A linear operator with a closed graph that is not bounded.

Question

Let $D$ be a linear operator on the subspace $ S=\{(x_n):\sum_n n^2 \vert x_n \vert^2 < \infty \} \subset l_2 $ such that $ D(x_n) = (n x_n) $. How to prove that the graph of $D$ is closed but is not bounded.

Answer 1

$S+I$ is closed iff $S$ is closed. $S+I$ has a bounded inverse $(S+I)^{-1}$, which is closed because it is bounded. So $S+I$ is closed because its graph is the transpose of the graph of $(S+I)^{-1}$. Therefore $S$ is closed.