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This question is informed by a lemma that is needed in pursuit of an ergodic theory result specifically to do with weak mixing.

We first need two definitions before we can state the proposition in question:

Upper and Lower Density: Given a set $A\subseteq \mathbb{Z}$, define it's upper density as, $$ \overline{\delta}(A) := \limsup_{n\to \infty}\frac{|A\cap \{1,2,\cdots, n\}|}{n}.$$ We define the lower density of $A$, write $\underline{\delta}(A)$, in the same way by just replacing $\limsup$ above by $\liminf$.

Density Convergence: A real valued sequence $(x_n)$ and a fixed $x\in\mathbb{R}$. The sequence is said to converge in density if for every open neighbourhood of $x$, say $V_x$, $$ \overline{\delta}\left(\{n\in \mathbb{N}: x_n \in V_x\}\right) =1. $$ We write, $$ D-\lim_{n\to\infty}x_n = x.$$

Proposition: Given a real valued sequence $(x_n)$. Then we have that, $$D-\lim_{n\to\infty}x_n >0 $$ if and only if there exists a set $S\subseteq \mathbb{N}$ with $\underline{\delta}(S)>0$ such that $x_n>0$ for all $n\in S$.

$(\impliedby)$

Assume that there is a set $S\subseteq \mathbb{N}$ with positive lower density such that $x_n>0$ for all $n\in S$. Suppose for the sake of a contradiction that $D-\lim_{n\to\infty}x_n=0$. There exists some $\epsilon>0$ such that $x_n>\epsilon>0$ for a set $S'\subseteq \mathbb{N}$ with positive lower density. By the supposition that $D-\lim_{n\to\infty}x_n=0$, we have that, $$ \overline{\delta}\left(\{n\in \mathbb{N}: x_n\in (-\epsilon, \epsilon)\}\right) =1. $$As a consequence, \begin{align*} 0 = \overline{\delta}\left(\{n\in \mathbb{N}: x_n>\epsilon \}\right)\geq \underline{\delta}(\{n\in \mathbb{N}: x_n>\epsilon \}) =\underline{\delta}(S')>0. \end{align*}Which clearly is a contradiction. It follows that $D-\lim_{n\to\infty}x_n>0$.

$(\implies)$ Assume that $D-\lim_{n\to\infty}x_n =c>0$. Fixing $\epsilon = c/2$, we know that, $$ \overline{\delta}\left(\{n\in \mathbb{N}: x_n \in (-c/2+c, c+c/2)\}\right) =1. $$ Thus, the set $\{n\in \mathbb{N}: x_n> c/2>0\}$ will also have upper density $1$.

At this point it is enough to show that any set $A\subseteq \mathbb{N}$ with upper density $1$ will necessarily have positive lower density. Although I am not sure if this is true. I am having trouble finding an alternative route.

Any suggestions are welcome, thanks in advance!

Walt van Amstel
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  • Is the proposition true? How about $x_n = 1/n$? – Junnan Mar 02 '18 at 03:44
  • @Junnan No, the proposition is a lemma needed for a larger result. The sequence $(1/n)$ won't work here since the standard limit (and hence the density limit) goes to zero. – Walt van Amstel Mar 02 '18 at 06:11
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    Exactly. Let $S = \mathbb{N}$, then $\underline{\delta}(S)=1$ and $x_n > 0$ for all $n\in S$. But $x_n \to 0$. So at least the sufficiency part is wrong. – Junnan Mar 02 '18 at 06:24
  • Oh goodness, I didn't see that. Thanks very much! This has been bugging me for a week – Walt van Amstel Mar 02 '18 at 06:26
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    Glad I could help :) – Junnan Mar 02 '18 at 06:27
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    If it helps, there is a result that a sequence is statistically convergent to the limit $x$ if and only if there is a subsequence on a subset of full density which converges to $x$. This result can be found, for example, in Šalát's paper referenced in this answer. – Martin Sleziak Oct 30 '18 at 11:16
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    If you are interested in condition that the sequence is positive on a big set, the notion of statistical limit superior might be interesting for you (although "big set" has here much weaker meaning than positive lower density). Maybe some references from this post might be useful, too: Where has this common generalization of nets and filters been written down? I will specifically mention my notes http://thales.doa.fmph.uniba.sk/sleziak/texty/rozne/trf/iconv/notions.pdf and a paper where I am a co-author http://dml.cz/dmlcz/132906. – Martin Sleziak Oct 30 '18 at 11:24

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