Let $I = (0, 1)$ and $H_0^2 (I)$ the closure of $C_c^\infty (I)$ in $W^{2, 2} (I)$. Consider $$a : H_0^2 (I) \times H_0^2 (I) \to \mathbb{R}$$ defined by $$a(u, v) = \underset{I}{\int} u''(x) v''(x) dx.$$ Show that a is coercive, i. e. there is a constant c such that $$a (u, u) \geq c \|u\|^2_{H_0^2 (I)}, \;\; \forall u \in H_0^2 (I).$$ We have $$u \in H_0^2 (I) : u \in W^{2, 2} (I) \;\; u = u' = 0 \;\; \text{on} \;\; \partial I $$ and $$a(u, u) = \|u ''\|^2_{L^2(I)}$$ $$\|u\|^2_{H_0^2 (I)} = \|u\|^2_{L^2(I)} + \|u'\|^2_{L^2(I)} + \|u''\|^2_{L^2(I)} = \|u\|^2_{L^2(I)} + \|u'\|^2_{L^2(I)} + a(u, u). $$ Also we have the Poincar$\acute{e}$ inequality, which says that there is a constant c such that $$\|u\|_{W^{1, 2}(I)} = \|u\|^2_{L^2(I)} + \|u'\|^2_{L^2(I)} \leq c \|u'\|_{L^2(I)}. $$
Can someone please give me a hint on how to conclude that a is coercive?
Thank you!
Since $u' \in H_0^1(I)$, you can apply Poincaré to $u'$. This gives $$\|u'\|_{H^1} \le C \, \|u''\|_{L^2}.$$ Can you conclude?