variational question Let $\Omega$ a bounded domain, connexe and regular, and let $f \in L^2(\Omega).$ Let the variational problem: Find $u \in H^1(\Omega)$ such 
$$\int_{\Omega} \nabla u \nabla v dx + (\int_{\Omega} u dx)(\int_{\Omega} v dx) = \int_{\Omega} f v dx, \forall v \in H^1(\Omega)$$
1- Prouve that this variational problem admits a unique solution in $H^1(\Omega).$
2- Deduce the boundary problem associate to this variational problem ( study the two cases $u \in  H^2(\Omega)$ and $u \notin H^2(\Omega)$.
I dont't understand in the question 2, why we mus study the two cases $u \in H^2$ and $u \notin H^2$?
 A: Case 1: $u\in H^2(\Omega)$
In this case, we have that $$\tag{1}\int_\Omega \nabla u\nabla v=-\int_\Omega v\Delta u+\int_{\partial\Omega}\frac{\partial u}{\partial\nu}v,\ \forall\ v\in H^1(\Omega)$$
From $(1)$, we conclude that $$\tag{2}-\int_\Omega v\Delta u+\int_{\partial\Omega}\frac{\partial u}{\partial\nu}v+\int_\Omega (\int_\Omega u)v=\int f v,\ \forall\ v\in H^1(\Omega)$$
If we take $v\in C_c^\infty(\Omega)$ in $(2)$, we can conclude by the Fundamental Lemma of Calculus of Variation that $$\tag{3}-\Delta u(x)+\int_\Omega u=f(x),\ a.e.\ x\in\Omega$$
By using $(3)$, we conclude from $(2)$ that $$\tag{4}\int_{\partial\Omega} \frac{\partial u}{\partial\nu}v=0,\ \forall\ v\in H^1(\Omega)$$ 
$(4)$ implies that $\frac{\partial u}{\partial\nu}=0$ in $\partial\Omega$, so your boundary balue problem is 
$$
 \left\{ \begin{array}{rl}
 -\Delta u+\int_\Omega u=f, &\mbox{ in $\Omega$} \\
 \frac{\partial u}{\partial\nu}=0  &\mbox{ in $\partial\Omega$}
       \end{array} \right.
$$
Case 2: $u\notin H^2(\Omega)$
I dont know how to treat the case. It is worth to note that the solution is unique, so I think that it is possible to prove that $u\in H^2(\Omega)$, by using difference quotient methods and then we are on the first case again, but this is just a guess.
Update: Suppose that $u\in H^1(\Omega)$ satisfies $$\tag{5}\int_\Omega \nabla u\nabla v+\left(\int_\Omega u\right)\left(\int_\Omega v\right)=\int_\Omega fv,\ \forall\ v\in H^1(\Omega)$$
Take $v=1$ in $(5)$ and get $$\tag{6}\int_\Omega u=\frac{1}{|\Omega|}\int_\Omega f$$
From $(5)$ and $(6)$ we concude that $$\tag{7}\int_\Omega\nabla u\nabla v=\int_\Omega\left(\frac{1}{|\Omega|}\int_\Omega f-f\right)v,\ \forall\ v\in H^1(\Omega)$$
As you can verify in Brezis's book chapter 9, equation $(7)$ implies that $u\in H^2(\Omega)$, so the same argument as above can be used.
Remark: After the update, we note that the first part of the proof could be carried without distinguishing two distinct cases.
