I am reading the proof of Theorem 3.24 in Functional Analysis, Sobolev Spaces and Partial Differential Equations, by Haim Brezis.
We have $E$ and $F$ two Banach spaces which are reflexive, and $A: D(A) \subset E \rightarrow F$ a densely defined and closed unbounded linear operator.
That is, D(A) is dense in E and the graph of A, $G(A)= \{(x, Ax) | x \in D(A) \} $, is closed in $E \times F$.
It is needed in a part of the proof to strictly separate a point $(0, \varphi) \notin G(A)$ from the set $G(A)$ by a closed hyperplane. I assume Hahn Banach (Second Geometric form) theorem is needed in order to do so. Both sets are closed and compact, so no problems here. The point $(0, \varphi)$ is also convex, but the theorem also needs the set $G(A)$ to be convex.
Maybe this is trivial, but I don't know how to prove it.
From the linearity of $A$ you get quite easily that $G(A)$ is a subspace, hence convex.