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 $$a(u, v) = \int\limits_I u''(x) v''(x) \, dx.$$ Can someone please give me a hint on how to show that $$\{u \in H_0^2(I) : |a(u, v)| \leq c \|v\|_{L^2(I)}, \forall v \in H_0^2(I)\} = H^4(I) \cap H_0^2(I)?$$ (This set is the domain of the introduced operator.)
Thank you!
Use integration by parts and the Riesz representation theorem.
For the one inclusion, if $u\in H^4(I) \cap H^2_0(I)$, we can integrate by parts twice to obtain
$$a(u,v) = \int\limits_I u''(x)v''(x)\,dx = \int\limits_I u^{(4)}(x)v(x)\,dx$$
and see that
$$\lvert a(u,v)\rvert \leqslant \lVert u^{(4)}\rVert_{L^2(I)}\cdot \lVert v\rVert_{L^2(I)}.$$
For the other inclusion, if $u\in H^2_0(I)$ is such that $v\mapsto a(u,v)$ is continuous in the $L^2(I)$-norm, then the Riesz representation theorem asserts the existence of a $w\in L^2(I)$ such that
$$a(u,v) = \int\limits_I w(x)v(x)\,dx$$
for all $v\in H_0^2(I)$. Then it remains to see that $w = u^{(4)}$ (as the distributional derivative). For that, consider $v\in C_c^\infty(I)$ and integrate by parts.