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 $\struct {Y, \norm \cdot_X}$ 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:


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

Since $T$ is surjective, we have $\map T X = Y$, and so:


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

Also known as
This theorem is also known as the open mapping theorem.