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It is known that $\zeta(s)$ has an infinity of zeros in the strip $0<\sigma<1$ and that those zeros become closer together as $t\rightarrow\infty$. More precisely, Littlewood showed that there is a zero $\rho=\beta+i\gamma$ with $|\gamma-t|\in O(1/\log\log\log t)$.

A positive proportion of these zeros are known to have $\beta=1/2$, but $\zeta(1/2+it)$ is unbounded. In fact, $\zeta(\sigma+it)$ is bounded as $t\rightarrow\infty$ only if $\sigma>1+\delta$, $\delta>0$ fixed.

On the other hand, there seem to be good reasons to expect that $$f(t)=\min_{0\leq\sigma\leq 1} |\zeta(s)|$$ approaches zero as $t\rightarrow\infty$. One reason is that zeros of $\zeta(s)$ are rather differently distributed than those of $\zeta(s)-z$ for any other value of $z$.

I would like to argue that $f(t)\rightarrow 0$ here, but my approach is incomplete. In fact, I would like to conjecture that, for every $\epsilon>0$, we have $$f(t)\in O (\log^{\Theta(t) -1+\epsilon}t),$$ where $\Theta(t)=\max\{\beta:\gamma<t\} $.

My questions are as follows: is anything at all non-trivial known about upper bounds on $f(t)$ and/or who has studied it? Does anyone have any computational data supporting this conjecture? Thanks!

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  • $\begingroup$ I would be surprised if this were true. For certain values of t zeta(sigma+ i t) is almost real for 0<sigma<1 and take large values >1. For example for t=87568424.95851 the minimum of the absolute value is >6 (more or less). $\endgroup$ – juan Jul 17 '14 at 20:46
  • $\begingroup$ I have no proof, it is only a guess after seeing many x-ray plots, T. Kotnik has given many of these ordinates. $\endgroup$ – juan Jul 17 '14 at 20:53
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This is not exactly an answer, but too long for a comment.

Perhaps I have not been sufficiently clear in the comments. In the paper: T. Kotnik, Computational estimation of the order of $\zeta(1/2+i t)$, Math. of Comp. 73 (2003) 949-956 it is shown how to get values $t$ where $|\zeta(1/2+it)|$ is large.

Many (almost all) of these values of $t$ are near a Gram point. The x-ray shows that these Gram points are places where a real line (one where zeta is real) hit the critical line. Precisely the real line goes from $\sigma=+\infty$. On these lines $\zeta(s)$ is real and $>1$. What the x-ray's show is that the points signaled by Kotnik are those in which these lines are almost parallel to the real line. So that at these height $t$ the value $\zeta(\sigma+it)$ is almost real for $0<\sigma<1$. By the election of the lines $|\zeta(1/2+it)|>>1$. This is the what Kotnik is searching, and we know that $\Re\zeta(\sigma+it) >1$, because these points are almost on the real line.

Therefore we expect that $$\limsup_{t\to\infty} f(t)\ge 1.$$ Of course this contradicts the conjecture above, except if you assume that $\Theta(t)\to1$, which contradicts the RH.

For example I take one of the values given by Kotnik (and he has means to get these values apparently without limit).
I take $t=21559062801.941668$ where he reported $Z(t)=-192.996$. And compute the values $$\zeta(it)=410676.057+11579.014 \cdot i$$ $$\zeta(0.1+it)=67487.871+ 1567.236\cdot i$$ $$\zeta(0.2+it)=12335.222+225.398\cdot i$$ $$\zeta(0.3+it)=2584.380+37.231\cdot i$$ $$\zeta(0.4+it)=640.801+7.986\cdot i$$ $$\zeta(0.5+it)=192.982+2.341\cdot i$$ $$\zeta(0.6+it)=71.315+0.841\cdot i$$ $$\zeta(0.7+it)=32.010+0.315\cdot i$$ $$\zeta(0.8+it)=17.005+0.101\cdot i$$ $$\zeta(0.9+it)=10.355+0.010\cdot i$$ $$\zeta(1+it)=7.013-0.027\cdot i$$

As I said the values are 'almost real' and the absolute value $|\zeta(1/2+it)|>7$. But I have taken any of the values of Kotnik.

