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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.

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  • $\begingroup$ If you like my answer, please accept it officially (so that it turns green). Thanks in advance! $\endgroup$
    – GH from MO
    Jan 31 at 11:51
  • $\begingroup$ @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! $\endgroup$
    – user494572
    Jan 31 at 16:05
  • $\begingroup$ 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. $\endgroup$
    – GH from MO
    Jan 31 at 16:39

1 Answer 1

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$F(H)$ 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*}

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    $\begingroup$ 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? $\endgroup$
    – Qiaochu Yuan
    Jan 29 at 8:50
  • $\begingroup$ Thank you for this example! Of course, the setting that I had in mind is more natural, like what @QiaochuYuan suggested. Do you think that under some nice hypothesis, one could write down a convolution-type formula? (But don't assume that f and g are arithmetic, as that is already known!) $\endgroup$
    – user494572
    Jan 29 at 16:58
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    $\begingroup$ @QiaochuYuan Under extra hypotheses things might go better, but my feeling is that monotonicity will not be enough. $\endgroup$
    – GH from MO
    Jan 29 at 19:52

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