# Inequality for the modulus of Riemann zeta on horizontal lines and alleged partial result of Maple

According to a conjecture p.4 $$|\zeta(\frac12 -\Delta + it))| > |\zeta(\frac12 + \Delta + i t|$$ for $$0 < \Delta < \frac12$$ and $$|t| > 2 \pi +1$$. Since $$\zeta(\overline{s}) = \overline{\zeta(s)}$$, this is equvalent to $$|\zeta(s)| > |\zeta(1-s)|$$ for $$0 < \sigma < \frac12$$ and $$t$$ large enough.

Set $$F(s) = {\frac {\Gamma \left( 1/2-1/2\,s \right) {\pi }^{-1/2+1/2\,s}{\pi }^{ 1/2\,s}}{\Gamma \left( 1/2\,s \right) }}$$

Then from the functional equation $$\frac{\zeta(s)}{\zeta(1-s)} = F(s)$$ and

$$\frac{|\zeta(s)|}{|\zeta(1-s)|} = |F(s)|$$

$$|F(s)| > 1$$ for $$0 < \sigma < \frac12$$ and $$t$$ large enough would imply the conjecture unless $$s$$ is a zero of zeta off the critical line.

In Maple 13 using with(MultiSeries); and assuming $$0 < \sigma < 1/2$$ we get:

$$\lim_{t \to \infty} |F(\sigma+ it)| = \infty , \, 0 < \sigma < 1/2$$

Looks like if Maple's result is correct this would mean the conjecture is true at infinity, unless $$s$$ is a zero off the critical line.

Is Maple's result true?

Proof that $$|F(s)| > 1$$ for $$t$$ large enough (except at zeros)?

For fixed $t>12$, let us consider for $0\leq\sigma\leq \frac{1}{2}$ the function

$$G(\sigma):=|\pi^{-(\sigma-it)/2}\Gamma(\sigma+it)|^2 = \pi^{-\sigma}|\Gamma(\sigma+it)|^2.$$

Following the accepted answer here, we see that

$$\frac{d}{d\sigma}\log G(\sigma)=-\sigma\log\pi + \psi(\sigma+it) + \psi(\sigma-it)$$

$$\geq -\frac{1}{2}\log\pi + 2(1 - \gamma) - \sum_{n=1}^{\infty} \frac{2}{n^2 + t^2} > 0.27-\frac{\pi}{t}>0,$$

whence $G(\sigma)$ is increasing on $[0,\frac{1}{2}]$.

It follows that $|F(s)|>1$ for $0<\Re(s)<\frac{1}{2}$ and $|\Im s|>24$.

• Thanks GH. Your answer means the conjecture about the inequality is true in the relevant ranges except at hypothetical zeros off the critical line? – joro Jul 16 '13 at 5:03
• @joro: Yes, I think so. – GH from MO Jul 16 '13 at 16:51

The function you are calling $F(s)$ is often called $\chi(s)$ in the theory of the Riemann zeta-function (e.g. in Titchmarsh's book The Theory of the Riemann Zeta-Function). Standard asymptotic estimates for the gamma function imply that $$\chi(s) = \Big( \frac{2\pi}{t} \Big)^{\sigma+it-1/2} e^{i(t+\pi/4)} \Big\{ 1 + O\Big(\frac{1}{t}\Big) \Big\}$$ in any fixed strip $\alpha \le \sigma \le \beta$ as $t \to \infty$ (for $s=\sigma+it$), which also answers your question -- though not as precisely as in GH's answer.
• Actually this formula does not tell us what happens when $\sigma$ is very close to $\frac{1}{2}$, this is why I used the approach above. – GH from MO Jul 15 '13 at 15:49