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 $\{m_n\}_{n=0}^\infty $ be a positive definite sequence, i.e. the  Hankel matrix $\{m_{i+j}\}$ is positive definite. Due to Hamburger theorem we have $$m_n=  \int\limits_{\mathbb{R}}x^n\,d\mu(x)\qquad (*)$$ for a positive measure $\mu  .$  The sequence $m_n$ is called indeterminate if the measure $\mu$ is not uniquely determined.
Fix an indeterminate sequence $m_n.$ Then there are so  called $N$-extremal measures satisfying $(*).$
Such measures have discrete unbounded supports, which are disjoint and cover the real line.
Moreover the polynomials are dense in $L^2(\mu).$
Due to indeterminacy the operator $M_x$ defined on polynomials by $$M_xp(x)=xp (x)$$ is not essentially self-adjoint, hence its closure is not self-adjoint.
Let $\mu$ be one of such measures satisfying  $0\notin {\rm supp}\,\mu.$
The adjoint operator $M_x^*$ is then  injective as the range of $M_x$ is dense in $L^2(\mu).$
