Analysis of function using fixed point

Question

Let $f$ be function defined by
$$f(x)=\frac{x^3+1}{3}$$ has 3 fixed points say $\alpha ,\beta ,\gamma$ where $-2 < \alpha < -1$,$0 < \beta < 1$,$1 <\gamma <2$

I want to show the following :

for arbitray choice of $x_1$, define $x_n$ as $x_{n+1}=f(x_n)$

a) If $x_1$ < $\alpha$ then $x_n \to -\infty$ as $n\to \infty$

b) If $\alpha < x_1 < \gamma$ then $x_n \to \beta $ as $n\to \infty$

c) If $x_1$ > $\gamma$ then $x_n \to \infty$ as $n\to \infty$

My Attempt:

$$g(x)=f(x)-x$$

I can show using Intermediate value property there exist 3 fixed as per given condition. And $f'(x)>0$ therefore function is increasing on $\mathbb{R}$. I am not able to proceed further. Any help will be appreciated

Answer 1

As regards (c), note that for $t>\gamma$, $f'(t)=t^2>\gamma^2>1$. Hence if $x_n>\gamma$ then by the Mean Value Theorem, $$ x_{n+1}-\gamma=f(x_n)-f(\gamma)=f'(t)(x_n-\gamma)>\gamma^2(x_n-\gamma)$$ which implies that if $x_1>\gamma$ then the sequence $x_n$ is strictly increasing and it goes to infinity: $$x_{n+1}>\gamma+\gamma^{2n}(x_1-\gamma)\to +\infty.$$

What about (a) and (b)?

Answer 2

Notations

Let

$a_1=x_1$

$a_2=f\left(a_1\right)$

$a_3=f\left(a_2\right)$

$a_4=f\left(a_3\right)$

.

.

.

. In general

$a_n=f\left(a_{n-1}\right)$ This is a sequence of numbers $a_1,a_2,a_3,a_4....a_n......$.If it converges to, say $x$.then because we have $a_n=\frac{a_{n-1}^3+1}{3}$

applying limits $lim_{\to \infty}$ both sides we get $x=\frac{x^3+1}{3}$ and since you assumed that $\alpha,\beta$ and $\gamma$ are fixed points.Then the values of x is nothing but $\alpha,\beta$ and $\gamma$ only

You can use induction to show that it is bounded. To be clear if $a_n<\alpha$ then $a_{n+1}<\alpha$ etcetra.

Now you are ready to use induction to prove that the sequence is increasing or decreasing, according to wether $x_1<\alpha$ or $\alpha <x_1<\beta$ or $\beta <x_1<\gamma$ or $\gamma <x_1$.

Once you had shown that it is monotonic in the intervals, use induction to show that it is bounded above if the sequence is increasing or it is bounded below if the sequence is decreasing.

In the cases when it diverges you need to show that it is not bounded.

I am giving you idea how to show its monotonic. In the same way you can show that it is bounded. Hint

$a_{n+1}-a_n=\frac{a_{n}^3+1}{3}-a_n=\frac{a_{n}^3-3a_n+1}{3}=\frac{\left(x-\alpha\right)\left(x-\beta\right)\left(x-\gamma\right)}{3}$