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today I read book Sobolev space, PDEs of H.Brezis, and when I read chapter 8, I don't know why following remark is easy:

Remark 11 (page 214). Let $I$ be a bounded interval, let $1\leq p\leq \infty$, and et $1\leq q\leq \infty$, then

$$|||u|||=||u'||_p+||u||_q$$

is equivalent to the norm of $W^{1,p}(I)$.

I know $$||u||_{W^{1,p}(I)}=||u||_p+||u'||_p$$ By Sobolev embedding: there exists a constant $C$ such that $$||u||_q=\left(\int_I |u|^q\right)^\frac{1}{q}\leq \left(\int_I ||u||_\infty^q\right)^\frac{1}{q}=I^\frac{1}{q} ||u||_\infty\leq CI^\frac{1}{q} ||u||_{W^{1,p}(I)}$$ and $$||u'||_p\leq ||u||_{W^{1,p}(I)}$$ so $$|||u|||=||u'||_p+||u||_q\leq (CI^\frac{1}{q} +1) ||u||_{W^{1,p}(I)}$$

But I can't find $\alpha$ such that $$|||u|||\geq \alpha ||u||_{W^{1,p}(I)}$$ Can anyone help me?

José Carlos Santos
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Muniain
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1 Answers1

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We can (for example) consider two cases:

Case 1 - $q> p$

We have that $\|u\|_p^p=\int_I |u|^p$. Because $q>p$, we have that $\frac{q}{p}>1$, hence we can apply Holder inequality, to conclude that $\|u\|_p^p\leq C \|u\|_q^{p}$, where $C>0$ is a positive constant (that can change in every line). From it we have that $\|u\|_{1,p}=\|u\|_p+\|u'\|_p\leq C\|u\|_q+\|u'\|_p\leq C\||u\||$

Case 2- $q\leq p$

In this case $W^{1,p}(I)$ is embedded in $W^{1,q}(I)$, hence $\|u\|_p=\Big(\int_I|u|^p\Big)^{1/p}\leq C\|u\|_\infty\leq C\|u\|_{1,q}$. From here you can conclude.

Tomás
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