Is the symmetric adjoint of a densely defined symmetric operator essentially self-adjoint?

Question

Let $T^{*}$ be the adjoint of a densely defined, symmetric operator $T$. Suppose that the operator $T^{*}$ is a symmetric operator. Why does it follow that $T$ is essentially self-adjoint?

Thanks in advance!

Answer 1

Here is a proof using the definitions.

First check that if $T$ is symmetric and densely defined, that then $D(T^*)\supseteq D(T)$ and $T^*\lvert_{D(T)}= T$. For if $x \in D(T)$ then: $$|\langle x, Ty\rangle| = |\langle Tx,y\rangle| ≤ \|Tx\| \ \|y\|$$ for all $y\in D(T)$ and $x\in D(T^*)$ follows. Further $T^*(x)$ is defined via: $$\langle T^* x, y\rangle := \langle x, Ty\rangle = \langle Tx,y\rangle$$ holding for all $y\in D(T)$ (which is dense). It follows that $T^*x=Tx$ for all $x\in D(T)$.

Now if $T^*$ is also symmetric this very same step gives you $D(T^{**})\supseteq D(T^*)$ and $T^{**}\lvert_{D(T^*)}=T^*$. The only thing we are still interested in proving is that $D(T^*)\supseteq D(T^{**})$. So let $x\in D(T^{**})$. This means that $$|\langle x,T^*y\rangle| ≤ C_x\|y\|$$ for all $y\in D(T^*)$. In particular since $D(T^*)\supseteq D(T)$ it holds for all $y\in D(T)$ and you get the desired $x\in D(T^{*})$.