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Prove that for $1<\sigma_1 \leq \sigma_2$ is true that $$\frac{\zeta(\sigma_2)}{\zeta(\sigma_1)} \leq \left|\frac{\zeta(\sigma_2+it)}{\zeta(\sigma_1+it)}\right|\leq \frac{\zeta(\sigma_1)}{\zeta(\sigma_2)}$$

uniform for $t \in \mathbb{R}.$

Any ideas?

Thanks.

EDIT: Notation: We assume that $s=\sigma + it$, where $\sigma, t\in \mathbb{R}$.

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You do know that the complex numbers do not form an ordered field, right? Are $\sigma_1,\sigma_2$ real numbers? –  Asaf Karagila Jan 9 '13 at 12:37
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up vote 2 down vote accepted

Ok, I think this works. The key fact is that $\zeta(s)$ is non zero in the half plane, so there is an analytic branch of the logarithm. Then with $s_j=\sigma_j+i t$ we have that $$ \log|\zeta(s_2)|-\log|\zeta(s_1)|=\text{Re}\left(\int_{s_1}^{s_2}\frac{\zeta^\prime}{\zeta}(z)\, dz\right). $$ Since $s_1$ and $s_2$ have the same imaginary part, the path of integration is a horizontal line and with $dz=dx+idy$, we have $dy=0$. The real part of the integral then reduces to $$ \int_{\sigma_1}^{\sigma_2}\text{Re}\left(\frac{\zeta^\prime}{\zeta}(z)\right)\, dx $$ With $\zeta^\prime/\zeta(z)=\sum_n-\Lambda(n)n^{-z}$, we have the real part is $\le$ the absolute value $\sum_n\Lambda(n)n^{-\sigma}=-\zeta^\prime/\zeta(\sigma)$. Put this in the integral to see that our expression is $$ \le-\int_{\sigma_1}^{\sigma_2}\frac{\zeta^\prime}{\zeta}(\sigma)\, d\sigma=\log(\zeta(\sigma_1))-\log(\zeta(\sigma_2)). $$ Exponentiating gives one inequality. For the other, replace "the real part is $\le$ the absolute value" with "the negative of the real part is $\le$ the absolute value".

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Thank you @stopple, really nice solution. –  Cortizol Jan 10 '13 at 14:49
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