From 28407333ffceca9b99fae721c30e8ae146a863da Mon Sep 17 00:00:00 2001
From: Justin Gassner <justin.gassner@mailbox.org>
Date: Wed, 14 Feb 2024 07:24:38 +0100
Subject: Update

---
 .../uniform-boundedness-theorem.md                 | 44 ++++++++++++++--------
 1 file changed, 28 insertions(+), 16 deletions(-)

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

diff --git a/pages/functional-analysis-basics/the-fundamental-four/uniform-boundedness-theorem.md b/pages/functional-analysis-basics/the-fundamental-four/uniform-boundedness-theorem.md
index 13460da..47ddd3f 100644
--- a/pages/functional-analysis-basics/the-fundamental-four/uniform-boundedness-theorem.md
+++ b/pages/functional-analysis-basics/the-fundamental-four/uniform-boundedness-theorem.md
@@ -3,27 +3,35 @@ title: Uniform Boundedness Theorem
 parent: The Fundamental Four
 grand_parent: Functional Analysis Basics
 nav_order: 2
-description: >
-  The
-# spellchecker:words preimages pointwise
 ---
 
 # {{ page.title }}
 
 Also known as *Uniform Boundedness Principle* and *Banach–Steinhaus Theorem*.
 
-{: .theorem-title }
-> {{ page.title }}
-> {: #{{ page.title | slugify }} }
->
-> If $\mathcal{T}$ is a set of bounded linear operators
-> from a Banach space $X$ into a normed space $Y$ such that
-> $\braces{\norm{Tx} : T \in \mathcal{T}}$
-> is a bounded set for every $x \in X$, then
-> $\braces{\norm{T} : T \in \mathcal{T}}$
-> is a bounded set.
+{% definition Pointwise and Uniform Boundedness %}
+Let $X$, $Y$ be normed spaces.
+We say that a collection $\mathcal{T}$ of bounded linear operators
+from $X$ to $Y$ is
+{: .mb-0 }
+- *pointwise bounded* if the set $\braces{\norm{Tx} : T \in \mathcal{T}}$ is bounded for every $x \in X$,
+- *uniformly bounded* if the set $\braces{\norm{T} : T \in \mathcal{T}}$ is bounded.
+{% enddefinition %}
+
+Clearly, every uniformly bounded collection of operators is pointwise bounded since $\norm{Tx} \le \norm{T} \norm{x}$.
+The converse is true, if $X$ is complete:
+
+{% theorem * Uniform Boundedness Theorem %}
+If a collection of bounded linear operators
+from a Banach space into a normed space
+is pointwise bounded,
+then it is uniformly bounded.
+{% endtheorem %}
 
 {% proof %}
+Suppose $X$ is a Banach space, $Y$ is a normed space
+and $\mathcal{T}$ is a pointwise bounded collection
+of bounded linear operators from $X$ to $Y$.
 For each $n \in \NN$ the set
 
 $$
@@ -38,9 +46,8 @@ the set $\braces{\norm{Tx} : T \in \mathcal{T}}$ is bounded by assumption.
 This means that there exists a $n \in \NN$
 such that $\norm{Tx} \le n$ for all $T \in \mathcal{T}$.
 In other words, $x \in A_n$.
-Thus we have show that $\bigcup A_n = X$.
-XXX Apart from the trivial case $X = \emptyset$,
-the union $\bigcup A_n$ has nonempty interior.
+Thus, we have shown that $\bigcup A_n = X$.
+In particular, $\bigcup A_n$ has nonempty interior.
 Now, utilizing the completeness of $X$, the
 [Baire Category Theorem]({% link pages/general-topology/baire-spaces.md %})
 implies that there exists a $m \in \NN$ such that $A_m$ has nonempty interior.
@@ -74,3 +81,8 @@ $$
 $$
 
 If $X$ is not complete, this may be false.
+
+TODO:
+- strong operator convergence
+- Kreyszig 4.9-5
+- Haase 15.6
-- 
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