From e7f7dfc8e3dc7a6f5b15421991abf441799528f4 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Tue, 27 Jan 2026 15:43:23 -0500 Subject: [PATCH] Fixed typo. --- src/fa/lc/hahn-banach.tex | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/src/fa/lc/hahn-banach.tex b/src/fa/lc/hahn-banach.tex index 6096c3b..7543afe 100644 --- a/src/fa/lc/hahn-banach.tex +++ b/src/fa/lc/hahn-banach.tex @@ -56,7 +56,7 @@ 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. - (1): Given that $\rho$ is a seminorm, for any $u \in \hom(E; \real)$, $u \le \rho$ if and only if $\abs{u} \le \rho$. + (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 \ref{proposition:polarisation-linear}. 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 \ref{proposition:polarisation-linear}. In addition, for any $x \in E$, if $\alpha = \overline{\sgn(\Phi(x))}$, then \[ @@ -82,7 +82,7 @@ For any $y \in A \cap (-A)$, \[ - \dpb{y, \phi}{E} \le [y]_A \le 1 \quad -\dpb{y, \phi}{E} \le [-y]_A < 1 + \dpb{y, \phi}{E} \le [y]_A < 1 \quad -\dpb{y, \phi}{E} \le [-y]_A < 1 \] so $\phi \in E^*$.