It seems to be that the simplicity of all the zeros is quite widely accepted as a working hypotheses, and it is known that a positive proportion are as such.

Titchmarsh explains in the last chapter on RH that the weak Mertens hypothesis, $M(x)=O(x^{1/2})$, implies that all zeros are simple. Though I think this is conjectured to be false by many authors. Is there a weaker bound whose optimality is known to imply that not all zeros are simple?

Given the classical estimate $N(\sigma, T)=O(T^{4\sigma(1-\sigma)+\epsilon})$ combined with the fact that the number of solutions to $\zeta(s)=z$ in any rectangle of height $T$ contained in $1/2\leq \sigma\leq 1$ is proportional to $T$, for any $z\neq 0$, simplicity of the zeros suggests that the derivative must vanish vastly more frequently in each such rectangle else one could presumably tie the number of such solutions to the number of zeros up to height $T$, for all sufficiently small $|z|$, by Rouche's theorem. One might therefore conjecture that $ \inf|\rho_{\zeta}-\rho_{\zeta'}|=0$ (is this known perhaps?).

An old theorem of MacDonald (Whittaker and Watson, p131) says that the number of zeros of an analytic function on the interior of a simple closed curve on which the modulus of $f$ is constant exceeds that of it's derivative by one. It can also be seen that the derivative necessarily vanishes on the largest such curve by the open and inverse mapping theorems. In the case of $\zeta(s)$ I'd expect that the largest of such curves around each zero would become small for large $T$, so I wonder if something in this direction is known?

Anyway, as Titchmarsh points out, $\Omega$-theorems for $M(x)$ seem to be more difficult to obtain than those for the prime counting functions and, since a significant distinction between the problems is that of the order of the zeros, I would like to enquire specifically about anything that is known in either direction, on any other hypotheses.

  • 1
    $\begingroup$ Perhaps I misunderstand, but $\inf|\rho-\rho'|=0$ is immediate from the fact that the number of zeros up to height $T$ is $\gg T\log T$. $\endgroup$ – GH from MO Feb 19 '14 at 15:16
  • $\begingroup$ The prime denotes zeros of the derivative- pardon, sloppy notation. $\endgroup$ – Kevin Smith Feb 19 '14 at 16:06
  • 5
    $\begingroup$ If $\zeta(s)$ has a multiple zero, then $M(x)=\Omega_\pm(\sqrt{x} \log x)$. So $M(x) = o(\sqrt{x}\log x)$ implies all zeros of $\zeta(s)$ are simple. $\endgroup$ – Micah Milinovich Feb 20 '14 at 19:16
  • $\begingroup$ Micah, can you give a reference for this? Is the same known for $\psi$? $\endgroup$ – Mark Lewko Feb 20 '14 at 21:39
  • $\begingroup$ Thank you Micah, I'm very interested in how this is proved. I can see that the factor comes from a residue of a double pole, but not the overall influence. Perhaps you could indicate how in an answer? $\endgroup$ – Kevin Smith Feb 21 '14 at 9:15

Quite a lot is known in the literature about zeros of $\zeta(s)$ and $\zeta^{\prime}(s)$, and in particular one can show that zeros of $\zeta^{\prime}(s)$ do get close to zeros of $\zeta(s)$. However, I am skeptical that these results have any implications for $M(x)$.

The number of zeros of $\zeta^{\prime}(s)$ with ordinates in $(0,T)$ is $$ = \frac{T}{2\pi} \log \frac{T}{4\pi e} + O(\log T). $$ This is due to Bruce Berndt, and is an application of the argument principle. Note that it is slightly different from the corresponding formula for $\zeta(s)$. The zeros of $\zeta^{\prime}$ are all expected to lie in the half plane $\sigma \ge 1/2$ -- this is equivalent to RH (proved first by Speiser). Levinson and Montgomery (Acta Math. 1974) investigated this further, and their work was the precursor to Levinson's famous result on the critical zeros of $\zeta(s)$. Levinson and Montgomery showed that most of the zeros of $\zeta^{\prime}(s)$ (with ordinate in $(0,T)$) lie in the region $|\sigma -1/2| \le w(T) \log \log T/\log T$ for any function $w(T)$ tending to infinity with $T$. In fact, most of the zeros of $\zeta^{\prime}$ probably lie closer to the critical line, on the scale of $1/\log T$ from $1/2$. For work in this direction, see papers of Conrey and Ghosh, Soundararajan, Yitang Zhang, Farmer and Ki, and most recently Radziwill (http://arxiv.org/abs/1301.3232, which will give the other references).

