I'm reading Section 2.6 An Introduction to Unbounded Linear Operators in Brezis' book of Functional Analysis
Let $E, F$ be Banach spaces and $A:D(A)\subseteq E \to F$ a densely defined unbounded linear operator. Let $A^\star : D(A^\star) \subseteq F^\star \to E^\star$ be the adjoint operator of $A$.
The graphs of $A$ and $A^{\star}$ are related by a very simple orthogonality relation: Consider the isomorphism $I: F^{\star} \times E^{\star} \rightarrow E^{\star} \times F^{\star}$ defined by $ I([v, f])=[-f, v]. $ Then $ I\left[G\left(A^{\star}\right)\right]=G(A)^{\perp}. $
Indeed, let $[v, f] \in F^{\star} \times E^{\star}$, then
$$\begin{align} [v, f] \in G\left(A^{\star}\right) & \iff \langle f, u\rangle=\langle v, A u\rangle \quad \forall u \in D(A) \\ & \iff -\langle f, u\rangle+\langle v, A u\rangle=0 \quad \forall u \in D(A) \\ & \iff [-f, v] \in G(A)^{\perp}. \end{align}$$
Because the graph $G(A) \subseteq E \times F$, so by definition of orthogonal complement in Section 1.3 The Bidual $$G(A)^\perp := (G(A))^\perp := \{T \in (E\times F)^\star \mid \forall (x,y) \in G(A),T(x,y) =0\}.$$
How $[-f, v] \in G(A)^{\perp}$ for $[v, f] \in F^{\star} \times E^{\star}$? I meant how can $[-f, v]$ belong to $(E\times F)^\star$.
Consider the map $$T: (E \times F)^\star \to E^\star \times F^\star, g \mapsto (g_1, g_2) :=(g\restriction E \times \{0_F\}, g\restriction \{0_E\} \times F).$$
It's easy to check that $g(x, y) := g_1(x)+g_2(y)$ and that $T$ is an isomorphism, i.e., $(E \times F)^\star \cong E^\star \times F^\star$. That's why we can identify $(E \times F)^\star$ with $E^\star \times F^\star$.