223 lines
9.9 KiB
TeX
223 lines
9.9 KiB
TeX
\section{Complexification}
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\label{section:complexification}
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\begin{definition}[Complexification]
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\label{definition:complexification}
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Let $E$ be a vector space over $\real$, then there exists a pair $(\complex(E), \iota)$ such that:
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\begin{enumerate}
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\item $\complex(E)$ is a vector space over $\complex$.
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\item $\iota: E \to \complex(E)$ is a $\real$-linear map.
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\item[(U)] For any pair $(F, T)$ satisfying (1) and (2), there exists a unique $\complex$-linear map $\complex(T): \complex(E) \to F$ such that the following diagram commutes:
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\[
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\xymatrix{
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\mathbb{C}(E) \ar@{->}[r]^{\mathbb{C}(T)} & F \\
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E \ar@{->}[u]^{\iota} \ar@{->}[ru]_{T} &
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}
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\]
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\item $\complex(E) = \iota(E) \oplus i\iota(E)$ as a vector space over $\real$. For each $z \in \complex(E)$ with $z = x + iy$, $x = \text{Re}(x)$ and $y = \text{Im}(y)$ are the \textbf{real} and \textbf{imaginary parts} of $z$.
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\end{enumerate}
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The pair $(\complex(E), \iota)$ is the \textbf{complexification} of $E$, and
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\begin{enumerate}
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\item[(F)] For any vector space $F$ over $\real$ and $\real$-linear map $T: E \to F$, there exists a unique $\complex$-linear map $\complex(T): \complex(E) \to \complex(F)$ such that the following diagram commutes:
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\[
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\xymatrix{
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\mathbb{C}(E) \ar@{->}[r]^{\mathbb{C}(T)} & \mathbb{C}(F) \\
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E \ar@{->}[u]^{\iota} \ar@{->}[r]_{T} & F \ar@{->}[u]_{\iota}
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}
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\]
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which is given by
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\[
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\complex(T)(x + iy) = Tx + iTy
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\]
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\end{enumerate}
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\end{definition}
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\begin{proof}
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(1): Let $\complex(E) = E \times E$ with coordinate-wise addition. For each $a, b \in \real$ and $x, y \in E$, let
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\[
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(a + bi)(x, y) = (ax - by, bx + ay)
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\]
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then $\complex(E)$ is a vector space over $\complex$.
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(2): Let $\iota: E \to \complex(E)$ be defined by $\iota(x) = (x, 0)$, then $\iota$ is $\real$-linear.
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(U): Let
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\[
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\complex(T): \complex(E) \to F \quad (x, y) \mapsto Tx + iTy
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\]
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then $\complex(T)$ is the unique $\complex$-linear map such that the given diagram commutes.
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(F): By (U) applied to $\iota \circ T$.
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\end{proof}
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\begin{definition}[Complex Conjugation]
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\label{definition:complex-conjugation}
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Let $E$ be a vector space over $\complex$ and $*: E \to E$ be a $\real$-linear map, then $*$ is a \textbf{complex conjugation} if:
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\begin{enumerate}
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\item[(C1)] For each $\lambda \in \complex$, $(\lambda x)^* = \ol \lambda x^*$.
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\item[(C2)] For each $x \in E$, $x^{**} = x$.
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\end{enumerate}
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In which case, $\text{Re}(E) = \bracs{x \in E| x^* = x}$ is the \textbf{real part} of $E$.
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\end{definition}
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\begin{proposition}
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\label{proposition:complex-conjugation-properties}
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Let $E$ be a vector space over $\complex$ and $*: E \to E$ be a complex conjugation, then:
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\begin{enumerate}
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\item $E = \complex(\text{Re}(E))$.
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\item For each $x \in E$,
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\[
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\text{Re}(x) = \frac{x + x^*}{2} \quad \text{Im}(x) = \frac{x - x^*}{2i}
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\]
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\item For each $x \in E$, $x^* = \text{Re}(x) - i\text{Im}(x)$.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(2): By properties of the complex conjugation, $\text{Re}(x), \text{Im}(x) \in \text{Re}(E)$.
