Topological isomorphisms in normed spaces Let $X$ be a complete normed space and let $T$ and $S$ be fixed bounded linear operators from $X$ into $X$. Suppose that $S$ is a topological isomorphism and that $\|T-S\|< \frac{1}{\|S^{-1}\|}$. Show that $T$ is a topological isomorphism.
I want to show that $T$ is surjective and that there are constant $A,a>0$ that satisfy $a\|x\| \leq \|Tx\| \leq A\|x\|$. Equivalently I could just show that both $T$ and $T^{-1}$ are bounded/continuous. I am having a hard time getting started with this. Of course, I can say that $\|Tx\| \leq \|T\|\|x\|$ and this proves the existence of such an $A$. The other direction is more difficult to show through. Also, I am not even sure I can prove that $T$ must be surjective in this case. Any help is welcomed. Thanks in advance.
Krull.

$\|S^{-1}\|\|T-S\|<1$ implies that $\|S^{-1}(T-S)\| \leq \|S^{-1}\|\|T-S\|<1$.
We deduce that $\|S^{-1}T-I\|<1$.
This implies that $I+(S^{-1}T-I)=S^{-1}T$ is invertible, (its inverse is $\sum_{n\geq 0}(-1)^n(S^{-1}T-I)^n$ which exists since $X$ is complete) and $T=S(S^{-1}T)$ is invertible.
