# Inequality involving $|u|^{p-1} u$

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}

Thanks.

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the first inequality is well explained here math.stackexchange.com/questions/9960/… – 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

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, $$r^{p-1} - 1 \leq (p-1)(r-1) \max(r^{p-2},1)\leq (p-1)(r-1) (r^{p-1}+1)$$