Difference of powers inequality

Question

Is there any inequality involving difference of powers? I want to estimate say $x^p - y^p$, or $|x^p -y^p|$ hopefully in terms of $|x-y|^p$

Edit : $0 < p <\infty$, and I'm sure that there will be cases depending on $p$

Surely not in terms of $x^p-y^p$ only, since for $x=y+1$, $x^2-y^2$ is big when $x$ is big. — André Nicolas, Mar 11 '13 at 07:29
You can create one, if you know your bounds well and prove. Otherwise, the closest I know is, minkowski inequality, $|f|_p+|g|_p\ge|f+g|_p$. You can see Hagen Von Eitzen's comment though. — , Mar 11 '13 at 07:39
@HagenvonEitzen Won't we need $x,y>1$ for your inequality? Otherwise take $p=2$, $x=10^{-4}$, $y=10^{-5}$. — Jyotirmoy Bhattacharya, Mar 11 '13 at 09:30
@HagenvonEitzen Something like that would be ideal... Is there a proof? — Euler....IS_ALIVE, Mar 11 '13 at 20:37
This is like what I wanted http://math.stackexchange.com/questions/165520/is-ub-10-rightarrow-mathbbr-beta-holder-continuous-given-by-ux-x — Euler....IS_ALIVE, Mar 12 '13 at 00:28

xpaul · Accepted Answer · 2013-04-15T12:43:06.807

2

In fact, we have the following inequalities:

If $x,y>0$, then $|x-y|^p\le p|x-y|(x^{p-1}+y^{p-1}), \text{ for }p\ge 1$,
If $x\ge y>0$, then $|x-y|^p\le \frac{p}{2}x^{p-2}|x-y|(x^2+y^2), \text{ for }p\ge 2$.
(Mazur Inequality) For any real $x,y$, then $2^{1-p}|x-y|^p\le\big|x|x|^{p-1}-y|y|^{p-1}\big|\le|x-y|(|x|^{p-1}+|y|^{p-1}), \text{ for }p\ge 1$.

edited Apr 15 '13 at 12:43

answered Apr 14 '13 at 16:51

xpaul

Can you point me towards book/paper where Mazur Inequality is discussed. I googled it myself, but couldn't find any comprehensive explanation (especially for real number case). – TheGrandDuke Oct 10 '19 at 10:43
(3) is missing a $p$ on the right side, is it not? – Nate Eldredge Oct 27 '21 at 03:29

score 0 · Answer 2 · answered Mar 11 '13 at 09:35

0

$$|x^p-y^p|=|x-y||x^{p-1}+x^{p-2}y+\cdots+xy^{p-2}+y^{p-1}|$$ If $|x|<M$, $y<M$ this gives us $$|x^p-y^p|<pM^{p-1}|x-y|$$

answered Mar 11 '13 at 09:35

2 Answers2