What is the norm of $H^4(0, 1) \cap H_0^2(0, 1)$?

Question

Let $I = (0, 1)$ and $H_0^2(I) = \{u \in L^2(I) : u', u'' \in L^2(I), u = u' = 0 \;\; \text{on} \;\; \partial I\}$.

What is the norm of $$H^4(I) \cap H_0^2(I)?$$ $\|u\|_{H^4(I)} = \Big[ \|u\|_{L^2(I)} + \|u'\|_{L^2(I)} + \|u''\|_{L^2(I)} + \|u'''\|_{L^2(I)} + \|u^{(4)}\|_{L^2(I)} \Big]^\frac{1}{2}$ $\|u\|_{H^2(I)} = \Big[ \|u\|_{L^2(I)} + \|u'\|_{L^2(I)} + \|u''\|_{L^2(I)} \Big]^\frac{1}{2}$

$\| u \|_{H^4(I) \cap H_0^2(I)} = ?$

Thank you!

Answer 1

Your examples are lacking squares (think of homogeneity), but apart from that they are both valid norms for $H^4(I) \cap H^2_0(I)$, as is the normal $L^2$-norm.

What you really want in most situations is not just any norm, but a norm with which the space is complete. This is not true for the $H^2$-norm: There holds by definition $$C^\infty_0(I) \subset H^4(I) \cap H^2_0(I) \subset H^2_0(I),$$ taking the closure of those spaces w.r.t. to the $H^2$-norm yields that $$\overline{H^4(I) \cap H^2_0(I)}^{\|\cdot\|_{H^2}} = H^2_0(I)$$ because $\overline{C^\infty_0(I)}^{\|\cdot\|_{H^2}} = H^2_0(I)$.

The space is complete with the $H^4$-norm on the other hand, because convergence in the $H^4$-norm implies convergence in the $H^2$-norm and $H^2_0(I)$ is a closed subspace of $H^2(I)$.