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.

That is, $T$ is a linear isomorphism.

Proof
Let $B_X^-$ be the closed unit ball of $X$.

Let $B_Y^-$ be the closed unit ball of $Y$.

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 Characterization of Open Linear Transformation between Normed Vector Spaces, there exists $\delta > 0$ such that:
 * $\delta B_Y^- \subseteq T \sqbrk {B_X^-}$

Then we have from Image of Subset under Mapping is Subset of Image and Image of Preimage under Mapping:
 * $T^{-1} \sqbrk {\delta B_Y^-} \subseteq B_X^-$

From Image of Dilation of Set under Linear Transformation is Dilation of Image, we have:
 * $T^{-1} \sqbrk {B_Y^-} \subseteq \delta^{-1} B_X^-$

From Norm of Bounded Linear Transformation in terms of Closed Unit Ball, $T^{-1}$ is therefore bounded.

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