Has this theorem (on existence of inverse) an analogous for unbounded operators?

Question

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.

Answer 1

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.

Answer 2

From the inequality $\lVert T-S\rVert \lVert T^{-1}\rVert < 1$, it follows that

$$R = (I - T^{-1}(T-S))$$

is an invertible bounded operator. Since $TR = T(I - T^{-1}(T-S)) = T - (T-S) = S$, we get the representation $S^{-1} = R^{-1}T^{-1}$ in the bounded case. Now we can verify that $R^{-1}T^{-1}$ is also the inverse of $S$ when $S$ is unbounded.

Since $T-S$ is bounded, we have $D(T) = D(S)$, and the identity $S = T - (T-S) = TR$ holds. Thus for every $x\in X$, we have

$$SR^{-1}T^{-1}x = TRR^{-1}T^{-1}x = TT^{-1}x = x,$$

and for $x \in D(S) = D(T)$,

$$R^{-1}T^{-1}Sx = R^{-1}T^{-1} TRx = R^{-1}Rx = x.$$