= Solution
\b[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> gives
$$
\boxed{f^n(x)=x^{2^n}.}
$$
For every finite $n$, these values tend to $1$ as $x\uparrow1$, so $\sup_{0<x<1}f^n(x)=1$. With $\varepsilon=1/2$, 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.
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