From 7c66b227a494748e2a546fb85317accd00aebe53 Mon Sep 17 00:00:00 2001
From: Justin Gassner <justin.gassner@mailbox.org>
Date: Thu, 15 Feb 2024 05:11:07 +0100
Subject: Update

---
 .../the-fundamental-four/hahn-banach-theorem.md            | 14 ++------------
 1 file changed, 2 insertions(+), 12 deletions(-)

(limited to 'pages/functional-analysis-basics/the-fundamental-four/hahn-banach-theorem.md')

diff --git a/pages/functional-analysis-basics/the-fundamental-four/hahn-banach-theorem.md b/pages/functional-analysis-basics/the-fundamental-four/hahn-banach-theorem.md
index 18cf64a..a2602ac 100644
--- a/pages/functional-analysis-basics/the-fundamental-four/hahn-banach-theorem.md
+++ b/pages/functional-analysis-basics/the-fundamental-four/hahn-banach-theorem.md
@@ -6,7 +6,6 @@ nav_order: 1
 ---
 
 # {{ page.title }}
-{: .no_toc }
 
 In fact, there are multiple theorems and corollaries
 which bear the name Hahn–Banach.
@@ -14,15 +13,6 @@ All have in common that
 they guarantee the existence of linear functionals
 with various additional properties.
 
-<details open markdown="block">
-  <summary>
-    Table of contents
-  </summary>
-  {: .text-delta }
-- TOC
-{:toc}
-</details>
-
 {% definition Sublinear Functional %}
 A functional $p$ on a real vector space $X$
 is called *sublinear* if it is
@@ -146,7 +136,7 @@ we define a functional $f_0$ by $f_0(\alpha x) = \alpha \norm{x}$ for $\alpha \i
 It is easy to check that $f_0$ is linear and bounded with norm $\norm{f_0} = 1$.
 By the Hahn–Banach Extension Theorem for Normed Spaces,
 there exists a bounded linear functional $f$ on $X$ extending $f_0$ with identical norm.
-Hence we have $f(x) = f_0(x) = \norm{x}$ and $\norm{f} = \norm{f_0} = 1$.
+Hence, we have $f(x) = f_0(x) = \norm{x}$ and $\norm{f} = \norm{f_0} = 1$.
 {% endproof %}
 
 Recall that for a normed space $X$ we denote its (topological) dual space by $X'$.
@@ -155,7 +145,7 @@ Recall that for a normed space $X$ we denote its (topological) dual space by $X'
 For every element $x$ of a real or complex normed space $X$ one has
 
 $$
-\norm{x} = \sup_{f \in X' \setminus \braces{0}} \frac{\abs{f(x)}}{\norm{f}} 
+\norm{x} = \sup_{f \in X' \setminus \braces{0}} \frac{\abs{f(x)}}{\norm{f}}
 $$
 
 and the supremum is attained.
-- 
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