Does $f(n\theta) \to 0$ for all $\theta>0$ and $f$ Darboux imply $f(x) \to 0$ as $x \to \infty$?

Solution 1:

Non-measurable example

By the axiom of choice there is a $\mathbb Q$-linear basis of $\mathbb R.$ This basis has the same cardinality as $\mathbb R$ so can be indexed as $a_r$ for $r\in\mathbb R.$ Define $f$ by setting $f(x)=r$ if $x$ is of the form $a_0+qa_r$ for some rational $q$ and real $r,$ and set $f(x)=0$ for $x$ not of this form. Then $f$ is Darboux because the set $\{a_0+qa_r\mid q\in\mathbb Q\}$ is dense for each $r.$ But for each $\theta>0,$ we can only have $f(q\theta)\neq 0$ for at most one rational $q$ - the reciprocal of the $a_0$ coefficient of $\theta.$ In particular $f(n\theta)\to 0$ as $n\to\infty$ with $n\in\mathbb N.$

Measurable example

For $n\geq 2$ let $b_n=n!(n-1)!\dots 2!.$ Each real has a unique "mixed radix" expression as $x=\lfloor x\rfloor + \sum_{n\geq 2}\frac{x_n}{b_n}$ where $x_n$ is the unique representative of $\lfloor b_n x\rfloor$ modulo $n!$ lying in $\{0,1,\dots,n!-1\}.$ For non-negative $x$ define $f(x)=\lim_{n\to\infty} \tfrac{1}{n}\sum_{m=2}^n x_m$ if this limit exists and $x_n\leq 1$ for all sufficiently large $n,$ and take $f(x)=0$ otherwise. For negative $x$ define $f(x)=f(-x).$ Note $f(x)\in[0,1].$ It is straightforward to see that $f$ takes all values in $[0,1]$ in every interval and is hence Darboux.

Now consider a real $x>0$ with $f(x)\neq 0$ and let $q<1$ be rational. We will show that $f(qx)=0.$ We know there exists $N$ such that $x_n\leq 1$ for all $n>N.$ Increasing $N$ if necessary we can assume that $qN$ is an integer. We also know that $x_n=1$ for infinitely many $n>N$ - otherwise we would have $\lim_{n\to\infty} \tfrac{1}{n}\sum_{m=2}^n x_m=0.$ Write $x=x'/b_{n-1}+1/b_n+\epsilon/b_{n+1}$ where $x'$ is an integer and $0\leq\epsilon< 2.$ So $qx b_{n+1}=qx'n!(n+1)!+q(n+1)!+q\epsilon.$ The first term is a multiple of $(n+1)!$ because $qn!$ is an integer, and the second term $q(n+1)!$ is an integer, and $q\epsilon<2.$ So $(qx)_{n+1}$ is either $q(n+1)!$ or $q(n+1)!+1$ (note this is less than $(n+1)!$). Since $q(n+1)!>1$ and there are infinitely many such $n,$ we get $f(qx)=0,$ .

This shows that for each $\theta>0,$ the sequence $f(n\theta)$ takes at most one non-zero value, and in particular $f(n\theta)\to 0.$

Remark: this $f$ appears to be a counterexample to https://arxiv.org/abs/1003.4673 Theorem 4.1.