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authorJustin Gassner <justin.gassner@mailbox.org>2024-02-15 05:11:07 +0100
committerJustin Gassner <justin.gassner@mailbox.org>2024-02-15 05:11:07 +0100
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tree9c649667d2d024b90b32d36ca327ac4b2e7caeb2 /pages/functional-analysis-basics/the-fundamental-four/hahn-banach-theorem.md
parent28407333ffceca9b99fae721c30e8ae146a863da (diff)
downloadsite-7c66b227a494748e2a546fb85317accd00aebe53.tar.zst
Update
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-rw-r--r--pages/functional-analysis-basics/the-fundamental-four/hahn-banach-theorem.md14
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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.