Set of Invertible Continuous Transformations is Open Subset of Continuous Linear Transformations in Supremum Operator Norm Topology

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Theorem

Let $X$ be a Banach space.

Let $\map {CL} X$ be the continuous linear operator space on $X$.

Let $\map {GL} X$ denote the set of all invertible continuous linear operators on $X$.


Then $\map {GL} X \subseteq \map {CL} X$ in the supremum operator norm topology.


Proof

Let $T_0 \in \map {GL} X$.

By definition:

$T_0^{-1} \in \map {CL} X$.

Let $T \in \map {B_\epsilon} {T_0}$ where $\map {B_\epsilon} x$ is an open ball in $\struct {\map {GL} X, \norm {\, \cdot \,} }$ topology.

By definition:

$\norm {T - T_0} < \epsilon$

We also have that:

\(\ds \norm {\paren {T - T_0} \circ T_0^{-1} }\) \(\le\) \(\ds \norm {T - T_0} \norm {T_0^{-1} }\) Supremum Operator Norm on Continuous Linear Transformation Space is Submultiplicative
\(\ds \) \(<\) \(\ds \epsilon \norm {T_0^{-1} }\)

Zero operator is not invertible.

Hence, $T_0^{-1} \ne \mathbf 0$ and $\norm {T_0^{-1} } \ne 0$.

Choose $\epsilon$ such that $\epsilon \norm {T_0^{-1} } < 1$, for example, $\epsilon = \dfrac 1 {2 \norm {T_0^{-1} } }$.

Then:

\(\ds \norm {\paren {T - T_0} \circ T_0^{-1} }\) \(<\) \(\ds \frac 1 2\)
\(\ds \) \(<\) \(\ds 1\)

By Neumann series theorem:

$I + \paren {T - T_0} \circ T_0^{-1} \in \map {GL} X$

However, $T_0 \in \map {GL} X$ too.

Hence:

\(\ds T\) \(=\) \(\ds T_0 + \paren {T - T_0}\)
\(\ds \) \(=\) \(\ds \paren {I + \paren {T - T_0}\circ T_0^{-1} } \circ T_0\)

We have that $T_0 \in \map {GL} X$ and $I + \paren {T - T_0}\circ T_0^{-1} \in \map {GL} X$.


Also, Composite of Bijections is Bijection and Bijection is Invertible.

Hence, their composition is also in $\map {GL} X$:

$T = \paren {I + \paren {T - T_0}\circ T_0^{-1} } \circ T_0 \in \map {GL} X$

$T$ was arbitrary.

So we have the implication that:

$\forall T \in \text{topology of } \struct {\map {GL} X, \norm {\, \cdot \,} } : \paren {T \in \map {GL} X} \implies \paren {T \in \map {CL} X}$

or in other words:

$\map {GL} X \subseteq \map {CL} X$

in the supremum operator norm topology.


$\blacksquare$


Sources

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