How do I prove this well known inequality?

Question

Let $u$ and $v$ be functions from $\mathbb{R}^{n+1}$ to $\mathbb{C}$.

How do I prove the following inequality: $$\big| |u|^{p-1}u - |v|^{p-1}v\big| \leq (|u|^{p-1}+|v|^{p-1})|u-v|?$$

This inequality is often used showing well-posedness of some non-linear PDE without proof.

EDIT: Some constant which is depending on $p$ and $n$ may be multiplied on RHS, i.e. $$\big| |u|^{p-1}u - |v|^{p-1}v\big| \leq C(n,p)(|u|^{p-1}+|v|^{p-1})|u-v|.$$

Answer 1

Observe \begin{align} \left||u|^{p-1}u-|v|^{p-1}v\right| =&\ \left||u|^{p-1}u-|u|^{p-1}v+|u|^{p-1}v -|v|^{p-1}v\right| \\ \le&\ |u|^{p-1}|u-v| + \left||u|^{p-1}-|v|^{p-1}\right||v|\\ \le&\ |u|^{p-1}|u-v|+(|u|^{p-2}|v|+|u|^{p-3}|v|^2+\ldots+|u||v|^{p-2}+|v|^{p-1})\left||u|-|v|\right|\\ \le&\ |u-v|\left(|u|^{p-1}+|u|^{p-2}|v|+\ldots +|u||v|^{p-2}+|v|^{p-1} \right). \end{align} Finally, we see that \begin{align} |u|^{a+b}+|v|^{a+b}-|u|^a|v|^b-|u|^b|v|^a =(|u|^a-|v|^a)(|u|^b-|v|^b)\geq 0 \end{align} which means \begin{align} \left||u|^{p-1}u-|v|^{p-1}v\right| \leq C|u-v|\left| |u|^{p-1}+|v|^{p-1}\right| \end{align} for some constant $C>0$.

Note that I have assumed $p$ is an integer. It's not hard to prove the statement for more general $p$.

Edit: Here is a more complete answer. For reference, one can consult Theorem 41 on page 39 of Inequalities by Hardy, Littlewood, and Polya.

There they obtained estimates of the form \begin{align} \left||u|^p-|v|^p \right|\leq p(|u|^{p-1}+|v|^{p-1})|u-v|. \end{align}

To relate to the desired estimate to the above estimate, observe that if $u$ and $v$ are of different sign then we have \begin{align} \left||u|^{p-1}u-|v|^{p-1}v\right| =& |u|^p+|v|^p\leq |u|^{p-1}(|u|+|v|) \\ =&\ |u|^{p-1}|u-v|\le (|u|^{p-1}+|v|^{p-1})|u-v|. \end{align} If $u$ and $v$ are of the same sign then this reduces to the situation considered in the book.

Answer 2

The inequality is false, at least for $p>2$. Let $u(x)=2$ and let $v(x)=1$. Then the left side equals $2^p-1$ and the right side equals $2^{p-1}+1$, however $$ (2^p-1)-(2^{p-1}+1)=2^{p-1}-2 $$ which is positive when $p>2$.

Answer 3

Proceeding with a multiplicative constant in light of the other 2 answers. I'll assume $u\in\mathbb R^d$, one can make an identification $\mathbb R^2 = \mathbb C$. These things like $f(a)-f(b)=g(a,b)|b-a|$ look like MVT which usually works out.

Set $f_j(u)=|u|^{p-1}u_j$, then $\partial_i f_j=(p-1)|u|^{p-2}\frac{u_j}{|u|}u_i + |u|^{p-1}\delta_{ij}$. Lazy calculations yield $$|\nabla f|=\|\partial_i f_j\|_{\ell^2_{ij}}\le C\|\nabla f\|_{\ell^\infty_{ij}} \le C[(p-1) |u|^{p-3}\|u_i\|^2_{\ell^\infty_{i}} + |u|^{p-1}] \le Cp|u|^{p-1},$$ since $\|u_i\|_{\ell^\infty_i} \le \|u_i\|_{\ell^2_i} = |u|$. I'd guess a smarter calculation would give $C=1$, instead we have $C=C(d)$.

Now MVT gives $$ | |u|^{p-1}u - |v|^{p-1}v | \le \sup_{\theta \in [0,1]} |\nabla f(\theta u + (1-\theta) v)| |u-v|$$ So the goal is just to bound the derivative on the line $[u,v]\subset \mathbb R^n$. $p=1$ is trivial. If $1<p<2$, then $s\mapsto s^{p-1}$ is concave, and $0^{p-1}=0$. This implies subadditivity, giving the result: \begin{align} |\nabla f(\theta u + (1-\theta) v)| &= C(n,d,p)|\theta u + (1-\theta) v|^{p-1} \\ &\le C(n,d,p)(|\theta u| + |(1-\theta)v|)^{p-1} \\ &\le C(n,d,p)(|\theta u|^{p-1} + |(1-\theta)v|^{p-1}) \\ &\le C(n,d,p)(|u|^{p-1} + |v|^{p-1}). \end{align} For $p\ge 2$, we can just directly use convexity, like the question body here and with some more consideration of constants here.