Banach-Schauder Theorem

== Theorem ==
Let $\struct {X, \norm \cdot_X}$ and $\struct {Y, \norm \cdot_Y}$ be Banach spaces.
Let $T : X \to Y$ be a surjective bounded linear transformation.
Then $T$ is an open mapping.
== Proof ==
For each $x \in X$ and $r > 0$, let $\map {B_X} {x, r}$ be the open ball in $\struct {X, \norm \cdot_X}$ with centre $x$ and radius $r$.
For each $y \in Y$ and $r > 0$, let $\map {B_Y} {y, r}$ be the open ball in $\struct {Y, \norm \cdot_Y}$ with centre $y$ and radius $r$.
Note that we can write: 
:$\ds X = \bigcup_{n \mathop = 1}^\infty \map {B_X} {0, n}$
From Image of Union under Mapping: General Result, we have: 
:$\ds T \sqbrk X = \bigcup_{n \mathop = 1}^\infty T \sqbrk {\map {B_X} {0, n} }$
Since $T$ is surjective, we have $T \sqbrk X = Y$, and so: 
:$\ds Y = \bigcup_{n \mathop = 1}^\infty T \sqbrk {\map {B_X} {0, n} }$
Since $\struct {Y, \norm \cdot_Y}$ is a Banach space, by the Baire Category Theorem it is also a Baire space.
From Baire Space is Non-Meager:
:$Y$ is non-meager.
So:
:$Y$ is not the countable union of nowhere dense subsets of $Y$.
So, we must have: 
:$T \sqbrk {\map {B_X} {0, m} }$ is not nowhere dense for some $m \in \N$.
So the topological closure of $T \sqbrk {\map {B_X} {0, m} }$ has non-empty interior.
So, there exists a open set $U \subseteq X$ such that: 
:$U \subseteq \paren {T \sqbrk {\map {B_X} {0, m} } }^-$
where $\paren {T \sqbrk {\map {B_X} {0, m} } }^-$ denotes the topological closure of $T \sqbrk {\map {B_X} {0, m} }$
Let $y \in U$.
Since $U$ is open set, there exists an open ball $\map {B_Y} {y, r}$ such that $r > 0$ and: 
:$\map {B_Y} {y, r} \subseteq U \subseteq \paren {T \sqbrk {\map {B_X} {0, m} } }^-$
=== Lemma 1 ===
=== Lemma 2 ===
Now let $V \subseteq X$ be an arbitrary open set.
We aim to show that $T \sqbrk V$ is open in $Y$.
Let $y \in T \sqbrk V$, then there exists $x \in V$ such that $y = T x$.
Since $V$ is open there exists $\delta > 0$ such that $\map {B_X} {x, \delta} \subseteq V$. 
We show that: 
:$\ds \map {B_Y} {T x, \frac \delta {2 m} r} \subseteq T \sqbrk V$
Let:
:$\ds v \in \map {B_Y} {T x, \frac \delta {2 m} r}$
Then: 
:$\ds \norm {v - T x}_Y < \frac \delta {2 m} r$ 
so:
:$\ds \norm {\frac {2 m} \delta \paren {v - T x} }_Y < r$ 
So:
:$\ds \frac {2 m} \delta \paren {v - T x} \in \map {B_Y} {0, r}$
Then, from Lemma 2, we have: 
:$\ds \frac {2 m} \delta \paren {v - T x} \in T \sqbrk {\map {B_X} {0, 2 m} }$
So: 
:$\ds \frac {2 m} \delta \paren {v - T x} = \map T {2 m u} = 2 m T u$
for some $u$ with $\norm u_X < 1$. 
So:
:$\ds \frac {v - T x} \delta = T u$
Then: 
:$v - T x = \map T {\delta u}$
with $\norm {\delta u}_X < \delta$.
Then: 
:$v = \map T {x + \delta u}$
with $x + \delta u \in \map {B_X} {x, \delta}$.
So: 
:$v \in T \sqbrk {\map {B_X} {x, \delta} }$
Then $v = T x'$ for some $x' \in \map {B_X} {x, \delta}$.
Since $\map {B_X} {x, \delta} \subseteq V$, we have $x' \in V$.
So $v \in T \sqbrk V$.
So:
:$\ds \map {B_Y} {T x, \frac \delta {2 m} r} \subseteq T \sqbrk V$
That is: 
:$\ds \map {B_Y} {y, \frac \delta {2 m} r} \subseteq T \sqbrk V$
So $T \sqbrk V$ is open in $Y$.
We have now shown that for each open set $V \subseteq X$, $T \sqbrk V$ is open in $Y$. 
So $T$ is an open mapping.
== Also known as ==
This theorem is also known as the open mapping theorem.
== Sources ==
* : Chapter XV, $\S1$. The Open Mapping Theorem
* $23.1$: The Open Mapping and Inverse Mapping Theorems
Category:Linear Transformations on Banach Spaces
Category:Banach-Schauder Theorem
Category:Open Mappings