I have this problem:
Let $k>0$ (integer) and $1 \leq p < \infty$. Show that the norms
$$ \|u\|_{W^{k,p}(U)} = \bigg( \sum_{|\alpha|\leq k}\|D^{\alpha}u\|_{L^{p}(U)}^{p}\bigg)^{\frac{1}{p}} $$
and
$$ \||u|\|_{W^{k,p}(U)} = \sum_{|\alpha|\leq k}\|D^{\alpha}u\|_{L^{p}(U)}$$
are equivalent norms on $W^{k,p}(U).$
So i have to show something like this: $ \beta|||u|||\leq||u||\leq\gamma|||u|||. $
I showed the second inequality. I looked for the powers $p$ of the norm ($p\geq 1$). So we have
$$ \big(\|u\|_{W^{k,p}(U)}\big)^p =\sum_{|\alpha|\leq k}\|D^{\alpha}u\|_{L^{p}(U)}^{p} \leq \bigg(\sum_{|\alpha|\leq k}\|D^{\alpha}u\|_{L^{p}(U)}\bigg)^p = \big(\||u|\|_{W^{k,p}(U)}\big)^p $$
as we know that $\sum_{i=1}^n a_i^{p} \leq \big(\sum_{i=1}^n a_i\big)^p$ (proof with the help of binomial theorem). Therefore $$ \|u\|_{W^{k,p}(U)} \leq 1 \cdot \||u|\|_{W^{k,p}(U)} $$
and $\gamma = 1.$ But the lower bound is troubling me. I would be really grateful for some tips.
Best Answer
Using H$\ddot{\mbox{o}}$lder's inequality, it is clear that $$\sum_{k=1}^n|x_i|\leq\left(\sum_{k=1}^n|x_i|^p\right)^{1/p}n^{1/q},$$ for any numbers $x_i$, whatsoever. That should help. Moreover, you should realize that the equivalence of Sobolev norms is really a question of the equivalence of the p-norms on a finite dimensional space.