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(1): For any $x, y \in \text{Re}(x)$ with $x = iy$, $x = -iy$ as well by (2) of the complex conjugation, so $x = y = 0$. Thus if $z = x + iy = x' + iy'$, then $x = x'$ and $y = y'$, and the decomposition is unique.
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\end{proof}
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\begin{definition}[Hermitian]
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\label{definition:hermitian-functional}
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Let $E$ be a vector space over $\real$, $*: \complex(E) \to \complex(E)$ be the canonical complex conjugation map, and $\phi \in \hom(\complex(E); \complex)$, then the following are equivalent:
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\begin{enumerate}
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\item $\phi|_E \in \hom(E; \real)$.
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\item For each $x \in E$, $\dpn{x, \phi}{\complex(E)} = \ol{\dpn{x^*, \phi}{\complex(E)}}$.
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\end{enumerate}
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If the above holds, then $\phi$ is \textbf{Hermitian}.
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\end{definition}
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\begin{definition}[Complexification of Topological Vector Space]
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\label{definition:complexification-tvs}
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Let $E$ be a TVS over $\real$, then there exists a pair $(\complex(E), \iota)$ such that:
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\begin{enumerate}
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\item $\complex(E)$ is a TVS over $\complex$.
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\item $\iota: E \to \complex(E)$ is a continuous $\real$-linear map.
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\item[(U)] For any pair $(F, T)$ satisfying (1) and (2), there exists a unique continuous $\complex$-linear map $\complex(T): \complex(E) \to F$ such that the following diagram commutes:
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\[
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\xymatrix{
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\mathbb{C}(E) \ar@{->}[r]^{\mathbb{C}(T)} & F \\
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E \ar@{->}[u]^{\iota} \ar@{->}[ru]_{T} &
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}
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\]
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\item $\complex(E) = \iota(E) \oplus i\iota(E)$ as a TVS over $\real$.
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\end{enumerate}
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The pair $(\complex(E), \iota)$ is the \textbf{complexification} of $E$ as a topological vector space, and
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\begin{enumerate}[start=4]
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\item If $E$ is locally convex, then so is $\complex(E)$.
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\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:
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\[
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\xymatrix{
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\mathbb{C}(E) \ar@{->}[r]^{\mathbb{C}(T)} & \mathbb{C}(F) \\
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E \ar@{->}[u]^{\iota} \ar@{->}[r]_{T} & F \ar@{->}[u]_{\iota}
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}
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\]
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which is given by
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\[
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\complex(T)(x + iy) = Tx + iTy
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\]
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Moreover, if $E$ and $F$ are normed, then $\norm{\complex(T)}_{L(\complex(E); \complex(F))} = \norm{T}_{L(E; F)}$.
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\end{enumerate}
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\end{definition}
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\begin{proof}
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(1), (2): Let $(\complex(E), \iota)$ be the complexification of $E$ as a vector space, and equip it with the \hyperref[direct sum]{definition:tvs-direct-sum} topology.
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(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}.
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(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.
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(F): Existence of $\complex(T)$ is given by (U) applied to $\iota \circ T$.
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For the isometry,
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\begin{align*}
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\norm{\complex(T)}_{L(\complex(E); \complex(F))} &= \norm{\complex(T)}
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\end{align*}
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\end{proof}
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\begin{definition}[Complexification of Normed Spaces]
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\label{definition:complexification-of-normed-spaces}
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Let $E$ be a normed vector space over $\real$, then
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\[
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\norm{\cdot}_{\complex(E)}: \complex(E) \to [0, \infty) \quad (x, y) \mapsto \sup_{\theta \in [0, 2\pi]}\norm{\cos(\theta)x + \sin(\theta)y}_E
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\]
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is a norm on $\complex(E)$ such that the inclusion map $\iota: E \to \complex(E)$ is isometric.
