condition for uniform convergence of $\sum_{n=1}^{\infty} a_n \max\{0,1-|x-2n|\}$
Now slightly modified and figure out necessary and sufficient condition for uniform continuity of $ \sum_{n=1}^{\infty} a_n \operatorname{max}\{1-|x-2n|,0\}, \quad x \in [0,\infty)$.
i.e.,
Let $a_n$ be a non-negative real numbers and define \begin{align} f(x) = \sum_{n=1}^{\infty} a_n \operatorname{max}\{1-|x-2n|,0\}, \quad x \in [0,\infty). \end{align} I want to find necessary and sufficient conditions on $a_n$ for the function $f$ to be uniformly continuous on $[0,\infty)$.
Following are my trial
Recall the definition of uniformly continuous of $f$
$f : D \rightarrow \mathbb{R}$ is uniformly continuous if $\forall \epsilon >0$, $\exists \delta >0$ such that $\forall x ,y \in D$, $|x-y| < \delta \Rightarrow |f(x) - f(y) | < \epsilon.$
From the comment of @Kavi Rama Murthy, pick $x \in (2n-1, 2n+1)$ I noticed that $ \sum_{k=n}^{m} a_k \max\{1-|x-2k|,0\} \leq \max\{a_n, \cdots, a_m\}$ for $x\in (2n-1, 2n+1)$ and $n<m$, But since here I am considering uniform continuity, so $x,y \in [0,\infty)$, but since the value $f(x)$ does not depend on the choice of $x$ it seems $f(x) = f(y)$ which seems wrong.