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Let $A$ and $B$ be closed operators on a (separable complex) Hilbert space with dense domains $D(A)$ and $D(B)$ respecitvely. Then, we may define the operator $A+B$ on $D(A)\cap D(B)$. In general, we have no reason to believe that this operator will be closed, which begs the question, is it closable?

I hope I'm not being an idiot again. . . Any ideas?

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Are you assuming that $D(A) \cap D(B)$ is dense? But even then I don't see an immediate reason. A densely defined operator is closable if and only if its adjoint is densely defined and then its closure is $\bar{A} = A^{\ast\ast}$. Why would that be without additional assumptions? – t.b. Jul 26 '11 at 23:31
up vote 9 down vote accepted

On $\ell^2$, define $A$ and $B$ by $(Ax)_n = -(Bx)_n = n^2 x_n$ for $n > 1$, $(A x)_1 = \sum_{n=1}^\infty n x_n$ and $(B x)_1 = 0$, with $D(A) = D(B) = \{x: \sum_{n =1}^\infty n^4 |x_n|^2 < \infty \}$. Then if I'm not mistaken $A$ and $B$ are closed but $A + B$ is not closable, e.g. (with $e_n$ the standard unit vectors) $\lim_{n \to \infty} e_n/n =0$ while $(A + B) e_n/n = e_1$.

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I believe that's correct. Thanks much! – Jonathan Gleason Jul 27 '11 at 16:58
How do we say $A$ and $B$ are closed. Any references? – Vishal Gupta May 31 '13 at 12:21
@VishalGupta their inverses are bounded (in fact compact), and therefore closed – Omnomnomnom May 15 at 21:28

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