Suppose $f:\mathbb{N}\to [0,1]$ is a multiplicative function (i.e. $f(n,m)=f(n)f(m)$ 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.

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.