inverse of self-adjoint operator is self-adjoint Given a (unbounded) self-adjoint operator $T: D(T) \subset X \to X$. 
Assume that $T^{-1}$ exists. Is it true that $T^{-1}$ is self-adjoint?
My understanding is that for any $r,s \in X$, there exist some $u, v \in D(T)$ such that 
$$Tu = r, \quad Tv= s.$$
But then, 
$$(T^{-1}r, s) = (u, Tv) = (Tu, v) = (r, T^{-1}s),$$
so $T^{-1}$ must be self-adjoint. 
However, I got some feeling that it's not that easy. Any suggestions/references are welcome!

Here's one result:


If $T$ is self-adjoint on a Hilbert space $H$ and is injective, then $T^{-1},$ with dense domain $R(T)$, is self-adjoint.


Reference: functional analysis appendix of Michael Taylor's PDE I textbook
