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.
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.
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
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.