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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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    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?2011-04-10

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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, then there is a value $c$ such that $a 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.

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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    That fact that the binomial theorem works for non-integer $p$ was more complicated that needed, when I have the intermediate value thorem around.2011-04-10