TITLE: Sum of Closed Operators Closable?
QUESTION [10 upvotes]: 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?

REPLY [12 votes]: 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 $(A + B) e_n/n = e_1$.