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?
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.