If A is included in B, what can I say about their adjoint? I'm studying the theory of linear operators, but even if I tried to bruteforce my way I can't really continue without understanding this (I believe simple) statement:
Let A and B be linear operators and $D(A) \subset D(B)$, then $D(B*) \subset D(A*)$.
The proof I found is:
Let f $\in$ D(B*), then exists h $\in$ H such that 
$$(f, Bv) = (h, v) \,\, \forall v \in D(B)$$
then being $A \subset B$
$$ (f,Av) = (f,Bv) = (h,v) \,\, \forall v \in D(A) $$
Hence $D(B*) \subset D(A*)$.
It all makes sense, but I don't understand why this last line is right.
I was searching around about the theory of operators but I only find operators defined in the whole Hilbert space, where instead of inclusion symbols there are equalities, like A = A** instead of A $\subseteq$ A**, and my notes don't have a demonstration. And I don't want only to memorize.
 A: One characterization of $\mathcal{D}(A^{\star})$ is as the set of $y$ for which $x\mapsto (Ax,y)$ is a bounded linear functional on $\mathcal{D}(A)$. That is, $y$ is in the domain of $A^{\star}$ iff there exists a constant $M$ such that
$$
                      |(Ax,y)| \le M\|x\|, \;\;\; x\in\mathcal{D}(A).
$$
If $B$ extends $A$, then the condition that $x\mapsto (Bx,y)$ is a bounded linear functional is stronger than requiring $x\mapsto (Bx,y)=(Ax,y)$ be bounded on the smaller domain $\mathcal{D}(A)$.
A: What you have proven is that, given $f\in D(B^*)$, $\exists h\in H$ such that
\begin{equation}
(f,A\nu)=(h,\nu)\,\forall\nu\in D(A).
\end{equation}
Thus $f\in D(A^*)$, by the definition of adjoint you used just above.
A: Thanks to all. Your answers and comments leaded me to understand the nature of the problem. You see, I had very clear that 
$$ (f, Bv) = (h,v) \,\,\,\forall v \in D(B)  $$
and
$$ (f, Av) = (f, Bv) = (h, v) \,\,\,\forall v \in D(A) $$
Now, I know that for $h$ to be unique (if exists) the domain of the operators should be dense in $H$. But I was making a bad reasoning, thinking that being the domain dense in $H$ and being the domain of the adjoint closed, the domain of the adjoint needed to be the whole Hilbert space. I realized the domain of the adjoint can even be the zero vector.
Understanding that $f \in D(A^*)$ and using (intuitively) the fact that one equation implies the other, but not necessarily otherwise (strong conditions), by the definition of the adjoint operator I understood that the second equation says that the $D(A^*)$ are all f for which it have a representation. So, for saying
$$ (f, Ax) = (A^* f, x) $$
$$ (f, Ax) = (B^* f, x) $$
for all $x \in D(A)$ and all $f \in D(A^*)$. So the domain of B, given in the 4th, equation is necessarily a subset of the domain of A, which contain already any possible vector which satisfies the 3rd equation. But Keith McClary raised again another doubt: this subset can be proper or improper.
