From 7c66b227a494748e2a546fb85317accd00aebe53 Mon Sep 17 00:00:00 2001 From: Justin Gassner Date: Thu, 15 Feb 2024 05:11:07 +0100 Subject: Update --- .../the-fundamental-four/uniform-boundedness-theorem.md | 6 ++++-- 1 file changed, 4 insertions(+), 2 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 47ddd3f..1140e45 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 @@ -14,11 +14,13 @@ 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 $x \in X$, +- *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 $\norm{Tx} \le \norm{T} \norm{x}$. +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 %} -- cgit v1.2.3-54-g00ecf