> [!definition] > > Let $\cx$ be a [[Topological Vector Space|TVS]], then the space of **toplinear isomorphisms** $\text{Laut}(\cx)$ is the space of [[Continuity|continuous]] invertible [[Bounded Linear Map|linear maps]] whose inverses are continuous. > [!theorem] > > Let $\cx$ be a [[Banach Space|Banach space]] and $T \in L(\cx, \cx)$ such that $\norm{I - T} < 1$, then $T \in \text{Laut}(\cx)$. Moreover, $T^{-1} = \sum_{n = 0}^\infty(I - T)^n$, where the series converges in $L(\cx, \cx)$. > > *Proof*. Since $\norm{I^{-1}}^{-1} = 1$ and $\norm{I - T} < 1$, the series is absolutely convergent. > > Now, > $ > \begin{align*} > (I - T) \cdot \sum_{n = 0}^\infty(I - T)^n &= \sum_{n = 1}^\infty(I - T)^n \\ > &= \sum_{n = 0}^\infty(I - T)^n - I > \end{align*} > $ > so $T \cdot \sum_{n = 0}^\infty(I - T)^n = I$. Similarly, > $ > \braks{\sum_{n = 0}^\infty(I - T)^n} \cdot (I - T) = \sum_{n = 0}^\infty(I - T)^n - I > $ > so $\sum_{n = 0}^\infty(I - T)^n \cdot T = I$. Therefore $T$ is invertible. > [!theorem] > > Let $\cx$ be a Banach space, then $\text{Laut}(\cx)$ is [[Open Set|open]] in $L(\cx, \cx)$. > > *Proof.* Let $T \in \text{Laut}(\cx)$ and $S \in L(\cx, \cx)$ with $\norm{S - T} < \norm{T^{-1}}^{-1}$, then > $ > \begin{align*} > \norm{I - T^{-1}S} &\le \norm{T^{-1}} \cdot \norm{T - S} < 1 > \end{align*} > $ > Therefore $T^{-1}S \in \text{Laut}(\cx)$, and $S \in \text{Laut}(\cx)$ as well.