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
$$\langle (T-\lambda I)^{-1}x,y \rangle = \langle x, [(T-\lambda I)^{-1}]^* y \rangle = \langle x, (T^* - \overline \lambda I)^{-1} y \rangle = \langle x, (T-\lambda I)^{-1}y \rangle,$$
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!