Suppose $f(x)$ is a function such that $f(x)=O(x)$, then $$\sum_{n\leq x} f\left(\frac{x}{n}\right) =O\left( \sum_{n\leq x} \frac{x}{n}\right) = O\left(x\sum_{n\leq x}\frac{1}{n} \right) = O\left( x\log(x)\right)$$
Conversely, if $g(x)$ is an increasing function such that $$\sum_{n\leq x} g\left(\frac{x}{n}\right) \sim x\log(x)$$ Then, $$\sum_{n\leq x} g\left(\frac{x}{n}\right) - 2 \sum_{n\leq x/2} g\left(\frac{x}{2n}\right) \sim x\log(x)-2(x/2)\log(x/2) = 2\log(2) x =O(x)$$ Therefore, $$g(x)-g(x/2) \leq \sum_{n\leq x} g\left(\frac{x}{n}\right) - 2 \sum_{n\leq x/2} g\left(\frac{x}{2n}\right) =O(x)$$ Finally, $$g(x) = \sum_{i=0}^{\log_2(x)} g\left(\frac{x}{2^i}\right)-g\left(\frac{x}{2^{i+1}}\right) = O\left(x\sum_{i=0}^{\log_2(x)} \frac{1}{2^i} \right) = O(x)$$
Now, suppose we have an arbitrary, increasing, sequence of integers $a_1<a_2<\dots$. Then if $f(x)=O(x)$, we get $$\sum_{\substack{n \\ a_n\leq x}} f\left(\frac{x}{a_n}\right) = O\left(x \sum_{\substack{n \\ a_n\leq x}} \frac{1}{a_n} \right)$$
My question is, is the similar converse true as in the case $a_n=n$. That is, if we have an increasing function $g$ such that $$\sum_{\substack{n \\ a_n \leq x}} g\left(\frac{x}{a_n}\right) \sim x \sum_{\substack{n \\ a_n\leq x}} \frac{1}{a_n}$$ Then, must $g(x)=O(x)$?
Edit:
I say two function, $f_1,f_2$ satisfy the relation $f_1(x)\sim f_2(x)$ if $$\lim_{x\to\infty} \frac{f_1(x)}{f_2(x)} =1$$
