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I've been working on a problem involving fractals and iterated function systems...

Consider a contractive map $f \circ f : X \rightarrow X$.

Is there an easy way to prove/disprove that $f$ is contractive?

I've have been trying to solve using the contrapositive... i.e if $f$ is not contractive, is there an easy way to prove $f \circ f$ is not contractive.

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The answer to the title question is no. Let $f:\Bbb{R}\to\Bbb{R}$ be the characteristic function of the rationals (so $f(x)=1$ if $x$ rational and $0$ otherwise). Then, $f$ is not continuous (actually nowhere continuous), so it cannot be a contraction (contractions are even Lipschitz continuous). On the other hand, $f\circ f = 1$ is constant so it is a contraction.

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No, it is not true, even supposing $f$ smooth: let $f\colon\mathbb{R}^2\to\mathbb{R}^2$ the linear map defined by the matrix $$M=\begin{pmatrix} & a \\ b & \end{pmatrix},$$ then $f\circ f$ is defined by $$M^2=\begin{pmatrix} ab & \\ & ab\end{pmatrix}.$$

If you chose $|a|>1$ and $|b|<1/|a|$, then $f\circ f$ is a contraction while $f$ is not (in fact is a dilation in one direction).

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    $\begingroup$ I opened this tab a couple of hours ago in order to give essentially this answer, but got distracted by work before typing it up. (+1) $\endgroup$ – Xander Henderson May 20 at 2:19

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