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I'm trying to bound the distance of fixed points of two functions assuming there's some bound on the distance between the functions.

Specifically, assume $f_1, f_2:[0,1] \rightarrow [0,1]$ are two continuous functions (and assume any regularity conditions you would like) with unique fixed points on $[0,1]$.

Also assume that $\forall x \in [0,1]$, $|f_1(x)- f_2(x)| \le M$.

Denote the fixed points of $f_1$ and $f_2$ as $x^*_1$ and $x^*_2$ respectively.

I'm interested in results about the distance $|x^*_1 - x^*_2|$ and about $|f_1(x^*_1)- f_2(x^*_2)|$. Any measure for distance ($L_1$, $L_2$ and others) are fine.

I would appreciate references to anything similar, or to how this problem will be called.

Thanks,

  • Ron
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    Aft$e$r som$e$ discussion, some extra restrictions might seem to work. Examples that come to mind: If both are concave (or convex). $A$lso requiri$n$g continuous differentiability. $A$$n$y point$e$rs to releva$n$t r$e$sults?2012-10-19

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I don't now, if by assumptions on $X$ something can be said, but I doubt the for general $X$ you will get an interesting bound: Consider $X = [0,1]$, and for $\epsilon > 0$ the functions $f_1, f_2 \colon [0,1] \to [0,1]$ given by \begin{align*} f_1(x) &= \begin{cases} 0 & x \le \epsilon\\ x-\epsilon & x \ge \epsilon \end{cases}\\ f_2(x) &= \begin{cases} x+\epsilon & x \le 1-\epsilon\\ 1 & x \ge 1 - \epsilon \end{cases} \end{align*} Then $f_1$ and $f_2$ are continuous, fulfill $\|f_1 - f_2\|_{\infty} \le 2\epsilon$, and have unique fixed points $x_1^* = 0$, $x_2^* = 1$. So the distance between the fixed points is as large as possible, but the distance between the functions can be made arbitrary small.

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    That's an excellent example, thanks. The question is what other restrictions on $f_1$ and $f_2$ might help to give some sort of other bound. Something about their derivatives, or any other condition? There might be something to be assumed on $X$, but at this point, sticking to $[0,1]$ is actually a good idea.2012-10-19
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Consider $f_1(x) = x + \epsilon \tanh(x - x_1)$ and $f_2(x) = x + \epsilon \tanh(x - x_2)$ on $\mathbb R$, where $\epsilon > 0$ is arbitrary. These have unique fixed points $x_1$ and $x_2$ and $|f_1(x) - f_2(x)| < 2 \epsilon$. So you'll need to base any bound on more than just $M$.

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    Wish I could upvote (probably later today). Thanks - this is excellent. So enough concavity/convexity should also work.2012-10-19