We have the problem: $$\begin{cases}div(\alpha\nabla u)+\gamma u=f & \text{in } \Omega\\u=0 & \text{in } \Gamma_1\subset\partial \Omega\\\alpha \nabla u\cdot \vec{n}=0 & \text{in } \Gamma_2 \subset \partial \Omega\end{cases}$$
So the variational formulation for the problem is: find $u \in V:=\{v\in H^1(\Omega): v\vert_{\Gamma_1}=0\}$ such that
$$\int_{\Omega}\alpha \nabla u \nabla v+\gamma uv \;dx=\int_{\Omega}fv \;dx$$ for all $v\in V$.
We suppose that $\alpha>a_0>0$ is a funtion such that $\alpha \in L^{\infty}(\Omega)$, $\gamma\in L^{\infty}(\Omega)$ is a positive function and $f \in L^2(\Omega)$. I want to prove that the bilinear form associated to the variational form is bicontinous and coercive, in order to apply Lax Milgram theorem. These things are going to be proved in $V$. Is that correct?
If the answer is yes, how can I prove these? I suppose that the bilinear form associated is $$a(u,v)=\int_{\Omega}\alpha \nabla u \nabla v+\gamma uv \;dx$$
and then I define a norm in $V$ as $\vert v \vert_{V}=\lVert \nabla v \rVert_{L^2(\Omega)}$. How can I prove bicontinuity from this new norm?
And finally, Is there a simplier way to demonstrate Lax Milgram theorem?
Thanks a lot.