9.6 The Hahn-Banach Theorem

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

Let

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

Proof. (1): Let $x_{0} \in E \setminus F$, then by Lemma 9.6.1, there exists $\lambda \in \real$ such that if

then $\phi_{x_0, \lambda}(x) \le \rho(x)$ for all $x \in F + \real x_{0}$. Thus for any $F \subsetneq E$, $\phi$ admits an extension to a subspace that strictly contains $F$.

Let

For any $\Phi, \Phi' \in \mathbf{F}$, define $\Phi \le \Phi'$ if $\Phi'$ is an extension of $\Phi$. Let $\cf \subset \mathbf{F}$ be a chain. For any $\Phi \in \cf$, let $\cd(\Phi)$ be its domain. Since $\bigcup_{\phi \in \cf}\cd(\Phi)$ is a subspace and $\cf$ is a chain, by Lemma 2.0.12, there exists $\Phi \in \hom\paren{\bigcup_{\Phi' \in \cf}\cd(\Phi'); \real}$ such that $\Phi|_{\cd(\Phi')}= \Phi'$ for all $\Phi' \in \cf$. Thus $\Phi \in \mathbf{F}$ is an upper bound of $\cf$.

By Zorn’s lemma, $\mathbf{F}$ admits a maximal element $\Phi$. If $\cd(\Phi) \subsetneq E$, then $\Phi$ is not maximal by the preceding discussion. Therefore $\cd(\Phi) = E$ and $\Phi$ is a desired extension.

(2): Given that $\rho$ is a seminorm, for any $u \in \hom(E; \real)$, $u \le \rho$ if and only if $\abs{u}\le \rho$.

Let $u = \re{\phi}$, then $u \in \hom(E; \real)$ by Proposition 8.4.1. By (1), there exists $U \in \hom(E; \real)$ such that $\abs{U}\le \rho$ and $U|_{F} = u$. For each $x \in E$, let $\Phi(x) = U(x) - iU(ix)$, then $\Phi \in \hom(E; \complex)$ and $\Phi|_{F} = \phi$ by Proposition 8.4.1. In addition, for any $x \in E$, if $\alpha = \overline{\sgn(\Phi(x))}$, then

so $\abs{\Phi}\le \rho$.$\square$

Proof. By translation, assume without loss of generality that $0 \in A$. In which case, $A \in \cn^{o}(0)$ is convex.

Let $[\cdot]_{A}: E \to [0, \infty)$ be the gaugeg of $A$, then $[\cdot]_{A}$ is a sublinear functional on $E$. For any $y, z \in E$ and $t > 0$ with $y, z \in tA$,

Hence $[\cdot]_{A}$ is continuous on $E$.

Let $\phi_{0}: \real x \to \real$ be defined by $\lambda x \mapsto \lambda$. Since $x \not\in A$, $[x]_{A} > 1$. Hence $\phi_{0}|_{Kx}\le [\cdot]_{A}|_{Kx}$. By the Hahn-Banach Theorem, there exists $\phi \in \hom(E; \real)$ such that $\phi|_{\real x}= \phi_{0}$ and $\phi(y) \le [y]_{A}$ for all $y \in E$.

For any $y \in A \cap (-A)$,

so $\phi \in E^{*}$.

Finally, let $y \in A$, then $\dpb{y, \phi}{E}\le [y]_{A} < 1 = \dpb{x, \phi}{E}$.$\square$

Proof. Let $C = A - B$, then $C \subset E$ is an open convex set by Lemma 9.1.3. Since $A \cap B = \emptyset$, $0 \not\in C$.

By Lemma 9.6.3, there exists $\phi \in E^{*}$ such that $C \subset \bracs{\phi < 0}$. In which case, for any $a \in A$ and $b \in B$, $\dpb{a - b, \phi}{E}< 0$ implies that $\dpb{a, \phi}{E}< \dpb{b, \phi}{E}$. Let $\alpha \in [\sup_{a \in A}\dpb{a, \phi}{E}, \inf_{b \in B}\dpb{b, \phi}{E}]$, then $A \subset \bracs{\phi \le \alpha}$ and $B \subset \bracs{\phi \ge \alpha}$.$\square$

Proof. Let $C = A - B$, then $C$ is closed by Proposition 8.1.8 with $0 \not\in C$. Since $E$ is locally convex, there exists $U \in \cn^{o}(0)$ convex such that $U \cap C \ne \emptyset$.

By Theorem 9.6.4, there exists $\phi \in E^{*} \setminus \bracs{0}$ and $\alpha \in \real$ such that $U \subset \bracs{f \le \alpha}$ and $C \subset \bracs{f \ge \alpha}$. In which case, for any $u \in U$, $a \in A$, and $b \in B$,

As $\phi \ne 0$ and $U \in \cn^{o}(0)$, $r = \sup_{u \in U}\dpb{u, \phi}{E}> 0$. Thus

Proof. (1): Let $\rho_{M}: E \to [0, \infty)$ be the quotient of $\rho$ by $M$, then $\rho_{M} \le \rho$ is a continuous seminorm on $E$ by Definition 9.3.1. Let $\phi_{0}: Kx \to K$ be defined by $\lambda x \mapsto \lambda \rho_{M}(x)$. By the Hahn-Banach theorem, there exists $\phi \in \hom{E; K}$ such that $\dpb{x, \phi}{E}= \rho_{M}(x)$ and $\phi \le \rho_{M} \le \rho$.

(2): By (1) applied to $M = \bracs{0}$.

(3): By (2) applied to $x - y$.$\square$