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I am trying to prove that given $p>1$ there exists a constant $C=C(p,n)$ such that $\big||x|^px-|y|^py\big|\leq C\big(|x|^p+|y|^p\big)|x-y|$ for all $x,y\in\mathbb{R}^n$. It seems useful to consider the inequality $|x+y|^p\leq C\big(|x|^p+|y|^p\big)$ but I don´t know how to follow.

Any help will be appreciated!

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2 Answers 2

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Let $f(x)=|x|^p\,x$. Then $f$ has a continuous derivative given by $f'(x)=(p+1)\,|x|^p$. By the mean value theorem, there exists a point $\xi$ between $x$ and $y$ such that $$ |x|^p\,x-|y|^p\,y=(p+1)\,|\xi|^p\,(x-y). $$ Since $\xi$ is between $x$ and $y$ $$ |\xi|^p\le\max(|x|^p,|y|^p)\le|x|^p+|y|^p. $$ Then $$ \bigl|\,|x|^p\,x-|y|^p\,y\,\bigr|\le(p+1)\,(|x|^p+|y|^p)\,|x-y|. $$ Note that the argument is valid for $p>0$.

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  • $\begingroup$ Thank you very much, nice proof. $\endgroup$
    – John F.
    Apr 8, 2013 at 18:04
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A different approach than that consider by Julián go as follows: Note that

\begin{eqnarray} |x|^px-|y|^py &=& \int_0^1\frac{d}{dt}|y+t(x-y)|^{p}(y+t(x-y))dt \nonumber \\ &=& (x-y)\int_0^1|y+t(x-y)|^p+ \\ &&p\int_0^1 |y+t(x-y)|^{p-2}\langle y+t(x-y),x-y\rangle (y+t(x-y))dt \nonumber \end{eqnarray}

Also, by using the hint, we have that \begin{eqnarray} |x-y|\int_0^1|y+t(x-y)|^p &\leq& |x-y|\int_0^1 C(t^p|x|^p+(1-t)^p|y|^p) \nonumber \\ &\leq& C|x-y|(|x|^p+|y|^p) \nonumber \end{eqnarray}

On the other hand

\begin{eqnarray} \int_0^1 |y+t(x-y)|^{p-2}\langle y+t(x-y),x-y\rangle (y+t(x-y))dt &\leq& \int_0^1 |y+t(x-y)|^p|x-y| \nonumber \end{eqnarray}

and you can use the same reasoning as above. From here you can conclude

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  • $\begingroup$ This also includes the vectorial case, great! $\endgroup$
    – John F.
    Apr 8, 2013 at 18:49

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