0
$\begingroup$

Suppose $t_{n}$ is a sequence of positive real numbers such that $c_{1}\geq \lim \sup_{n\to \infty}t_{n}/n\geq \lim \inf_{n\to \infty}t_{n}/n\geq c_{2}>0$ where $c_{1}\geq c_{2}>0$ are positive constants.

Does it follow that $\lim \inf_{N\to \infty}\dfrac{m(\cup^{N}_{n=1}[t_{n}-1,t_{n}+1])}{N}>0$ where $m$ is Lebesgue measure.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

Define $t_n$ by $t_n = 2^k$ for $2^{k-1} < n \le 2^k$, so that it begins $$1, 2, 4,4, 8,8,8,8, 16,16,16,16,16,16,16,16,32, \dots$$ Then we have $n \le t_n \le 2n$ for all $n$ so your hypothesis is satisfied with $c_1 = 2$, $c_2 = 1$. But there are only about $\log_2 N$ distinct values among $t_1, \dots, t_N$, so $m(\cup^{N}_{n=1}[t_{n}-1,t_{n}+1]) \approx 2 \log_2 N$ and the desired liminf equals zero.

Improve this answer
$\endgroup$
1
  • $\begingroup$ @YellowPig: Try a sequence where you take a number $n_k$ and repeat it $\sqrt{n_k}$ times, then take the number $n_{k+1} = n_k+\sqrt{n_k}$ and repeat it $\sqrt{n_{k+1}}$ times, etc. If my back-of-the-envelope calculation is right, you should get $c=1$, but the first $N$ terms contain about $N^{2/3}$ distinct values. $\endgroup$
    – Nate Eldredge
    Commented Jul 27, 2017 at 19:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .