Let $T:D(T)\to H$ be an unbounded densely defined operator on a Hilbert space $H$. Suppose that $T^{-1}$ is continuous, i.e. that $0$ belongs to the resolvent set $\rho(T)$ of $T$.
As $T^{-1}$ exists, this means that $T$ is injective, isn't it? So $T^{-1}$ is a left inverse (but not necessarily a right one, correct?).
Alternatively, suppose that I have the following informations:
By definition, $T^{-1}f = u\iff f = Tu$. If for any $f\in H$, there exists a unique $u\in D(T)$ such that the previous holds, this exactly means that the operator $T^{-1}$ is continuous, no? Actually, this overall means that $T$ is surjective, which seems to imply that $T$ is bijective since being invertible means that it is injective.
My main questions:
- Is the reasoning about the continuity of $T$ right?
- Where is the flaw in my reasoning about the injectivity and surjectivity. This would mean that any operator with $0$ in its resolvent set is bijective from $D(T)$ to $T$ which seems to be surprising for some operators such as the laplacian on an appropriate domain.
1) Your reasoning about continuity is wrong. Best is to read a bit about continuity.
2) Yes, the map is a bijection