Krdytkyu

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 $gleft( x_{n+1} right) < g left( x_n right) < 0$, 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 $x_n rightarrow -infty$.