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Some time ago, I managed to show that $sin(x)$ and $cos(x)$ were both contractive almost everywhere and then I noted that $cos$ was the derivative of $sin$ so there might be a deeper connection.

Let's suppose $F:X \rightarrow X$ is a differentiable contractive map where $(X,d)$ is a metric space. This implies that there exists $ K < 1$,

$$d(F(x),F(y)) \leq K d(x,y) \forall x,y \in X\tag{*}$$

  1. Now, I wonder whether there is a general theorem which provides some conditions on $F$ such that $F'$ is also a contractive mapping. I can think of some particular examples but a general theorem eludes me.
  2. A related question is whether a mapping $G$ defined on $X$ that is contractive almost everywhere, behaves like a contractive mapping. I believe this is true.

Note 1: I'm mainly interested in Riemannian manifolds but if there's a more general theorem which answers these particular questions for Riemannian geometry I'd definitely be interested to hear about it.

Note 2: When I say that $sin$ and $cos$ are both contractive almost everywhere I mean that $\{x,y\in \mathbb{R}^2\ :|cos(x)-cos(y)|=|x-y|\}$ is a set of measure zero.

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    What definition of derivative are you using for a general metric space? – 2017-01-02
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    Usually contractive means that $|f(x)-f(y)| \le \lambda |x-y$ for some $\lambda <1$. In this regard, neither $\sin$ not $\cos$ are contractive. – 2017-01-02
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    True. But, they behave like contractive mappings almost everywhere. – 2017-01-02
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    @copper.hat I added a point on this in my question. – 2017-01-02
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    Not a full answer: $F(x)=\sqrt{x^2+1}$ is $1$-Lipschitz, and satisfies $|F'(x)|<1$ for all $x$, but is a useful counterexample for certain intuitions we might have about functions satisfying $|F(x)-F(y)|<|x-y|$. For example, if $a_{n+1}=F(a_n)$, $a_n \to \infty$ for any starting values of $a_1$. – 2017-01-02
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    Notice that $sin(x)$ and $cos(x)$ take value in $[-1,1]$, therefore you can narrow down your domain into $[-1,1]$ where $|sin(x)|,|cos(x)|$ are strictly less than 1. – 2017-01-02
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    @SteveD I didn't make this clear when I first asked the question but I'm particularly interested in Riemannian geometry. – 2017-01-02

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