variational problem-exercice Let $\Omega$ an open bounded connexe and regular, and let $f \in L^2(\Omega)$
We consider the variational problem: find $u \in H^1(\Omega)$ such:
$$\displaystyle\int_{\Omega} A \nabla u \cdot \nabla v dx + \left(\displaystyle\int_{\Omega} u dx\right)\left(\displaystyle\int_{\Omega} v dx\right) = \displaystyle\int_{\Omega} f v dx, \forall v \in H^1(\Omega)$$
with: $$\exists \alpha > 0, A(x) \xi \xi \geq \alpha |\xi|^2, \forall \xi \in \mathbb{R}^n, \exists \beta > 0, |A(x) \xi| \leq \beta |\xi|, \forall \xi \in \mathbb{R}^n$$
-The question is: prove that this variational problem admits a unique solution $u \in H^1(\Omega)$.
My solution is: we put 
$$a(u,v) = \displaystyle\int_{\Omega} A \nabla u \cdot \nabla v dx + \left(\displaystyle\int_{\Omega} u dx\right)\left(\int_{\Omega} v dx\right)$$
$$L(v) = \displaystyle\int_{\Omega} f v dx$$
We begin to prouve that $a$ is coecive. For this, we have to prouve that $$\exists \lambda > 0, a(v,v) \geq \lambda ||v||^2_{H^1(\Omega)}$$
We have:
$a(v,v)= \displaystyle\int_{\Omega} A \nabla v \cdot \nabla v dx + (\displaystyle\int_{\Omega} v dx)\left(\int_{\Omega} v dx\right)$.
we prove the coercivity of $a$ by absurd. We suppose that $a$ isn't coercitive, then: $$\forall \lambda > 0, \exists v \in H^1(\Omega): a(v,v) < \lambda ||v||^2_{H^1}$$
then, in particular, for $\lambda = \dfrac{1}{n}$, we have:
$$\forall n \in \mathbb{N}, \exists v_n \in H^1: a(v_n,v_n) < \dfrac{1}{n} ||v_n||^2_{H^1}$$
and we can choice $v_n$^such that $||v_n||=1,$ then, we obtain: 
$$\forall n \in \mathbb{N}, \exists v_n \in H^1(\Omega): ||v_n||_{H^1}=1 , a(v_n,v_n) < \dfrac{1}{n}$$
as $\Omega$ is bounded and régular, we have by Rellich theorem that there exist a subsequence $v_n$ who converge to $v \in L^2(\Omega)$. Then, $(v_n)$ is Cauchy sequence in $L^2(\Omega)$ and as $a(v_n,v_n)=\displaystyle\int_{\Omega} A \nabla v_n \cdot \nabla v_n dx + (\displaystyle\int_{\Omega}v_n dx)^2$ converges to 0, and as $a(v_n,v_n)$ is a sum of positif termes, we deduce that
$\displaystyle\int_{\Omega} A |\nabla v_n|^2 dx$ converge to 0 ans $\displaystyle\int_{\Omega} v_n dx$ converge to 0.Then $\nabla v_n$ converge to 0 in $L^2(\Omega)$. So, $v_n$ is a Cauchy sequence in $H^1(\Omega)$ and because of $H^1(\Omega)$ is an Hilbert space, he is a complet space, i.e. All Cauchy sequence converge, then $v_n$ converge in $H^1(\Omega)$ to $v$. 
Further,
$||\nabla v||_{L^2} = \lim_n ||\nabla v_n||_{L^2} = 0$^and $||v_n||_{L^2}$ converge to 0.
and $||v||_{L^2} = \lim_n ||v_n||_{L^2} = 1$
Then, $v=0$ and $||v||_{L^2}=1$ and there is a contradiction. So $a$ is coercitive.
My solution is it correct? please
 A: I do agree with you, until the part where you prove that $$\tag{1}\|\nabla v\|_2=\lim \|\nabla v_n\|_2=0$$
Then you conclude that $\|v_n\|_2\to 0$ without proof. This part seem to me to be wrong, but we can fix it. 
Also, as you have concluded, we have that $$\tag{2}\|v_n\|_{1,2}=\|v_n\|_2+\|\nabla v_n\|_2=1$$
From $(1)$ and $(2)$ we conclude that $\|v_n\|_2\to \|v\|_2=1$. Moreover, from $(1)$ we have that $v$ is a constant function, $\|v\|_2=1$ implies that $v=c$ where $c\neq 0$.
To conclude first your have to prove that $a$ is continuous. After this, you have that $a(v_n,v_n)\to a(v,v)$. But $a(v_n,v_n)\to 0$ and $a(v,v)\neq 0$, because $v$ is constant and not zero.
A: Okay, so i hope to prouve the coercivity of $a$ in a logical way.
We process by absurde, so, we suppose that that $$\forall \nu > 0, \exists v \in H^1(\Omega), a(v,v) \geq \nu ||v||^2_V$$
In partical, for $_nu=\dfrac{1}{n},$ we have $$\forall n \in \mathbb{N}, \exists v_n \in H^1(\Omega):a(v_n,v_n) < \dfrac{1}{n} ||v_n||^2_V$$
from this point I do not know how to think logically and naturally to find the right arguments to prouve the coercivity.
