Exercise 5.23

The fixed points of $f$ are the zeros of $g(x)=f(x)-x$, of which there are at most three. Since $g(-2)=-1∕3$, $g(-1)=1$, $g(0)=1∕3$, $g(1)=-1∕3$, and $g(2)=1$, there are three fixed points of $f$, lying in the intervals $(-2,-1)$, $(0,1)$, and $(1,2)$, as asserted.

Note that $\mathrm{\Delta }{x}_n=x_n-x_{n-1}=f(x_{n-1})-x_{n-1}=g(x_{n-1})$. For $x<\alpha$, $g(x)<0$, and $g$ decreases monotonically to $-\infty$ as $x\to -\infty$. Hence if $x_1<\alpha$, then $g(x_{n-1})=\mathrm{\Delta }{x}_n<0$, so $x_n<x_{n-1}$ and $\mathrm{\Delta }{x}_{n+1}=g(x_n)<\mathrm{\Delta }{x}_n$. Hence $x_n\to -\infty$ as $n\to \infty$.

Similarly, for $x>\gamma$, $g(x)>0$, and $g$ increases monotonically to $+\infty$ as $x\to +\infty$. Hence if $x_1>\gamma$, then $g(x_{n-1})=\mathrm{\Delta }{x}_n>0$, so $x_n>x_{n-1}$ and $\mathrm{\Delta }{x}_{n+1}=g(x_n)>\mathrm{\Delta }{x}_n$. Hence $x_n\to +\infty$ as $n\to \infty$.

For $x\in (\alpha ,\beta )$ we have $g(x)>0$, and for $x\in (\beta ,\gamma )$ we have $g(x)<0$. I want to compare $g$ with the linear function $\beta -x$, which also has a zero at $x=\beta$ and has a slope of $-1$. Since $g^{\prime }(\beta )={\beta }^2-1>-1$, we have $g(x)<\beta -x$ for $x\in (\alpha ,\beta )$ and $g(x)>\beta -x$ for $x\in (\beta ,\gamma )$.

Putting this together, we get that if $x_n\in (\alpha ,\beta )$, then $0<\mathrm{\Delta }{x}_{n+1}<\beta -x_n$ so that $x_{n+1}\in (x_n,\beta )$. Similarly, if $x_n\in (\beta ,\gamma )$, then $\beta -x<\mathrm{\Delta }{x}_{n+1}<0$ so that $x_{n+1}\in (\beta ,x_n)$.

That is, if $x_1\in (\alpha ,\beta )$, then $\{x_n\}$ is a monotonically increasing sequence with upper bound $\beta$, and if $x_1\in (\beta ,\gamma )$, then $\{x_n\}$ is a monotonically decreasing sequence with lower bound $\beta$. In either case they must have a limit ${\beta }^{\prime }$. This must be a zero of $g$ (and so a fixed point of $f$) since otherwise $g({\beta }^{\prime })\ne 0$, so for $x_n$ close to ${\beta }^{\prime }$, the value of $|\mathrm{\Delta }{x}_n|$ would be larger than $|{\beta }^{\prime }-x_n|$ which would contradict ${\beta }^{\prime }$ being a limit point of the monotonic sequence $\{x_n\}$. Hence ${\beta }^{\prime }=\beta$.