Evans, chapter 6 exercise 2 Consider the following question

I know I'm supposed to show boundedness and coercivity but haven't made much progress.
Do I need to integrate by parts? Also I'm really interested in seeing the coercivity step
The norm I am considering is the $H^1_{0}$ norm
 A: To prove that is bounded is straightforward as soon as the coefficients are bounded. For the coercivity, I guess you are assuming that the matrix $a$ of the coefficients is uniformly elliptic. Then you just have to use Poincaré inequality.
I just focus on coercivity. Call $A$ the uniformly elliptic matrix whose entries are the $a_{ij}$. The fact that $A$ is uniformly elliptic means that there exits $\lambda >0$ such that $A v\cdot v\geq \lambda |v|^2$ for every $v\in\mathbb{R}^n$. You have
$$
\int_\Omega A \nabla u\cdot \nabla u +c u^2 \geq \int_\Omega \lambda |\nabla u|^2 +c u^2 \geq \int_\Omega \frac{\lambda}{2} |\nabla u|^2 + \frac{\lambda}{2}|\nabla u|^2 +c u^2\;.
$$
Therefore, by the Poincaré inequality, i.e.,
$$
\int_\Omega |\nabla u|^2 \geq \frac{1}{c_P^2}\int_\Omega |u|^2 \, ,$$
applied to one of the $\lambda/2$ terms, you have
$$
\int_\Omega \frac{\lambda}{2} |\nabla u|^2 + \frac{\lambda}{2}|\nabla u|^2 +c u^2
\geq
\int_\Omega \frac{\lambda}{2} |\nabla u|^2 + (c+\frac{\lambda}{2 c_P^2})|u|^2\; .$$
Therefore it is sufficient to take the $\alpha$ you were looking for as
$$
\alpha:=\frac{\lambda}{2 c_P^2}
$$
because the $|u|^2$ term is nonegative as soon as $c+\frac{\lambda}{2 c_P^2}\geq 0$, i.e., as soon as $c\geq =-\frac{\lambda}{2 c_P^2}=:- \alpha$.
A: Since Kosh has proved the coercitivity condition, I will prove the boundedness one.
The bilinear form associated to $\mathcal L$ is
$$B(u,v)=-\int_\Omega\sum_{i,j}^n(a_{ij}u_{x_i})_{x_j}v+\int_\Omega\sum_{i,j=1}^ncuv=\int_\Omega\sum_{i,j=1}^na_{ij}u_{x_i}v_{x_j}+\int_\Omega cuv$$
where in the second equality we used the integration by parts formula, toghether with the fact that $v$ vanishes in $\partial\Omega$ (recall that functions in $W_0^{1,p}(\Omega)$ are precisely those whose trace is equal to zero in the boundary $\partial\Omega$).
Now
\begin{align}
   |B(u,v)|&\leq\int_\Omega\sum_{i,j=1}^n|a_{ij}||u_{x_i}||v_{x_j}|+\int_\Omega |c||u||v|\\
&\leq C_1\left(\int_\Omega|u_{x_i}|v_{x_j}|+\int_\Omega |u||v|\right)\\
&\leq C_1\left(\int_\Omega|\nabla u||\nabla v|+\int_\Omega|u||v|\right)\\
&\leq C_1(\lVert\nabla u\rVert_{L^2(\Omega)}\lVert\nabla v\rVert_{L^2(\Omega)}+\lVert u\rVert_{L^2(\Omega)}\lVert v\rVert_{L^2(\Omega)})\\
&\leq C_2\lVert\nabla u\rVert_{L^2(\Omega)}\lVert\nabla v\rVert_{L^2(\Omega)}\leq C_2\lVert\nabla u\rVert_{H_0^1(\Omega)}\lVert\nabla v\rVert_{H_0^1(\Omega)}
\end{align}
where in the first inequality we used the usual inequality involving the norm of an integral, and the integral of a norm, toghether with the triangular inequality, in the second we used the coefficients $a_{ij}$ and $c$ are bounded in $\Omega$ (we are supposing that they belong to $L^\infty(\Omega)$), in the third we simply use that
$$\sqrt{|\nabla u|}=\sqrt{u_{x_1}^2+\dots+u_{x_n}^2}\geq|u_{x_i}|$$
for all $i=1,\dots,n$, In the forth we are using Hölder's inequality, and in the fifth we are using Poincaré's inequality. Finally, the last inequality follows from the definition of the norm in the Sobolev spaces $W^{k,p}$.
