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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?2010-09-16
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    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 $n$.2010-09-16
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    Couldn't your first $|f_n(x)|$ be rewritten as $1_{[0,0.5)}$? What am I missing?2010-09-16
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    @Jens the interval is partitioned into 2*n intervals.2010-09-16
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    @alext87 you are right. edited.2010-09-16
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    In your second example how are you defining $f_n(\frac{1}{2})$?2010-09-16
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    @user957: Yes, it is. But since for each $n$ the intervals are all disjoint and their union is $[0, 0.5)$, $f_n(x)$ is in $\\{-1,1\\}$ if $x<0.5$ or $0$. The absolute then kills the sign.2010-09-16
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    The definition of A_n completely escapes me. What notation is that?2010-09-16
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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

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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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    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?2010-09-16
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    Btw, what would be the $\limsup$ and $\liminf$ for these cases?2010-09-16
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    @user957: You are absolutely right! I made the same error I corrected you for! :)2010-09-16
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    @alext87: Perhaps you will be kind enough to explain the definition of A_n? Seems like user957 has ignored my request.2010-09-16
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    @Moron - when did you ask me? $A_n$ is a set of all $x$ values in $[0,1]$ satisfying given condition for $n$.2010-09-16
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    Yes, $A_n\subseteq [0,1]$. The condition on $x\in A_n$ is that $|f_n(x)|\geq p$ with $p\geq0$ constant. $|f_n(x)|=1$ for $x\in[0,1]$. Thus the definition of $f_n(x)$ using characteristic functions is just a distraction. Hence $A_n(p)=\emptyset$ if $p>1$ and $A_n(p)=[0,1]$ if $p\leq 1$.2010-09-16
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    @user957: Not in those exact words, but see my comment to main question.2010-09-16