# Naive conjecture about zeros and local extrema of $\Re \zeta(\sigma+i t)$ (resp. $\Im \zeta(\sigma+ it)$) for $0 \le \sigma \le \frac12$

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).

For Siegel $$Z$$ function on the critical line RH implies this for $$t$$ large enough.
Maybe can be generalized to $$\sigma \le \frac12$$.
• "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 $\sigma>1/2$. – Stopple Aug 6 '13 at 17:31
• 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. – Wolfgang May 24 at 13:28