How can I prove that the adjoint of the adjoint is an extension of the operator?

Question

Let $H$ be a Hilbert space and $T:D_T\subset H\rightarrow\Delta_T\subset H$ an operator with $D_T$ dense in $H$. Let $T^*$ be the adjoint of $T$ and $T^{**}$ be the adjoint of $T^*$. How can I prove that $T^{**}$ is an extension of $T$?

Answer 1

Recall that $x \in D(T^*)$ with $T^*x = y$ if and only if $$\langle x, Tz \rangle = \langle y, z \rangle \qquad \forall z \in D(T).$$

To prove that $T^{**}$ is an extension of $T$, you want to show that if $x \in D(T)$ with $Tx = y$ then $x \in D(T^{**})$ with $T^{**}x = y$. Let's unpack what this means.

By the definition, $x \in D(T^{**})$ with $T^{**}x = y$ if and only if $$\langle x, T^*z \rangle = \langle y, z \rangle \qquad \forall z \in D(T^*).$$ So suppose $x \in D(T)$ and $Tx = y$ and $z \in D(T^*)$. Then by the definition of $T^*$ we have that $$\langle z, y \rangle = \langle z, Tx \rangle = \langle T^*z, x \rangle $$ which implies that $$\langle y, z \rangle = \langle x, T^*z \rangle$$ as desired so $T^{**}$ extends $T$.