The Eigenspace

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/Part 3: Functional Analysis/Chapter 4: Locally Convex Space

4.4 Hahn-Banach Theorem

Lemma 4.4.1.label Let $E$ be a vector space over $\R$, $\rho:E\to\R$ be a sublinear functional, $F\suf E$ be a subspace, $\phi\in \text{Hom}(F;\R)$ with $\phi(x)\leq \rho(x)$ for all $x\in F$. For each $x_{0}\in E\del F,\lam\in \R$ define the extension

\begin{align*}\phi_{x_0,\lam}:F+\R x_{0}\to \R&&y+tx_{0}\mto \phi(y)+\lam t\end{align*}

then there exists $\lam\in \R$ depending on $x_{0}$ such that $\phi_{x_0,\lam(x)}\leq\rho(x)$ for all $x\in F+\R x_{0}$.

Proof. Let $x,y\in F$, then

\begin{align*}\phi(x)+\phi(y)=\phi(x+y)&\leq\rho(x+y)\leq \rho(x-x_{0})+\rho(y+x_{0})\\ \phi(x)-\rho(x-x_{0})&\leq \rho(y+x_{0})-\phi(y)\\ \SUP{x\in F}\braks{\phi(x)-\rho(x-x_0)}&\leq\INF{x\in F}\braks{\rho(x+x_0)-\phi(x)}\end{align*}

Let

\begin{align*}\lam\in \braks{\SUP{x\in F}\braks{\phi(x)-\rho(x-x_0)},\INF{x\in F}\braks{\rho(x+x_0)-\phi(x)}}\end{align*}

then for any $x\in F,t>0$

\begin{align*}\phi_{x_0,\lam}(x+tx_{0})&=\phi(x)+\lam t=t \braks{\phi(t\inv x)+\lam}\\&\leq t \braks{\rho(t\inv x+x_0)-\phi(t\inv x)+\phi(t\inv x)}\\&=t\rho(t\inv x+x_{0})=\rho(x+tx_{0})\\ \phi(x-tx_{0})&=\phi(x)-t\lam=t \braks{\phi(t\inv x)-\lam}\\&\geq t \braks{\rho(t\inv x-x_0)+\phi(t\inv x)-\phi(t\inv x)}\\&=t \rho(t\inv x-x_{0})=\rho(x-tx_{0})\end{align*}

$\square$

Proposition 4.4.2 (Polarisation).label Let $E$ be a vector space over $\com$ then the map

\begin{align*}\hom{\com}{E}{\com}\to \hom{\R}{E}{\R}&&\phi\mto \Re(\phi)=:u\end{align*}

is a bijection whose inverse is

\begin{align*}\hom{\R}{E}{\R}\to \hom{\com}{E}{\com}&&u\mto (x\mto \inner{x}{u}_{E}-i\inner{ix}{u}_{E})=:\phi\end{align*}

If $E$ is normed then $\norm{\phi}{}=\norm{\Re(\phi)}{}$

Proof. Let $\phi\in \hom{\com}{E}{\com},x,y\in E,\lam\in \R$ then

\begin{align*}\Re(\inner{\lam x+y}{\phi}_{E})&=\Re(\lam \inner{x}{\phi}_{E}+\inner{y}{\phi}_{E})=\lam \Re(\inner{x}{\phi}_{E})+\Re(\inner{y}{\phi}_{E})\\ \Im(\inner{x}{\phi}_{E})&=\Re(-i\inner{x}{\phi}_{E})=-\Re(\inner{ix}{\phi}_{E})=-\inner{ix}{\Re(\phi)}_{E}\end{align*}

showing $\Re(\phi)\in \hom{\R}{E}{\R}$ and $\inner{x}{\phi}_{E}=\inner{x}{u}_{E}-i\inner{ix}{u}_{E}$.

