Orthogonality & Adjoint Operator I am trying to prove this simple statement left to the reader in Brézis's book. 
Let $A \colon D(A)\subset E \longrightarrow F$ be an unbounded operator. Let $G:=\operatorname{Graph}(A)$ and $L=E \times \{0\}$.
Prove that $G^{\perp}+L^{\perp}=R(A^*)\times F^*$, where $A^*$ is the adjoint operator.
Well, it is obvious that $L^{\perp}=(E\times\{0\})^{\perp}=\{0\}\times F^*$.
I think I should show $G^{\perp}=R(A^*) \times (something)$, but I haven't been able, so far.
Help me, please. Many Thanks!

I'm assuming that $E,F$ are Hilbert spaces with some scalar products $\langle \cdot , \cdot \rangle_{E}$ and $\langle \cdot , \cdot \rangle_{F}$.
Let $\begin{pmatrix} e \\ f \end{pmatrix} \in G^\perp$, i.e. $\left\langle \begin{pmatrix} e \\ f \end{pmatrix}, \begin{pmatrix} x \\ Ax \end{pmatrix}\right\rangle_{E\times F}=0$ 
for all $x\in D(A)$. Using the definition of the scalar product in $E\times F$, this means $\langle f,Ax \rangle_F = \langle -e,x \rangle_E$, which is equivalent to $f \in D(A^*)$ and $A^*f = -e$, by (the algebraic) defintion of the adjoint operator. This means $e\in R(A^*)$, and furthermore, every $e\in A^*$ fulfills the equations above.
