1
$\begingroup$

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.

Remark: Yesterday I asked whether this was true assuming only $\lim \inf_{n\to \infty}t_{n}/n>0$. The answer to this question was no, as answered by DanielWainfleet here: Does a neighborhood of a sequence of positive density have positive measure?

$\endgroup$
0
0
$\begingroup$

While the factorial grows too quickly to be useful here, the idea behind DanielWainfleet's answer to your previous question still works. Define $t_{2^n+k}=2^n$ for $n\ge0$ and $1\le k\le2^n$. Observe that $\frac12n\le t_n\le n$ for all $n$, so certainly your condition holds. However,

$$m\left(\bigcup_{j=1}^{2^n+k}[t_j-1,t_j+1]\right)=m\left(\bigcup_{i=0}^n[2^i-1,2^i+1]\right)\le2(n+1)\le C\log(2^n+k)$$

and so $\frac1Nm\left(\bigcup_{n=1}^N[t_n-1,t_n+1]\right)\to0$.

Using the same principle, you can show that it's not even enough for $\frac{t_n}n\to c$. In that case, let $t_n=k^2$ for $k^2\le n<(k+1)^2$. Then $$\frac1{N^2}m\left(\bigcup_{n=1}^{N^2}[t_n-1,t_n+1]\right)=\frac1{N^2}m\left(\bigcup_{k=1}^N[k^2-1,k^2+1]\right)\le\frac 2N.$$ The problem is we need something like the the following condition: there exists $k$ and $\delta>0$ such that $\max\{t_{n+1},\ldots,t_{n+k}\}\ge t_n+\delta$ for all $n$ sufficiently large. In all examples provided so far, no such $k$ exists. Assumptions on $\{t_n\}$ that imply this condition likely require finer asymptotics, such as $t_n=cn+o(1)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.