Has this theorem (on existence of inverse) an analogous for unbounded operators? Let $S,T:X\to X$ be bounded linear operators, where $X$ is a Banach space. It's a consequence of the Banach Fixed Point Theorem that if $T$ is invertible and $\|T-S\|\|T^{-1}\|<1$ than $S$ is invertible.
I would like to know if is there a similar result for unbounded operators that are defined just in a dense subset of $X$. More precisely, let $T,S:Y\subset X\to X$ be linear operators, where $X$ is a Banach space and $Y$ is dense in $X$ (you can suppose $X$ Hilbert if you want). Supose that:  
(1) $T$ is invertible;
(2) $T^{-1}$ and $T-S$ are bounded;
(3) $\|T-S\|\|T^{-1}\|<1$.
Is $S$ invertible?
Thanks.

IF $T-S$ is bounded on $Y$, and if $T^{-1} : X\rightarrow Y$ is bounded, then
$(T-S)T^{-1}=I-ST^{-1}$ is bounded on $X$, and $\|I-ST^{-1}\| < 1$ by your assumptions. Therefore, $ST^{-1}=I-(I-ST^{-1})$ is invertible in $\mathcal{L}(X)$. And $(ST^{-1})^{-1}ST^{-1}=ST^{-1}(ST^{-1})^{-1}=I$. It follows that $S$ is surjective from the second equality. And, because $T^{-1}$ must have a range equal to $Y$, then $(ST^{-1})^{-1}ST^{-1}=I$ implies $S$ is injective. So $S$ is invertible, and $T^{-1}(ST^{-1})^{-1}=S^{-1}$ must be bounded.
