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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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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. – Olivier Bégassat Apr 9 '11 at 20:49
up vote 4 down vote accepted

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.

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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 $D(B)$ . – TaQ Apr 11 '11 at 9:21
Thanks, I correct the typo. – András Bátkai Apr 11 '11 at 13:06

See Caradus's paper entitled: Semiclosed operators just the abstract or Messirdi's paper: Almost closed operator.

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