= Solution
\b[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 $n>M$,
$$
\sup_{|x|\le M}|f_n(x)-f(x)|=M/n\longrightarrow0.
$$
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.
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