# How to prove the following inequality?

Show that there exist constant $C= C(n,p)$ depending only on $n$ dimension and $p \in \mathbb{N}$ such that $$C|a-b|^{p} \le \langle |a|^{p-2} a - |b|^{p-2}b, a-b \rangle$$ for any $a,b \in \mathbb{R^{n}}$. Thank you.

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Vector inequalities of this type are constantly used in the study of the $p$-Laplace equation. Chapter 10 of Notes on the p-Laplace equation has enough of them to answer this question, see (I) and (VII) therein. However, I don't like Lindqvist's approach via integrals: I'd much rather see a clean algebraic derivation.

1. The dimension $n$ is irrelevant. Taking the span of $a$ and $b$, you can assume $n=2$.
2. The stated inequality is false for $p=1$.
3. There is no need for $p$ to be an integer. However, you do need $p\ge 2$ to have an inequality as stated.
4. The way in which the inequality is stated is suboptimal. Here is a better one: $$c|a-b|^2(|a|+|b|)^{p-2}\le \langle |a|^{p-2}a-|b|^{p-2}b,a-b \rangle \le C|a-b|^2(|a|+|b|)^{p-2}$$ which is true for $1<p<\infty$ (the constants depend on $p$).

I wish I had a clean proof of 4 up my sleeve, but I don't right now. I do have a proof of the following fact: the inner product is comparable to the product of norms. I actually inserted the following computations, elementary as they are, in a paper upon referee's request.

Let $S_{\alpha}(x)=|x|^{\alpha-1}x$, $0<\alpha<\infty$. (So, $\alpha=p-1$ in the above notation.)

Claim. The mapping $S_{\alpha}\colon\mathbb R^n\to\mathbb R^n$ satisfies the inequality $$\langle S_{\alpha}(x)-S_{\alpha}(y),x-y\rangle\ge \delta|S_{\alpha}(x)-S_{\alpha}(y)||x-y|$$ with $$\delta=\begin{cases}\alpha/(2-\alpha), \quad &0<\alpha\le 1;\\ 1/(2\alpha-1),\quad &1\le\alpha<\infty.\end{cases}$$

Proof. Since $S_{1/\alpha}=S_{\alpha}^{-1}$, it suffices to consider the case $0<\alpha\le 1$. The homogeneity of $S_{\alpha}$ reduces our task to proving the inequality $$\langle{x-S_{\alpha}(y),x-y}\rangle\ge \frac{\alpha}{2-\alpha}|x-S_{\alpha}(y)||x-y|$$ under the assumptions $|x|=1$, $|y|\le 1$. We estimate both sides as follows. $$\langle x-S_{\alpha}(y),x-y\rangle =|{x-y}|^2+\langle (1-|y|^{\alpha-1})y,x-y \rangle \ge |x-y|^2+(1-|y|^{\alpha-1})|y||x-y| \\ =|x-y|(|x-y|+|y|-|y|^{\alpha})$$ and $$|x-S_{\alpha}(y)|\le |x-y|+|y-|y|^{\alpha-1}y|=|x-y|+|y|^{\alpha}-|y|.$$ The desired inequality will follow once we prove that $$|x-y|+|y|-|y|^{\alpha}\ge\frac{\alpha}{2-\alpha}(|x-y|+|y|^{\alpha}-|y|).$$ The latter is equivalent to $$(1-\alpha)|x-y|\ge |{y}|^{\alpha}-|y|.$$ Note that $|y|^{\alpha}\le 1+\alpha(|y|-1)$ because the function $t\mapsto t^{\alpha}$ is concave for $t\ge 0$. Hence $$|y|^{\alpha}-|{y}|\le (1-\alpha)(1-|y|)\le (1-\alpha)|x-y|,$$ which is what we wanted. QED

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Thank you very much. – user29999 Jun 15 '12 at 21:21