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Let $(A,\rho)$ be a compact metric space and let $f: A \to A$ be a function satisfying $$ \forall \quad x,y\in A, \ x \ne y \ \ \implies \rho(f(x),f(y))< \rho(x,y). $$

Now define the function $\chi: A \to \mathbb{R}$ by $$\chi(x) = \rho(x,f(x)), \qquad x \in A.$$

Then I have to show that $\chi$ is uniformly continuous.

I have to show for every $\epsilon >0$, there exists $\delta>0$ such that for every $x,y \in A$ with $\rho(x,y) < \delta$ we have $\rho(\chi(x),\chi(y)) <\epsilon$.

Can anyone explain in detail?

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Why do you have to do that? – Jonas Meyer Dec 14 '12 at 15:44
$ \chi(x) \in \mathbb{R}$, thus you have to prove that $| \chi(x) - \chi (y)| < \varepsilon.$ $\rho$ is a metric on the space $A$ that a priori can not be $\mathbb{R}$. – user29999 Dec 14 '12 at 15:52


Note that $\rho(x,f(x))-\rho(y,f(y))$ is less than or equal to (using the triangle inequality and the contraction assumption)

$$\begin{aligned}\rho(x,f(y))+\rho(f(x),f(y))-\rho(y,f(y)) & \leqslant \rho(x,y)+\rho(y,f(y))+\rho(x,y)-\rho(y,f(y))\\ &=2\rho(x,y)\end{aligned}$$

apply the same calculating to get that $\rho(y,f(y))-\rho(x,f(x))\leqslant 2\rho(x,y)$ and so $|\rho(x,f(x))-\rho(y,f(y))|\leqslant 2\rho(x,y)$.

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