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  • $\begingroup$ This is very interesting. Thank you. $\endgroup$ – Kevin Smith Jul 19 '14 at 9:35
  • $\begingroup$ Do you mean that there is a way of determining arbitrarily large t for which a curve on which zeta is real and increasing (from right to left) is sufficiently nearby to guarantee that the lower bound holds? $\endgroup$ – Kevin Smith Jul 20 '14 at 10:41
  • $\begingroup$ Not exactly. (a) Kotnik explains a procedure by which he obtains the points $t_k$ where $|Z(t_k)|$ has a record. This records are infinite because $Z(t)$ is not bounded. (b) I observed that at these points a real line of zeta cross almost perpendicular to the critical line. I know that at these lines zeta is real and $>1$. (c) the two previous points are reasons to suspect that $\limsup f(t) > 1$. $\endgroup$ – juan Jul 20 '14 at 15:53
  • $\begingroup$ My experience with x-rays of zeta. Difficult to explain without graphs and a common language(I am not English speaker) make me suspect that (b) is not a coincidence. The near parallell line is associated to big values of $Z(t)$ and some other things. $\endgroup$ – juan Jul 20 '14 at 15:56
  • $\begingroup$ Finally. This is all based on the computation of $Z(t)$ that are possible only for $t$ small. I may be very wrong, (as those that assume the RH), because we ignore how zeta behave at higher values. Say when $S(t)$ is of the order of 1000. $\endgroup$ – juan Jul 20 '14 at 16:01
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The following paper deals with the behavior of the modulus of the Zeta function in the critical strip along horizontal lines:

  • MR1986257 (2004c:11152) Saidak, Filip ; Zvengrowski, Peter . On the modulus of the Riemann zeta function in the critical strip. Math. Slovaca 53 (2003), no. 2, 145--172.

See also the slightly related paper

  • MR2857985 (Reviewed) Srinivasan, Gopala Krishna ; Zvengrowski, P. On the horizontal monotonicity of |Γ(s)|. Canad. Math. Bull. 54 (2011), no. 3, 538--543.

See also arXiv:1205.2773

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Not a compete answer, but observations on the value of $\sigma$ for which the minimum occurs. Throughout I'm assuming the Riemann Hypothesis.

In "Zeros of $\zeta^\prime(s)$ and the Riemann Hypothesis", Spira proves that the Riemann Hypothesis holds if and only if $|\zeta(s)|$ is increasing as $\sigma$ decreases from $1/2$ to $-\infty$ for $t\gg 1$ (He seems to prove $t>165$.) So it suffice to consider the minimum over $1/2\le \sigma\le 1$.

From now on we assume $\zeta(1/2+it)\ne 0$ (else the minimum over $\sigma$ is already $0$ for that $t$.) Then $|\zeta(\sigma+i t)|>0$, and so $\ln|\zeta(\sigma+i t)|$ is defined. Since $\ln$ is increasing and preserves inequalities, the same $\sigma$ minimizes both $|\zeta(\sigma+i t)|$ and $\ln|\zeta(\sigma+i t)|$. There are three possibilities:

I: $\ln|\zeta(\sigma+i t)|$ is increasing on $1/2\le \sigma\le 1$ and the minimum occurs at $\sigma=1/2$.

II: $\ln|\zeta(\sigma+i t)|$ is decreasing on $1/2\le \sigma\le 1$ and the minimum occurs at $\sigma=1$.

III: $\ln|\zeta(\sigma+i t)|$ has a local minimum on $(1/2,1)$. (Conceivably, the global minimum may still occur at an endpoint.) This is the case in which $$ \frac{d}{d\sigma}\ln|\zeta(s)|=0 \Leftrightarrow \frac{\partial}{\partial \sigma} \text{Re}(\log(\zeta(s)))=0 \Leftrightarrow \text{Re}\left(\frac{\zeta^\prime(s)}{\zeta(s)}\right)=0. $$

This latter happens when either

IIIa: $\zeta^\prime(s)=0$, or

IIIb: $\arg\left(\frac{\zeta^\prime(s)}{\zeta(s)}\right)=\pm \pi/2.$

The zeros of $\zeta^\prime(s)$ occur discretely (just as do the zeros of $\zeta(s)$). Meanwhile the level curves for $\arg=\pm \pi/2$ will connect the zeros of $\zeta^\prime/\zeta$ (i.e. the zeros of $\zeta^\prime(s)$) with the poles of $\zeta^\prime/\zeta$ (i.e. the zeros of $\zeta(s)$ on the critical line.)

