Suppose $f:\mathbb{N}\to [0,1]$ is a multiplicative function (i.e. $f(nm)=f(n)f(m)$ whenever $m$ and $n$ are coprime). Suppose $f$ has non-zero mean, which means $$ \lim_{N\to\infty}\frac{1}{N} \sum_{n=1}^N f(n) >0. $$ Also, lets assume $f(n)\neq 0$ for all $n$. Then, can we say something about the set of all $n$ such that $f(n)\leq \varepsilon$? What is the density (or upper density) of this set? In particular, does this density go to zero as $\varepsilon\to 0$?

Thank you very much for your help.

  • $\begingroup$ Yes, thank you Hurkyl. I have edited my question. $\endgroup$
    – Karl
    Aug 19 '15 at 23:54

Yes, the density goes to zero as $\epsilon \to 0$, but the convergence to zero can be arbitrarily slow (depending on the choice of the function $f$). To see this, first note that the condition that the mean value of $f$ is strictly positive is equivalent to $$ \sum_p \frac{1-f(p)}{p}<\infty, $$ which is also the same as $$ \sum_{p^k} \frac{1-f(p^k)}{p^k} <\infty, $$ where the sum is over all prime powers $p^k$. This is a simple case of results of Wirsing, Halasz, Delange, and follows from the estimate (for non-negative multiplicative functions bounded by $1$) $$ \sum_{n\le x} f(n) \ll \frac{x}{\log x} \sum_{n\le x} \frac{f(n)}{n} \ll \frac{x}{\log x} \exp\Big(\sum_{p\le x} \frac{f(p)}{p}\Big). $$ (See for example Theorem 2 of Halberstam and Richert).

Next, for any $1> \delta >0$, put $$ {\mathcal F}(\delta)= \sum_{\substack{p^k \\ f(p^k) \le \delta} } \frac{1}{p^k}, $$ the convergence of the sum for any $\delta <1$ being guaranteed by our previous observation. We claim that ${\mathcal F}(\delta) \to 0$ as $\delta \to 0$; this step crucially uses the assumption that $f(n)$ is strictly positive for all $n$. For each $\delta >0$ let $p(\delta)$ denote the smallest prime power $p^k$ with $f(p^k)<\delta$. Then the assumption that $f(n)>0$ gives that $p(\delta) \to \infty$ as $\delta \to 0$. Therefore, for $\delta \le 1/2$ say,
$$ {\mathcal F}(\delta) \le \sum_{\substack {p^k \\ p^k \ge p(\delta) \\ f(p)\le 1/2}} \frac{1}{p^{k}}, $$ and the RHS is the tail of a convergent series, and therefore goes to zero as $p(\delta)\to \infty$, or in other words as $\delta \to 0$. This proves our claim.

Now suppose $\epsilon >0$ is given (and suitably small), and we want to bound the density of $n$ with $f(n) \le \epsilon$. Put $\delta=\exp(-(\log 1/\epsilon)^{\frac 12})$, which is bigger than $\epsilon$ but still goes to zero with $\epsilon$. Suppose first that $p^k \Vert n$ for some prime power $p^k$ with $f(p^k) <\delta$. The density of such $n$ is clearly at most ${\mathcal F}(\delta)$. Now suppose that if $p^k \Vert n$ then $f(p^k) \ge \delta$. Note here that (using $x \ge \exp(2\log (1/\delta) (x-1))$ for $\delta <x\le 1$) $$ \epsilon > f(n) = \prod_{p^k \Vert n} f(p^k) \ge \exp\Big( 2\log \frac{1}{\delta} \sum_{p^k \Vert n} (f(p^k)-1) \Big), $$
and so, for such $n$, $$ \sum_{p^k \Vert n} (1-f(p^k)) \ge \frac{1}{2} \Big(\log \frac {1}{\epsilon}\Big)^{\frac 12}. \tag{1} $$ Since $$ \sum_{n\le N} \sum_{p^k \Vert n} (1-f(p^k)) \ll N \sum_{p^k}\frac{1-f(p^k)}{p^k} \ll N, $$ we conclude that the density of $n$ satisfying (1) is $\ll (\log \frac {1}{\epsilon})^{-\frac 12}$. Thus our desired density is $$ \ll {\mathcal F}(\delta) + \Big(\log \frac{1}{\epsilon}\Big)^{-\frac 12}, $$ which goes to $0$ as $\epsilon \to 0$.


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