Conversely let $u\in\hom{\R}{E}{\R},x,y\in E,\lam\in \com$ then

\begin{align*}\inner{\lam x+y}{\phi}_{E}&=\inner{\lam x+y}{u}_{E}-i\inner{i(\lam x+y)}{u}_{E}\\&=\lam(\inner{x}{u}_{E}-i\inner{ix}{u}_{E})+(\inner{y}{u}_{E}-i\inner{iy}{u}_{E})\\&=\lam\inner{x}{\phi}_{E}+\inner{y}{\phi}_{E}\end{align*}

In particular taking $y=0, \lam=i$ we have

\begin{align*}\inner{ix}{\phi}_{E}=\inner{ix}{u}_{E}-i\inner{i(ix)}{u}_{E}=\inner{ix}{u}_{E}-i\inner{-x}{u}_{E}=i\inner{x}{\phi}_{E}\end{align*}

Since $\abs{\Re(\inner{x}{\phi}_E)}\leq \abs{\inner{x}{\phi}_E}$ for any $x\in E$ we have $\norm{\phi}{}=\norm{u}{}$. On the other hand, if $\inner{x}{\phi}_{E}\neq0$ then let $\al=\cl{\text{sgn}\inner{x}{\phi}_E}$. Since

\begin{align*}\abs{\inner{x}{\phi}_E}=\al\inner{x}{\phi}_{E}=\inner{\al x}{\phi}_{E}=\inner{\al x}{u}_{E}\leq \norm{u}{}\norm{\al x}{}=\norm{u}{}\norm{x}{}\end{align*}

whence $\norm{\phi}{}\leq\norm{u}{}$.$\square$

Theorem 4.4.3 (Hahn-Banach, Analytic Form).label Let $E$ be a vector space over $K\in \curl{\R,\com},\rho:E\to\R$ be a sublinear functional, and $F\sub E$ be a subspace, then

  1. (1)

    For any $\phi\in \text{Hom}(F;\R)$ with $\phi\leq \rho|_{F}$, there exists $\Phi\in \hom{}{F}{\R}$ such that $\Phi\leq \rho$ and $\Phi|_{F}=\phi$.

  2. (2)

    If $\rho$ is a seminorm, then for any $\phi\in \hom{}{F}{K}$ with $\abs{\phi}\leq\rho|_{F}$, there exists $\Phi\in \hom{}{E}{K}$ such that $\abs{\Phi}\leq \rho$ and $\Phi|_{F}=\phi$.

Proof.

  1. (1)

    Let $x_{0}\in E\del F$, then by Lemma 4.4.1, there exists $\lam\in\R$ such that if

    \begin{align*}\phi_{x_0,\lam}:F+\R x_{0}\to \R&&(x+tx_{0})\mto x+\lam t\end{align*}

    then $\phi_{x_0,\lam}(x)\leq \rho(x)$ for all $x\in F+\R x_{0}$. Thus for any $F\sub E$, $\phi$ admits an extension to a subspace that strictly contains $F$. Let

    \begin{align*}\textbf{F}=\curl{\Phi\in\hom{}{F'}{\R}:F'\supf,\Phi|_F=\phi,\Phi\leq \rho|_{F'}}\end{align*}

    For any $\Phi,\Phi'\in \textbf{F}$, define $\Phi\leq\Phi'$ if $\Phi'$ is an extension of $\Phi$. For any $\Phi\in\textbf{F}$, let $\cali{D}(\Phi)$ be its domain. Let $\cali{F}\suf \textbf{F}$ be a chain then $\cups{}{\Phi\in\cali{F}}\cali{D}(\Phi)$ is a subspace, hence there exists $\Phi\in\hom{}{\cups{}{\Phi'\in \cali{F}}\cali{D}(\Phi')}{\R}$ such that $\Phi|_{\cali{D}(\Phi')}=\Phi'$ for all $\Phi'\in\cali{F}$. Thus $\Phi\in \textbf{F}$ is an upper bound of $\cali{F}$. By Zorn’s lemma, $\textbf{F}$ admits a maximal element $\Phi$. If $\cali{D}(\Phi)\sub E$, then $\Phi$ is not maximal by Lemma 4.4.1. Therefore $\cali{D}(\Phi)=E$ and $\Phi$ is a desired extension.

  2. (2)

    By absolute homogeneity of $\rho$, for any $u\in \hom{}{E}{\R}$, $u\leq \rho$ if and only if $\abs{u}\leq\rho$. By (1) we assume WLOG $K=\com$.

    By Proposition 4.4.2 $\Re(\phi)\in \hom{}{E}{\R}$. By (1), there exists $u\in\hom{}{E}{\R}$ such that $\abs{u}\leq\rho$ and $u|_{F}=\Re(\phi)$ so by Proposition 4.4.2 we are done.

$\square$

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The Eigenspace

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