Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For $u,v \in L^q(\Omega)$ with $q \ge p \ge 1$, how does one show that: $$ \begin{aligned} \||u|^{p-1}u - |v|^{p-1}v\|_{L^{p/q}} & \le C\,\|(|u|^{p-1} + |v|^{p-1})\,|u-v|\,\|_{L^{p/q}}\\ & \le C\,(\|u\|^{p-1}_{L^q} + \|v\|^{p-1}_{L^q})\,\|u-v\|_{L^q} \end{aligned} $$


share|cite|improve this question
the first inequality is well explained here… – uforoboa Sep 19 '12 at 10:12
... and the second one is Hölder with $\frac 1q + \frac 1\alpha = \frac 1{\frac pq}$ giving $\alpha = \frac pq + \frac 1{p-1}$. – martini Sep 19 '12 at 10:15
Ok, I nearly see it now. Except for the mean-value theorem step... – David Chapel Sep 19 '12 at 10:21

Just to put something in this box: the "mean-value theorem step" is

For all real $r\ge 1$ one has $r^{p-1} - 1 \leq c_p(r - 1)(r^{p-1} + 1)$

Indeed, applying MVT to $f(x)=x^{p-1}$ on the interval $[1,r]$ we get $$f(r)-f(1)=f'(\xi)(r-1)=(p-1)\xi^{p-2}(r-1),\qquad \exists \xi\in (1,r)$$ Here $\xi^{p-2}\le \max(r^{p-2},1)$ where we take $\max $ because $p-2$ could be either negative or positive. Hence, $$r^{p-1} - 1 \leq (p-1)(r-1) \max(r^{p-2},1)\leq (p-1)(r-1) (r^{p-1}+1)$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.