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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!

stephen
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    Write $f'=f'\cdot 1$ and use the fact that $f'\in L^p$. – Giuseppe Negro Dec 12 '12 at 01:45
  • I thought about that. Then doesn't the constant $L$ that you get depend on $x$ and $y$? – stephen Dec 12 '12 at 01:47
  • Oh if you take $L = \int_{\mathbb R} f'(t) dt$, then I think it works applying Holder. – stephen Dec 12 '12 at 01:49
  • 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 02:42
  • 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 02:44
  • @Jonas. It is not directly part of my definition, but it follows from the fundamental theorem of calculus for Lebesgue integration. – stephen Dec 14 '12 at 03:41
  • @Jonas. Is it important in this case? – stephen Dec 14 '12 at 03:42
  • @stephen: The definition of absolutely continuous is definitely important if you are trying to prove something about absolutely continuous functions. Since there is no widespread standard for what absolutely continuous means when the domain is $\mathbb R$, it would be appreciated if you wouldn't mind stating it for our benefit. If you are asking whether it is important that $f'$ is in $L^1$, that depends on whether you want to use such a property; in your comment it seemed you did. I do not know what you mean by "it follows from the fundamental theorem of calculus." – Jonas Meyer Dec 14 '12 at 03:58
  • stephen: Where is the problem from? I would enjoy looking up the context myself if it is an available reference. – Jonas Meyer Dec 14 '12 at 04:13
  • @Jonas: The problem comes from a problems set that our professor gave us to practise for the final exam. I am not sure where she took them from. The definition of absolutely continuous function that I use is: For every $\epsilon>0$, there exists $\delta > 0$ such that for any finite collection of disjoint intervals $(a_i, b_i)_{i=1}^n$, $\sum_i |F(b_i)-F(a_i)|<\epsilon$ whenever $\sum_i (b_i-a_i) < \delta$. This is the definition from Folland's real analysis. The Fundamental theorem of calculus states that $F$ is absolutely continuous if and only if $F'$ exists almost everywhere and... – stephen Dec 15 '12 at 02:04
  • ...is integrable, and we have $F(b)-F(a)= \int_a^b F'(x)dx$. I will post an answer to my question so that you will understand the discussion between Giuseppe and I. – stephen Dec 15 '12 at 02:05
  • stephen: Thank you for clarifying. The things you have written now make sense, but your earlier comment referencing $\int_\mathbb R f'(t)dt$ still does not, which is why I was having trouble figuring out the underlying assumptions. (Maybe this was a typo, and you meant $|f'|_p$?.) Anyway, to be clear, that condition implies (and is stronger than) the condition that $f$ is absolutely continuous on bounded intervals (which would suffice for your problem), but certainly does not imply that $f'$ is integrable on $\mathbb R$. – Jonas Meyer Dec 15 '12 at 05:05
  • stephen: While your answer looks good except for the loss of $1/q$, and your definition of AC works here, I will note more explicitly: (1) That definition of AC does not imply that $F'$ is integrable on $\mathbb R$: e.g., consider $F(x)=x$. (2) That definition of AC is strictly stronger than the FTC holding on bounded intervals, as the example $F(x)=x^2$ shows. AC on bounded intervals suffices for your problem and is a property I have seen used more often than the definition you gave here. – Jonas Meyer Dec 15 '12 at 05:25
  • @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! – stephen Dec 15 '12 at 16:06

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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.

stephen
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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 02:37
  • @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 05:14
  • @stephen: You should have $\alpha=1/q$, not $1$, correct? – Jonas Meyer Dec 15 '12 at 05:15
  • @Jonas. Yes thank you! – stephen Dec 15 '12 at 16:04