I'm working on this problem studying for an applied analysis qualifying exam.
Let $T$ and $S$ be densely defined linear operators on a Hilbert space, $H$. Prove that $$T \subset S \Rightarrow S^* \subset T^*.$$
In order to do this, we need to show:
$D(S^*) \subset D(T^*)$
$S^* v = T^*v$ for all $v \in D(S^*).$
Assuming the first, I can show the second, but I'm having a hard time working out part 1.
Proof of 2 Assuming 1:
Choose $v \in D(S^*) \subset D(T^*)$.
By the definition of $v \in D(T^*)$, we have that $<Tu, v> = <u, T^*v>$ for all $u \in D(T)$. However, as $Tu = Su$ on $D(T)$, we then have that $<u, T^*v> = <Su, v>$ for all $u \in D(T)$.
Finally, we also have that $v \in D(S^*)$ which implies that $<Su, v> = <u, S^*v>$ for all $u \in D(S)$, which clearly implies that $<Su, v> = <u, S^*v>$ for all $u \in D(T)$.
Combing these, we than have that $<u, T^*v> = <u, S^*v>$ for all $u \in D(T)$, which, as $T$ is densely defined, implies that $S^*v = T^*v$ for $v \in D(S^*)$.