Is $T$ self-adjoint if it is closed, symmetric and $\ker T=\ker T^*$?

Question

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.

Answer 1

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 $$\mathcal{N}(T)=\mathcal{N}(T^*)=\{0\}.$$

Answer 2

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 $L^2(\mu).$