# Distance between fixed points of functions

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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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

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.
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 $M$.
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