Show that if \(f\) is a differentiable contraction, \(f\) is uniformly continuous.
Solution
By definition, a contraction satisfies \[ d(f(x),f(y))\le c\,d(x,y) \] for some constant \(0\le c<1\) and for all \(x,y\) in its domain. Thus \(f\) is Lipschitz. Given \(\varepsilon>0\), if \(c=0\) then \(f\) is constant; otherwise take \(\delta=\varepsilon/c\). Whenever \(d(x,y)<\delta\), \[ d(f(x),f(y))\le c\,d(x,y)<\varepsilon. \] Therefore \(f\) is uniformly continuous. Differentiability is not needed.