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Let $f:\mathbb{R}\to \mathbb{R}$ be a measurable function and $(h_j)_j$ be a sequence of nonzero real numbers converging to zero as $j\to \infty$. Is it true that for almost every $x\in \mathbb{R}$: $\frac{1}{h_j}(\cos(f(x+h_j)-f(x))-1)\to 0\quad\mathrm{as}~~ j\to \infty~~?$

It feels that the Lebesgue differentiation theorem or maybe some other measure-theoretic theorem must be used to prove it, if it is true at all.

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    The function $f$ does not have to be Lipschitz for it to work. If $f$ is continuous, then it works iff $(f(x+h_j)-f(x))/|h_j|^{1/2}\to0$. So it works, for example, if $f$ is locally $\alpha$-Hölder with \alpha>1/2. And yes, there are continuous functions where this not only fails on a set of positive measure, but in fact fails almost everywhere.2011-07-08

4 Answers 4

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Here's another approach; hopefully it adds something to the answers already given. Suppose there is a Borel set $E$, with $m(E)>0$, and a preassigned sequence $\{h_j\}$ such that $\chi_E(x+h_j)\not\to \chi_E(x)$ for a.e. $x\in E$. Then $f(x)=\chi_E(x)$ provides a counterexample. In the spirit of @Ben Derret's original answer, I'm trying to show that a much weaker conclusion fails to hold, even in the context of characteristic functions of Borel sets.

As for the existence of $E$, I think a fat Cantor set will work if the sequence $\{h_j\}$ is chosen carefully. More precisely, let $E=\bigcap_kE_k$, where $E_0=[0,1]$, and $E_{k+1}$ is obtained from $E_k$ by removing an open segment of length $\frac{1}{2}\cdot\frac{1}{3^{k+1}}$ from the center of each of the $2^{k}$ disjoint intervals whose union is $E_k$. It's not hard to show that $m(E)=\frac{1}{2}$.

Suppose $x\in E$, and let $I=[a,b]$ be the interval housing $x$ at the $k$th stage in the construction; i.e., $I$ is one of the $2^k$ disjoint intervals (of common length) whose union is $E_k$. The length of $I$ is then

$\frac{1}{2^k}\left(1-2^{-1}\sum_{n=1}^k\frac{2^{n-1}}{3^n}\right)=\frac{3^k+2^k}{6^k}<\left(\frac{5}{6}\right)^k\qquad(\text{assuming }k>1).$

So $x\in\left(b-\frac{i}{6^k},b-\frac{i-1}{6^k}\right]$ for some $i\leq 5^k$.

For $i=0,\dots,5^k$, define $p_{i,k}=\frac{i}{6^k}$, and let $\{h_j\}$ be an enumeration of the $p_{i,k}$ satisfying $h_j\to0$. (For instance, $p_{0,1},\dots,p_{5,1};p_{0,2},\dots,p_{25,2};\dots.$) If $x\in E$ and $I=[a,b]$ is as above, choose $p_{i,k}$ so that

$b

The complement $E_k^c$ of $E_k$ is the disjoint union of segments $S$, each with length at least $\frac{1}{2}\cdot\frac{1}{3^k}>\frac{1}{6^k}$. So $p_{i,k}+x$ lies in one of the $S\subset E_k^c$ (namely, the one that has $b$ for a left endpoint).

There is a $j$ for which $p_{i,k}=h_j$. Then $x+h_j\notin E_k\supset E$. This can be done for all $k$, so $x+h_j\notin E\,$ infinitely often, and, if I haven't made an error, $\chi_E(x+h_j)\not\to\chi_E(x)$ whenever $x\in E$.

