# Absolutely continuous functions with derivatives in $L^p$

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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Write $f'=f'\cdot 1$ and use the fact that $f'\in L^p$. –  Giuseppe Negro Dec 12 '12 at 1:45
I thought about that. Then doesn't the constant $L$ that you get depend on $x$ and $y$? –  stephen Dec 12 '12 at 1:47
Oh if you take $L = \int_{\mathbb R} f'(t) dt$, then I think it works applying Holder. –  stephen Dec 12 '12 at 1:49
Exactly. To be more precise, you apply Hölder first and then estimate $$\left(\int_x^y \lvert f'(t)\rvert^p\, dt\right)^{1/p} \le \left(\int_{-\infty}^\infty \lvert f'(t)\rvert^p\, dt\right)^{1/p}.$$ (provided $x<y$) –  Giuseppe Negro Dec 12 '12 at 2:42
stephen: Is it part of your definition of absolutely continuous that $f'\in L^1$? Since there are several ways to generalize "absolutely continuous" from bounded to unbounded intervals, it would help to clarify precisely what that means here. (I have seen in some contexts "locally AC", which would imply $f(y)-f(x)=\int_x^y f'(t)dt$, but not generally $f'\in L^1$. The $\varepsilon$-$\delta$ version of the definition is stronger than locally AC, but still does not imply $f'\in L^1$.) I might not have asked, but you mention "$\int_\mathbb R f'(t)dt$" in a comment. –  Jonas Meyer Dec 14 '12 at 2:44

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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The issue here is that something like $x^2$ is absolutely continuous, but its derivative is not in $L^1$. –  JSchlather Dec 15 '12 at 2:37
@Jacob: Although $x^2$ is AC on bounded intervals, it is not AC according to the definition which stephen is using (although this definition was only given in a comment a few hours ago). It is the $\varepsilon$-$\delta$ definition usually applied to bounded intervals, but instead applied to the real line. In particular, it implies uniform continuity, hence $x^2$ is not $AC$. (And in any case, there is the additional assumption that $f'\in L^p$ which $x^2$ wouldn't satisfy.) –  Jonas Meyer Dec 15 '12 at 5:14
@stephen: You should have $\alpha=1/q$, not $1$, correct? –  Jonas Meyer Dec 15 '12 at 5:15
@Jonas. Yes thank you! –  stephen Dec 15 '12 at 16:04