Why is a densely defined symmetric operator $T$ extended by its adjoint $T^*$?

Question

This is a result I've seen stated a few times, but I can't seem to come up with a proof!

Suppose $T$ is a densely defined linear operator with domain $D(T)\subset H$, where $H$ is a Hilbert space then if $T$ is symmetric i.e. $\langle Tx,y\rangle = \langle x,Ty\rangle$ for all $x,y \in D(T)$, it follows that the adjoint $T^*$ extends $T$, i.e. $T \subset T^*$.

What I'm having trouble showing is why $D(T)\subset D(T^*)$. Where $$D(T^*)=\{y\in H:x\mapsto \langle Tx,y\rangle \text{ is a bounded linear functional on } D(T)\}.$$

Answer 1

For $y \in D(T)$, we have, due to the symmetry of $T$,

$$x \mapsto \langle Tx,y\rangle = x \mapsto \langle x, Ty\rangle,$$

and the latter is easily seen to be continuous, hence $D(T) \subset D(T^\ast)$.