Show that a Bilinear form is Coercive I'm reading through Brezis' book on functional analysis, Sobolev spaces and PDE, and I'm having trouble showing that the Bilinear form: $a(u,v) = \int_{0}^{2}u'v'dx+\left(\int_{0}^{1}u dx\right) \left(\int_{0}^{1}v dx\right)$
is coercive.(Problem 8.25 #2).
The book offers the following hint:
(Argue by contradiction and assume that there is a sequence $(u_n)$ in $H^1$ such that $a(u_n,u_n)\rightarrow 0$ and $||u_n||_{H^1}=1$. Let $(u_{n_k})$ be a subsequence such that $u_{n_k} \rightharpoonup u$ in $H^1$ and $||u_{n_k}||_{L^2}\rightarrow||u||_{L^2}$ for some limit u. Show that u=0.
From the previous question I have:
$|a(u,v)|\leq ||u||_{H^1}||v||_{H^1}~~~~~~~$     and $~~~~~~~a(u,u)=0 ~~~\Rightarrow~~~ u=0$
I've been throwing ideas at this for most of today and I've failed to come up with anything leading to the contradiction. Any suggestions for the right place to start?
 A: Suppose ad absurdum that there is a sequence $u_n\in H^1$ such that $$a(u_n,u_n)\to 0,\ \ \|u_n\|_{H^1}=1$$
In other words $$\tag{1} \int_0^2 |u_n'|^2+\left(\int_0^1 u_n\right)^2\to 0,\ \int_0^2u_n^2+\int_0^2|u_n'|^2=1$$
We conclude from $(1)$ that $$\int_0^2 |u'_n|\to 0,\ \int_0^2u_n^2\to 1\tag{2}$$
Assume without loss of generality that $u_n\rightharpoonup  u$ in $H^1$, hence, $u_n'\rightharpoonup u'$ in $L^2$ which implies that $$\|u'\|_{L^2}\leq\liminf \|u'_n\|_{L^2}=0$$
We conclude from the last inequality that $u=c$ where $c$ is a constant. Moreover, because $u_n\to u$ in $L^2$ (compact embedding), we have that $$\int_0^2 u_nf\to\int_0^2 uf,\ \forall \ f\in L^2$$
Take $f$ as the characteristic function of the interval $(0,1)$, hence $$\int_0^1 u_n\to \int_0^1 u\tag{3}$$
We combine $(3)$ and $(1)$ to conclude that $u=0$ in $[0,1]$. Because $u$ is continuous (every function in $H^1(0,2)$ is continuous) we have that $u=0$ in $[0,2]$. To finish the argument, use $(2)$ to conclude that 
$$\int_0^1 u^2=1$$
which is an absurd.
