# Prob. 23, Chap. 5 in Baby Rudin: Fixed points of $(x^3+1)/3$

Here is Prob. 23, Chap. 5 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

The function $$f$$ defined by $$f(x) = \frac{ x^3 + 1}{3}$$ has three fixed points, say $$\alpha$$, $$\beta$$, $$\gamma$$, where $$-2 < \alpha < -1, \qquad 0 < \beta < 1, \qquad 1 < \gamma < 2.$$ For arbitrarily chosen $$x_1$$, define $$\left\{ x_n \right\}$$ by setting $$x_{n+1} = f \left( x_n \right)$$.

(a) If $$x_1 < \alpha$$, prove that $$x_n \to - \infty$$ as $$n \to \infty$$.

(b) If $$\alpha < x_1 < \gamma$$, prove that $$x_n \to \beta$$ as $$n \to \infty$$.

(c) If $$\gamma < x_1$$, prove that $$x_n \to +\infty$$ as $$n \to \infty$$.

Thus $$\beta$$ can be located by this method, but $$\alpha$$ and $$\gamma$$ cannot.

My Attempt:

Let the function $$g$$ be defined on $$\mathbb{R}^1$$ by $$g(x) = f(x) - x = \frac{ x^3 - 3x +1 }{3}.$$ Then we note that $$g(-2) = -\frac{1}{3} < 0 < 1 = g(-1),$$ $$g(0) = \frac{1}{3} > 0 > -\frac{1}{3} = g(1),$$
and $$g(1) = -\frac{1}{3} < 0 < 1 = g(2),$$ and $$g$$ is continuous on all of $$\mathbb{R}^1$$. Hence applying the intermediate-value theorem to $$g$$ on the intervals $$[-2, -1]$$, $$[0, 1]$$, and $$[1, 2]$$, respectively, we can conclude the existence of points $$\alpha \in (-2, -1)$$, $$\beta \in (0, 1)$$, and $$\gamma \in (1, 2)$$ at which $$g$$ vanishes (and these points are fixed points of $$f$$).

Moreover, $$g^\prime(x) = \frac{3x^2 - 3 }{3} = x^2-1.$$ So $$g^\prime(x) \ \ \begin{cases} > 0 \ & \ \mbox{ if } x < -1, \\ =0 \ & \ \mbox{ if } x = -1, \\ < 0 \ & \ \mbox{ if } -1 < x < 1, \\ = 0 \ & \ \mbox{ if } x = 1, \\ > 0 \ & \ \mbox{ if } x > 1. \end{cases}$$ This implies that $$g \ \begin{cases} \mbox{ is strictly increasing} & \mbox{on } \ (-\infty, -1], \\ \mbox{ has a local maximum} & \mbox{at } \ x= -1, \\ \mbox{ is strictly decreasing} & \mbox{on } \ [-1, 1], \\ \mbox{ has a local minimum} & \mbox{at } \ x = +1, \\ \mbox{ is strictly increasing} & \mbox{on } \ [1, +\infty). \end{cases}$$

Part (a):

Now since $$-2 < \alpha < -1$$, $$g$$ is strictly increasing on $$(-\infty, -1]$$, and $$g(\alpha) = 0$$, so if we choose $$x_1 < \alpha$$, then we must have $$g \left(x_1 \right) < g(\alpha) = 0$$, that is, we must have $$x_2 < x_1 < \alpha$$.

Now suppose that $$n$$ is a natural number such that $$x_{n+1} < x_n < \alpha$$. Then $$f \left( x_n \right) < x_n$$, which implies that $$g \left( x_n \right) < 0$$, But as $$g$$ is strictly increasing on $$(-1, -1]$$ and as $$x_{n+1} < x_n < -1$$, so we must have $$g\left( x_{n+1} \right) < g \left( x_n \right) < 0$$, which implies that $$g \left( x_{n+1} \right) < 0$$ and hence $$f \left( x_{n+1} \right) < x_{n+1}$$, that is, $$x_{n+2} < x_{n+1}$$.

Therefore by induction we can conclude that $$x_{n+1} < x_n$$ for all $$n \in \mathbb{N}$$. And as $$g(x) \to -\infty$$ as $$x \to -\infty$$, so $$\left\{ x_n \right\}$$ is unbounded from below. Hence $$x_n \to -\infty$$ as $$n \to \infty$$.

Part (c):

If $$\gamma < x_1$$, then $$0 = g(\gamma) < g(x_1)$$, that is, $$x_1 < x_2$$. That is, $$\gamma < x_1 < x_2$$.

Suppose that, for some $$n \in \mathbb{N}$$, we have $$\gamma < x_n < x_{n+1}$$. Then $$0 < g \left( x_n \right) < g \left( x_{n+1} \right)$$, which implies that $$x_{n+1} < f \left( x_{n+1} \right)$$ and hence $$x_{n+1} < x_{n+2}$$.

Therefore the sequence $$\left\{ x_n \right\}$$ is strictly increasing and $$g(x) \to +\infty$$ as $$x \to +\infty$$. So $$x_n \to +\infty$$ as $$n \to \infty$$.

Are these proofs correct? If so, then is my argument rigorous enough for Rudin? If not, then where are the issues? How can it be made more sound and rigorous?

How to tackle Part (b)?

Of course, I would like to have a proof of each of Parts (a), (b), and (c) that uses only what Rudin has established until this point in his book.

$\frac{x^3+1}{3}=x$ iff $f(x)=x^3-3x+1=0$. Observe that $f(-2)=-1<0$ and $f(-1)=3>0$, $f(0)=1>0$ and $f(1)=-1<0$ and $f(2)=3>0$. By Darboux Theorem $\frac{x^3+1}{3}=x$ has precisely three fixed points, one in each of intervals $(-2,-1)$, $(0,1)$ and $(1,2)$.
For $(a)$, first note that $x_n<\alpha$ for every $n$ because $f'(x)>0$ for $x<\alpha$. Applying the mean value theorem we get $x_{n+1} - \alpha = \frac{f(x_n) - f(\alpha)}{x_n - \alpha}(x_n - \alpha)=f'(c_n)(x_n - \alpha)$ for some $x_n < c_n < \alpha$. By induction $x_{n+1} - \alpha = f'(c_n) \ldots f'(c_1)(x_1 - \alpha)$. Since $x_1 - \alpha < 0$ and $f'(c_i) = c_i^2 > \alpha ^ 2 > 1$ for each $i$, we have $x_n \rightarrow -\infty$.