Titchmarsh shows that $\zeta(s)\ne 0$ for $\sigma>3$, and zeros with $\sigma>1$ do exist. But Spira shows in "Zeros of $\zeta^\prime(s)$ in the critical strip" that 'most' zeros are close to $\sigma=1/2$ (in the sense that the number of zeros with $\sigma>1/2+\delta$, $t<T$ is only O(T).) Lots of research has been done on the clustering of the zeros of $\zeta^\prime$ near the critical line.

Edit: For $t\gg1$, possibility I above can not occur. For if it did, then combined with Spira's result above, we would have that $\sigma=1/2$ is a local minimum. But the formula for the logarithmic derivative $$ \frac{\zeta^\prime(s)}{\zeta(s)}=\log(2\pi)-1-C/2-\frac{1}{s-1}-\frac{\Gamma^\prime(s/2+1)}{2\Gamma(s/2+1)}+\sum_{\rho}\left(\frac{1}{s-\rho}+\frac{1}{\rho}\right), $$ along with Stirling's formula, gives that $$ \text{Re}\frac{\zeta^\prime}{\zeta}(1/2+it)\sim -\log(t/2)/2, $$ and in particular is not $0$.

Case II above should correspond to zeros of $\zeta^\prime(s)$ with $\sigma>1$, or consecutive zeros of $\zeta$ with no zero of $\zeta^\prime$ in between (for otherwise the horizontal line $t=\text{const.}$ should cross a level curve $\arg=\pm \pi/2$.) This latter phenomenon happens at intervals on average of $2\pi/\log(2)\approx 9.064$, due to the different asymptotics of the zeros of $\zeta$ and $\zeta^\prime$.

Final edit: the graphic below shows $t$ on the vertical axis, and, on the horizontal axis, the Hardy function $Z(t)$ (in purple) and the value of $\sigma$ which minimizes $f(t)$. Also marked on the vertical axis are the zeros of $Z(t)$ (in black). The $t$ which correspond to zeros of $\zeta^\prime(s)$ are marked in red when $1/2<\sigma<1$, and in green when $\sigma\ge1$. [Thanks to Ricky Farr for computation of zeros of $\zeta^\prime$.] The data illustrates that minima occuring at $\sigma=1$ tend to be associated to consecutive zeros of $\zeta$ with either no intervening zero of $\zeta^\prime$, or a zero of $\zeta^\prime$ with $\sigma\ge 1$.

    alt text (source)

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  • $\begingroup$ I'll make this a comment since it's speculative. I think the $t$ of type II above will be associated to large gaps between zeros, and large values of $|\zeta(1/2+it)|$. Titchmarsh Theorem 8.12 specializes to say that for any $\alpha<1/2$, there exist arbitrarily large values of $t$ for which $|\zeta(1/2+it)|>\exp(\log(t)^\alpha))$. Now for those same $t$, $|\zeta(1+it)|$ will be smaller, but I don't think it's plausible that these values will tend to $0$. $\endgroup$ – Stopple Aug 27 '14 at 19:47
  • $\begingroup$ Somehow I didn't know you had given this answer until just now. On first reading I think it will take me a while to digest, but for the moment I would like to thank you. $\endgroup$ – Kevin Smith Sep 23 '14 at 21:35
  • $\begingroup$ I can email you the graphics if you want a closer look. $\endgroup$ – Stopple Sep 23 '14 at 22:01
  • $\begingroup$ Please do- my email address is k.p.q.smith@gmail.com. $\endgroup$ – Kevin Smith Sep 24 '14 at 5:26

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