Suppose $f(x)$ is a positive continuous function on $[0,\infty)$ and that $f(x+u)-f(x)\to 0$ as $x\to\infty$ for every given $u\in[0,\infty)$. Prove that, given any $a>0$, $f(x+u)-f(x)\to 0$, as $x\to\infty$ uniformly for $u$ over $[0,a]$.


$f$ is actually called regularly (slowly) varying function. The usual assumption is that $f$ is measurable or a Baire function. I would like to see how the stronger condition on $f$ simplifies the proof of the conclusion.



Local Uniform Convergence
Fix $\varepsilon>0$. For any positive integer $n$ consider the (closed) set $E\_n$ of $u\in [0,4a]$ such that $|f(x+u)-f(x)|\leq \varepsilon$ for all $x\geq n$. Then $[0,4a]=\cup\_n E\_n$, thus there exists $n$ such that $\mu(E\_n)>3a$. Then for any $b\in [0,a]$ we have $\mu(E\_n\cap [0,3a])>2a$, hence sets $B=E\_n\cap [0,3a]$ and $b+B$ must have a common point. Take $c\in B\cap (b+B)$. For $x>n$ we have $|f(x+b)-f(x)|\leq |f(x+c)-f(x)|+|f(x+b)-f(x+c)|\leq 2\varepsilon$ since $c,c-b\in E\_n$. 


What we actually use here is that $E\_n$ are measurable and that inequalities hold pointwise, not a.e. Both problems may be routinely fixed for measirable $f$. 