From above one knows that there are plenty of zeros of $\zeta^{\prime}(s)$ lying very close to the half line. From zero density results, we also know that there are plenty of zeros of $\zeta(s)$ lying close to the half line. For example, we can guarantee that there are at most $T/\log T$ zeros of $\zeta(s)$ with $|\sigma- 1/2| \ge 100 \log \log T/\log T$, say. The only thing is to rule out a conspiracy separating ordinates of such zeros of $\zeta$ and $\zeta^{\prime}$. A straightforward way to do this is to use Littlewood's theorem that the gaps between consecutive ordinates of zeros of $\zeta(s)$ tends to zero (at the rate of $1/\log \log \log |\gamma|$). First find, using the results above, $\gg T$ zeros of $\zeta^{\prime}$ (with ordinates in $[T,2T]$ say) such that these zeros all lie near the critical line, and their ordinates are spaced a distance $1$ apart. Next by Littlewood, for each such zero of $\zeta^{\prime}$, find a zero of zeta whose ordinate is close to it. Lastly, note that by the zero density theorem for zeta, at most $T/\log T$ of these corresponding zeros of zeta can be far from the $1/2$ line.

That completes the proof. The lazy argument above shows that $\min |\rho_{\zeta} -\rho_{\zeta^{\prime}}| \ll 1/\log \log \log T$ for zeros around ordinate $T$. By working harder one should be able to get $\ll 1/\log T$, and one probably cannot prove anything better. It is believed that zeros of $\zeta$ can get very close to each other, and so plausibly the truth here is that $|\rho_{\zeta} -\rho_{\zeta^{\prime}}|$ can get as small as $T^{-\frac 13+o(1)}$. This is very speculative, and even with this I don't see any implication for $M(x)$.

| cite | improve this answer | |
  • $\begingroup$ Thanks for this- it is helpful that you have included a brief summary of relevant results. The connection with the conjecture that $M(x)\notin O(x^{1/2})$ is far from obvious, but hopefully I will be able to explain it. $\endgroup$ – Kevin Smith Feb 27 '14 at 19:48

Here is a good reason why $\inf | \rho - \rho'| =0$ should be true.

Fact 1: It follows from Voronin's universality theorem applied to $\zeta'$, that given $\epsilon>0$, there number of zeros $\rho'$ of $\zeta'$ with $|\Re \rho' -1/2| < \epsilon$ and $0 \leq \Im \rho \leq T$ grow at least like $c_\epsilon T$.

Here is a suggestion how to prove it. Consider $\epsilon_0> 0$ and a ball $B$ around $1/2 + \epsilon_0$ of radius $\epsilon_0/2$. Consider an entire function $f$ vanishing inside of $B$, but not on the boundary. Let $\lambda(T)$ be the Lebesgue measure of the set $T>t>0$ with $\sup_{s \in B}| \zeta'(s+it) - f(s)| < \inf_{s \in \partial B} |f(s)|$, which implies $f$ and $\zeta'(s+it)-f+f$ have the same number of zeros in $B$ (Rouché). We have that $\lim\inf_{T \rightarrow \infty} \lambda(T)/T>0$ (Voronin).

Fact 2 (Selberg?): Now a positive proportion of the zeros lies on $\Re s =1/2$, that being, the number of zeros $\rho$ with $\Re \rho =1/2$ and $0<\Im \rho < T$ grow like $c T \log T$.

In a previous version, I thought this would imply $\inf | \rho - \rho'|=0$, but it does not trivially.

