Bijective Continuous Linear Operator is not necessarily Invertible

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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

\(\ds \forall \mathbf x, \mathbf y \in \ell^\infty : \forall \lambda \in \mathbb F: \, \) \(\ds \map A {\mathbf x} + \lambda \map A {\mathbf y}\) \(=\) \(\ds \tuple {x_1, \frac {x_2} 2, \frac {x_3} 3, \ldots} + \lambda \tuple {y_1, \frac {y_2} 2, \frac {y_3} 3, \ldots}\)
\(\ds \) \(=\) \(\ds \tuple {x_1 + \lambda y_1, \frac {x_2} 2 + \lambda \frac {y_2} 2, \frac {x_3} 3 + \lambda \frac {y_3} 3, \ldots}\) 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 $\ds \sequence {\lambda_i}_{i \mathop \in \N_{> 0} } = \sequence {\frac 1 i}_{i \mathop \in \N_{> 0} }$.

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

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

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
\(\ds \) \(=\) \(\ds 1 \cdot \norm {\mathbf x}_\infty\)
\(\ds \) \(=\) \(\ds \norm {\mathbf x}_\infty\)


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

$\Box$

$A$ 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:

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

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:

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

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

This is a contradiction.

$\Box$

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

$\blacksquare$

Sources

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