Based on limited numerical evidence, I suspect this conjecture.

Conjecture: Fix $ 0 \le \sigma \le \frac12$ and let $t > 0$. Between consecutive local extrema of $\Re \zeta(\sigma+i t)$ (resp. $\Im \zeta(\sigma+ it)$), there is always a zero of $\Re \zeta(\sigma+i t)$ (resp. $\Im \zeta(\sigma+ it)$).

Verification for several random $\sigma$ and $ 0 < t < 30000$ and for a few random intervals didn't show any counterexamples.

For $\sigma > \frac12$ it is false and on the other hand this appears counterintuitive to me.

Counterexamples? (please check for closely spaced zeros that might look like a single local minimum on a large plot).

Does this contradict something?

Even if it is true, a conditional proof probably will be hard yet welcome.

For Siegel $Z$ function on the critical line RH implies this for $t$ large enough.

Maybe can be generalized to $\sigma \le \frac12$.

Plot of a random interval:

  • $\begingroup$ It seems that the zeros of the function you plotted are abscissas of inflexion points. $\endgroup$ Jul 29 '13 at 11:17
  • $\begingroup$ @SylvainJULIEN I am not sure they are really inflexion points, though they are very close to them. Here is a plot of the second derivative: s11.postimg.org/woe6fbzhv/re_zeta_0_123_inflexion.png The plot legend shows the function. $\endgroup$
    – joro
    Jul 29 '13 at 12:03
  • 2
    $\begingroup$ "For $\sigma>1/2$ it is false and on the other hand this appears counterintuitive to me." This may be related to the fact that RH is equivalent to $\zeta^\prime(s)$ has no zeros in $\sigma\le 1/2$. But $\zeta^\prime(s)$ does have zeros in $\sigma>1/2$. $\endgroup$
    – Stopple
    Aug 6 '13 at 17:31
  • 1
    $\begingroup$ Note that under RH, as it has been proved for $\sigma = \frac12$ in the other thread, if ever there is a counterexample for $\sigma <\frac12$, this implies by continuity that there must be a $\sigma_0 < \frac12$ and $t$ such that $\Re \zeta(\sigma_0+i t)$ or $\Im \zeta(\sigma_0+i t)$ has a double zero. Not that this simplifies the thing, but I guess it is an interesting fact. $\endgroup$
    – Wolfgang
    May 24 '19 at 13:28

This question has been on my mind from time to time since it was posted seven years ago. Joro says "Maybe can be generalized to $\sigma\le 0$. In fact, outside the critical strip is actually easier, and that's what this post addresses.

Write $\zeta=u+iv$, $\zeta^\prime=u^\prime+iv^\prime=u_\sigma+iv_\sigma=v_t-iu_t$. Inspired by Arias-de-Reyna's paper X-ray of the Riemann zeta function, we look at the level curves $u=0$, $v=0$, $u^\prime=0$, $v^\prime=0$.

First, for $\sigma>1$ $$ u=1+\sum_{n=2}^\infty n^{-\sigma}\cos(t\log n) $$ and $$ v=-\sum_{n=2}^\infty n^{-\sigma}\sin(t\log n). $$

The sum for $u$ is bounded by $\zeta(\sigma)-1$, so for $\sigma\ge 2$ is bounded by $\pi^2/6-1<1$, which tells us that $u(t)\ne 0$ for $\sigma\ge 2$. On the other hand, $$ 2^\sigma v=-\sin(t\log 2)+o_\sigma(1), $$ so $v=0$ when $t\approx (2k\pi \pm \pi/2)/\log 2$.

Similarly, $$ 2^\sigma u_t=-\log 2\sin(t\log 2)+o_\sigma(1) $$ $$ 2^\sigma v_t=-\log 2\cos(t\log 2)+o_\sigma(1). $$

So, in a right half plane $u$ is never $0$, while the level curves for $v=0$ and $u_t=0$ are very close and approximately horizontal lines. The level curves for $v_t=0$ are also approximately horizontal line lying halfway between those of $v=0$. So in a right half plane, along a vertical line $u$ is never $0$ but alternately increasing and decreasing, while $v$ has an extreme point between consecutive zeros.

Next we consider a half plane to the left of the critical strip. We have the functional equation $$ \zeta(s)=\pi^{s-1/2}\frac{\Gamma((1-s)/2)}{\Gamma(s/2)}\zeta(1-s). $$ Now $\zeta(1-s)\approx 1$, and for bounded $\sigma$, as $t\to+\infty$, Stirling's formula shows the argument of the remaining terms is asymptotic to $-t\log(t/2\pi)-t$. In particular $\zeta(s)$ is purely imaginary ($u=0$), or real ($v=0$) along approximately horizontal lines (since the asymptotic for the argument does not depend on $\sigma$.). It's interesting to observe that because of the $\log(t/2\pi)$ factor, there are more level curves in the left half plane.

Now since $u=0$ is in particular constant approximately along a line, $-v_t=u_\sigma=0$ along a curve which is approximately this same line. And similarly with $v=0$ and $u_t=0$.

One can make this a little more precise by computing the logarithmic derivative of the functional equation, which gives

$$ \zeta^\prime(s)=\zeta(s)\cdot(\log(2\pi)+\pi/2\tan(\pi(1-s)/2)-\Gamma^\prime/\Gamma(1-s)-\zeta^\prime/\zeta(1-s)). $$ For $1-\sigma\gg 1$, $\zeta(1-s)=1+o_\sigma(1)$ is very close to real, and the Stirling's formula asymptotic of the Psi function say the same is true for the remaining terms in parenthesis. Thus for $\sigma$ in a bounded strip and $t$ large, the level curves $v_t=0$ and $-u_t=0$ are just slight perturbations of the level curves $u=0$ and $v=0$.

Thus in a left half plane sufficiently far to the left of $\sigma=0$, along a vertical line $u$ has an extreme point between pair of consecutive zeros, and the same for $v$.

Mathematica can easily plot multiple level curves with the ContourPlot command. The graphic show the level curves for $u=0$ in solid blue, $v=0$ in solid red, $u_t=0$ in dashed blue, and $v_t=0$ in dashed red. The figure shows the region $-12\le\sigma\le 13$, $0\le t\le 50$.

enter image description here

Update: One can imagine why Joro's conjecture might follow from the Riemann Hypothesis (which was assumed to prove the conjecture on the critical line in the linked question). If the conjecture were to fail, say for $u$, in the region $\sigma<1/2$, one would need a level curve $u=0$ (solid blue) and a level curve $u_t=0$ (dashed blue) to swap places. In particular, they need to cross. To the extent that the level curves $u=0$ and $v_t=0$ (dashed red) are close, there will be an intersection of $u_t=0$ and $v_t=0$ nearby. Such are the zeros of $\zeta^\prime$. On the Riemann Hypothesis these points must lie to the right of the critical line.


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