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).

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:

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It seems that the zeros of the function you plotted are abscissas of inflexion points. –  Sylvain JULIEN Jul 29 '13 at 11:17
@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. –  joro Jul 29 '13 at 12:03
"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