Set of Invertible Continuous Transformations is Open Subset of Continuous Linear Transformations in Supremum Operator Norm Topology
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:
\(\ds \norm {\paren {T - T_0} \circ T_0^{-1} }\) | \(<\) | \(\ds \frac 1 2\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds 1\) |
- $I + \paren {T - T_0} \circ T_0^{-1} \in \map {GL} X$
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
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous): Chapter $\S 2.4$: Composition of continuous linear transformations