Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 2 / 3G (Analysis II)

The uniform convergence condition on a set $X$ isThe index $N$ is independent of $x$; this is stronger than pointwise convergence.

Fix $x_0\in \mathbb{R}$ and $ϵ>0$. Choose one $N$ such that $|f_N(x)-f(x)|<ϵ/3$ for every $x$. The continuity of $f_N$ at $x_0$ gives $\delta >0$ with $|f_N(x)-f_N(x_0)|<ϵ/3$ whenever $|x-x_0|<\delta$. The triangle inequality then givesSince $x_0$ was arbitrary, $\boxed{f\text{ is continuous on }\mathbb{R}}$, the uniform limit theorem. Continuity of each approximant need not be uniform continuity.

No: local uniform convergence need not be global uniform convergence. For a bounded counterexample use the continuous functions $f_n(x)=\min (1,|x|/n)$ and $f(x)=0$. On any fixed interval $[-M,M]$, for $n>M$,Thus there is locally uniform convergence to zero. On all of $\mathbb{R}$, however, ${\sup }_x|f_n(x)-f(x)|=1$ for every $n$, so uniform convergence fails. The cutoff needed to control the tails cannot be chosen independently of the interval.