Moreover, it is conjectured that the zeros of $\zeta$ are distributed uniformly, see e.g. here: http://empslocal.ex.ac.uk/people/staff/mrwatkin//zeta/bump-gue.htm. I also must admit that besides these facts, I don't know much about the spacings of the zeros of $\zeta$ on $\Re s=1/2$ or $\zeta'$, so maybe the two above facts are enough to yield the result.

| cite | improve this answer | |
  • $\begingroup$ I had thought a proof might come from a universality argument, but I didn't know the derivative had this property- thank you. I would be very interested if you have an idea of how to find an upper bound on the rate at which the distance goes to zero, because I think this ought to furnish us with a means to disproove $M(x)=O(x^{1/2})$. $\endgroup$ – Kevin Smith Feb 21 '14 at 12:36
  • $\begingroup$ This would require an effective version of the universality depending upon $\epsilon_0$. To my knowledge, there are none or only horrible estimates known in that direction (Steuding's book discusses this or something similar). $\endgroup$ – Marc Palm Feb 21 '14 at 15:15
  • $\begingroup$ Btw the universality theorem for $\zeta'$ immedeiately follows from that for $\zeta$ plus you can drop that vanishing condition on $f$ encountered in the statement for $\zeta$. Note that on a simply connected set you will always have a primitive $F$ for $f$ ($F'= f$), then add a sufficiently big constant. $\endgroup$ – Marc Palm Feb 21 '14 at 15:55
  • $\begingroup$ @Marc: I don't see how you apply Rouché's theorem. The fact that $\sup_{s\in B}|g(s)|<\sup_{s\in B}3|f(s)|/2$ certainly does not imply that $g(s)$ vanishes in $B$. Sorry if I am missing the obvious! $\endgroup$ – GH from MO Feb 23 '14 at 17:11
  • 1
    $\begingroup$ @GH Thanks, I memorized the statement in a wrong manner. Should have checked Wiki. I have fixed the mistake. $\endgroup$ – Marc Palm Feb 24 '14 at 10:29

I think the following argument settles the conjecture that $\inf |\rho_{\zeta}-\rho_{\zeta'}|=0$, assuming RH. Basically, it follows because the zeros can be arbitrarily close.

Littlewood proved that $$\inf \gamma_n-\gamma_{n-1}=0.$$ On RH this implies that $$\inf \rho_n-\rho_{n-1}=0.$$ So, for every $\epsilon>0$ one can find a pair of zeros between which $$\max\{|\zeta(1/2+it)|:\gamma_{n(\epsilon)-1}<t<\gamma_{n(\epsilon)}\}=\epsilon.$$

By the theorem of Macdonald (cited in my question above) and the inverse mapping theorem, the derivative must vanish somewhere on the boundary of the closed curve enclosing $\rho_{n(\epsilon)}$ defined by the condition that $|\zeta(s)|=\epsilon$, and not inside it. By the maximum principle, as $\epsilon\rightarrow 0$ the modulus on and throughout the interior of the curve tends to zero, so such a curve must tend to a point as $\epsilon\rightarrow 0$. Since the derivative vanishes on the boundary, the result follows.

EDIT. This argument is not quite right. Actually the curves described above are not necessarily closed, and therefore need not tend to a point. However, the derivative certainly vanishes on the boundary of the largest closed curve enclosing $\rho$ and on which $|\zeta(s)|=c$, and $c$ is less than or equal to $\epsilon$. The conclusion is the same, because these curves are closed.

| cite | improve this answer | |
  • $\begingroup$ I can't see why two zeros are close, then the maximal distance on the curve is close. That seems not to be true. Can you elaborate what you mean "by the max princip". $\endgroup$ – Marc Palm Feb 26 '14 at 15:47
  • $\begingroup$ I have edited the answer. I hope this helps. I think the key point is that the curve is always closed, but in hindsight actually I am wondering if it is the case. Perhaps this is wrong. $\endgroup$ – Kevin Smith Feb 26 '14 at 17:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.