jerrylicious/garden
1
0
Fork 0
Code Issues Pull Requests Actions Packages Projects Releases Wiki Activity

Reworked the order chapter.

Browse Source
This commit is contained in:
Bokuan Li
2026-07-07 14:40:27 -04:00
parent f4e5004e8c
commit f5ebcd7979
5 changed files with 132 additions and 65 deletions
Download Patch File Download Diff File Expand all files Collapse all files

44
src/fa/tvs/complexify.tex
View File

@@ -119,8 +119,7 @@
The pair $(\complex(E), \iota)$ is the \textbf{complexification} of $E$ as a topological vector space, and
\begin{enumerate}[start=4]
\item If $E$ is locally convex, then so is $\complex(E)$.
\item If $E$ is normed, then $\complex(E)$ is normable, and there exists a norm $\norm{\cdot}_{\complex(E)}: \complex(E) \to [0, \infty)$ such that $\iota: E \to \complex(E)$ is isometric.
\item[(F)] For any vector space $F$ over $\real$ and continuous $\real$-linear map $T: E \to F$, there exists a unique continuous $\complex$-linear map $\complex(T): \complex(E) \to \complex(F)$ such that the following diagram commutes:
\item[(F)] For any topological vector space $F$ over $\real$ and continuous $\real$-linear map $T: E \to F$, there exists a unique continuous $\complex$-linear map $\complex(T): \complex(E) \to \complex(F)$ such that the following diagram commutes:
\[
\xymatrix{
\mathbb{C}(E) \ar@{->}[r]^{\mathbb{C}(T)} & \mathbb{C}(F) \\
@@ -132,6 +131,8 @@
\[
\complex(T)(x + iy) = Tx + iTy
\]
Moreover, if $E$ and $F$ are normed, then $\norm{\complex(T)}_{L(\complex(E); \complex(F))} = \norm{T}_{L(E; F)}$.
\end{enumerate}
\end{definition}
@@ -141,25 +142,56 @@
(U): By (U) of the \hyperref[complexification]{definition:complexification}, there exists a $\complex$-linear map $\complex(T): \complex(E) \to F$ such that the given diagram commutes. Since $T \circ \iota$ and $iT \circ \iota$ are continuous, $T$ is continuous by (U) of the \hyperref[direct sum]{definition:tvs-direct-sum}.
(4): By \autoref{proposition:finite-lc-product}, the direct sum and product of finitely many locally convex spaces coincide. By \autoref{proposition:lc-projective-topology}, this topology is locally convex.
(5): Let $\norm{\cdot}_E: E \to [0, \infty)$ be the norm of $E$, and define
(F): Existence of $\complex(T)$ is given by (U) applied to $\iota \circ T$.
For the isometry,
\begin{align*}
\norm{\complex(T)}_{L(\complex(E); \complex(F))} &= \norm{\complex(T)}
\end{align*}
\end{proof}
\begin{definition}[Complexification of Normed Spaces]
\label{definition:complexification-of-normed-spaces}
Let $E$ be a normed vector space over $\real$, then
\[
\norm{\cdot}_{\complex(E)}: \complex(E) \to [0, \infty) \quad (x, y) \mapsto \sup_{\theta \in [0, 2\pi]}\norm{\cos(\theta)x + \sin(\theta)y}_E
\]
then for any $\phi \in [0, 2\pi]$ and $x, y \in E$,
is a norm on $\complex(E)$ such that the inclusion map $\iota: E \to \complex(E)$ is isometric.
Moreover, for any normed vector space $F$ over $\real$ and $T \in L(E; F)$, $\norm{\complex(T)}_{L(\complex(E); \complex(F))} = \norm{T}_{L(E; F)}$.
\end{definition}
\begin{proof}
For any $\phi \in [0, 2\pi]$ and $x, y \in E$,
\begin{align*}
\normn{e^{i \phi}(x, y)}_{\complex(E)} &= \normn{(\cos(\phi)x - \sin(\phi)y, \sin(\phi)x + \cos(\phi)y)}_{\complex(E)} \\
&= \sup_{\theta \in [0, 2\pi]}\norm{\cos(\theta - \phi)x + \sin(\theta - \phi)y}_E \\
&= \norm{(x, y)}_{\complex(E)}
\end{align*}
so $\norm{(x, y)}_{\complex(E)}$ is a norm. For any $x \in E$,
so $\norm{\cdot}_{\complex(E)}$ is a norm on $\complex(E)$. For any $x \in E$,
\[
\norm{\iota x}_{\complex(E)} = \sup_{\theta \in [0, 2\pi]}\norm{\cos(\theta)x}_E = \norm{x}_E \\
\]
Therefore $\iota: E \to \complex(E)$ is isometric.
(F): By (U) applied to $\iota \circ T$.
Now, let $F$ be a normed vector space over $\real$ and $T \in L(E; F)$, then
\begin{align*}
\norm{\complex(T)}_{L(\complex(E); \complex(F))} &= \sup\bracsn{\norm{\complex(T)(x,y)}_{\complex(F)}|(x, y) \in \complex(E), \norm{(x, y)}_{\complex(E)} = 1} \\
&\ge \sup\bracsn{\norm{Tx}|x \in E, \norm{x}_E = 1} = \norm{T}_{L(E; F)}
\end{align*}
On the other hand, let $(x, y) \in \complex(E)$, then there exists $\theta \in [0, 2\pi]$ such that
\begin{align*}
\normn{\complex(T)(x, y)}_{\complex(F)} &= \normn{(Tx, Ty)}_{\complex(F)} = \norm{\cos(\theta)Tx + \sin(\theta)Ty}_F \\
&\le \norm{T}_{L(E; F)} \cdot \norm{\cos(\theta)x + \sin(\theta)y}_E \\
&\le \norm{T}_{L(E; F)} \cdot \sup_{\phi \in [0, 2\pi]}\norm{\cos(\phi)x + \sin(\phi)y}_E \\
&= \norm{T}_{L(E; F)} \cdot \norm{(x, y)}_E
\end{align*}
\end{proof}