Problem

One of my best friends asked me to think about the following problem:

Suppose a sequence $\{a_n\}_{n=1}^\infty$ satisfies $\lim_{n\to\infty}a_{\lfloor\alpha^n\rfloor}=0$ for each $\alpha>1$. Is it true that $\lim_{n\to\infty}a_n=0$?

He told me that the preceding proposition (if true) implies Kolmogorov's strong law of large numbers in probability theory. I don't know how, but it's irrelevant.

Thoughts

Suppose $E\subseteq(1,+\infty)$ is countable (for example, $E=(1,+\infty)\cap\mathbb Q$) and $\lim_{n\to\infty}a_{\lfloor\alpha^n\rfloor}=0$ for each $\alpha\in E$, we cannot conclude that $\lim_{n\to\infty}a_n=0$. In fact, we can choose an infinite set $S\subseteq\mathbb Z_{>0}$ such that $S\cap\{\lfloor\alpha^n\rfloor\colon n\in\mathbb Z_{>0}\}$ is finite for each $\alpha\in E$ as follows:

Suppose $E=\{\alpha_1,\alpha_2,\dotsc\}$. We choose $S$ inductively. Suppose $T_0=\mathbb Z_{>0}$. Given $T_{n-1}$, we set $s_n=\min T_{n-1}$ and $T_n=(T_{n-1}\setminus\{\lfloor\alpha_n^k\rfloor\colon k\in\mathbb Z_{>0}\})\setminus\{s_n\}$. By a density argument, it's easy to see that $T_n$ are infinite therefore the process doesn't terminate. Let $S=\{s_n\colon n\in\mathbb Z_{>0}\}$. It's apparent that $\#(S\cap\{\lfloor\alpha_n^m\rfloor\colon m\in\mathbb Z_{>0}\})\le n$.

Given $S$, we set $a_n=1/n$ if $n\not\in S$, and $a_n=1$ if $n\in S$, then $\lim_{n\to\infty}a_{\lfloor\alpha^n\rfloor}=0$ for each $\alpha\in E$, but $\lim_{n\to\infty}a_n$ doesn't exist.

Here we choose $S$ by a diagonal process, therefore we cannot mimic the construction when $E$ is uncountable. In fact, the falsehood of the original statement is equivalent to the existence of $S$, therefore it's essential combinatorial, or related to some topological structure of $\mathbb Z_{>0}$ (say, compactness or Baire category, etc). I have no idea on the general case. Any idea? Thanks!

Solution 1:

Here is an attempt. Hopefully, this is correct.

Fix $\varepsilon >0$. We are looking for an integer $N$ such that $\vert a_n\vert<\varepsilon$ for every $n\geq N$.

For any $k\in\mathbb N$, set $$ A_k:=\left\{ \alpha>1;\; \forall l\geq k\;:\; \vert a_{\lfloor \alpha^l\rfloor}\vert<\varepsilon \right\} .$$

Then $(1,\infty)=\bigcup_{ n\in\mathbb N} A_k$ by assumption. By the Baire category theorem, it follows that at least on $A_k$ is not nowhere dense. So one can fix $1<u<v$ and $k_0\in\mathbb N$ such that $A_{k_0}\cap (u,v)$ is dense in $(u,v)$.

Now, choose an integer $K\geq k_0$ such that $v^k> u^{k+1}$ for all $k\geq K$; this is possible since $v>u$. Then $\bigcup_{k\geq K} (u^k,v^k)$ is the interval $(u^K,\infty)$ (the intervals overlap by the definition of $K$).

Finally, choose an integer $N>u^K$. Let us check that this $N$ works.

Take any integer $n\geq N$. Then one can write $n=\beta^k$ for some $\beta\in (u,v)$ and some integer $k\geq K$. Now, since $A_{k_0}$ is dense in $(u,v)$, one can find a point $\alpha\in A_{k_0}$ such that $\alpha >\beta$ and $\alpha$ is very close to $\beta$. Then, since the floor function is upper semi-continuous and $n=\beta^k$, we have $n=\lfloor \alpha^k\rfloor$. By the definition of $A_{k_0}$ and since $k\geq K\geq k_0$, it follows that $\vert a_n\vert<\varepsilon$, as required.