bilinear continuous, coercive form Let $k\in \mathbb{R}, k\neq 1$, consider the space 
$$ V = \{u\in H^1(0,1): u(0) = ku(1)\}$$
Let
$$a(u,v) = \int_0^1 (u'v'+ uv)\; dx - \left(\int_0^1 u\; dx\right) \left(\int_0^1 v\; dx\right)\quad\;\forall\; u,v\in V$$
Show that $a$ is coercive.
I can't find such a way to estimate term $-\left(\int_0^1 u\; dx\right)\left(\int_0^1 v\; dx\right)$ 
 A: The answer uses a argument of compacity, see this answer here and the comments therein.
First note that by Holder inequality $$\left(\int_0^1 u\right)^2\leq \int_0^1 u^2,\ \forall u\in L^2,$$
hence, $a(u,u)\ge \|u\|_{1,2}^2-\int_0^1 u^2$ for all $u\in V$. We conclude that it is sufficient to find a positive constant $C$ such that $$\tag{1} \|u\|_{1,2}^2-\int_0^1 u^2\ge C \|u\|_{1,2}^2,\ \forall\ u\in V.$$
Assume that $(1)$ is not true, i.e. assume that there is a sequence $u_n\in V$ such that $$\|u_n\|_{1,2}^2-\int_0^1 u^2_n<\frac{1}{n}\|u_n\|^2_{1,2},\ \forall n\in\mathbb{N}\tag{2}$$
Let $v_n=\frac{u_n}{\|u_n\|_{1,2}}$. Then, we get from $(2)$ that $$1-\int_0^1 v^2_n<\frac{1}n,\ \forall n\tag{3},$$
which is equivalent to $$\int_0^1 v^2_n\to 1,\ \|v_n\|_{1,2}=1\tag{4}$$
Once $(4)$ is satisfied, we can assume without loss of generality the existence of $v\in V$ such that $v_n$ weakly converges to $v$ in $H^1$, $v_n$ strongly converges to $v$ in $L^2$. Because $\|v_n\|_2\to 1$ we conclude that $\|v'_n\|_2\to 0$ which implies that $\|v'\|_2=0$, hence, $v$ is constant a.e. 
The condition $v(0)=kv(1)$ for $k\neq 1$ implies that $v=0$, which is an absurd, because $\|v\|_2=1$, therefore, $(1)$ is true. 
