Show that there is an equilibrium point

Question

Show that if $x_{k+1}=f(x_{k})$, where $f$ is continuous, has two stable equilibrium points, then there's a third equilibrium point between then.

I'm trying to approach this problem using the Intermediate Value Theorem. but I'm having no success. Any ideas?

Any help is quite welcome!! Thanks

Answer 1

Let $a<b$ be the stable equilibrium points. Consider $g(x)= f(x)-x$, $g'(a)=f'(a)-1<0, g'(b)=f'(b)-1<0$ since $a,b$ are stable, thus $\mid f'(a)\mid <1$ and $\mid f'(b)\mid <1$.

This implies there exists $c>a$ such that the restriction of $g'$ on $[a,a+c]$ is strictly negative thus $g$ decreases on $[a,c]$ and $g(a)=0>g(c)$. On the other hand, there exists $d>c$ such that the restriction of $g'$ on $[d,b]$ is negative, thus $g$ decreases on $[d,b]$, $g(d)>g(b)=0$.

We have an interval $[c,d]$ with $g(c)<0, g(d)>0$, since $g$ is continue, there exists $e\in [c,d]$ such that $g(e)=0$, this is equivalent to say that $f(e)=e$.

Answer 2

Let's say that $a<b$ are stable fixed points and assume that there are no others. Then for any $\varepsilon>0$ sufficiently small, we have $f(a+\varepsilon)<a+\varepsilon$ and $f(b-\varepsilon)>b-\varepsilon$ (because $a$ and $b$ are stable fixed points). Hence, the function $g(x)=f(x)-x$ must have a zero between $a+\varepsilon$ and $b-\varepsilon$. This contradiction shows that there are other fixed points.