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Moreover, for any normed vector space $F$ over $\real$ and $T \in L(E; F)$, $\norm{\complex(T)}_{L(\complex(E); \complex(F))} = \norm{T}_{L(E; F)}$.
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\end{definition}
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\begin{proof}
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For any $\phi \in [0, 2\pi]$ and $x, y \in E$,
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\begin{align*}
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\normn{e^{i \phi}(x, y)}_{\complex(E)} &= \normn{(\cos(\phi)x - \sin(\phi)y, \sin(\phi)x + \cos(\phi)y)}_{\complex(E)} \\
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&= \sup_{\theta \in [0, 2\pi]}\norm{\cos(\theta - \phi)x + \sin(\theta - \phi)y}_E \\
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&= \norm{(x, y)}_{\complex(E)}
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\end{align*}
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so $\norm{\cdot}_{\complex(E)}$ is a norm on $\complex(E)$. For any $x \in E$,
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\[
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\norm{\iota x}_{\complex(E)} = \sup_{\theta \in [0, 2\pi]}\norm{\cos(\theta)x}_E = \norm{x}_E \\
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\]
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Therefore $\iota: E \to \complex(E)$ is isometric.
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Now, let $F$ be a normed vector space over $\real$ and $T \in L(E; F)$, then
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\begin{align*}
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\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} \\
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&\ge \sup\bracsn{\norm{Tx}|x \in E, \norm{x}_E = 1} = \norm{T}_{L(E; F)}
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\end{align*}
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On the other hand, let $(x, y) \in \complex(E)$, then there exists $\theta \in [0, 2\pi]$ such that
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\begin{align*}
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\normn{\complex(T)(x, y)}_{\complex(F)} &= \normn{(Tx, Ty)}_{\complex(F)} = \norm{\cos(\theta)Tx + \sin(\theta)Ty}_F \\
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&\le \norm{T}_{L(E; F)} \cdot \norm{\cos(\theta)x + \sin(\theta)y}_E \\
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&\le \norm{T}_{L(E; F)} \cdot \sup_{\phi \in [0, 2\pi]}\norm{\cos(\phi)x + \sin(\phi)y}_E \\
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&= \norm{T}_{L(E; F)} \cdot \norm{(x, y)}_E
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\end{align*}
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\end{proof}
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\begin{proposition}
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\label{proposition:hermitian-functional-norm}
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Let $E$ be a normed vector space over $\complex$, $*: E \to E$ be a complex conjugation map such that $\norm{x}_E = \normn{x^*}_E$ for all $x \in E$, and $\phi \in E^*$ be a Hermitian functional, then
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\[
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\norm{\phi}_{E^*} = \sup\bracsn{\dpn{x, \phi}{E}|x \in E, x = x^*, \norm{x}_E = 1}
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\]
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\end{proposition}
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\begin{proof}
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Since $\bracsn{x \in E|x = x^*} \subset E$, $\norm{\phi}_{E^*} \ge \sup\bracsn{\dpn{x, \phi}{E}|x \in E, x = x^*, \norm{x}_E = 1}$.
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On the other hand, let $x \in E$ with $\norm{x}_E = 1$. Assume without loss of generality that $\dpn{x, \phi}{E} \in \real$, then
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\begin{align*}
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\dpn{x, \phi}{E} &= \dpn{\text{Re}(x), \phi}{E} + \underbrace{i\dpn{\text{Im}(x), \phi}{E}}_{\in \real} =
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\dpn{\text{Re}(x), \phi}{E} \\
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&\le \norm{\text{Re}(x)}_E \cdot \sup\bracsn{\dpn{y, \phi}{E}|y \in E, y = y^*, \norm{y}_E = 1}
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\end{align*}
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where $\norm{\text{Re}(x)}_E = \norm{{(x + x^*)}/{2}}_E \le \norm{x}_E$. As the above holds for all $x \in E$,
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\[
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\norm{\phi}_{E^*} \le \sup\bracsn{\dpn{x, \phi}{E}|x \in E, x = x^*, \norm{x}_E = 1}
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\]
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\end{proof}
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