A coercive bilinear form Let $\alpha > 0$ and $X = [H^{2}(\Omega)\cap H^{1}_0(\Omega)] \times H^{1}_{0}(\Omega)$.
Find $\lambda_0 > 0$ for which the bilinear form $B: X \times X \rightarrow \mathbb{R}$ given by
$$
B((u,v),(\phi,\psi)) = \int_{\Omega}(\lambda_0 u \phi - v \phi + (\lambda_0 + \alpha) v \psi - \Delta u \psi)
$$
is coercive, that is, there exist $\beta > 0$ such that
$$
B((u,v),(u,v)) \geq \beta ||(u,v)||_{X}^{2}, \forall (u,v) \in X.
$$
I don't know exactly which norm I should use in X, because the intersection on definition of $X$ is confusing me.
What I have tried: For $(u,v) \in X$ we have
$$
B((u,v), (u,v)) = \int_{\Omega}(\lambda_0 u^2 - v u + (\lambda_0 + \alpha) v^2) - \int_{\Omega}(\Delta u) v
$$
Now I consider the general Green's formula:
$$
\int_{\Omega} \nabla u \cdot \nabla v = - \int_{\Omega} (\Delta u)v + \int_{\partial \Omega} v \frac{\partial u}{\partial \nu}, \forall u \in H^2(\Omega), v \in H^{1}(\Omega)
$$
In our case, as $v \in H^{1}_{0}(\Omega)$, we get
$$
B((u,v), (u,v)) = \int_{\Omega}(\lambda_0 u^2 - v u + (\lambda_0 + \alpha) v^2) + \int_{\Omega} \nabla u \cdot \nabla v.
$$
Still we have,
$$
B((u,v), (u,v)) = \lambda_0 ||u||_{L^2}^2 - \langle u,v\rangle_{L^2}  + (\lambda_0 + \alpha) ||v||_{L^2}^2 + \langle u,v\rangle_{H^{1}_{0}}
$$
At this point I tried to use the Cauchy-Schwarz inequality, but nothing worked. So I dont know what to do and I don't know which norm I should use in X.
The context:
I have to find a $\lambda_0 > 0$ for which $Im(\lambda_0 I - A) = Y$, where $Y = H^1_0(\Omega) \times L^2(\Omega)$ and
$$
A : X = [H^2(\Omega) \cap H^{1}_{0}(\Omega)] \times H^{1}_{0}(\Omega) \rightarrow Y
$$
is difined by $A(u,v) = (v, \Delta u - \alpha v)$. I'm trying to solve this using Lax-Milgram theorem.
 A: @Jose27's comment does not help you conclude what you want - it proves that it is false.
Let $u\in H^2(\Omega) \cap H^1_0(\Omega)$ be arbitrary and let $v= -u$. Then $$B((u,v),(u,v))= (2\lambda_0+\alpha +1) \| u \|_{L^2(\Omega)}^2 - \| \nabla u \|_{L^2(\Omega)}^2. $$ If $B((u,v),(u,v)) \geqslant 0$ then this shows that there exists a constant $C>0$ such that $$\| \nabla u \|_{L^2(\Omega)} \leqslant C\| u \|_{L^2(\Omega)} \tag{$\ast$}$$ for all  $u\in H^2(\Omega) \cap H^1_0(\Omega)$. The inequality ($\ast$) is known to be false in general. For example, take $\Omega = B_1$ and $$ u_k(x) = k^{n/4} \big ( e^{-k\vert x\vert^2}-e^{-k}\big ).$$ One can check that $u_k \in H^2(B_1) \cap H^1_0(B_1)$, $\| u_k \|_{L^2(B_1)} \leqslant C_0$ with $C_0$ independent of $k$, but $\| \nabla u \|_{L^2(B_1)} \to \infty $ as $k \to \infty$ which contradicts ($\ast$).
Thus, it is not true that $B((u,v),(u,v)) \geqslant 0$ for all $(u,v) \in X$ so there is no way that this bilinear form can be coercive with respect to any norm on $X$.
