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. $\endgroup$ – Sylvain JULIEN 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
  • $\begingroup$ 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 at 13:28

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