 Solution

No: pointwise convergence need not be uniform convergence. The continuous function $f(x)=x^2$ maps $(0,1)$ into itself and is strictly below the identity. Its iteration of a map givesFor every finite $n$, these values tend to $1$ as $x\uparrow 1$, so ${\sup }_{0<x<1}{f}^n(x)=1$. With $\epsilon =1/2$, choosing $x>2^{-1/2^n}$ gives $f^n(x)>1/2$. Thus no single iterate works for every starting point, although each fixed starting point tends to zero.