Is $T$ self-adjoint if it is closed, symmetric and $\ker T=\ker T^*$? Let $T$ be a densely defined, closed, unbounded, and symmetric linear operator (i.e., $T\subset T^*$) defined in a Hilbert space, with domain $D(T)$.
Is it true that if we further suppose $\ker T=\ker T^*$, then $T$ is self-adjoint, i.e., $T=T^*$?
I know this is true for the class of quasinormal operators. Recall also, that we always have $\ker T\subset \ker T^*$ when $T$ is symmetric.
I hope this is not too obvious!
Thanks a lot.

Let $\mathcal{H}=L^2[0,\infty)$ and let $T : \mathcal{D}(T) \subset \mathcal{H}\rightarrow\mathcal{H}$ be defined as $Tf=if'$ on the domain $\mathcal{D}(T)$ consisting of all absolutely continuous functions $f\in L^2[0,\infty)$ such that $f(0)=0$ and $f'\in L^2[0,\infty)$. $T$ is a closed, densely-defined, symmetric linear operator. However, $T^*$ is not the same as $T$ because the domain of $T^*$ includes functions $f$ that do not vanish at $0$, unlike the functions in the domain of $T$. The closed symmetric operator $T$ is not self-adjoint, even though $$\mathcal{N}(T)=\mathcal{N}(T^*)=\{0\}.$$
