% % satzvongelfand.tex % % (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule % \bgroup \begin{frame}[t] \setlength{\abovedisplayskip}{0pt} \setlength{\belowdisplayskip}{0pt} \setbeamercolor{block body}{bg=blue!20} \setbeamercolor{block title}{bg=blue!20} \frametitle{Satz von Gelfand} {\usebeamercolor[fg]{title}Behauptung:} $\varrho(A)=\pi(A)$\uncover<2->{, $A(\varepsilon) = \displaystyle\frac{A}{\varrho(A)+\varepsilon}$}\uncover<3->{, $\varrho(A(\varepsilon))=\displaystyle\frac{\varrho(A)}{\varrho(A)+\varepsilon} \uncover<4->{=\frac{1}{1+\varepsilon/\varrho(A)}}$} \uncover<5->{% %{\usebeamercolor[fg]{title}Beweisidee:} %$\displaystyle\pi\biggl(\frac{A}{\varrho(A)+\epsilon}\biggr) %= %\frac{\pi(A)}{\varrho(A)+\epsilon}$ berechnen \vspace{-5pt} \begin{columns}[t,onlytextwidth] \begin{column}{0.48\textwidth} \begin{block}{$\varepsilon < 0$} \vspace{-10pt} \begin{align*} \uncover<6->{ \varrho(A(\varepsilon))&>1}\uncover<7->{\quad\Rightarrow\quad \|A(\varepsilon)^k\|\to \infty} \\ \uncover<8->{\|A(\varepsilon)^k\| &\ge m\varrho(A(\varepsilon))^k} \\ \uncover<9->{\|A(\varepsilon)^k\|^{\frac1k} &\ge m^{\frac1k} \varrho(A(\varepsilon))} \\ \uncover<10->{\pi(A) &\ge \lim_{k\to\infty}m^{\frac1k}\varrho(A(\varepsilon))} \\ &\uncover<11->{= \varrho(A(\varepsilon))}\uncover<12->{ > 1} \\ \uncover<13->{\frac{ \pi(A(\varepsilon))}{\varrho(A)+\varepsilon} &> 1} \\ \uncover<14->{ \pi(A) &> \varrho(A)+\varepsilon } \end{align*} \end{block} \end{column} \begin{column}{0.48\textwidth} \begin{block}{$\varepsilon > 0$} \vspace{-10pt} \begin{align*} \uncover<16->{ \varrho(A(\varepsilon)) &<1} \uncover<17->{\quad\Rightarrow\quad \|A(\varepsilon)^k\| \to 0} \\ \uncover<18->{\|A(\varepsilon)^k\| &\le M\varrho(A(\varepsilon))^k} \\ \uncover<19->{ \|A(\varepsilon)^k\|^{\frac1k} &\le M^{\frac1k}\varrho(A(\varepsilon)) } \\ \uncover<20->{ \pi(A(\varepsilon)) &\le \varrho(A(\varepsilon)) \lim_{k\to\infty} M^{\frac1k} } \\ &\uncover<21->{= \varrho(A(\varepsilon))} \uncover<22->{ < 1} \\ \uncover<23->{\frac{\pi(A)}{\varrho(A)+\varepsilon}&< 1} \\ \uncover<24->{\pi(A)&< \varrho(A) + \varepsilon} \end{align*} \end{block} \end{column} \end{columns}} \uncover<15->{% \vspace{2pt} {\usebeamercolor[fg]{title}Folgerung:} $\varrho(A)-\varepsilon < \pi(A) \uncover<25->{< \varrho(A)+\varepsilon}\quad\forall\varepsilon>0 \uncover<26->{ \qquad\Rightarrow\qquad \varrho(A)=\pi(A)}$ } \end{frame} \egroup