# 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 $\mathbb F$.

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

### $A$ is a linear operator

 $\ds \forall \mathbf x, \mathbf y \in \ell^\infty : \forall \lambda \in \mathbb F: \,$ $\ds \map A {\mathbf x} + \lambda \map A {\mathbf y}$ $=$ $\ds \tuple {x_1, \frac {x_2} 2, \frac {x_3} 3, \ldots} + \lambda \tuple {y_1, \frac {y_2} 2, \frac {y_3} 3, \ldots}$ $\ds$ $=$ $\ds \tuple {x_1 + \lambda y_1, \frac {x_2} 2 + \lambda \frac {y_2} 2, \frac {x_3} 3 + \lambda \frac {y_3} 3, \ldots}$ Space of Bounded Sequences with Pointwise Addition and Pointwise Scalar Multiplication on Ring of Sequences forms Vector Space $\ds$ $=$ $\ds \tuple {x_1 + \lambda y_1, \frac {x_2 + \lambda y_2} 2, \frac {x_3 + \lambda y_3} 3, \ldots}$ $\ds$ $=$ $\ds \map A {\mathbf x + \lambda \mathbf y}$

By definition, $A$ is a linear operator.

$\Box$

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

### $A$ is a continuous operator

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

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

Then:

 $\ds \forall \mathbf x \in c_{00}: \,$ $\ds \norm {\map A {\mathbf x} }_\infty$ $\le$ $\ds \norm A \norm {\mathbf x}_\infty$ Supremum Operator Norm as Universal Upper Bound $\ds$ $=$ $\ds 1 \cdot \norm {\mathbf x}_\infty$ $\ds$ $=$ $\ds \norm {\mathbf x}_\infty$

$\Box$

### $A$ is an injection

Suppose:

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

Then:

 $\ds \map A {\mathbf x}$ $=$ $\ds \map A {\mathbf y}$ $\ds \leadsto \ \$ $\ds \mathbf 0$ $=$ $\ds \map A {\mathbf x} - \map A {\mathbf y}$ $\ds \leadsto \ \$ $\ds \mathbf 0$ $=$ $\ds \map A {\mathbf x - \mathbf y}$ $A$ is a linear operator $\ds \leadsto \ \$ $\ds \tuple {0, 0, 0, \ldots}$ $=$ $\ds \tuple {x_1 - y_1, \frac {x_2 - y_2} 2, \frac {x_3 - y_3} 3, \ldots}$ $\ds \leadsto \ \$ $\ds \forall i \in \N_{>0}: \,$ $\ds x_i$ $=$ $\ds y_i$ $\ds \leadsto \ \$ $\ds \mathbf x$ $=$ $\ds \mathbf y$

By definition, $A$ is injective.

$\Box$

### $A$ is a surjection

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

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

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:

 $\ds \forall \mathbf y \in c_{00} : \exists \mathbf x \in c_{00}: \,$ $\ds \map A {\mathbf x}$ $=$ $\ds \map A {\tuple {1 \cdot y_i, 2 \cdot y_2, 3 \cdot y_3, \ldots} }$ $\ds$ $=$ $\ds \tuple {\frac {1 \cdot y_i} 1, \frac {2 \cdot y_2} 2, \frac {3 \cdot y_3} 3, \ldots}$ $\ds$ $=$ $\ds \tuple {y_1, y_2, y_3, \ldots}$ $\ds$ $=$ $\ds \mathbf y$

By definition, $A$ is surjective.

$\Box$

By definition, $A$ is bijective.

### $A$ is not invertible

Aiming for a contradiction, suppose 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:

 $\ds \forall m \in \N_{> 0}: \,$ $\ds 1$ $=$ $\ds \norm {\mathbf e_m}_\infty$ $\ds$ $=$ $\ds \norm {B \circ \map A {\mathbf e_m} }_\infty$ $\ds$ $\le$ $\ds \norm B \norm {\map A {\mathbf e_m} }_\infty$ $\ds$ $=$ $\ds \norm B \cdot \frac 1 m$ $\ds \leadsto \ \$ $\ds \forall m \in \N_{> 0}: \,$ $\ds m$ $\le$ $\ds \norm B$ $\ds \leadsto \ \$ $\ds \norm B$ $=$ $\ds \infty$ Definition of Infinity

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

$\Box$
Hence, $A$ is bijective, but not invertible.
$\blacksquare$