I found this question on another forum, and after processing it a bit, I didn't find a good answer. The question is:
Is the sum of two closed operators closable? If not, give an example of two closed operators such that their sum is not closable.
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asked Apr 9, 2011 at 19:43
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$\begingroup$ a quick google search produced following paper : On the sums and products of unbounded linear operators in Hilbert space, M.J.J. Lennon. In the first paragraph the author states that the answer to your question is no. I don't think he gives a counter example in his paper, it seems to be a well known fact, but you might find one in his references. $\endgroup$– Olivier BégassatCommented Apr 9, 2011 at 20:49
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2 Answers
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I might change your question a little bit: Let $X=L^2[0,1]$, $Af:=f''$ with $D(A)=H^2[0,1]$ and let $Bf=f'(0)\cdot\mathbb{1}$, with $D(B)=H^2[0,1]$. Then $B$ is not closable (easy exercise from the definition), but $B$ is relatively $A$-bounded with $A$-bound zero. Hence, $A+B$ is closed (see Kato, Thorem IV.1.1).
This example also answers your question.
answered Apr 9, 2011 at 21:40
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$\begingroup$ Shouldn't $D(B)$ be some set of functions having well-defined derivatives at $0$ , e.g. $C^1([0,1])$ . $H^1[0,1]$ is not appropriate as $D(B)$ . $\endgroup$– TaQCommented Apr 11, 2011 at 9:21
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$\begingroup$ I don't see why the $A$-bound of $B$ is $0$ ? I can see that $B$ is relatively $A$-bounded due to the continuity of the trace operator for the derivative. However it seems that the constant cannot be made arbitrarily close to $0$. ... Sorry to unearth this post after so many years $\endgroup$– ThelebCommented May 29, 2023 at 15:38
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$\begingroup$ See for ecxample Engel-Nagel, One-Parameter Semigroups... Example III.2.2. (page 170) : math.uni-tuebingen.de/de/forschung/agfa/members/… $\endgroup$ Commented May 30, 2023 at 8:08
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See Caradus's paper entitled: Semiclosed operators just the abstract or Messirdi's paper: Almost closed operator.
