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/open-mapping-theorem.md   | 30 +++++++++++-----------
 1 file changed, 15 insertions(+), 15 deletions(-)

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

diff --git a/pages/functional-analysis-basics/the-fundamental-four/open-mapping-theorem.md b/pages/functional-analysis-basics/the-fundamental-four/open-mapping-theorem.md
index 53da008..b191bb2 100644
--- a/pages/functional-analysis-basics/the-fundamental-four/open-mapping-theorem.md
+++ b/pages/functional-analysis-basics/the-fundamental-four/open-mapping-theorem.md
@@ -3,7 +3,6 @@ title: Open Mapping Theorem
 parent: The Fundamental Four
 grand_parent: Functional Analysis Basics
 nav_order: 3
-# cspell:words surjective bijective
 ---
 
 # {{ page.title }}
@@ -13,12 +12,10 @@ where $X$ and $Y$ are topological spaces,
 is called *open* if the image under $T$ of each open set of $X$
 is open in $Y$.
 
-{: .theorem-title }
-> {{ page.title }}
-> {: #{{ page.title | slugify }} }
->
-> A bounded linear operator between Banach spaces is open
-> if and only if it is surjective.
+{% theorem * Open Mapping Theorem %}
+A bounded linear operator between Banach spaces is open
+if and only if it is surjective.
+{% endtheorem %}
 
 {% proof %}
 Let $X$ and $Y$ be Banach spaces
@@ -91,19 +88,22 @@ Conversely, suppose that $T$ is open. TODO
 
 ---
 
-XXX injective
 For a bijective mapping between topological spaces, to say that it is open,
 is equivalent to saying that its inverse is continuous.
 The inverse of a bijective linear map between normed spaces is automatically linear
 and thus continuous if and only if it is bounded.
 As a corollary to the {{ page.title }} we obtain the following:
 
-{: .corollary-title }
-> Bounded Inverse Theorem
-> {: #bounded-inverse-theorem }
->
-> If a bounded linear operator between Banach spaces is bijective,
-> then its inverse is bounded.
-XXX relax to injective
+{% corollary * Bounded Inverse Theorem %}
+If a bounded linear operator between Banach spaces is bijective,
+then its inverse is bounded.
+{% endcorollary %}
 
 Also known as *Inverse Mapping Theorem*.
+
+{% corollary %}
+Let $T: X \to Y$ be a bounded linear operator between Banach spaces
+and suppose that $T$ is injective, so that the inverse $T^{-1} : R(T) \to X$
+is defined on the range of $T$.
+The linear operator $T^{-1}$ is bounded if and only if $R(T)$ is closed in $X$.
+{% endcorollary %}
-- 
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