Conditions necessary for a function to contain multiple attractive fixed points

Question

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?

Answer 1

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.

Answer 2

Suppose $f:\mathbb{R} \to \mathbb{R}$ is continuously differentiable. If $f(x)=x$ and $|f'(x)|<1$ then $x$ is an attractive fixed point. One can prove this as follows.

There is $\delta>0$ such that for all $y\in I=(x-\delta,x+\delta)$, $|f'(y)|<1$ because $f'$ is continuous. Also, $f(I)$ is an interval about $x$ because $f(x)=x$ and $f$ continuous. Now take $y_1, y_2 (>y_1) \in I$. Then, by the mean value theorem there exists $t\in (y_1,y_2)$ such that $$|f(y_1)-f(y_2)|=|f'(t)||y_1-y_2|<|y_1-y_2|$$ Becuase this is true for all $y_1,y_2$ we can conclude that $f(I)$ is mapped strictly inside $I$. (Fix one to be $x$ then it is easier to see). From this it is easy to see that, $$\bigcap_{n \in \mathbb{N}}f^n(I)=\{x\}$$So and hence $\lim_{n \to \infty}f^n(y)=x$ for all $y \in I$.

But this is a local behaviour. Points outside $I$ may not converge to $x$ under repeated application of $f$. Also, we need not assume differentiability of $f$ on the whole real line! We need it just near the fixed point $x$.

Given a set of points $S=\{x_i | i\in \mathbb{N}\}$ with no limit points, one can construct a function $f$ whose attracting fixed points are exactly the points in set $S$. Because $S$ is discrete there exists $\epsilon>0$ such that the intervals $I_i=(x_i-\epsilon,x_i+\epsilon)$ are disjoint. Now on each of these interval define $f$ linearly with $f(x_i)=x_i$ and the derivative either $1/2$ or $-1/2$ alternating between intervals. Now define the function in the rest of the real line also linearly in such a way that the resulting function is continuous (shrinking or expanding $I_i$ as necessary). Because we chose derivatives to alternate sign on $I_i$'s it is easy to see that the only points at which this function crosses the line $y=x$ are $x_i$'s and hence they are the only fixed points. Because the derivative is less than $1$ in absolute value at each of $x_i$ they are also attracting fixed points.