Question 2.18 from Brezis: Would someone help me to understand a solution given to this question? 
I underlined parts of the solution where I'm struggling with. Would someone help me with this?

*

*In (Part 1) I did not understand the equality in (1), because from
the theory, for me  $$N(A^*)=\{v\in D(A^*):  A^*v=0 \in E^*)\}\subset
   F^*$$ So, how can I find the equality in part 1, from this equality
above?


*In (Part 2), the doubts in (2) are: First of all, why $A$ being closed implies that $G(A)$ is a Banach subspace of $E\bigoplus F$?(I tried to explain using the closed graph theorem, but it did not work). Secondly, I did not understand why there exists a continuous linear functional such that $f|_{G(A)}\equiv 0$? Finally, I did not understand why $f(v,0)=1$?


*In topic (3) underlined, the doubts are: why $f\longrightarrow (0,f)$ is a bounded linear map? (Who is the domain and the codomain of $f$?) Also, why $f(0, Av_n)\longrightarrow 0?$


*At topic (4) underlined, I did not understand why $\displaystyle\lim_{n\rightarrow \infty} f(v_n, Av_n)-f(0, Av_n)$
is going to be zero?
 A: *

*I think from your definition you just have to test against all $u\in E$  the equality $A^*v=0$, unless I am mistaken


*Doesn't A closed mean that G(A) is a closed subset ? If so, it is enough to conclude.
For your secound question, it use the Hahn-Banach theorem on it's geometric form : since a banach is locally convexe, $\{(v,0)\}$ is compact and G(A) is a closed convex disjoint of $\{(v,0)\}$, there existe an affine hyperplan separating $\{(v,0)\}$ and G(A). We can reformulate this as there is an $f\in (E\oplus F)^*$ and $\alpha \in R$ such that $$\forall w \in G(A), f(w)<\alpha<f((v,0))$$
Then using that G(A) is a vectorial subset, we can show that in fact $G(A) \in ker(f)$, and up to a multiplication by $1/f((v,0))$ we can assume that $f((v,0))=1$.


*That's a conflict of notation, I would rewrite it as : Since the inclusion $i : F\to F\oplus E$ defined as $i : x \in F \mapsto (0,x)$ is continuous between banach space, then $f\circ i \in F^*$. So by what is just above, with $w=f\circ i$, we have that $0=lim (f\circ i,Av_n)=lim f(0,Av_n)$


*Since $(v_n,Av_n) \in G(A)$, we have that $f(v_n,Av_n)=0$ by construction of $f$. The limit given in 3) allow to conclude
