Erdős Problem 239: mean values of real multiplicative functions
1) Statement and scope
Every multiplicative function $f:\mathbb N\to\{-1,1\}$ has a Cesaro mean. This result was proved by Wirsing and generalized by Halász. We give the deduction from a standard form of Halász's theorem, together with a direct treatment of the absolutely convergent case.
2) An absolutely convergent convolution
Suppose first that $\sum_p(1-f(p))/p<\infty$. Put $g=f*\mu$, so $f=1*g$, $g$ is multiplicative, and \[
g(p^j)=f(p^j)-f(p^{j-1})\quad(j\geq1).
\] Here $f(1)=1$ follows from multiplicativity and the nonzero range. For $j=1$, $|g(p)|=1-f(p)$; for $j\geq2$, $|g(p^j)|\leq2$. It follows that \[
\sum_p\sum_{j\geq1}\frac{|g(p^j)|}{p^j}
\leq\sum_p\frac{1-f(p)}p+
2\sum_p\frac1{p(p-1)}<\infty.
\] The Euler product with nonnegative coefficients now shows that $\sum_d|g(d)|/d<\infty$. Hence \[
\frac1N\sum_{n\leq N}f(n)
=\sum_{d\leq N}g(d)\frac{\lfloor N/d\rfloor}{N}
\longrightarrow\sum_{d\geq1}\frac{g(d)}d
=\prod_p\left[\left(1-\frac1p\right)
\sum_{j\geq0}\frac{f(p^j)}{p^j}\right].
\] Dominated convergence is justified by $|g(d)|/d$. A local Euler factor is allowed to vanish; this presents no difficulty.
3) The divergent case
We use the following external consequence of Halász's theorem: if $|f(n)|\leq1$ and \[
\sum_p\frac{1-\operatorname{Re}(f(p)p^{-it})}{p}=\infty
\quad\hbox{for every fixed }t\in\mathbb R,
\] then $\sum_{n\leq N}f(n)=o(N)$. Suppose $\sum_p(1-f(p))/p=\infty$. This verifies the criterion for $t=0$. For $t\ne0$ and $u=t\log p$, \[
1-f(p)\cos u\geq1-|\cos u|
\geq\frac{1-\cos(2u)}4.
\] We next verify the divergence of the resulting nonnegative series. For $\sigma>1$, the Euler product for $\zeta$ gives \[
\sum_p p^{-\sigma}\cos(2t\log p)
=\log|\zeta(\sigma+2it)|+O(1).
\] The error from prime powers of exponent at least two is bounded uniformly as $\sigma\downarrow1$. Since $\zeta$ is holomorphic near $1+2it$, the right side is bounded above there. On the other hand $\sum_p p^{-\sigma}\to\infty$. Therefore \[
\sum_p p^{-\sigma}\left(1-\cos(2t\log p)\right)\longrightarrow\infty.
\] If the corresponding sum with weight $1/p$ were finite, it would dominate these sums, a contradiction. Halász's criterion applies, and the mean is zero.
4) Why the real-valued hypothesis matters
For comparison, $f(n)=n^i$ is completely multiplicative and has modulus one. Comparison with its integral, using $|(x^i)'|=1/x$, yields \[
\frac1N\sum_{n\leq N}n^i
=\frac{N^i}{1+i}+O\left(\frac{\log N}{N}\right),
\] which has no limit. Thus an analogous assertion for arbitrary complex values on the unit circle would be false.
5) Sources and scope
Sources:
https://www.erdosproblems.com/239; A. Granville, A. J. Harper and K. Soundararajan,
A new proof of Halász's Theorem, and its consequences,
https://arxiv.org/abs/1706.03749. Halász's theorem and the standard analytic continuation and Euler product of $\zeta$ are external dependencies. The two-case deduction above proves the full assertion from them; no original resolution or local formal verification is claimed.
6) Completion estimate
COMPLETION: 100%