If a closed symmetric operator $T$ has at least one real number in its resolvent set, then does it follow that $T$ is self adjoint? Since $T$ is closed and symmetric, we know that it is extended by its adjoint. We just need to show if we assume that for some $\lambda \in \mathbb{R}$ the operator $T-\lambda I$ is invertible, then this means $T^*$ is also extended by $T$ and hence $T=T^*$.
I do not know how to proceed however, is this supposeed to be a simple exercise?
Claim: if $T$ is dense, symmetric and there exists $\lambda \in \mathbb C$ such that $\lambda - T : D(T) \rightarrow H$ and $\overline \lambda - T : D(T) \rightarrow H$ are onto, then $T$ is self adjoint.
Proof: We have $T \subset T^*$ and we are left to show that $T^* \subset T$, or equivalently $D(T^*) \subset D(T)$. So let $x \in D(T^*)$, use the hypothesis to get $z \in D(T)$ such that $(\lambda - T)z = (\lambda - T^*)x$ and then write $$ (x, (\overline{\lambda}-T)y)_H = ((\lambda - T^*)x,y)_H = ((\lambda - T)z,y)_H = (z,(\overline{\lambda}-T)y)_H,\quad \forall y \in D(T). $$ Use the surjectivity of $\overline{\lambda}-T$ to get $$ x=z \in D(T). $$