MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

When one looks at the quotient of Euler products $$\prod_p\frac{\sum_{\alpha=0}^{\infty}f(p^{\alpha})p^{-\alpha s}}{1+f(p)p^{-s}}$$ with $|f|\leq 1$, it is observed that the resulting expression contains no $p^{-s}$ terms, so the quotient converges absolutely for $\sigma\geq 1/2+\epsilon$, $\epsilon>0$.

Roughly speaking, for $\delta\geq \epsilon$, this amounts to saying that the statements $$S(x)=\sum_{n\leq x}|\mu(n)| f(n)=O(x^{1/2+\delta})$$ and $$N(x)=\sum_{n\leq x}f(n)=O(x^{1/2+\delta})$$ are equivalent. This is just because an absolutely convergent Euler product is bounded above and away from zero in absolute value.

I had at first guessed that this kind of inference would break down if we asked for sharper information. For instance, say $S(x)=O(x^{1/2}g(x))$ for some $g(x)=o(x^{\epsilon})$, then I had supposed that it would not necessarily follow that the same must be said of $N(x)$.

This guess was based on the observation that, if we could evaluate the inverse Mellin transforms by residues, then the distinction between the two sums would be governed by behaviour in the closed half plane $\sigma\leq1/2$, and this is the region in which the influence of the quotient of the Euler products is not apparent.

On further reflection, I am yet to be convinced. If we consider the Fourier transform $$2\pi e^{-\sigma y}N(e^y)=\int_{\mathbb{R}}\frac{F(\sigma+it)e^{iyt}dt}{\sigma+it},$$ with $\sigma>1/2$ say, then it appears that the finer behaviour may be determined by considering the integral on vertical lines close to $\sigma=1/2$. For example, choosing $\sigma=1/2+y^{-1}\log y$ we are looking at $N(x)/(x^{1/2}\log x)$, and for any finite $x$ this integral is within the region where the Mellin transforms have the same large and small values (because the quotient Euler product is bounded above and away from zero there).

I would like to know about how the integrals might be compared in order to extend the influence of the Euler product to the finer shades of the behaviour of the transform, if that is possible. In particular, I would like to be able to say something like "If $x^{-1/2}N(x)=O(g(x))$ then the same is true of $S(x)$".

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.