Characterization of Invertible Bounded Linear Transformations

Theorem

Let $\struct {U, \norm \cdot_U}$ and $\struct {V, \norm \cdot_V}$ be normed vector spaces.

Let $A : V \to U$ be a linear transformation with inverse $A^{-1} : U \to V$.

Then $A^{-1}$ is a bounded linear transformation if and only if:

there exists a real number $c > 0$ such that $\norm {A x}_U \ge c \norm x_V$ for all $x \in V$.

That is, $A$ is invertible as a bounded linear transformation if and only if:

there exists a real number $c > 0$ such that $\norm {A x}_U \ge c \norm x_V$ for all $x \in V$.

Proof

From Inverse Mapping of Linear Transformation is Linear Transformation, we have:

$A^{-1}$ is a linear transformation.

So we are interested in determining when $A^{-1}$ is bounded.

Necessary Condition

Suppose that $A^{-1}$ was bounded.

Then:

there exists a real number $M > 0$ such that $\norm {A^{-1} y}_V \le M \norm y_U$ for all $y \in U$.

Let $x \in V$ and set $y = A x \in U$.

Then:

$\norm {A^{-1} y}_V \le M \norm y_U$

We have:

\(\ds \norm {A^{-1} y}_V\)

\(=\)

\(\ds \norm {A^{-1} A x}_V\)

\(\ds \)

\(=\)

\(\ds \norm x_V\)

and:

$\norm y_U = \norm {A x}_U$

giving:

$\norm x_V \le M \norm {A x}_U$

So, we have:

$\norm {A x}_U \ge \dfrac 1 M \norm x_V$

for all $x \in V$.

Then, setting $c = 1/M$, we have that:

there exists a real number $c > 0$ such that $\norm {A x}_U \ge c \norm x_V$ for all $x \in V$.

$\Box$

Sufficient Condition

Suppose that:

there exists a real number $c > 0$ such that $\norm {A x}_U \ge c \norm x_V$ for all $x \in V$.

Let $y \in U$ and let $x = A^{-1} y \in V$.

Then:

$\norm {A x}_U \ge c \norm x_V$

We have:

\(\ds \norm {A x}_U\)

\(=\)

\(\ds \norm {A A^{-1} y}_U\)

\(\ds \)

\(=\)

\(\ds \norm y_U\)

and:

$\norm x_V = \norm {A^{-1} y}_V$

so:

$\norm y_U \ge c \norm {A^{-1} y}_V$

So, we have:

$\norm {A^{-1} y}_V \le \dfrac 1 c \norm y_U$

for all $y \in U$.

So:

$A^{-1}$ is bounded.

$\blacksquare$