# Is there a "convolution" of asymptotic growth?

Suppose that I have two asymptotic counts given by $$\#\{x \in [0,H] \cap \mathbb Z: f(x) \leq H\} \sim F(H)$$ and also $$\#\{x \in [0,H] \cap \mathbb Z: g(x) \leq H\} \sim G(H).$$

From these two counts, is there any way to deduce the asymptotic growth of $$\#\{x \in [0,H] \cap \mathbb Z: f(x) + g(x) \leq H\}?$$

It seems that there are some ways to approximate this if $$f, g$$ are arithmetic functions, and in theory it seems to me that the answer should in some way be related to the convolution of $$F(H)$$ and $$G(H)$$, but I don't readily see it.

• If you like my answer, please accept it officially (so that it turns green). Thanks in advance! Jan 31 at 11:51
• @GHfromMO I learned from your answer, but it's not the answer that I was looking for (I was looking for the right conditions on $f$ and $g$ so that some type of convolution would work, rather than a pathology) so I'm going to keep the question open in case someone else comes along. Thanks for your answer!
– user494572
Jan 31 at 16:05
• I think that $f$ and $g$ have to be very regular for a positive answer, and natural arithmetic functions (like multiplicative functions) are not like that. Jan 31 at 16:39

$$F(H)$$ and $$G(H)$$ do not determine the asymptotic growth of the third display. Indeed, consider the following two functions from $$\mathbb{Z}_{\geq 0}$$ to $$\mathbb{Z}_{\geq 0}$$: $$f(x):=\begin{cases}0,&\text{x is even;}\\x^2,&\text{x is odd;}\end{cases}\qquad\qquad g(x):=\begin{cases}x^2,&\text{x is even;}\\0,&\text{x is odd;}\end{cases}$$ Then \begin{align*} \#\{x \in [0,H]: f(x) \leq H\}& \sim H/2,\\ \#\{x \in [0,H]: g(x) \leq H\}& \sim H/2, \end{align*} but \begin{align*} \#\{x \in [0,H]: f(x)+g(x) \leq H\}& \sim \sqrt{H},\\ \#\{x \in [0,H]: f(x)+f(x) \leq H\}& \sim H/2. \end{align*}
• I was wondering if a counterexample along these lines are possible. Things might go better if you ask for $f$ and $g$ to be monotonic? Jan 29 at 8:50