If $\lim_{n \to +\infty} \ f(\alpha n)$ exists , does this imply $\lim_{x \to +\infty} f(x)$ exists?

Question

Let $f:\mathbb{R} \to \mathbb{R}$. Suppose $f$ has the following property: for all $\alpha >0$, the sequence $\{f(\alpha n)\}_{n \geq 0}$ converges either to a finite or infinite extended real number as $n \to +\infty$. In other words, we assume

$$\forall \alpha \in \mathbb{R}_+,\lim_{n \to +\infty} f(\alpha n) = L(\alpha) $$ where $L(\alpha)$ is a value depending on $\alpha$. Note that $L(\alpha)$ may be $\pm \infty$.

Does it follow that $$\lim_{x \in \mathbb{R}, x \to +\infty} f(x)$$ exists as a finite or infinite extended real number?

Note that if the above claim is true, then $L(\alpha)$ is in fact independent of $\alpha$. However, we do not assume this.

We can also play around with this to get stronger and weaker statements. Does it hold if we just assume $a \in \mathbb{Q}_+$? Does it hold if we assume $f$ is continuous?

I haven't made too much progress. However, this is quite similar to a classic little theorem which states that for continuous $f$, if for all $\alpha>0$, $f(\alpha n)\xrightarrow[n \in \mathbb{N}, n\to+\infty]{} 0$ then $f(x)\xrightarrow[x \in \mathbb{R}, x\to+\infty]{} 0$ which is proven with the Baire category theorem. However, my 'conjecture' only assumes the existence of the limit (rather than giving an explicit value, like $0$) and also does not assume continuity. I can't quite modify that proof to make it work here.

Answer 1

It is true if we further assume that $f$ is continuous!

First we shall use the following Lemma:

Lemma. Let $I_n=[a_n,b_n]\subset(0,\infty)$, $n\in\mathbb N$, where $a_n<b_n$ and $a_n\to\infty$, and $K$ be an infinite subset of $\mathbb N$. We set $$ S_K=\Big\{x\in(0,\infty) : \text{$nx\in \bigcup_{k\in K} I_k$ for infinitely many $n\in\mathbb N$}\Big\}. $$ If $K,L$ are infinite subsets of $\mathbb N$, then $S_K\cap S_L$ is dense in $(0,\infty).$

We postpone the proof of this Lemma. If $\lim_{x\to\infty}\,f(x)$ does NOT exist, then we can pick $\,A,B \in\mathbb R$, such that $$ \liminf_{x\to\infty}\, f(x) < A < B < \limsup_{x\to\infty}\, f(x). $$ Due to the continuity of $f$ it is possible to define intervals $I_n=[a_n,b_n]$ and $J_n=[c_n,d_n]$, $n\in\mathbb N$, such that $\,a_n<b_n<c_n<d_n<a_{n+1}$, and $$ f\,\big|_{I_n} <A\quad\text{while}\quad f\,\big|_{J_n} >B, \quad \text{for all $n\in\mathbb N$}. $$ Due to the Lemma, there exists a set of points $x$, dense in $(0,\infty)$, with the property that, for infinitely many $n$'s the multiple $nx$ belongs to $\bigcup_{k\in\mathbb N} I_k$, and for infinitely many $n$'s the multiple $nx$ belongs to $\bigcup_{k\in\mathbb N} J_k$.

This in turn implies that $\lim_{n\to\infty}f(nx)$ does not exist.

Proof of the Lemma. Let $[c,d]\subset(0,\infty)$. We shall prove that there exists an $x\in [c,d]$, such that for infinitely many $n$ the multiple $nx$ belongs to an interval of the form $I_k$, $\,k\in K$, and for infinitely many $n$ the multiple $nx$ belongs to an interval of the form $I_\ell$, $\ell\in L$. This is based on the observation that, if $a_n$ is sufficiently large, then $$ [c',d']=\frac{1}{N}[a_n,b_n]\cap[c,d]\quad\text{is a nontrivial interval}, $$ for $N=\lfloor a_n/d\rfloor+1$. This allows us to recursively define a sequence of nontrivial closed intervals $J_n=[c_n,d_n]$, where $[c_0,d_0]=[c,d]$, $[c_1,d_1]=[c',d']$, and we pick the $a_n$'s, so that the first one is in $K$, the next in $L$, the next in $K$ and so on. In this way we have a sequence of closed intervals $$ J_0\supset J_1\supset\cdots\supset J_n\supset J_{n+1}\supset\cdots. $$ Clearly $S=\bigcap_{n\in\mathbb N} J_n\ne\varnothing$, and for each $x\in S$, infinitely many multiples belong to $I_k$'s, $k\in K$, while infinitely many multiples belong to $I_\ell$'s, $\ell\in L$.

