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Define $\xi\in C^1([-1,1]\times[-1,1])$ such that $$ \int\limits_{-1}^1 \xi(x,y)\,dy = 1 $$ for all $x\in[-1,1]$ and $\xi\geq 0$. Put $A_0 = [0,1]$ and $$A_{n+1} = \left\{x\in A_n:\int\limits_{A_n}\xi(x,y)\,dy = 1\right\}.$$

Are there methods to find the rate of convergence of $\lambda(A_{n}\setminus A_{n+1})$ where $\lambda$ is a Lebesgue measure? Maybe you can refer me to the relevant literature?

There are at least two kinds of situation:

  1. for some $N$ we have $A_N = \emptyset$, then $A_{N+1} = A_N$ and $\lambda(A_n\setminus A_{n+1}) = 0$ for $n\geq N$.

  2. for some $N$ we have $A_N = A_{N+1}$, then again $\lambda(A_n\setminus A_{n+1}) = 0$ for $n\geq N$.

Can you help me to construct an example for the third case, namely when $A_n\neq A_{n+1}$ for all $n\geq0$ (of course if such an example exists)? We can also assume for this example that $\xi\in \operatorname{Lip}([-1,1]\times[-1,1])$, not necessary differentiable.

I am mostly interested if it is possible to find a Lipschitz $\xi$ such that $A_n$ coincide with iterations of Smith-Volterra-Cantor set (Fat Cantor Set)?

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Am I misunderstanding something? Doesn't $A_n = [0,1]$ for all $n$? –  Tournabout Jun 24 '11 at 15:41
    
@Tournabout: you're certainly right, edited. –  Ilya Jun 24 '11 at 15:48
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1 Answer

up vote 5 down vote accepted
+125

Choose any strictly increasing function $g:[-1,1]\to[-1,1)$ and define

$$ \xi(x,y)= \begin{cases} 0&y<g(x)\;,\\ 3\frac{(y-g(x))^2}{(1-g(x))^3}&y\ge g(x)\;.\\ \end{cases} $$

The complication is just for achieving $C^1$ continuity; basically all we need is a properly normalized function with support in $y\ge g(x)$.

Then $A_n=[a_n,1]$ with $a_0=0$ and $a_{n+1}=g^{-1}(a_n)$ as long as $g^{-1}(a_n)>a_n$, and otherwise $a_{n+1}=a_n$. This will converge towards a fixed point with $a_n=g^{-1}(a_n)$ for suitable $g$, and you can make the convergence rate anything you like by choosing $g$ accordingly.

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