Let $E$ and $F$ be two Banach spaces and let $A: D(A) \subset E \to F$ be a closed densely defined unbounded operator. Does it follow that $N(A) = R(A^*)^\perp$?
Notation. Let $E$ and $F$ be two Banach spaces. An unbounded linear operator from $E$ into $F$ is a linear map $A: D(A) \subset E \to F$ defined on a linear subspace $D(A) \subset E$ with values in $F$. The set $D(A)$ is called the domain of $A$. We also have$$\text{range of }A = R(A) = \{Au : u \in D(A)\} \subset F,\text{ kernel of }A = N(A) = \{u \in D(A) : Au = 0\} \subset E.$$Progress. We probably want to argue by contradiction somehow? Maybe suppose there is some $u \in R(A^*)^\perp$ such that$$[u, 0] \notin \{[u, Au] : u \in D(A)\} \subset E \times F$$and apply Hahn-Banach? But I am not sure how to proceed any further.
You don't need such heavy machinery as Hahn-Banach. This is a completely elementary fact:
Let $J\colon E\times F\to F\times E,\,J(x,y)=(-y,x)$. It is a (more or less) immediate consequence of the definiton of $A^\ast$ that $G(A^\ast)=(J G(A))^\perp$, where $G(T)$ denotes the graph of the operator $T$.
Then we have $x\in N(A)$ iff $(x,0)\in G(A)$ iff $(0,x)\in JG(A)=(JG(A))^{\perp\perp}=G(A^\ast)^\perp$ iff $x\in R(A^\ast)^\perp$.
Observe that $G(A)$ is closed since $A$ is closed, and therefore $JG(A)$ is also closed und thus $(JG(A))^{\perp\perp}=JG(A)$.