Show that bilinear form is $H^1(0,l)$-elliptic/coercive Let
$$a(u,v) := \int_0^l \partial u(x) \partial v(x) + cu(x)v(x)\, \mathrm{d}x + \alpha u(l)v(l)$$
Show that $a(\cdot,\cdot)$ is $H^1(0,l)$-elliptic if either $c > 0$ or $\alpha > 0$.
My attempt at a solution:
The case where $\alpha$ is $0$ is straightforward. I am stuck on the $c =0$ and $\alpha > 0$ case. I need to show that 
\begin{align}
\int_0^l (\partial v(x))^2 \, \mathrm{d}x + \alpha u(l)v(l) \ge \int^l_0 (\partial v(x))^2 + v(x)^2 \, \mathrm{d}x.
\end{align}
 I've been comparing the terms of the $||v||_{H^1}$ term by term to the bilinear form - clearly, $a(v,v) \ge \int_0^l (\partial v(x))^2$, so that term is fine. But I can't figure out why $a(v,v) \ge c_0 \int_0^lv(x)^2$. The only estimate that I found to work with is that $|u(l)| \le 2l^{1/2} ||\partial u||_{L^2} + 2l^{-1/2}||u||_{L^2}$, and that's not getting me anywhere.
 A: This is a consequence of Friedrich's inequality, which, when specialized to your case, states that
\begin{align}
\left\|u\right\|_{L^{2}[0,l]}^{2} \le l^{2}\left\|\partial u\right\|_{L^{2}[0,l]}^{2}
\end{align}
Obviously,
\begin{align}
a(u,u) &:= \int_{0}^{l} \partial u \partial u \, \mathrm{d}x + \alpha u(l)^{2} \ge \left\|\partial u\right\|_{L^{2}[0,l]}^{2}. 
\end{align}
Then
\begin{align}
\left\|\partial u\right\|_{L^{2}[0,l]}^{2} 
&= \left(1 - \frac{1}{l^{2}} \right)\left\|\partial u\right\|_{L^{2}[0,l]}^{2}  + \frac{1}{l^{2}}\left\|\partial u\right\|_{L^{2}[0,l]}^{2} \\
&\ge \frac{1}{l^{2}}\left\| u\right\|_{L^{2}[0,l]}^{2}  + \frac{1}{l^{2}}\left\|\partial u\right\|_{L^{2}[0,l]}^{2}  - \frac{1}{l^{2}}\left\|\partial u\right\|_{L^{2}[0,l]}^{2} \\
\left(1+\frac{1}{l^2}\right)\left\|\partial u\right\|_{L^{2}[0,l]}^{2} 
&\ge \frac{1}{l^{2}}\left\|u\right\|_{H^{1}[0,l]}^{2} 
\end{align}
so that $a(u,u) \ge \frac{1}{1+l^2}\left\|u\right\|_{H^{1}[0,l]}^{2}$.
I have a proof of Friedrich's inequality sitting around if you want it; but it seemed off-topic here . . .
