Suppose we have a function $f$ that has a fixed point $\tau$. Let's also consider a real number $x_0$, that, when we infinitely apply the function $f$ to it, it converges towards $\tau$.
Also, $x_0$ is inside the biggest monotonic interval of $f$ that contains $\tau$, such that $f(x_0)$ is also inside that interval, and so on.
I just wonder if, $\forall n\in\mathbb{N}, |f^{n+1}(x_0)-\tau|<|f^{n}(x_0)-\tau|$
With the initial conditions. Or if it isn't, what are the restrictions so that it is true?
Consider the function $f : [0,1) \to [0,1)$ defined by $$f(t) = \begin{cases} 0&t = 0\\ \frac1k & \frac 1{k+1} \le t < \frac1k, k\text{ odd}\\ \frac1{k+3} & \frac 1{k+1} \le t < \frac1k, k\text{ even} \end{cases}$$
Note that for any $x\in(0,1), f^n(x)$ converges to $0$, but it is never decreasing. Instead it alternates between smaller increases and larger decreases.
This is just one simple example. You can vary the idea to make $f$ continuous, or even analytic, and to be defined on all $\Bbb R$.