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In Remark 11 on page 214 ( Functional analysis, Sobolev spaces and partial differential equations written by Haim Brezis): Let $I$ be an bounded interval, let $1 \le p \le \infty,$ and $1 \le q \le \infty.$ From Theorem 8.2 and (5), it can be shown easily that the norm $|||u|||=\|u'\|_p+\|u\|_q$ is equivalent to the norm $\|u\|_{1,p}=\|u'\|_p+\|u\|_p$ of $W^{1,p}(I).$ Note that $\|u\|_\infty \le \|u\|_{1,p}, \forall 1 \le p \le \infty$ $\cdots (5)$ and Theorem 8.2 says the existence of continuous representative $\tilde u$ on $\bar I$ for $u \in W^{1,p}(I)$ and $\tilde u(y)-\tilde u(x)=\int_x^y u'(t)dt~\forall x,y \in \bar I.$

Since $I$ is a bounded interval, $\|u\|_b \le C \|u\|_a$ if $b \le a,$ it is easy to prove that $\exists$ $C_1>0$ such that $\|u\|_{1,p} \ge C_1 |||u|||$ for all $u \in W^{1,p}(I).$ Also, if $q \ge p,$ it's also easily prove that $\exists$ $C_2>0$ such that $|||u||| \ge C_2 \|u\|_{1,p}$ for all $u \in W^{1,p}(I).$

I cannot prove $\exists$ $C_2>0$ such that $|||u||| \ge C_2 \|u\|_{1,p}$ for all $u \in W^{1,p}(I)$ for the case $q <p.$ Would you tell me the reason why it's true.

04170706
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1 Answers1

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I'll only consider the case when $q \leq p$. Take $u\in W^{1,p}(I)$, and note that $u\in W^{1,q}(I)$ as well. Since $I$ is bounded we have that $$\|u\|_{L^p}\leq c\|u\|_{L^\infty}.$$ Now the result follows: \begin{align*} \newcommand{\vertiii}[1]{{\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert #1 \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}} \|u\|_{W^{1,p}} & = \|u'\|_{L^p} + \|u\|_{L^p} & \qquad & \\ & \leq \|u'\|_{L^p} + c\|u\|_{L^\infty} & & \\ & \leq \|u'\|_{L^p} + c\|u\|_{W^{1,q}} & & \text{(by Theorem 8.8 since $u\in W^{1,q}(I)$)}\\ & = \|u'\|_{L^p} + c\|u'\|_{L^q} + c\|u\|_{L^q} & & \\ & \leq (1+c)\|u'\|_{L^p} + c\|u\|_{L^q} & & \\ & \leq c\vertiii{u}. \end{align*}

Note that I'm being a bit sloppy with the constant $c$ changing from line to line.

Yousuf Soliman
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