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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/uniform-boundedness-theorem.md
parent28407333ffceca9b99fae721c30e8ae146a863da (diff)
downloadsite-7c66b227a494748e2a546fb85317accd00aebe53.tar.zst
Update
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-rw-r--r--pages/functional-analysis-basics/the-fundamental-four/uniform-boundedness-theorem.md6
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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 %}