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:


 * $\ds Y = \bigcup_{n \mathop = 1}^\infty T \sqbrk {\map {B_X} {0, n} }$

Assuming the axiom of dependent choice, the Baire Category Theorem can be used.

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 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.