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 $-\left(\int_0^1 u\; dx\right)\left(\int_0^1 v\; dx\right)$
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It holds $a(u,u) \ge |u'|_{L^2(0,1)}^2$ trivially. You need to prove $\int_0^1 u^2 dx \ge c (\int_0^1 u dx)^2$ with $c\in(0,1)$. – daw Apr 11 '14 at 16:34
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Why $\int_0^1 u^2;dx \geq c(\int_0^1 u; dx)^2$? can you explain ? – Apr 11 '14 at 16:50
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That would be sufficient for coercivity, I have no idea how to prove this :( – daw Apr 11 '14 at 16:56
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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.
