I´m working on a book for my seminar in the university and I am stuck in the argumentation of one Theorem.
So the Theorem says Zeta has no zeta in $\sigma>1$.
To proof we say $\zeta(1+i\tau_0)=0$, so $\tau_0$ can´t be $0$ and $zeta$ is holomorphic in some neighbourhood of $1+i\tau_0$. Therefore
$\zeta(\sigma+i\tau_0)\ll\sigma-1~~~(\sigma>1)$
on the other hand
$\zeta(\sigma)\ll(\sigma-1)^{-1},~\zeta(\sigma+2i\tau_0)\ll 1~~~~(\sigma>1)$
the rest of the proof is quite clear for me, but where do these three estimates come from? Anyone got a clue?
We want to show that the functions $\zeta(\sigma)(\sigma-1), \frac{\zeta(\sigma + i\tau_0)}{\sigma-1}$ and $\zeta(\sigma + 2i\tau_0)$ are bounded as $\sigma \to 1^+$.
Since $\zeta(s)$ has a pole at $s=1$, the function $\zeta(s)(s-1)$ is holomorphic at $s=1$, and therefore $\zeta(\sigma)(\sigma-1)$ is bounded as $\sigma \to 1^+$.
Now $\zeta(1+ i\tau_0) = 0$, so again by holomorphicity $$\frac{\zeta(\sigma + i\tau_0)}{\sigma-1} = \frac{\zeta(\sigma + i\tau_0) - \zeta(1 + i\tau_0)}{(\sigma + i\tau_0)-(1+i\tau_0)}$$ is bounded as $\sigma \to 1^+$.
And finally $\zeta(\sigma + 2i\tau_0)$ converges to $\zeta(1+ 2i\tau_0)$ as $\sigma \to 1^+$, so it is also bounded.