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.

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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!

  • 1
    $\begingroup$ 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 $\overline{\Gamma(T)} = \overline{\Gamma(T\lvert_{D_1})}$) $\endgroup$
    – s.harp
    Nov 19, 2020 at 15:55
  • $\begingroup$ @s.harp I submitted accidentally, now you can see the whole question. $\endgroup$ Nov 19, 2020 at 16:23

1 Answer 1


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 $$\Gamma(T)\subseteq \overline{\Gamma(T\lvert_{D_1})}$$

  • $\begingroup$ God it turns out that easy... Thank you so much!! $\endgroup$ Nov 20, 2020 at 3:44

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