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My lecturer wrote that there exists $c>0$ such that if $\sigma\geq 1-c/\log(|t|+4)$ then $\zeta(s)\neq 0$. He explained the proof from p. 173. I understood most of it. But there are some points that I do not understand. I had no chance of asking him some questions. What I do not understand is,

  • What is the goal of the proof? I understand that $\zeta(s)\neq 0$ for $\sigma>1$. I suppose the proof starts with looking at the region $0<\sigma\leq 1$. Is the goal something about the behaviour when letting $\delta\to 0^+$ in the inequality $$\Re\left ( -3\frac{\zeta(1+\delta)}{\zeta(1+\delta)} \right ) +\Re\left ( -4\frac{\zeta(1+\delta+i\gamma_0)}{\zeta(1+\delta+i\gamma_0)} \right )+\Re\left ( -\frac{\zeta(1+\delta+2i\gamma_0)}{\zeta(1+\delta+2i\gamma_0)} \right )\geq 0$$ so that we can find out how far any zero $\rho_0=\beta_0+i\gamma_0$ would coincide a point $s$ horizontally from right (ie imaginary part)?
  • I understand that the proof uses some restrictions, $5/6\leq \beta_0\leq 1$ and $|\gamma_0|\geq 7/8$ for estimating the two last functions above. The bottom of the proof says $$1-\beta_0\gg \frac{1}{\log(|\gamma_0|+4)}.$$ What does this (conclusion) mean? Does it mean that if $\zeta(\rho_0)=0$ then $1-\beta_0\gg \frac{1}{\log(|\gamma_0|+4)}?$
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    You need some Fourier/harmonic analysis intuition : $\sum_{n < x} a_n = \frac{1}{2i\pi}\int_{\sigma-i\infty}^{\sigma+i \infty} F(s)\frac{x^s}{s} ds= \frac{1}{2i\pi}\int_{\sigma-iT}^{\sigma+i T} F(s)\frac{x^s}{s} ds+ \mathcal{O}(....)$ and $\sigma$ has to be in the domain of convergence, but with the residue theorem you can move it to the left, and the more you move the contour on the left, the more $x^s$ becomes small. For example [see there p.69-70](http://www.dms.umontreal.ca/~koukoulo/documents/notes/primes.pdf) – 2017-01-09
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    When you say 'the goal' are you wondering about the goal moving forward (error estimate for the prime number theorem) or are you asking about motivation for the steps in the proof? – 2017-01-09

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