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

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?

• Are you assuming that $T$ is symmetric and densely defined as well? Commented Mar 18, 2020 at 12:42
• Yes, I correct that now.
– user525192
Commented Mar 18, 2020 at 12:48

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 $$x\in D(T^{*})$$.

• What definition of essential-self-adjointness are you using?
– user525192
Commented Mar 18, 2020 at 13:48
• I had some of the things mixed up: The general statement is that if $T$ is symmetric then $T^{**}$ is the closure of $T$. This is always true and a calculation can be found in the appendix of these notes for example. What is done here shows that $T^*$ is self-adjoint, hence $T^{**}=T^*$ is self-adjoint giving that the closure of $T$ is self-adjoint. Commented Mar 18, 2020 at 14:13
• Thank you very much!
– user525192
Commented Mar 18, 2020 at 14:14