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A sequence of sets is defined as $A_n=\{x \in [0,1] : |\sum_{i=0}^{n-1} 1_{[\frac{i}{2n},\frac{2i+1}{4n})} - 1_{[\frac{2i+1}{4n},\frac{i+1}{2n})}| \geq p\}$ for some positive $p\geq0$. What is $\limsup_{n\to\infty}A_n(p)$ and $\liminf_{n\to\infty}A_n(p)$?

For any $x$, the function $f_n(x) = \sum_{i=0}^{n-1} 1_{[\frac{i}{2n},\frac{2i+1}{4n})} - 1_{[\frac{2i+1}{4n},\frac{i+1}{2n})}$ is $1$ or $-1$ based on number of paritions $n$. So, $|f_n(x)|$ is always $1$ for any given $x$. Hence, if $p > 1$, $A_n$ is empty set for all $n$. Hence both $\limsup$ and $\liminf$ is empty. If $p \leq 1$, then $A_n = [0,1]$ for all $n$. Therefore, both $\limsup$ and $\liminf$ is $[0,1]$. Is this correct?

I experimented further by changing $f_n(x)$ definition as follows:

$f_n(x) = \sum_{i=0}^{n-1} 1_{[\frac{i}{2n},\frac{2i+1}{4n})} - 1_{[\frac{2i+1}{4n},\frac{i+1}{2n})} ~~~~\mbox{if } 0 \leq x < 1/2$

$f_n(x) = x^n ~~~~\mbox{if } 1/2 \leq x \leq 1$

What is $\limsup_{n\to\infty}A_n(p)$ and $\liminf_{n\to\infty}A_n(p)$?

I am still trying to solve this.

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P.S. how do I insert multi line latex block in question? – user957 Sep 16 '10 at 10:53
From your working if $p\leq 1$ then $|f_n(x)|=1\geq p$ so haven't you got the cases mixed up. Shouldn't it be if $p<1$ then $A_n=[0,1]$ for all $n$. – alext87 Sep 16 '10 at 10:59
Couldn't your first $|f_n(x)|$ be rewritten as $1_{[0,0.5)}$? What am I missing? – Jens Sep 16 '10 at 11:16
@Jens the interval is partitioned into 2*n intervals. – user957 Sep 16 '10 at 12:19
@alext87 you are right. edited. – user957 Sep 16 '10 at 12:22
up vote 1 down vote accepted

I believe you have got the first example correct now. For the second example:

If $p>1$ then since $f_n(x)\leq 1$ for all $x\in(\frac{1}{2},1]$ we have $A_n=\emptyset$. If $p\leq 1$ then $f_n(x)\geq p$ if and only if $x\geq p^{\frac{1}{n}}$. Thus $A_n=[0,\frac{1}{2})\cup [p^{\frac{1}{n}},1]$

For clarity for $p=1$: I take the set $[1,1]=\{1\}$ so this agrees with what you suspected.

For $p\geq 1$, $A_n$ doesn't depend on $n$ so the $\liminf$ and $\limsup$ are easy.

For $p<1$ we have $A_{n+1}\subset A_n$. So $\liminf=\limsup=\lim=[0,\frac{1}{2})\cup\{1\}$.

You can always get this from $\liminf A_n=\bigcup_{n=1}^{\infty}\bigcap_{m=n}^{\infty} A_m$ and similarly $\limsup A_n=\bigcap_{n=1}^{\infty}\bigcup_{m=n}^{\infty} A_m$

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if $p=1$, then $A_n = [0,1/2) \cup \{1\}$, isn't it? If $p>1$, then $A_n$ should be empty set as $f_n(x)$ is never greater than 1. Your answer for $p<1$ looks correct. Do you agree? – user957 Sep 16 '10 at 18:18
Btw, what would be the $\limsup$ and $\liminf$ for these cases? – user957 Sep 16 '10 at 18:37
@user957: You are absolutely right! I made the same error I corrected you for! :) – alext87 Sep 16 '10 at 18:43
@alext87: Perhaps you will be kind enough to explain the definition of A_n? Seems like user957 has ignored my request. – Aryabhata Sep 16 '10 at 18:59
@Moron - when did you ask me? $A_n$ is a set of all $x$ values in $[0,1]$ satisfying given condition for $n$. – user957 Sep 16 '10 at 19:09

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