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Using Reverse Triangle Inequality, one can write for $x,y\in R^1$

$$ ||x|-|y||\leq |x-y| $$

Is there any suitable inequality doing the following $$ ||x|^p-|y|^p|\leq f_p(|x-y|) $$

for $1 \leq p < \infty$

Thanks for any advice.

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Presumable, you want an $f_p$ which depends on $p$, not a general $f$? A general $f$ can't be found because you can choose $x=1-\epsilon$ and $y=1+\epsilon$ and make $||x|^p-|y|^p||$ abitrarily large when $|x-y|$ is very small. –  Thomas Andrews Apr 10 '11 at 8:48
    
Sure, I want $f_p$ which depends on $p$. –  Jirka cigler Apr 10 '11 at 8:52
    
Never mind, it shouldn't be possible even for $p=2$, because if $x$ and $y$ are positive, $x>y$, then $|x^2-y^2| = x^2-y^2 = (x-y)(x+y) < 2y|x-y|$. So you can find |x-y| small but $|x^2-y^2|$ is arbitrarily large. I think this can be done for $p>1$, just easier to show it with $p=2.$ –  Thomas Andrews Apr 10 '11 at 8:54
    
Well, this makes sense. We can relax the general problem and assume that the variables are constrained by a-priory known lower and upper bounds i.e. $x_{lb} \leq x \leq x_{ub}$ and $y_{lb} \leq y \leq y_{ub}$, would it be possible to parametrize it in this case? –  Jirka cigler Apr 10 '11 at 9:10
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1 Answer

This can't be done when $p>1$ because the derivative of $x^p$, $px^{p-1}$ is unbounded.

I've forgotten the name of the theorem, but we'll use the following fact: If $g(x)$ is differentiable, and $a<b$, then there is a value $c$ such that $a<c<b$ and $g^\prime(c)=\frac{f(b)-f(a)}{b-a}$

So, if we take $x>0, y=x+1$, and $g(x)=x^p$, then: $||y|^p-|x|^p| = (x+1)^p - x^p = pc^{p-1}$ for some $c$ with $x<c<x+1$.

Now, if $f$ exists, then $||y|^p-|x|^p| \leq f(1)$ in this case. But $pc^{p-1}>px^{p-1}$ is unbounded.

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It's really that $x^p$ is not uniformly continuous. Also, you might just cite the binomial theorem instead of using the intermediate value theorem. –  Douglas Zare Apr 10 '11 at 10:00
    
That fact that the binomial theorem works for non-integer $p$ was more complicated that needed, when I have the intermediate value thorem around. –  Thomas Andrews Apr 10 '11 at 10:56
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