Banach Isomorphism Theorem

Theorem
Let $\struct {X, \norm \cdot_X}$ and $\struct {Y, \norm \cdot_Y}$ be Banach spaces.

Let $T : X \to Y$ be a bijective bounded linear transformation.

Then the inverse of $T$ is a bounded linear transformation.

Proof
Let $\map {B_X} {x, r}$ denote the open ball in $X$ centered at $x \in X$ with radius $r$.

Let $\map {B_Y} {x, r}$ denote the open ball in $Y$ centered at $y \in Y$ with radius $r$.

Let $T^{-1} : Y \to X$ be the inverse of $T$.

From Inverse of Linear Transformation is Linear Transformation, $T^{-1} : Y \to X$ is a linear transformation.

It remains to show that $T^{-1}$ is bounded.

Since $T$ is bijective, it is surjective.

So, by the Banach-Schauder Theorem:


 * $T$ is an open mapping.

From Open Ball is Open Set in Normed Vector Space, we have:


 * $\map {B_X} {0, 1}$ is open in $X$.

So:


 * $\map T {\map {B_X} {0, 1} }$ is is open in $Y$.

Since $0 \in \map {B_X} {0, 1}$ and $\map T 0 = 0$, there therefore exists $r > 0$ such that:


 * $\map {B_Y} {0, r} \subseteq \map T {\map {B_X} {0, 1} }$

We now show that:


 * $\map {T^{-1} } {\map {B_Y} {0, 1} } \subseteq \map {B_X} {0, r^{-1} }$

Let:


 * $x \in \map {T^{-1} } {\map {B_Y} {0, 1} }$

Then:


 * $\norm {T x}_Y < 1$

So, from linearity, we have:


 * $\norm {\map T {r x} }_Y < r$

Then:


 * $\map T {r x} \in \map {B_Y} {0, r}$

so:


 * $\map T {r x} \in \map T {\map {B_X} {0, 1} }$

So there exists $x' \in \map {B_X} {0, 1}$ such that:


 * $\map T {r x} = T x'$

Since $T$ is a bijection, we have $x' = r x$ and so:


 * $r x \in \map {B_X} {0, 1}$

That is:


 * $\norm {r x}_X < 1$

so:


 * $\norm x_X < r^{-1}$

So:


 * $x \in \map {B_X} {0, r^{-1} }$

showing that:


 * $\map {T^{-1} } {\map {B_Y} {0, 1} } \subseteq \map {B_X} {0, r^{-1} }$

Note that we now have:


 * $\norm {T^{-1} y}_X < r^{-1}$

for all $y \in \map {B_Y} {0, 1}$.

We aim to augment this bound to the whole of $Y$.

Note that for all $y \in Y$ with $y \ne 0$, we have:


 * $\ds \norm {\frac y {2 \norm y_Y} }_Y = \frac 1 2 < 1$

That is:


 * $\ds \frac y {2 \norm y_Y} \in \map {B_Y} {0, 1}$

So:


 * $\ds \norm {\map {T^{-1} } {\frac y {2 \norm y_Y} } }_X < r^{-1}$

so:


 * $\norm {T^{-1} y}_X < 2 r^{-1} \norm y_Y$

for all $y \in Y$ with $y \ne 0$.

For $y = 0$ we have:


 * $\norm {T^{-1} y}_X \le 2 r^{-1} \norm y_Y$

So $T^{-1}$ is bounded.