Say $A, B : H \supset D \to H$ are essentially selfadjoint operators on the dense common domain $D$. $H$ is some Hilbert space. Does it hold that $A + B$ is also essentially selfadjoint? If not, can you please give me a counterexample?
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2$\begingroup$ Why don't you just leaf through one of the hundreds of books on spectral theory and/or theory of operators (e.g., Reed and Simon's treatise)? a large part of the theory is devoted to answer precisely this question $\endgroup$– Piero D'AnconaJun 4, 2011 at 8:57
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Let $D=H^4(0,1)\cap H^2_0(0,1)$, $Au=u''''$, $Bu=u''''+u''$.

2$\begingroup$ It may be helpful to note that the reason counterexamples exist is that the intersection of the two domains may be smaller than each of the domains; and if $A$ is ess. selfadjoint with domain $D$ and $D_0\subset D$ is dense, the closure of the restriction of $A$ to $D_0$ may not coincide with $A$. (If the closure of the restriction of $A$ to $D_0$ is $A$ one sometimes says that $D_0$ is a core for $A$). So your $A+B$ will be ess. selfadjoint if you assume that there is a common core for $A$ and $B$. $\endgroup$ Jun 4, 2011 at 7:43

1$\begingroup$ In the example above D is a core for A and B. It is not a core for A+B. $\endgroup$ Jun 4, 2011 at 10:41