How to imagine (or visualize) the convergence of $x,\; f(x),\; f(f(x)),\; f(f(f(x))), \dots\; $?

Question

I found a problem (Calculus 3rd edition by Michael Spivak, Chapter 22 question 21) that goes as follows:

21. (a) Suppose that $f$ is continuous on $[0,1]$ and that $0\leq f(x) \leq 1$ for all $x$ in $[0,1]$. Problem 7-11 shows that $f$ has a fixed point. If $f$ is increasing, a much stronger statement can be made:
For any $x$ in $[0,1]$, the sequence $$x,\; f(x),\; f(f(x)),\dots$$ has a limit (which is necessarily a fixed point by problem 20). Prove this assertion, by examining the behavior of the sequence for $f(x)>x$ and $f(x)<x$ or by looking at Figure 10. [...]

I managed to prove the assertion but can't seem to visualize the concept. Could someone explain Figure 10 and its implications? Is $\varphi_n(x)=f^n(x)$ (where $n$ is the number of compositions) getting as close as desired to $id_{[0,1]}(x)=x$ given a sufficiently large $n$ ($\forall x\in [0,1]$)? What is happening to $\varphi_n(x)$ as $n$ gets larger? What about the case $f=id_{[0,1]}$?

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Proof of the assertion
Let $x\in\{\alpha\in [0,1]\mid 1\geq f(\alpha)>\alpha\}=A$, then, since $f$ is increasing, $f(f(x))>f(x)$ and $f(f(f(x)))>f(f(x))>f(x)>x$ so the sequence is increasing $\forall x\in A$ but $f(x)\leq 1$ so it's also bounded and thus a limit exists. For $x$ such that $f(x)<x$ the proof is similar $\square$.

Answer 1

Let $x_0\in[0,1]$. You have already proved that $\varphi(x_0)$ given by $\varphi_n(x_0) = f^n(x_0)$ converges to, let's say, $l$. Therefore, you may write $$l=\lim_{n\to\infty}\varphi_n(x_0) = \lim_{n\to\infty}\varphi_{n+1}(x_0) = \lim_{n\to\infty} f(\varphi_n(x_0)) = f\left(\lim_{n\to\infty}\varphi_n(x_0)\right) = f(l),$$ where we have used the continuity of $f$. Hence, $l$ is a fixed point of $f$.

What Figure 10 is displaying can be seen as the following iteration:

1. You begin with $q_0 = \left(x_0,0\right)$.
2. At step $n$, apply $f$ to move vertically from $q_n$ to $p_n = \left(\varphi_n(x_0),f\left(\varphi_{n}(x_0)\right) = \left(\varphi_n(x_0),\varphi_{n+1}(x_0)\right)\right)$.
3. Move horizontally from $p_n$ to $q_{n+1} = \left(\varphi_{n+1}(x_0),\varphi_{n+1}(x_0)\right)$.
4. Update $n = n + 1$ and go to step 2.

Notice that a fixed point will be reached in the limit as $$p_n = \left(\varphi_{n}(x_0), f\left(\varphi_{n}(x_0)\right)\right) = \left(\varphi_{n}(x_0), \varphi_{n+1}(x_0)\right) \to \left(l,l\right).$$

Answer 2

Basically, if you go along a vertical line until you hit the curve, then you get to the point $(x, f(x))$.

Now you go along the horizontal line until you hit the line $y = x$, then you get to the point $(f(x), f(x))$.

Going again along the vertical line, you get to $(f(x), ff(x))$.

Again along the horizontal line, you get to $(ff(x), ff(x))$.

The procedure continues forever. Repeating these two steps is thus equivalent to iterations of $f$.

The convergence should then be obvious from tracing the picture.