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

---
 .../the-fundamental-four/closed-graph-theorem.md   | 43 ++++++++++++++--------
 1 file changed, 27 insertions(+), 16 deletions(-)

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

diff --git a/pages/functional-analysis-basics/the-fundamental-four/closed-graph-theorem.md b/pages/functional-analysis-basics/the-fundamental-four/closed-graph-theorem.md
index f8b8254..f6a9783 100644
--- a/pages/functional-analysis-basics/the-fundamental-four/closed-graph-theorem.md
+++ b/pages/functional-analysis-basics/the-fundamental-four/closed-graph-theorem.md
@@ -3,29 +3,40 @@ title: Closed Graph Theorem
 parent: The Fundamental Four
 grand_parent: Functional Analysis Basics
 nav_order: 4
-# cspell:words
 ---
 
 # {{ page.title }}
 
-{: .theorem-title }
-> {{ page.title }}
-> {: #{{ page.title | slugify }} }
->
-> An (everywhere-defined) linear operator between Banach spaces is bounded
-> iff its graph is closed.
+{% theorem * Closed Graph Theorem %}
+An (everywhere-defined) linear operator between Banach spaces is bounded
+iff its graph is closed.
+{% endtheorem %}
 
 We prove a slightly more general version:
 
-{: .theorem-title }
-> {{ page.title }}
-> {: #{{ page.title | slugify }}-variant }
->
-> Let $X$ and $Y$ be Banach spaces
-> and $T : \dom{T} \to Y$ a linear operator
-> with domain $\dom{T}$ closed in $X$.
-> Then $T$ is bounded if and only if
-> its graph $\graph{T}$ is closed.
+{% theorem * Closed Graph Theorem (Variant) %}
+Let $X$ and $Y$ be Banach spaces
+and $T : \dom{T} \to Y$ a linear operator
+with domain $\dom{T}$ closed in $X$.
+Then $T$ is bounded if and only if
+its graph $\graph{T}$ is closed.
+{% endtheorem %}
 
 {% proof %}
+Let us assume first that $T$ is bounded.
+Let $(x_n,Tx_n)_n$ be a sequence in $\graph{T}$ that converges to some element $(x,y) \in X \times Y$.
+This means that $x_n \to x$ and $Tx_n \to y$ for $n \to \infty$.
+The continuity of $T$ implies $Tx_n \to Tx$.
+Since a convergent series in a Hausdorff space has a unique limit,
+it follows that $Tx = y$; hence $(x,y)$ lies in $\graph{T}$.
+This shows that $\graph{T}$ is closed.
+
+Conversely, suppose that $\graph{T}$ is a closed subspace of $X \times Y$.
+Note that $X \times Y$ is a Banach space with norm $\norm{(x,y)} = \norm{x} + \norm{y}$.
+Therefore $\graph{T}$ is itself as Banach space in the restricted norm $\norm{(x,Tx)} = \norm{x} + \norm{Tx}$.
+The canonical projections $\pi_X : \graph{T} \to X$ and $\pi_Y : \graph{T} \to Y$ are bounded.
+Clearly, $\pi_X$ is bijective, so its inverse $\pi_X^{-1} : X \to \graph{T}$ is a bounded operator by the
+[Bounded Inverse Theorem](/pages/functional-analysis-basics/the-fundamental-four/open-mapping-theorem.html#bounded-inverse-theorem).
+Consequently the composition, $\pi_Y \circ \pi_X^{-1} : X \to Y$ is bounded.
+To complete the proof, observe that $\pi_Y \circ \pi_X^{-1} = T$.
 {% endproof %}
-- 
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