analytic number theory - Partial sums of multiplicative functions

This reference contains the best result of this kind currently known for $mu(n)$:

Tadej Kotnik and Herman te Riele The Mertens Conjecture revisited. Algorithmic number theory, 156--167, Lecture Notes in Comput. Sci., 4076, Springer, Berlin, 2006.

They prove that $limsup_{x rightarrow +infty}M(x)/sqrt{x} geq 1.218$ and that $liminf_{x rightarrow +infty}M(x)/sqrt{x} leq -1.229$. Here

$M(x) = sum_{n leq x}mu(n)$

is the conventional notation for the summatory function of the Möbius function. Their proof is a mixture of analytic number theory and large scale computations. They also have a survey of what is known and what is conjectured about the size of $M(x)$.

Now on to the Liouville function $lambda(n)$ and its summatory function $L(x)$. The latter is very closely connected with $M(x)$, for

$sum_{n leq sqrt{x}}mu(n)Lleft(frac{x}{n^2}right) = sum_{n leq sqrt{x}}mu(n)sum_{m leq x/n^2}lambda(m) = sum_{N leq x}sum_{mn^2 = N}mu(n)lambda(m) = sum_{N leq x}mu(N) = M(x).$

Thus

$|M(x)| leq sum_{n leq sqrt{x}}|Lleft(frac{x}{n^2}right)|$

and so the assumption

$L(x) = oleft(frac{sqrt{x}}{log^{1 + epsilon}(x)}right)$

for example, leads to a contradiction with the Kotnik-te Riele result (or earlier results) for any $epsilon > 0$.

My guess is that if one looks up the old (pre-computer) results on $|M(x)|$ from below, one might prove that $limsup_{x rightarrow +infty}|L(x)|/sqrt{x} > 0$. This may even be in the literature somewhere.

Alternatively

$sum_{n = 1}^{infty}lambda(n)n^{-s} = prod_{p}sum_{k=0}^{infty}lambda(p^k)p^{-ks} = prod_{p}sum_{k=0}^{infty}(-1)^kp^{-ks} = prod_{p}(1 + p^{-s})^{-1} = frac{zeta(2s)}{zeta(s)}$

by the Euler product formula, and from here on it is more elementary than it was with $M(x)$ in the argument that David Speyer gave, because we don't need the zeros on the critical line. For $zeta(2s)$ has a pole at $s = 1/2$ which is not canceled by a pole of $zeta(s)$ at the same point. Thus $L(x) = O(x^{alpha})$ is impossible with $alpha < 1/2$ by partial summation.

For multiplicative functions of modulus $1$, the situation is much less clear. For simplicity, assume that $f(n)$ is a totally multiplicative function (multiplicative and $f(p^k) = f(p)^k$) with $|c_p| = 1$ where $c_p = f(p)$. The Liouville function is the case
$c_p equiv -1$. Then

$sum_{n = 1}^{infty}f(n)n^{-s} = prod_{p}frac{1}{1 - c_pp^{-s}}{quad},{quad}A(x) = sum_{n leq x}f(n)$

by the Euler product formula. The basic principle is that if $A(x) = O(x^{alpha})$, then the Dirichlet series on the left hand side is convergent in the half plane $sigma > alpha$, so the sum is holomorphic in that half plane. If we can find a singularity $s_0$ of the product on the right hand side with $mathrm{Re}(s_0) = sigma_0$, that tells us that
$A(x) = O(x^{alpha})$ with $alpha < sigma_0$ is impossible. The bad thing now is that the product may diverge at a point without having a singularity there, because the product may diverge to zero, and a holomorphic function may be zero at a point without being singular there.

But it is straightforward to show that if $mathrm{Re}(c_p) geq delta > 0$ for all $p$, then $A(x) = O(x^{alpha})$ is impossible for any $alpha < 1$, by showing that the series
$sum_{p}c_pp^{-sigma}$
goes to infinity as $sigma rightarrow 1^{+}$.

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