# 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.

• This might be a stupid question, but is there any reason why $T$ and $T^*$ have the same domain? If not, what does it mean that $\ker T = \ker T^*$? Aug 5, 2022 at 22:27
• I think you should replace $T$ by its closure or assume that $T$ i closed. Otherwise the conclusion is not true even for bounded operators. Aug 6, 2022 at 2:28
• @Yanko The domains of $T$ and ,$T^*$ are equal if $T$ is self-adjoint. Actually that is exactly the definition of self-adjointness. Aug 6, 2022 at 2:32
• The claim is not true. There are symmetric closed operators such that $\ker T=\{0\}$ and $\ker T^*\neq \{0\}.$ The explanation I know makes use of the theory of indeterminate moment problem. Perhaps someone finds an elementary solution. Aug 6, 2022 at 2:54
• Thank you for your comments...I have as an assumption that $\ker T=\ker T^*$, and also that $T$ is closed and symmetric. I think that when the range of $T$ is closed, then this is perhaps true (I need to check the details). But, I want to give this a try without assuming that $T$ has a closed range. Aug 6, 2022 at 6:15

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

• Thank you, Disintegrating By Parts. Aug 7, 2022 at 6:51
• @MohammedHichemMortad : You're welcome. If the answer is acceptable to you, please check the mark to the left of the answer in order to accept the answer. Aug 21, 2022 at 4:55

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 $$L^2(\mu).$$

• Thanks, Ryszard Szwarc. Aug 7, 2022 at 6:52
• You welcome, although I prefer the answer given by @Disintegrating By Parts as it is much more elementary Aug 7, 2022 at 7:35
• Yes, I agree...I have also found another example... Aug 7, 2022 at 10:57