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The Kato-Rellich Theorem is a classical result stating that if $A,B$ are unbounded operators on a Hilbert space with $A$ self-adjoint, $B$ symmetric, $\mathcal D (A)\subset \mathcal D(B)$ and $$ \|Bx\|\leq a\|A x\|+b\|x\|,\;\;x\in\mathcal D(A) $$ for positive constants $a,b$, with $a<1$, then $A+B$ is self-adjoint on the domain of $A$. The infimum of those $a$ for which the above inequality holds is called the relative bound of $B$. Thus, $A+B$ is self-adjoint if $B$ is relatively bounded with bound $a<1$.

I'm interested in non-examples where $A+B$ is not essentially self-adjoint even though $A$ is self-adjoint, $B$ is symmetric and $\mathcal D(A)\subset\mathcal D(B)$.

I have tried considering multiplication operators to no avail. Let $M_\phi$ denote the multiplication operator $$ M_\phi h=\phi h,\;\;\mathcal D(M_{\phi})=\{ h|\phi h\in L_2\}, $$ and let $A=M_f$, $B=M_g$ with $f,g$ real functions, finite almost everywhere. Since $A$ and $B$ are closed, the assumption $\mathcal D(A)\subset\mathcal D(B)$ implies the bound $$ \|B h\|\leq C(\|A h\| +\|h\|) $$ for some $C>0$. But then, for $h$ non-zero only on the set $\{x||f(x)|\leq M\}$, one obtains $$ \|B h\|\leq C(1+M)\|h\|, $$ which implies that $|g(x)|\leq C(1+M)$ almost everywhere on $\{x||f(x)|\leq M\}$. This in turn implies $|f(x)+g(x)|\leq (M+ CM + C)$ on that same set. But then $\mathcal D(A)$ is a core for $\mathcal D(M_{f+g})$, which implies that $A+B$ is essentially self-adjoint.

Next up were Schrödinger operators. 'Unfortunately', Theorem XIII.96 in Methods of Modern Mathematical Physics vol 4 by Reed and Simon implies that (on $\mathbb R^d$, with $d\leq 3$) the inclusion $\mathcal D(-\Delta)\subset\mathcal D(V)$ is sufficient in order to conclude that $V$ is relatively $(-\Delta)$-bounded with bound $0$.

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  • $\begingroup$ Thanks for sharing, if you could please elaborate why the assumption $\mathcal D(A)\subset\mathcal D(B)$ implies the bound $$ \|B h\|\leq C(\|A h\| +\|h\|) $$ for some $C>0$. $\endgroup$
    – Harish
    Mar 20, 2015 at 12:20
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    $\begingroup$ @Harish Let H denote the Hilbert space that A,B are operators in. You want to show that B is bounded as an operator from the domain of A (with the graph norm) into H. It suffices, by the closed graph theorem, to show that B is closed as an operator from the domain of A (with the graph norm) into H. Now use that, on the domain of A, convergence with respect to the graph norm implies convergence in H. Do you see how to finish the argument? $\endgroup$ Mar 20, 2015 at 14:50

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Suggestion: start with a symmetric operator $C$ that is not essentially self adjoint, and $A$ with ${\mathcal D}(A) \subset {\mathcal D}(C)$ that is self adjoint. Take $B = C - A$.

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    $\begingroup$ Ah, of course. Thank you! So an example might be $C=-i\partial_x$ on $\mathcal D(C)=H_0^1([0,1])$. Then $C$ is closed and symmetric, $C\neq C^*$, $C^*C$ is self-adjoint with $\mathcal D(C^*C)\subset\mathcal D(C)$ and $C$ is relatively $C^*C$-bounded with $C^*C$-bound $0$. It follows that $C^*C+t(C-C^*C)$ is self-adjoint for all $t\neq 1$, while $C$ is clearly not essentially self-adjoint. $\endgroup$ Jul 11, 2014 at 17:00

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