# Essential self-adjointness of restriction of symmetric operator

Suppose $$T$$ is a (unbounded) symmetric operator on Hilbert space $$\mathscr{H}$$, defined on $$D$$. Let $$D_1 \subseteq D$$ is a linear subset dense in $$D$$, $$T|_{D_1}$$ be the restriction of $$T$$ on $$D_1$$. Show that if $$T|_{D_1}$$ is essentially self-adjoint, then $$T$$ is essentially self-adjoint, and $$\overline{T} = \overline{T|_{D_1}}$$. I have the answer following.

It suffices to show that $$\overline{\Gamma (T)} \subseteq \overline{\Gamma (T|_{D_1})}$$, then I choose a sequence in $$D_1$$ converge to x $$\in D$$, using diagnal technique.Simply let $$x_n(\in D) \rightarrow x$$, while $$T(x_n) \rightarrow y$$ and let $$x_k^{(n)}(\in D_1)\rightarrow x_n$$ then $$x^{(n)}_n \rightarrow x$$, but how can I show that $$T(x^{(n)}_n) \rightarrow y$$? Thanks a lot in advance!

• What is the question? Are you asking whether your proof is correct? The proof is correct, although it is possible (but not necessary) to add more details (like why $\overline{\Gamma(T)} = \overline{\Gamma(T\lvert_{D_1})}$) Commented Nov 19, 2020 at 15:55
• @s.harp I submitted accidentally, now you can see the whole question. Commented Nov 19, 2020 at 16:23

Let $$(x, Tx)$$ lie in $$\Gamma(T)$$. Then for all $$y\in D_1$$ you have:
$$|\langle x, T\lvert_{D_1}y\rangle| = |\langle Tx,y\rangle| ≤\|Tx\|\,\|y\|$$
and $$x$$ lies in the domain of $$(T\lvert_{D_1})^*$$ and that $$(T\lvert_{D_1})^*x = Tx$$.
Since $$T\lvert_{D_1}$$ is essentially self-adjoint you have that $$(T\lvert_{D_1})^*=\overline{T\lvert_{D_1}}$$, so we have just seen $$(x, Tx)\in\overline{\Gamma(T\lvert_{D_1})}$$, ie $$\Gamma(T)\subseteq \overline{\Gamma(T\lvert_{D_1})}$$