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Let $f(x) = \frac{\pi}{2}\mathbf{1}_\mathbb{Q}(x)$

Take $x\in \mathbb{R}\backslash\mathbb{Q}$. Then $f(x)=0$. Let $q_n$ be a sequence of rationals converging to $x$. Let $h_n=q_n-x$. Then

$\frac{1}{h_n}(\cos(f(x+h_n)-f(x))-1)= \frac{-1}{h_n}\nrightarrow 0$

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    @Gerry: Sorry about the confusion; I actually meant "for all h" but my comment was informal (I tried to indicate this by the word "somehow"). Please look at the question, which is now correctly stated (at least I think it is, except for the word "measurable" which might be ambiguous and should read Borel measurable). I decided to edit the question in a different way: the sequence of $h$es now has to be independent of $x$; that's why this answer no longer works.2011-07-07
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Let $f$ be a path of Brownian motion. Then $E [|f(x+h_j) - f(x)|^2] = |h_j|$ so $\frac{E [ \cos(f(x+h_j) - f(x)) - 1 ]}{h_j}$ does not converge to 0. I think it shouldn't be hard to convert the expected value statement to an "almost surely" statement, and then to go from that to an "almost all $x$" using Fubini's theorem.

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This is an elaboration on the answer of @Robert Israel. Let $B:\mathbb{R}\times\Omega\to\mathbb{R}$ be a (two-sided) Brownian motion defined on some probability space. Let $ X_j(x,\omega) = \frac{\cos(B(x + h_j,\omega) - B(x,\omega)) - 1}{h_j}. $ Let $S = \{(x,\omega): X_j(x,\omega) \to 0\}$, $S_x=\{\omega:(x,\omega)\in S\}$, and $S_\omega=\{x:(x,\omega)\in S\}$. Let $m$ denote Lebesgue measure. We claim that $ P(\{\omega: m(S_\omega) = 0\}) = 1. $ This claim implies that for $P$-a.e. $\omega$, the function $f(x)=B(x,\omega)$ serves as a (continuous) counterexample, showing that the answer to the original question is no. In fact, for such an $f$, the set of $x$ for which the quotient converges to zero is a set of measure zero.

To prove the claim, let us begin by defining $ Z_j(x,\omega) = \frac{B(x + h_j,\omega) - B(x,\omega)}{|h_j|^{1/2}}. $ Then \begin{align*} |X_j| &= \left|{\frac{\cos(|h_j|^{1/2}Z_j) - 1}{h_j}}\right|\\ &= \frac{\sin^2(|h_j|^{1/2}Z_j)}{|h_j||\cos(|h_j|^{1/2}Z_j) + 1|}\\ &= \left({\frac{\sin(|h_j|^{1/2}Z_j)}{|h_j|^{1/2}Z_j}}\right)^2 \frac1{|\cos(|h_j|^{1/2}Z_j) + 1|}Z_j^2. \end{align*} Since $x\mapsto B(x,\omega)$ is continuous, $|h_j|^{1/2}Z_j\to0$ as $j\to\infty$. Hence, $X_j\to0$ if and only if $Z_j\to0$, which implies $S = \{(x,\omega): Z_j(x,\omega) \to 0\}$.

Now fix $x\in\mathbb{R}$, and suppose that $P(S_x)>0$. By Blumenthal's $0$-$1$ law, this implies $P(S_x)=1$. Thus, $Z_j(x)\to0$ a.s. But each $Z_j(x)$ is a standard normal random variable. Thus, $E|Z_j(x)|^2=1$ for all $j$, which implies that the sequence $\{Z_j(x)\}_j$ is uniformly integrable. Hence, almost sure convergence implies convergence in $L^1$, giving $E|Z_j(x)|\to0$ as $j\to\infty$. But this contradicts the fact that each $Z_j(x)$ is a standard normal. Therefore, $P(S_x)=0$ for all $x\in\mathbb{R}$.

We now have $ \int_\mathbb{R}\int_\Omega 1_S(x,\omega)\,dP(\omega)\,dx = \int_\mathbb{R} P(S_x)\,dx = 0. $ By Fubini's theorem, $ \int_\Omega\int_\mathbb{R} 1_S(x,\omega)\,dx\,dP(\omega) = \int_\Omega m(S_\omega)\,dP(\omega) = 0. $ Hence, $m(S_\omega)=0$ a.s., which proves the claim.