Consider the function $f:\Bbb R\to \Bbb R$ given by $f(x)=\max\{0,1-|x|\}$, and let $(s_n)$ be a sequence of real numbers. I am asked to show to find a necessarily and sufficient condition for the series $h(x)=\sum_n s_nf(x-2n)$ is uniformly continuous.
It is obvious that $h(x)$ is well defined for all $x$, because $f(x-2n)$ is nonzero for at most one $n$. Also, it is easy to find sufficient conditions, for example, by letting $s_n=0$ for all but finitely many $n$. However, how should I find a necessarily and sufficient condition?
Partial answer. Look at $h(2m)$ and $h(2m+a_m)$ with $a_m \to 0$ to see that $s_m a_m \to 0$ as $m \to \infty$ for every sequence $a_m \to 0$. This implies that $s_m $ is bounded. Conversely if $s_m \to 0$ show that $|h(x)| \leq |s_m|$ on $2m-1\leq |x| \leq 2m+1$. This proves that $\lim_{x \to \pm \infty} h(x)=0$. Any continuous function on $\mathbb R$ with this property is uniformly continuous. I cannot find a necessary and sufficient condition.