Question about a proof. I was just reading over this proof here Adjoint identity .
I am just learn functional analysis, and I wonder, in the step where it says "Since $D(A)$ is a subspace we must have $f(v,Av)=0$", why this must be true? I cannot see how Hanh Banach gives such a property.
Thank you! Sorry if I have bad english.

Hahn-Banach gives you the separation, i.e. the existence of an $\alpha$ such that
$$f(u,0) < \alpha < f(v,Av) \quad \forall v \in D(A).$$
Suppose there were some $v\in D(A)$ such that $f(v,Av)\neq 0$. Using the fact that $D(A)$ is a subspace, hence closed under multiplication by scalars, and also of course using the fact that $f$ is linear, show that you'd then be able to scale this particular $v$ in such a way as to make the above inequality false.
