Closed graph theorem question?

Question

Let $H$ be a Hilbert space. Let $A:\operatorname{dom}A\to H$ has a closed graph, where $\operatorname{dom}A$ is dense in $H$. Let $S\subseteq \operatorname{dom}A$ be dense. Is it true $A_{|S}$ has a closed graph?

Answer 1

Let $H=L^{2}[0,2\pi]$. Let $f \in \mathcal{D}(A)$ iff $f$ is equal a.e. to an absolutely continuous periodic function $\tilde{f}$ on $[0,2\pi]$ such that $\tilde{f}'\in L^{2}[0,2\pi]$. Define $A : \mathcal{D}(A)\subset H\rightarrow H$ by $Af=-i\tilde{f}'$ for all $f\in\mathcal{D}(A)$. $A$ is closed, and the domain of $A$ is dense in $H$. ($A$ is, in fact, selfadjoint.)

Let $A_{p}$ be the restriction of $A$ to the subspace $\mathcal{D}(A_{p})$ consisting of polynomials in $e^{ix}$ and $e^{-ix}$. Notice that $A_{p}e^{inx}=ne^{inx}$ for $n=0,\pm 1,\pm 2,\ldots $. The trigonometric polynomials are dense in $H$ because $\{ e^{inx}\}_{n=-\infty}^{\infty}$ is a complete orthornomal basis of $H$. So $\mathcal{D}(A_{p})$ is dense in $H$. Furthermore, the graph of $A_{p}$ is dense in the graph of $A$ under the graph norm $\|x\|_{A}^{2}=\|x\|_{H}^{2}+\|Ax\|_{H}^{2}$. To see this, notice that that if $f \in \mathcal{D}(A)$, then $$ f_{n}=\sum_{n=-N}^{N}(f,e_{inx})e^{inx} $$ converges in $L^{2}$ to $f$, and $$ Af_{n} = \sum_{n=-N}^{N}in(f,e^{inx})e^{inx}=\sum_{n=-N}^{N}(Af,e^{inx})e^{inx} $$ converges in $L^{2}$ to $Af$. That is, $(f_{n},Af_{n})$ converges to $(f,Af)$ in $L^{2}\times L^{2}$. So the graph of $A_{p}$ is dense in the graph of $A$. That means that $A_{p}$ cannot be closed because its closure is $A$.