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 $\mathbb F \in \set {\R, \C}$.

Let $\map {c_{00} } {\mathbb F}$ be the space of almost-zero sequences on $\mathbb F$.

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

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.

$A$ is a linear operator
By definition, $A$ is a linear operator.

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

$A$ is a continuous operator
By definition, $A$ is a diagonal operator with $\ds \sequence {\lambda_i}_{i \mathop \in \N_{> 0} } = \sequence {\frac 1 i}_{i \mathop \in \N_{> 0} }$.

By Supremum Operator Norm of Diagonal Operator over Bounded Sequence Space:


 * $\ds \norm {A} = \sup_{i \mathop \in \N_{> 0} } \size {\lambda_i}$

Then:

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

$A$ is an injection
Suppose:


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

Then:

By definition, $A$ is injective.

$A$ is a surjection
Let $\mathbf y \in c_{00}$ and $N \in \N_{> 0}$ such that:


 * $\forall i \in \N_{> 0} : \paren {i \le N \implies y_i \in \R} \land \paren {i > N \implies \paren {y_i = 0} }$

Also, Real Multiplication is Closed.

Then:


 * $\exists \mathbf x \in c_{00} : \forall i \in \N_{> 0} : \paren {i \le N \implies \paren {x_i = i \cdot y_i \in \R} } \land \paren {i > N \implies \paren {x_i = 0} }$

Hence:

By definition, $A$ is surjective.

By definition, $A$ is bijective.

$A$ is not invertible
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:

However, by definition of supremum operator norm:


 * $\norm B = \map \sup {\norm {\map B {\mathbf x} }_\infty : \mathbf x \in c_{00}, \norm {\mathbf x}_\infty \le 1}$

where $\norm {\map B {\mathbf x}}_\infty \in \R$.

This is a contradiction.

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