Let $f : \mathbb R \to [0,\infty)$ be continuous and have $\int_{\mathbb R} f(x) dx < \infty$. Also let $I=[0,1]$. For some $c \geq 1$ define $F_n : I \to [0,\infty)$ for $n\in\mathbb N$ by $$ F_n(x) = \sum_{k=0}^n f(x + ck). $$ I'll let $F(x) = \lim_{n\to\infty} F_n(x)$. Basically I'm looking at picking a point $x\in I$ and summing all the values of $f$ on the translations $\{x + ck : k\in \mathbb N\}$.
My question: does $(F_n)$ converge uniformly over $I$?
I suspect so because $\int f < \infty$ means $f$ gets small quickly, so no matter my $x\in I$ I should be able to choose $N$ large enough to have $\sup_{x\in I}|F_n(x) - F(x)| < \varepsilon$ for $n\geq N$, but I haven't been able to prove this. I wonder if it's not true unless I add more regularity conditions on $f$, like maybe monotonicity?
No, $f$ need not get small quickly. On each interval $[n,n+1], n=1,2,\dots$ consider the isosceles triangle over the midpoint, of height $n$ and base $1/n^3.$ Define $f$ to be upper part of that triangle on each of those intervals. Everywhere else define $f=0.$ Then $f$ is continuous and nonnegative, and
$$\int_0^\infty f = \sum_{n=0}^{\infty}\frac{n}{2n^3}<\infty.$$
This should help in thinking about uniform continuity.