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I understand that if $M(N)=O(N^\sigma)$, then $\sum_{n=1}^\infty \frac{\mu(n)}{n^s}=\frac{1}{\zeta(s)}$ and therefore $$ \frac{1}{s\zeta(s)} = \int_0^\infty M(x) x^{-(s+1)} dx $$ for $s>\sigma$, and that having $\sigma=1/2+\epsilon$ for every $\epsilon>0$ will thus prove the RH.

The Wikipedia article on the Mertens Conjecture states that the reverse also holds, but I don't understand the details of argument:

Using the Mellin inversion theorem we now can express M in terms of 1/ζ as

$$M(x) = \frac{1}{2 \pi i} \int_{\sigma-i\infty}^{\sigma+i\infty} \frac{x^s}{s \zeta(s)}\, ds$$ which is valid for $1 < \sigma < 2$, and valid for $1/2 < \sigma < 2$ on the Riemann hypothesis.

Here is my first problem. It seems to me we can only extend this formula if $$\lim_{T \to \pm \infty} \frac{x^{\sigma+iT}}{(\sigma+iT)\zeta(\sigma+iT)}= 0$$ uniformly on the strip $1/2 < \sigma < 2$. If this is indeed necessary, is there an easy way to show this without invoking heavy machinery?

From this, the Mellin transform integral must be convergent,

Again this seems to need the limit above (in order for the Mellin inversion theorem to apply).

and hence $M(x)$ must be $O(x^e)$ for every exponent e greater than $1/2$.

This is also very unclear to me. There are certainly functions $f(x)$ such that $$\int_0^\infty f(x)x^{-(s+1)} dx$$ converges but $f(x)$ is not $O(x^e)$ for every $e>1/2$. Take for instance

$$f(x) = \left\{\begin{array}{cc}x &\mbox{if $\lfloor x \rfloor$ is a power of two} \\ 0 & \mbox{otherwise}\end{array}\right.$$

for which the integral is convergent, but $f(x)$ is not $O(x^e)$ for any $e<1$. By using that $|M(n)-M(n-1)| \le 1$, I've managed to prove that $M(x) = \mathcal{O}(x^e)$ for $e>0.75$ but I still need $e>0.5$.

Or am I misinterpreting this and should I get that from the vertical line Mellin integral?

From this it follows that

$$M(x) = O(x^{\frac12+\epsilon})$$

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  • $\begingroup$ Note you say, ".. then $\sum_{n=1}^\infty \frac{\mu(n)}{n^s}=\frac{1}{\zeta(s)}$." That is not a consequence of the first statement; it is true for all $s$ where the left side converges. $\endgroup$ – Thomas Andrews Aug 18 '15 at 22:37
  • $\begingroup$ Certainly, but the requirement that $M(N)=O(N^e)$ is sufficient for convergence of the LHS when $\mbox{Re}(s)>e$. $\endgroup$ – Sean Wentzel Aug 19 '15 at 10:04
  • $\begingroup$ it is detailed in Titchmarsh's book (the last chapter consequences of the RH), But at first you should look at page.60 to 66 where the same idea is applied to show $\zeta(s)$ has no zeros on $Re(s) = 1 \implies M(x) = o(x)$ $\endgroup$ – reuns Jul 16 '16 at 18:58
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You'll need Titchmarsh's book "The theory of the Riemann zeta-function" for this one, as pointed out by reuns:

http://plouffe.fr/simon/math/The%20Theory%20Of%20The%20Riemann%20Zeta-Function%20-Titshmarch.pdf

The argument you're looking for is given in ch.14.25, p.370, in Theorem 14.25(C), using the inverse Mellin transform on region $\sigma>1$, particularly with $\sigma=2$.

Best regards.

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