Are resolvents for self-adjoint operators themselves self-adjoint?

[AxiomOfChoice]

Let T be a (possibly unbounded) self-adjoint operator on a Hilbert space \mathscr H with domain D(T), and let \lambda \in \rho(T). Then we know that (T-\lambda I)^{-1} exists as a bounded operator from \mathscr H to D(T). Question: do we also know that (T-\lambda I)^{-1} is self-adjoint? Can someone prove or give a counterexample?

 

[AxiomOfChoice]

Note: I have reason to believe this is true in the case that T is a positive operator (i.e., (Tx,x) \geq 0 for all x\in D(T)). Whether it is true in general...well, I'm not sure!

 

[AxiomOfChoice]

I believe I have a quick and easy proof of this proposition, provided the resolvent has the form (T-\lambda I)^{-1} for \lambda \in \mathbb R. We have
<br /> \langle (T-\lambda I)^{-1}x,y \rangle = \langle x, [(T-\lambda I)^{-1}]^* y \rangle = \langle x, (T^* - \overline \lambda I)^{-1} y \rangle = \langle x, (T-\lambda I)^{-1}y \rangle,<br />
since \lambda = \overline \lambda and T^* = T by assumption. This also uses the fact that (T^*)^{-1} = (T^{-1})^* for T bounded and invertible. Now, if \lambda is complex, this argument is obviously bogus, and in fact I'm reasonably confident that it's just not true!