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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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@gary: By measurable I mean Borel-measurable (maybe I should add that to the question), that is, the preimage of each Borel set is Borel. Continuous functions are particular examples of Borel measurable functions, and Borel measurable functions are closed under composition. – Florian Jul 7 '11 at 13:43
I deleted my comment, since it is nonsense; I was thinking of something else; the composition of measurables is , of course, measurable. Sorry. – gary Jul 7 '11 at 14:13
To sum up, the statement is false. The convergence works for those points where $f$ is locally Lipschitz continuous (at least for the points which are captured by the sequence $h_j$) but it often fails if $f$ has a worse modulus of continuity or is even discontinuous, and there are functions $f$ where this happens on a set with positive measure. – Florian Jul 8 '11 at 11:46
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. – Jason Swanson Jul 8 '11 at 22:09
up vote 2 down vote accepted

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

$$b<p_{i,k}+x\leq b+\frac{1}{6^k}.$$

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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Thanks. Actually it's not what I intended; I somehow need to relax the condition "for all h" to "for almost all h". I'll ask a new question. – Florian Jul 7 '11 at 11:54
I edited the question rather than asking a new one, because I needed only to change a tiny bit. But this invalidates your answer; sorry about that. – Florian Jul 7 '11 at 12:16
When you write, "for almost all $h$," I think you mean, "for almost all $x$." Now, Ben claims his result for all irrational $x$. Well, that's almost all $x$, so isn't the answer still valid? – Gerry Myerson Jul 7 '11 at 13:44
@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. – Florian Jul 7 '11 at 14:07

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.

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