From 777f9d3fd8caf56e6bc6999a4b05379307d0733f Mon Sep 17 00:00:00 2001
From: Justin Gassner <justin.gassner@mailbox.org>
Date: Tue, 12 Sep 2023 07:36:33 +0200
Subject: Initial commit

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+---
+title: Open Mapping Theorem
+parent: The Fundamental Four
+grand_parent: Functional Analysis Basics
+nav_order: 3
+# cspell:words surjective bijective
+---
+
+# {{ page.title }}
+
+Recall that a mapping $T : X \to Y$,
+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.
+
+{% proof %}
+Let $X$ and $Y$ be Banach spaces
+and let $T : X \to Y$ be a bounded linear operator.
+Let $B_X$ and $B_Y$ denote the open unit balls in $X$ and $Y$, respectively.
+
+First, suppose that $T$ is surjective.
+The balls $m B_X$, $m \in \NN$, cover $X$.
+Since $T$ is surjective,
+their images $mTB_X$ cover $Y$.
+This remains true, if we take closures:
+$\bigcup \overline{mTB_X} = Y$.
+Hence, we have written the space $Y$,
+which is assumed to have a complete norm,
+as the union of countably many closed sets. It follows form the
+[Baire Category Theorem]({% link pages/general-topology/baire-spaces.md %})
+that $\overline{mTB_X}$ has nonempty interior for some $m$.
+Thus there are $q \in Y$ and $\alpha > 0$
+such that $q + \alpha B_Y \subset \overline{mTB_X}$.
+Choose a $p \in X$ with $Tp=q$.
+It is a well known fact, that in a normed space
+the translation by a vector and the multiplication with a nonzero scalar
+are homeomorphisms and thus compatible with taking the closure.
+We conclude $\alpha B_Y \subset \overline{T(mB_X-q)}$.
+Since $mB_X-q$ is a bounded set,
+it is contained in a ball $\beta B_X$ for some $\beta > 0$.
+Thus, $\alpha B_Y \subset \overline{T \beta B_X} = \beta \overline{TB_X}$.
+With $\gamma := \alpha / \beta > 0$ we obtain $\gamma B_Y \subset \overline{TB_X}$.
+
+Clearly, every $y \in \gamma B_Y$ is the limit of a sequence $(Tx_n)$,
+where $x_n \in B_X$.
+However, the sequence $(x_n)$ *may not converge*!
+We show that it is possible to find a *convergent* sequence $(s_n)$ in $4B_X$
+such that $Ts_n \to y$.
+To construct $(s_n)$, we recursively define a sequence $(y_k)$
+with $y_k \in 2^{-k} \gamma B_Y$ for $k \in \NN_0$.
+The sequence starts with $y_0 := y \in 2^0 \gamma B_Y$.
+Given $y_k \in 2^{-k} \gamma B_Y$, one has $y_k \in \overline{T 2^{-k} B_X}$.
+By the definition of closure, there exists a $x_k \in 2^{-k} B_X$
+such that $Tx_k$ lies in the open $2^{-(k+1)} \gamma$-ball about $y_k$.
+This means that $y_{k+1} := y_k - Tx_k \in 2^{-(k+1)}\gamma B_Y$.
+Now define $s_n$ as the $n$-th partial sum of the series $\sum_{k=0}^{\infty} x_k$.
+The series converges,
+because it converges absolutely (Here we use the completeness of $X$).
+The latter is true because $\sum \norm{x_k} \le  \sum 2^{-k} = 3$.
+This also shows that each $s_n$ and the limit $x := \lim s_n$ lie in $4B_X$.
+The auxiliary sequence $(y_n)$ converges to $0$ by construction.
+Therefore, in the limit $n \to \infty$
+
+$$
+Ts_n = \sum_{k=0}^{n} Tx_k = \sum_{k=0}^{n} y_k - y_{k+1}
+= y_0 - y_{n+1} \to y_0 = y,
+$$
+
+as desired.
+It follows from the continuity of $T$ that $Ts_n \to Tx$, thus $Tx = y$.
+
+In the preceding paragraph it was proven that $\gamma B_Y \subset 4TB_X$.
+Hence, $\delta B_Y \subset TB_X$ where $\delta := \gamma/4$.
+To show that $T$ is open, consider any open set $U \subset X$.
+If $y$ lies in $TU$, there exists a $x \in U$ such that $Tx=y$.
+Since $U$ is open, there is an $\epsilon > 0$ such that $x+\epsilon B_X \subset U$.
+Applying $T$, we find $y + \epsilon TB_X \subset TU$.
+Combine with $\delta B_Y \subset TB_X$ to see $y + \epsilon \delta B_X \subset TU$.
+Hence, $TU$ is open.
+This shows that $T$ is indeed an open mapping.
+
+Conversely, suppose that $T$ is open. TODO
+{% endproof %}
+
+---
+
+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
+
+Also known as *Inverse Mapping Theorem*.
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