is $\mathcal{D}(A)=\mathcal{D}(B)$ a sufficient condition for $(A+B)^*=A^*+B^*$ , where $A$ and $B$ are densely defined (not necessarily symmetric) operators on some Hilbert space?
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2Certainly not. Consider the case A=-B. – Michael Renardy Jun 02 '14 at 15:01
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Is it really a counterexample? – Jun 02 '14 at 15:06
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Or do you mean that we have to assume at least that $A^$ resp. $B^$ have dense domain? – Jun 02 '14 at 15:08
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1$(A+B)^=0$, $A^+B^=0|_{D(A^)}$. – Michael Renardy Jun 02 '14 at 15:20
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Yes, so could we save the statement by assuming that $\mathcal{D}(A^)$ and $\mathcal{D}(B^)$ are dense? – Jun 02 '14 at 15:24
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@user51527: No, if $A=-B$ then $(A+B)^$ is everywhere* defined, so unless $A^*$ is everywhere defined (i.e. bounded, by the closed graph theorem), the equality will not hold as written. – Nate Eldredge Jun 02 '14 at 15:50
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1Okay, but $A^-A^=0\upharpoonright_{\mathcal{D}(A^*)}$ is a bounded densely defined operator and therefore has a unique extension to $0$ on the whole space. What seems to go wrong, is that we lose closedness. One last try: what if $B$ is in addition $A$-bounded with $A$-bound strictly less than 1? – Jun 02 '14 at 15:59
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I guess I should have assumed that $A,B$ are closed in the first place. – Jun 02 '14 at 15:59
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1ADJOINT OF SUMS AND PRODUCTS OF OPERATORS INHILBERT SPACES https://arxiv.org/pdf/1507.08418.pdf This paper give condition for this to hold. – Manuel Dec 27 '19 at 17:57
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I know this is an old question, but OP was almost right in its last comments. If $A$, $B$ are closed operators, and we assume $B$ to be $A$-bounded and $B^$ to be $A^$-bounded both with relative bounds strictly less than one, then $(A+B)^=A^+B^*$. This is Corollary 1 in P. Hess, T. Kato, Perturbation of Closed Operators and Their Adjoints, Comment. Math. Helv., 45 (1970) 524-529. In particular, this property reduces to Kato-Rellich's theorem when $A$ is self-adjoint and $B$ is symmetric. – Davide Oct 25 '21 at 12:36