Completely multiplicative functions with values in $\{-1,1\}$ This question is from Eric Saias and myself:
Let $A$ be the set of abscissas of convergence of Dirichlet Series $\sum_{n\ge 1} \frac{f(n)}{n^s}$
where $f(n)$ is completely multiplicative and $f(n) \in \{-1,1\}$ for all $n\ge 1$.
Our question is: What is known about the set $A$ ?
a) We clearly have $A \subseteq [0,1]$.
b) By taking $f(n) = 1$, we have  $1 \in A$.
c) By taking $f$ to be the completely multiplicative function defined by $f(p) =
\chi(p)$ for $p>2$, where $\chi$ is the non-principal Dirichlet character modulo 4,
and $f(2) = 1$, we have $0 \in A$.
d) By taking $f$ to be Liouville's function ($f(p) = -1$), then, assuming the Riemann Hypothesis (RH), $1/2 \in A$.
Question: With or without RH, are there other numbers in $A$ besides $0, 1/2, 1$ ?
 A: One can show that every $\alpha \in [0,1]$ is attained.  This may be achieved by tweaking suitably a fixed Dirichlet character, for example the non-trivial character $\pmod 4$.  Let $\alpha \in (0,1)$ be given, and let ${\mathcal P}$ denote a set of primes all $\equiv 3\pmod 4$ with 
$$ 
\#\{ p\le x: p \in {\mathcal P} \} \sim x^{\alpha}. 
$$ 
Define the completely multiplicative function $f$ by taking $f(2)=1$, and setting $f(p)=1$ if $p\equiv 1\pmod 4$ or if $p\in {\mathcal P}$, and then taking $f(p)=-1$ on all the primes that are $\equiv 3\pmod 4$ but not in ${\mathcal P}$.  Thus we can write $f$ as the convolution of $\chi$ (the character $\pmod 4$) and a multiplicative function $g$ given by $g(2^k) =1$, and $g(p^k) =0$ unless $p\in {\mathcal P}$ in which case $g(p^k)=2$ (for all $k\ge 1$).  
Thus 
$$ 
F(s)=\sum_{n=1}^{\infty} \frac{f(n)}{n^s} = L(s,\chi) \sum_{n=1}^{\infty} \frac{g(n)}{n^s}, 
$$ 
and by our assumption on ${\mathcal P}$, it is easy to see that the series $F(s)$ converges for Re$(s)>\alpha$.  Now we note that the series does not converge at $\alpha$.  To see this note that 
$$ 
\sum_{n\le x} \frac{f(n)}{n^{\alpha}} =\sum_{d\le x} \frac{g(d)}{d^{\alpha}} \sum_{m\le x/d} \frac{\chi(m)}{m^{\alpha}}. 
$$ 
Since the inner sum is alternating with monotone decreasing coefficients it is always $\ge (1-1/3^{\alpha})$, and $g$ is non-negative, the above is 
$$ 
\ge \sum_{d\le x} \frac{g(d)}{d^{\alpha}} \Big(1-\frac{1}{3^{\alpha}}\Big) \gg \sum_{p\le x, p\in {\mathcal P}} \frac{1}{p^{\alpha}} \gg \log x,
$$ 
by our assumption on ${\mathcal P}$.  So $\alpha$ is the abscissa of convergence. 
