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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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    You asked how to make multiline LaTeX. Unfortunately it's impossible to have newlines in comments, so take a look at [this pastebin link](http://pastebin.com/htrJUvvP) instead. Ready and formatted for use on this site. Note the quadrouple backslash in the `cases` environment.2010-11-22

1 Answers 1

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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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    @user957: Not in those exact words, but see my comment to main question.2010-09-16