Invertible Continuous Linear Operator has Unique Inverse

Theorem

Let $\struct {X, \norm {\, \cdot \,} }$ be the normed vector space.

Let $\map {CL} X := \map {CL} {X, X}$ be a continuous linear transformation space.

Let $I \in \map {CL} X$ be the identity element.

Suppose $A \in \map {CL} X$ is invertible.

Then there is a unique $B \in \map {CL} X$ such that $A \circ B = B \circ A = I$.

Proof

Let $B_1, B_2 \in \map {CL} X$.

Suppose:

$A \circ B_1 = I = B_1 \circ A$

$A \circ B_2 = I = B_2 \circ A$

Then:

\(\ds B_1\)

\(=\)

\(\ds I \circ B_1\)

Definition of Identity Element

\(\ds \)

\(=\)

\(\ds B_2 \circ A \circ B_1\)

\(\ds \)

\(=\)

\(\ds B_2 \circ I\)

\(\ds \)

\(=\)

\(\ds B_2\)

Definition of Identity Element

$\blacksquare$

Sources

• 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 2.4$: Composition of continuous linear transformations