Let me remind you that I already have a counterexample when $T$ is not symmetric.

Also, it is plain that for a counterexample we must avoid a $T$ for which $T^2$ is densely defined.

I expect not. For a counter example I would consider a differential operator where the coefficients have differing differentiability conditions so the formal adjoint might not exist in a normal sense such as out of the definition of the adjoint associated with a ordinary operator where the coefficients have no smoothness and the formal adjoint is a "quasi differential operator" in the verbiage of Naimark. Naimark covered this topic in detail in his books Linear Differential Operators.

https://www.amazon.com/Linear-Differential-Operators-Elementary-Hilbert/dp/B095RLJNDY/ref=sr_1_27?Adv-Srch-Books-Submit.x=1918&Adv-Srch-Books-Submit.y=597&qid=1645741017&refinements=p_27%3AM.+A.%5CcNaimark&s=books&sr=1-27&unfiltered=1

That Naimark's work in differential equations was not better known is probably a victim of the Cold War. My wife at the time was a Russian linguist and I didn't have to depend on translations and not all of his better works were - but those two volumes above were.

In the thesis of G. C. Rota, a student of Jacob Schwartz, on the extension theory of ordinary differential operators he develops a theory which might point the way to the counter example you seek. His results were published in the Communications on Pure and Applied Mathematics in 1958. Another place to look is in the classic Dunford and Schwartz "Linear Operators", vol II where there is an detailed development of the extension theory of ordinary operators on Banach spaces (L^p).

My gut tells me without other conditions the conjecture is false and the counter example lies in differential operators.

Dear Professor T. Prevatt

Thank you very much for this detailed answer, which is also very informative.

In the first place, I thought a counterexample would be fairly simple, and I then thought that I just was not aware of it. Now, I think it has to be constructed in several steps. In Reed-Simon's (Vol. 2), the authors assumed somewhere $T^2$ is densely defined even for a symmetric operator $T$, and that has made me think that the answer to my question was probably negative (I have just emailed Prof. B. Simon who is probably aware of such a counterexample).

I will have a look at Naimark's paper and your other suggestions. If anything is found, I would certainly let you know.

If the answer to my question is positive (which is rather unlikely), then it would also be an interesting question to investigate the question for a closable and densely defined $T$?

Regards,

Hichem

This question spurred my interest. As for as closure of the formal adjoint - see Theorem 1.1 of Part II of Article Application of exponential dichotomies to asymptotic integra...

For background and if you cannot find Naimark's work see W. T. Reid's paper

https://www.ams.org/journals/tran/1957-085-02/S0002-9947-1957-0088625-6/S0002-9947-1957-0088625-6.pdf

is useful. When the coefficients of a differential operator are sufficient smooth so that D^j(ak(t) f(t)) can be calculated and the domains of definition of the formal adjoint are obvious. If not - then great care needs to be taken. The same would be true when one defines the square of such an operator if the coefficients are not sufficiently smooth. For a counter example, I expect one could find it in Strum-Louisville theory.

Great again! I will read them thoroughly.

I wanted to tell you that I have already found a densely defined (unclosable) T such that $D(T^2)=D(T^*)=D(TT^*)=D(T^*T)=\{0\}$. Perhaps this example may be used in some other way (e.g. using a matrix of operators) to obtain a symmetric operator with the very above required properties.

Preprint Some properties of the adjoint of the unbounded operators $T...

This study is very important.

Thanks Benamar...

I dont know, but my feeling is that it is not true. Best regards, K.S.

Thanks a lot for your comment Konrad Schmüdgen. Please have a look at

Preprint Some properties of the adjoint of the unbounded operators $T...

Regards,

Hichem

There are certain Chenroff like counterexamples.