Bijective Continuous Linear Operator is not necessarily Invertible

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 bijective.

Then $A$ is not necessarily invertible.

Proof
Let $c_{00}$ be the space of almost-zero sequences.

Let $\mathbf x = \tuple {x_1, x_2, \ldots, x_N, 0, \ldots} \in c_{00}$.

Let $A : c_{00} \to c_{00}$ be a mapping such that:


 * $\ds \map A {\tuple {x_1, x_2, x_3, \ldots}} = \tuple {x_1, \frac {x_2}{2}, \frac{x_3}{3}, \ldots}$

Let $\struct {\ell^\infty, \norm {\, \cdot \,}_\infty}$ be the normed vector space of bounded sequences.

We have that Space of Almost-Zero Sequences is Subspace of Space of Bounded Sequences:


 * $c_{00} \subseteq \ell^\infty$

Moreover, Space of Almost-Zero Sequences with Supremum Norm is Normed Vector Space.

Let $\norm {\, \cdot \,}$ be the supremum operator norm.

Then:

Also:

By definition, $A$ is a linear operator.

By Continuity of Linear Transformations, $A$ is continuous.

Suppose:


 * $\forall \mathbf x, y \in \ell^\infty : \map A {\mathbf x} = \map A {\mathbf y}$.

Then:

By definition, $A$ is injective.

Moreover, surjective.

By definition, $A$ is bijective.

there is $B \in \map {CL} {c_{00}}$ which is the inverse of $A$.

Let $\mathbf e_m = \tuple {\underbrace{0, \ldots, 0}_{m - 1}, 1, 0, \ldots}$.

Then:

This is a contradiction.

Hence, $A$ is bijective, but not invertible.