Answer 2

By indicator function I mean $f(x)=\left\{\begin{matrix} 1& x\in Q & \\ 0& x\in Q^c & \end{matrix}\right.$ This function is a counterexample.

Answer 3

Let me give a shorthand version of Yiorgos S. Smyrlis's answer. The following lemma will come handy:

Lemma. Let $U \subseteq (0,\infty)$ be open and unbounded. Then for any $N \geq 1$, the set $\bigcup_{n\geq N} \frac{1}{n}U $ is dense in $(0, \infty)$.

Proof. Let $W_N = \bigcup_{n\geq N} \frac{1}{n}U$. It suffices to prove that $W_N$ intersects every $(a, b) \subseteq (0, \infty)$. To this end, notice that $\frac{a}{b} < \frac{n}{n+1}$ for large $n$ and hence $$ (r, \infty) \subseteq \bigcup_{n \geq N} (na, nb) $$ holds for some $r > 0$. Since $U$ is unbounded, we can pick $x \in U \cap (r,\infty)$. Then $x \in (na, nb)$ for some $n \geq N$, which implies that $\frac{1}{n}x \in W_N \cap (a, b)$. $\quad \square$

Now we prove a proposition that is enough for our purpose. First, we introduce some notations. Fix a function $f:(0,\infty) \to \Bbb{R}$. Then we denote the set of limit points of $f$ as $x\to\infty$ by

$$ A := \{ \ell \in [-\infty, +\infty] : f(x_j) \to \ell \text{ for some } x_j \to \infty \} $$

Similarly, for each $\alpha > 0$ we denote the set of limit points of the sequence $(f(n\alpha) : n\geq 1)$ by

$$ A(\alpha) := \{ \ell \in [-\infty, +\infty] : f(n_j\alpha) \to \ell \text{ for some } n_j \to \infty \} $$

Proposition. Let $f : (0, \infty) \to \Bbb{R}$ be continuous. Then there exists $\alpha > 0$ such that $A = A(\alpha)$ holds. Moreover the set of all such $\alpha$ is dense in $(0,\infty)$.

Proof. It is clear that $A(\alpha) \subseteq A$ for all $\alpha \in (0, \infty)$. So it is enough to prove that $A \subseteq A(\alpha)$ for $\alpha$ on a dense subset of $(0,\infty)$. To this end, define the family $\mathcal{U}$ of open intervals by

$$ \mathcal{U} = \{ (a, b) : \text{ $a, b \in \Bbb{Q}\cup\{-\infty,\infty\}$ and $f^{-1}((a, b))$ is unbounded} \} $$

and notice that $\mathcal{U}$ is countable and

$$ \ell \in A \quad \Leftrightarrow \quad \begin{cases} \text{$(a,b) \in \mathcal{U}$ for all $a, b \in \Bbb{Q}$ with $a < \ell < b$}, & \text{if } \ell \in \Bbb{R} \\ \text{$(a,\infty) \in \mathcal{U}$ for all $a \in \Bbb{Q}$}, & \text{if } \ell = +\infty \\ \text{$(-\infty,b) \in \mathcal{U}$ for all $b \in \Bbb{Q}$}, & \text{if } \ell = -\infty \end{cases} $$

Then by the Baire category theorem and the lemma above,

$$ D = \bigcap_{U \in \mathcal{U}} \bigcap_{N\geq 1} \bigcup_{n\geq N} \frac{1}{n} f^{-1}(U) $$

is a dense subset of $(0, \infty)$. Moreover, for each $\alpha \in D$ and for each $U \in \mathcal{U}$, we have $f(n\alpha) \in U$ for infinitely many $n$. So each $\ell \in A$ is also a limit point of the sequence $f(n\alpha)$. Therefore we have $A \subseteq A(\alpha)$ for each $\alpha \in D$. $\quad \square$

For our problem, the conclusion follows by noticing that $A(\alpha)$ is a singleton $\{L(\alpha)\}$ for all $\alpha > 0$.