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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.

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  • $\begingroup$ 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 $\ker T = \ker T^*$? $\endgroup$
    – Yanko
    Aug 5 at 22:27
  • $\begingroup$ 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. $\endgroup$
    – Ryszard Szwarc
    Aug 6 at 2:28
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    $\begingroup$ @Yanko The domains of $T$ and ,$T^*$ are equal if $T$ is self-adjoint. Actually that is exactly the definition of self-adjointness. $\endgroup$
    – Ryszard Szwarc
    Aug 6 at 2:32
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    $\begingroup$ 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. $\endgroup$
    – Ryszard Szwarc
    Aug 6 at 2:54
  • $\begingroup$ 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. $\endgroup$
    – Mohammed Hichem Mortad
    Aug 6 at 6:15

2 Answers 2

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

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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 $L^2(\mu).$

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

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