closablity of unbounded operator

Question

Let $A$ be a Banach *-algebra, $\delta:D(\delta)\rightarrow A$ be a unbounded operator which is closed. Is that possible that a sequence $x_{n}\in D(\delta)$, converge to $x\in D(\delta)$, but $\delta(x_{n})$ does not converge to $\delta(x)$? Clearly $\delta(x_{n})$ cannot be a converging sequence if that is possible.

Answer 1

Yes it is possible. Take $A=C[0,1]$ with the supremum norm. It is a Banach algebra. Take $\delta : C^1 [0,1] \to C[0,1] $, $\delta (f) =f' .$ The operator $\delta $ is unbounded. We show that it is closed. So suppose that $f_n \to f$ and $f_n' \to g$ in the supremum norm, then $f_n \to f $ and $f_n'\to g$ uniformly hence $f$ is differentiable and $f'=g . $ So $\delta$ is closed. Now take $$ x_n (t) =\frac{\sin nt }{\sqrt{n}}$$ then $x_n \to 0$ in $A$ but $$\delta (x_n ) = \sqrt{n}\cos (nt) $$ does not coverge to $\delta (0) =0$ in $A.$