Analytic functions are continuous; 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), @@ -188,13 +185,12 @@ $$ This shows that $R$ is a bounded entire function. Now [Liouville's Theorem](/pages/complex-analysis/one-complex-variable/cauchys-integral-formula.html#liouvilles-theorem) (for vector-valued functions) implies that $R$ is constant. -This is contradictiory because XXX +This is contradictory because XXX {% endproof %} -{: .theorem-title } -> Gelfand–Mazur Theorem -> -> Every Banach algebra in which all nonzero elements are invertible is isometrically isomorphic to $\CC$. +{% theorem * Gelfand–Mazur Theorem %} +Every Banach algebra in which all nonzero elements are invertible is isometrically isomorphic to $\CC$. +{% endtheorem %} {% proof %} For any Banach algebra $A$, @@ -220,40 +216,39 @@ include: - $\mathcal{A}$ is a division algebra. - The underlying ring of $\mathcal{A}$ is a field. -{: .theorem-title } -> Spectral Radius Formula -> -> For every Banach algebra element $x$ the spectral radius is given by -> -> $$ -> r(x) = \lim_{n \to \infty} \norm{x^n}^{1/n}. -> $$ -> {: .katex-display .mb-0 } +{% theorem * Spectral Radius Formula %} +For every Banach algebra element $x$ the spectral radius is given by + +$$ +r(x) = \lim_{n \to \infty} \norm{x^n}^{1/n}. +$$ +{% endtheorem %} ## Gelfand’s Theory -Proposition +{% 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 $\mathcal{A}$. +{% endproposition %} -In other words, the mapping $\phi \mapsto \ker \phi$ is gives a bijection +In other words, the mapping $\phi \mapsto \ker \phi$ gives a bijection between the sets of nonzero multiplicative linear functionals and maximal ideals. +{% definition %} +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. +{% enddefinition %} -Definition +{% 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 +the weak* topology on the dual of $\mathcal{A}$ via the correspondence described above. {% enddefinition %} - +{% proposition %} +The *Gelfand space* of a unital commutative Banach algebra is a compact Hausdorff space. +{% endproposition %} -- cgit v1.2.3-54-g00ecf