I want to show that $\operatorname{Range}(A^*)^\perp \subset \operatorname{Null}(A)$ where $A:E \supset D(A) \to F$ is an unbounded closed linear operator densely defined in $E$, and $E$ and $F$ are Banach Spaces.
Notice this is trivial when $A$ is bounded. However, I am struggling to prove it with this hypothesis.
My idea so far has been to try a proof by contradiction:
Let $u \in \operatorname{Range}(A^*)^\perp$ be such that $(u,0) \notin \operatorname{Graph}(A)$. Then by Hahn Banach we can strictly separate $\{(u,0)\}$ from $\operatorname{Graph}(A)$ with a hyperplane $f \in (E \times F)^*$, say f(u,0) < \alpha < f(v,Av) \quad \forall v \in D(A).
I guess from this I should be able to find $g \in D(A^*)\subset F^*$ such that $g(Au) \neq 0$. This would contradict $u \in \operatorname{Range}(A^*)^\perp$ since $0=A^*(g)(u)=g(Au)$ for all $v\in D(A^*)$.