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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.


  • Ron
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After some discussion, some extra restrictions might seem to work. Examples that come to mind: If both are concave (or convex). Also requiring continuous differentiability. Any pointers to relevant results? – Ron Oct 19 '12 at 20:51
up vote 3 down vote accepted

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. – Ron Oct 19 '12 at 20:19

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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Thanks for the answer Robert. Please see my comment to @martini. Would appreciate your view as well. – Ron Oct 19 '12 at 20:23
As the example shows, the problem arises when both $f_1$ and $f_2$ are close to the identity function. You need to control that somehow. For example, suppose you know $|f_1(x) - x| > \epsilon$ for $|x - x_1| > r$. Then if $|f_1 - f_2| < \epsilon$, you'll certainly have $|x_2 - x_1| \le r$. – Robert Israel Oct 19 '12 at 20:54
Wish I could upvote (probably later today). Thanks - this is excellent. So enough concavity/convexity should also work. – Ron Oct 19 '12 at 21:03

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