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Is it true that for every completely multiplicative function $f:\mathbb N\to\mathbb C$ with $|f(n)|=1$ for all $n$, the logarithmic average $$\lim_{N\to\infty}\frac1{\log N}\sum_{n=1}^N\frac1nf(n)\quad\text{exists?}$$

In Elliott's book "Probabilistic number theory", a theorem attributed to Delange, Wirsing, Halasz (Theorem 6.3) describes exactly when a multiplicative function taking values in the unit disk has a (Cesaro) mean, and what is the value of the mean when it exists. It seems that the main obstacle to the existence of the (Cesaro) mean exist are the multiplicative functions $n^{it}$. On the other hand, it is easy to check that the logarithmic average of any such function exists (and is 0).

If the series $\sum_{p\text{ prime}}\frac{1-f(p)}p$ converges, then the answer is yes (this follows from the aforementioned theorem). On the other hand, an application of the Turan-Kubillius inequality shows that if for some $\epsilon>0$ the set $S:=\{p\text{ prime}:|f(p)-1|>\epsilon\}$ is divergent (in the sense that $\sum_{p\in S}\frac1p=\infty$), then again the answer to the question is yes (and the limit is $0$). However I don't see how to handle the general case.

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    $\begingroup$ If I understand it right, for the case when only finitely many $f(p)\ne 1$, the limit is $\prod\frac{1-\frac 1p}{1-\frac {f(p)}p}$. If $f(p)=e^{i\xi}$, this creates about $(1-\xi^2/p)$ in shrinking and about $\xi/p$ in rotation per single prime $p$, so you can rotate as much as you want without any noticeable shrinking by choosing $\xi$ small and then taking sufficiently many arbitrarily large primes. Now life becomes easy: throw in rotation by $\pi$ without shrinking, wait until the limit almost establishes, throw in another rotation by $\pi$ using much larger primes, and so on. $\endgroup$
    – fedja
    Commented Jun 20, 2019 at 23:08
  • $\begingroup$ Oh, I think you're right. So the answer is actually no in general... $\endgroup$
    – Joel Moreira
    Commented Jun 21, 2019 at 2:13

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