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Suppose you have an absolutely continuous function $f$, with derivative $f'\in L^p(\mathbb R)$ for some $p>1$. Then I would like to show that there exist constants $L$ and $\alpha$ such that

$|f(x)-f(y)| \leq L |x-y|^{\alpha}, \forall x,y.$

Since $f$ is absolutely continuous, we have that $f(x) - f(y) = \int_y^x f'(t) dt$. Then I should maybe use Hölder inequality, but I don't know how to apply it in this case. Any help would be appreciated!

Thanks!

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    @Jonas. Yes you are totally right. It is integrable on bounded intervals, and this is what I use for the proof. Sorry for my confusion!2012-12-15

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Since $f$ is absolutely continuous, we have that for any $x,y$,

$|f(y)-f(x)| = \left|\int_x^y f'(t) dt\right|.$

By applying Hölder inequality on $f'(t)\cdot 1$, we get,

$f(y)-f(x) \leq \left(\int_x^y|f'(t)|^p dt\right)^{1/p} \cdot \left(\int_x^y 1^q dt\right)^{1/q}$

$\leq \left(\int_{-\infty}^{\infty} |f'(t)|^p dt\right)^{1/p} \cdot |y-x|^{1/q},$

where $1/p + 1/q = 1$. Taking $L= (\int_{-\infty}^{\infty} |f'(t)|^p dt)^{1/p}$ and $\alpha = 1/q$, this proves the claim.

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    @Jonas. Yes thank you!2012-12-15