There exists a function $f: \mathbb{R} \to \mathbb{R}$ that satisfies $$ \sum_{n=1}^{\infty} |f^{[n]}(x) - f^{[n]}(y)| < \infty \quad \forall x,y \in \mathbb{R} $$ where $f^{[n]}(x) = f(f(f...(f(x)))$ iterated $n$ times, but there is no $\lambda < 1$ such that $|f(x) - f(y)| < \lambda |x - y|$. I think that a function that is not a contraction but $f(f(x))$ is, satisties my problem, but how to prove it? And how I find such function?
OBS: I prove that a function that satisfies this property has a unique fixed point
Consider the function \begin{align*} f: \mathbb{R} \rightarrow \mathbb{R}, f(x)=\begin{cases} 2^x \pi,& x\in \mathbb{N}, \\ 1,& x\in \mathbb{R} \setminus \mathbb{N}. \end{cases} \end{align*} We have that $f(f(x))=1$, thus, we have \begin{align*} \sum_{n=0}^\infty \vert f^{[n]}(x)-f^{[n]}(y) \vert = \vert f(x)-f(y) \vert <\infty. \end{align*} On the other hand, we have for $n\in \mathbb{N}$ $$ \vert f(n)-f(0) \vert = (2^n-2^0) \pi. $$ Assume that there exists $\lambda>0$ such that $\vert f(x)-f(y) \vert \leq \lambda \vert x-y \vert$ for all $x,y\in \mathbb{R}$ yields for $n\in \mathbb{N}$ $$ \lambda n \geq \vert f(n)-f(0) \vert = (2^n-2^0) \pi $$ and hence $$ \lambda \geq \frac{2^n-1}{n}. $$ However, the RHS tends to infinity for $n\rightarrow \infty$, which yields a contradiction.