diff --git a/src/op/c-star/state.tex b/src/op/c-star/state.tex index 1ee3f98..82ab7d2 100644 --- a/src/op/c-star/state.tex +++ b/src/op/c-star/state.tex @@ -90,6 +90,34 @@ \end{proof} +\begin{corollary} +\label{corollary:cstar-positive-property-probe} + Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then\footnote{The crude bound seems kind of tragic, but it wouldn't be true otherwise. } + \begin{align*} + \sigma_A(x) &\subset \bracs{\dpn{x, \phi}{A}|\phi \in P(A)} \\ + &\subset \bracs{\dpn{x, \phi}{A}|\phi \in S(A)} = \ol{\text{Conv}}(\sigma_A(x)) + \end{align*} + + + In particular, there exists $\phi \in P(A)$ such that $\norm{x}_A = |\dpn{x, \phi}{A}|$. +\end{corollary} +\begin{proof} + Let $\lambda \in \sigma_A(x)$. By \autoref{proposition:gelfand-transform-gymnastics}, there exists $\phi \in \Omega(A[x])$ such that $\dpn{x, \phi}{A[x]} = \lambda$. By \autoref{proposition:multiplicative-pure-state}, $\phi \in P(A[x])$. The \hyperref[pure state extension theorem]{theorem:cstar-pure-state-extension} implies that there exists $\Phi \in P(A)$ such that $\Phi|_{A[x]} = \phi$. Thus $\Phi$ is a pure state with $\dpn{x, \Phi}{A} = \lambda$, and $ \sigma_A(x) \subset \bracs{\dpn{x, \Phi}{A}|\Phi \in P(A)}$. + + Let $\Phi \in S(A)$ and $\phi = \Phi|_{A[x]}$, then $\phi \in S(A[x])$ as well. By the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark}, the \hyperref[Spectral Theorem]{theorem:spectral-c-star}, and the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon}, $\phi$ takes the form of a Radon probability measure $\mu$ on $\sigma_A(x)$. In which case, + \[ + \dpn{x, \Phi}{A} = \dpn{x, \phi}{A[x]} = \int_{\sigma_A(x)}\lambda \mu(d\lambda) \in \ol{\text{Conv}}(\sigma_A(x)) + \] + + Finally, since $S(A)$ is compact and convex by \autoref{proposition:state-space-compact-convex}, + \begin{align*} + \bracs{\dpn{x, \phi}{A}|\phi \in S(A)} &= \ol{\text{Conv}}(\bracs{\dpn{x, \phi}{A}|\phi \in P(A)}) \\ + &\subset \ol{\text{Conv}}(\sigma_A(x)) + \end{align*} + + by \autoref{proposition:compact-extensions} and \autoref{proposition:closure-of-image}. +\end{proof} + \begin{theorem} \label{theorem:cstar-state-existence} Let $A$ be a unital $C^*$-algebra, $x \in A$, and $\lambda \in \sigma_A(x)$, then there exists $\phi \in S(A)$ such that $\dpn{x, \phi}{A} = \lambda$. @@ -112,8 +140,15 @@ \item For each $x \in A$, $x = 0$ if and only if $\dpn{x, \phi}{A} = 0$ for all $\phi \in P(A)$. \item The linear span of $P(A)$ is weak*-dense in $A^*$. \end{enumerate} + + Moreover, for any $x \in A$, + \begin{enumerate}[start=2] + \item $x$ is self-adjoint if and only if $\dpn{x, \phi}{A} \in \real$ for all $\phi \in P(A)$. + \item $x$ is positive if and only if $\dpn{x, \phi}{A} \ge 0$ for all $\phi \in P(A)$. + \end{enumerate} + \end{corollary} -\begin{proof} +\begin{proof}[Proof, {{\cite[Theorem 13.9]{Zhu}}}. ] (1): Let $x \in A$ such that $\dpn{x, \phi}{A} = 0$ for all $\phi \in P(A)$. First suppose that $x$ is self-adjoint. By \autoref{theorem:cstar-state-existence}, $\sigma_A(x) = \bracs{0}$, and $\norm{x}_A = [x]_{sp} = 0$ by \autoref{theorem:c-star-normal-spectral-radius}. Now suppose that $x$ is arbitrary. In this case, for each $\phi \in P(A)$, @@ -124,40 +159,13 @@ because $\phi$ is Hermitian. Similarly, $\dpn{\text{Im}(x), \phi}{A} = 0$ as well. Thus $\text{Re}(x) = \text{Im}(x) = 0$, and $x = 0$ as well. (2): Since the linear span of $P(A)$ separates points in $A$, it is weak*-dense in $A^*$ by \autoref{lemma:duality-dense}. -\end{proof} - -\begin{corollary} -\label{corollary:cstar-positive-property-probe} - Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then\footnote{The crude bound seems kind of tragic, but it wouldn't be true otherwise. } - \begin{align*} - \sigma_A(x) &\subset \bracs{\dpn{x, \phi}{A}|\phi \in P(A)} \\ - &\subset \bracs{\dpn{x, \phi}{A}|\phi \in S(A)} = \ol{\text{Conv}}(\sigma_A(x)) - \end{align*} - - - In particular, - \begin{enumerate} - \item $x$ is self-adjoint if and only if $\dpn{x, \phi}{A} \in \real$ for all $\phi \in P(A)$. - \item $x$ is positive if and only if $\dpn{x, \phi}{A} \ge 0$ for all $\phi \in P(A)$. - \item There exists $\phi \in P(A)$ such that $\norm{x}_A = |\dpn{x, \phi}{A}|$. - \end{enumerate} -\end{corollary} -\begin{proof} - Let $\lambda \in \sigma_A(x)$. By \autoref{proposition:gelfand-transform-gymnastics}, there exists $\phi \in \Omega(A[x])$ such that $\dpn{x, \phi}{A[x]} = \lambda$. By \autoref{proposition:multiplicative-pure-state}, $\phi \in P(A[x])$. The \hyperref[pure state extension theorem]{theorem:cstar-pure-state-extension} implies that there exists $\Phi \in P(A)$ such that $\Phi|_{A[x]} = \phi$. Thus $\Phi$ is a pure state with $\dpn{x, \Phi}{A} = \lambda$, and $ \sigma_A(x) \subset \bracs{\dpn{x, \Phi}{A}|\Phi \in P(A)}$. - - Let $\Phi \in S(A)$ and $\phi = \Phi|_{A[x]}$, then $\phi \in S(A[x])$ as well. By the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark}, the \hyperref[Spectral Theorem]{theorem:spectral-c-star}, and the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon}, $\phi$ takes the form of a Radon probability measure $\mu$ on $\sigma_A(x)$. In which case, - \[ - \dpn{x, \Phi}{A} = \dpn{x, \phi}{A[x]} = \int_{\sigma_A(x)}\lambda \mu(d\lambda) \in \ol{\text{Conv}}(\sigma_A(x)) - \] - - Finally, since $S(A)$ is compact and convex by \autoref{proposition:state-space-compact-convex}, - \begin{align*} - \bracs{\dpn{x, \phi}{A}|\phi \in S(A)} &= \ol{\text{Conv}}(\bracs{\dpn{x, \phi}{A}|\phi \in P(A)}) \\ - &\subset \ol{\text{Conv}}(\sigma_A(x)) - \end{align*} - - by \autoref{proposition:compact-extensions} and \autoref{proposition:closure-of-image}. - - (1), (2): By \autoref{corollary:spectrum-characterisation-iff}. + + (3): Let $\phi \in P(A)$, then $\phi$ is Hermitian. If $x$ is self-adjoint, then $\dpn{x, \phi}{A} \in \real$. + + On the other hand, if $\dpn{x, \phi}{A} \in \real$, then $\dpn{x, \phi}{A} = \dpn{x^*, \phi}{A}$, and $\dpn{x - x^*, \phi}{A} =0 $. If this holds for all $\phi \in P(A)$, then $x - x^* = 0$ by (1), and $x$ is self-adjoint. + + (4): Let $\phi \in P(A)$, then $\phi$ is positive. Thus if $x$ is positive, $\dpn{x, \phi}{A} \ge 0$. + + On the other hand, if $\dpn{x, \phi}{A} \ge 0$ for all $\phi \in P(A)$, then $x$ is self-adjoint by (3). By \autoref{corollary:cstar-positive-property-probe}, $\sigma_A(x) \subset [0, \infty)$. As such, $x$ is positive by \autoref{corollary:spectrum-characterisation-iff}. \end{proof}