From 1bc1b17feee9c5c50b2caf3b2a72b550a9a29c0b Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Sat, 8 Aug 2026 19:39:18 -0400 Subject: [PATCH] Added elementary properties of adjoint maps. --- src/fa/duality/adjoint.tex | 75 ++++++++++++++++++++++++++++++++++++++ src/fa/duality/index.tex | 1 + src/op/vn/topologies.tex | 2 +- 3 files changed, 77 insertions(+), 1 deletion(-) create mode 100644 src/fa/duality/adjoint.tex diff --git a/src/fa/duality/adjoint.tex b/src/fa/duality/adjoint.tex new file mode 100644 index 0000000..89ff64a --- /dev/null +++ b/src/fa/duality/adjoint.tex @@ -0,0 +1,75 @@ +\section{Adjoint Maps} +\label{section:adjoint-maps} + +\begin{definition}[Adjoint Map] +\label{definition:adjoint-map} + Let $E, F$ be vector spaces over a field $K$, and $T \in \hom(E; F)$ be a linear map, then the mapping + \[ + T^*: F^* \to E^* \quad \dpn{x, T^*\phi}{E} = \dpn{Tx, \phi}{F} + \] + + is the \textbf{algebraic adjoint} of $T$. +\end{definition} + +\begin{proposition} +\label{proposition:adjoint-weak-continuous} + Let $\dpn{E, F}{\lambda}$ and $\dpn{G, H}{\mu}$ be dualities over $K \in \RC$ and $T \in \hom(E; G)$, then the following are equivalent: + \begin{enumerate} + \item $T$ is $\sigma(E, F)$-$\sigma(G, H)$ continuous. + \item $T^*(H) \subset F$. + \end{enumerate} + + If the above holds, then + \begin{enumerate}[start=2] + \item $T^*|_{H}$ is $\sigma(H, G)$-$\sigma(F, E)$ continuous. + \item $T^{**} = T$. + \end{enumerate} + + and the restriction of $T^*$ to $H$ is the \textbf{adjoint} of $T$ with respect to $\dpn{E, F}{\lambda}$ and $\dpn{G, H}{\mu}$. +\end{proposition} +% Proof omitted due to obviousness. + +\begin{proposition} +\label{proposition:adjoint-polar-gymnastics} + Let $\dpn{E, F}{\lambda}$ and $\dpn{G, H}{\mu}$ be dualities over $K \in \RC$, $T: E \to G$ be a $\sigma(E, F)$-$\sigma(G, H)$ continuous linear map, $A \subset E$, and $B \subset G$, then: + \begin{enumerate} + \item $T(A)^\circ = (T^{*})^{-1}(A^\circ)$. + \item If $T(A) \subset B$, then $T^{*}(B^\circ) \subset A^\circ$. + \end{enumerate} +\end{proposition} +\begin{proof} + (1): + \begin{align*} + T(A)^\circ &= \bracsn{\phi \in H| \text{Re}\dpn{Tx, \phi}{\mu} \le 1 \forall x \in A} \\ + &= \bracsn{\phi \in H| \text{Re}\dpn{x, T^*\phi}{\lambda} \le 1 \forall x \in A} = (T^{*})^{-1}(A^\circ) + \end{align*} + + (2): + \begin{align*} + T^*(B^\circ) &= T^*(\bracs{\phi \in H| \text{Re}\dpn{y, \phi}{\mu} \le 1 \forall y \in B}) \\ + &\subset T^*(\bracs{\phi \in H| \text{Re}\dpn{y, \phi}{\mu} \le 1 \forall y \in T(A)}) \\ + &= T^*(\bracs{\phi \in H| \text{Re}\dpn{Tx, \phi}{\mu} \le 1 \forall x \in A})\\ + &= T^*(\bracs{\phi \in H| \text{Re}\dpn{x, T^*\phi}{\lambda} \le 1 \forall x \in A}) \subset A^\circ + \end{align*} +\end{proof} + +\begin{corollary} +\label{corollary:adjoint-kernel-gymnastics} + Let $\dpn{E, F}{\lambda}$ and $\dpn{G, H}{\mu}$ be dualities over $K \in \RC$ and $T: E \to G$ be a $\sigma(E, F)$-$\sigma(G, H)$ continuous linear map, then: + \begin{enumerate} + \item $\ker(T^*) = T(E)^\perp = \bracs{\phi \in H| \dpn{y, \phi}{\mu} = 0 \forall y \in T(E)}$. + \item $T^*$ is injective if and only if $T(E)$ is $\sigma(G, H)$-dense in $G$. + \end{enumerate} +\end{corollary} + +\begin{proposition} +\label{proposition:adjoint-continuity} + Let $\dpn{E, F}{\lambda}$ and $\dpn{G, H}{\mu}$ be dualities over $K \in \RC$, $T: E \to G$ be a $\sigma(E, F)$-$\sigma(G, H)$ continuous linear map, $\sigma \subset 2^E$ be a saturated ideal of $\sigma(E, F)$-bounded sets, $\tau \subset 2^G$ be a saturated ideal of $\sigma(G, H)$-bounded sets, then the following are equivalent: + \begin{enumerate} + \item $T^*$ is continuous with respect to the $\tau$-uniform topology on $H$ and the $\sigma$-uniform topology on $F$. + \item $T(\sigma) \subset \tau$. + \end{enumerate} +\end{proposition} +\begin{proof} + (1) $\Rightarrow$ (2): Let $A \in \sigma$, then there exists $B \in \tau$ such that $T^*\phi(A) \subset \ol{B_K(0, 1)}$ for all $\phi \in H$ with $\phi(B) \subset \ol{B_K(0, 1)}$. In which case, $T^*(B^\circ) \subset A^\circ$. Assume without loss of generality that $A$ and $B$ are convex, circled, and closed. By \autoref{proposition:adjoint-polar-gymnastics} applied to $T^*$ and the \hyperref[Bipolar theorem]{theorem:bipolar}, $T(A) \subset B$. Therefore $T(\sigma) \subset \tau$. +\end{proof} diff --git a/src/fa/duality/index.tex b/src/fa/duality/index.tex index eca8a4f..3f127fa 100644 --- a/src/fa/duality/index.tex +++ b/src/fa/duality/index.tex @@ -4,5 +4,6 @@ \input{./definitions.tex} \input{./polar.tex} \input{./mackey.tex} +\input{./adjoint.tex} diff --git a/src/op/vn/topologies.tex b/src/op/vn/topologies.tex index 2abae7e..7b578b6 100644 --- a/src/op/vn/topologies.tex +++ b/src/op/vn/topologies.tex @@ -13,7 +13,7 @@ Depending on the topology placed on $H \otimes H$, and the corresponding complet Let $H$ be a complex Hilbert space, then the dual of $H \wh \otimes_\pi H$ is $B(H)$, and the $\sigma(B(H), H \wh \otimes_\pi H)$-topology is the \textbf{ultraweak}/\textbf{$\sigma$-weak} topology on $B(H)$. \end{definition} \begin{proof} - By \autoref{proposition:projective-tensor-product-dual}. + By \autoref{proposition:projective-tensor-product-dual} and the \hyperref[Riesz Representation Theorem]{theorem:riesz-hilbert}. \end{proof} Seeing that $B(H)$ is a dual Banach space, the following fact is immediate: