<s>[INST] Conditions necessary for a function to contain multiple attractive fixed points Given a function $f(x)$, what are the necessary conditions for $f(x)$ to contain multiple attractive fixed points?
Another key condition is I want the construction of the fixed point $x^*$ to be repeated application of $f$: $\lim\limits_{x \rightarrow n} f^n(x) = x^*$
One approach I looked at was forcing $f$ to be a contraction mapping therefore having a Lipshitz constant be less than one. This will force the function to contain a fixed point (via Banach's fixed point theorem), but I don't think a contraction map is capable of having multiple attractive fixed points. 
Any ideas on how to approach this construction?
 [/INST] All attracting fixed points must be fixed points first of all. So, the function $f(x)$ must have multiple fixed points, or solutions for $f(x^*)=x^*$. If function $f(x)$ is differentiable in open neighbourhoods of those fixed points $\{x^*\}$ and $|f'(x^*)|<1$, then those points are also attracting. Let's use these statements to construct a function $f(x)$ with multiple contracting fixed points. 
Consider $\left\{\frac{1}{4},\frac{2}{4},\frac{3}{4} \right\}\subset (0,1)$, the polynomial 
$$P_{3}(x)=\left(x-\frac{1}{4}\right)\left(x-\frac{2}{4}\right)\left(x-\frac{3}{4}\right)$$
and 
$$f(x)=x-P_{3}(x)$$
We have
$$f\left(\frac{1}{4}\right)=\frac{1}{4},f\left(\frac{2}{4}\right)=\frac{2}{4},f\left(\frac{3}{4}\right)=\frac{3}{4}$$
$$f'(x)=1-\left(x-\frac{2}{4}\right)\left(x-\frac{3}{4}\right)-\left(x-\frac{1}{4}\right)\left(x-\frac{3}{4}\right)-\left(x-\frac{1}{4}\right)\left(x-\frac{2}{4}\right)=\\
-3x^2+3x+\frac{5}{16}$$
And
$$\left|f'\left(\frac{1}{4}\right)\right|=\frac{7}{8}<1$$
$$\left|f'\left(\frac{2}{4}\right)\right|=\frac{17}{16}>1$$
$$\left|f'\left(\frac{3}{4}\right)\right|=\frac{7}{8}<1$$
So $f(x)$ has 2 attracting fixed points.
Now, try with $\left\{\frac{1}{6},\frac{2}{6},\frac{3}{6},\frac{4}{6},\frac{5}{6} \right\}\subset (0,1)$ ... and continue with $\left\{\frac{1}{2n},\frac{2}{2n},...,\frac{2n-1}{2n} \right\}\subset (0,1)$ to obtain a function with $n$ attracting fixed points.
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