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I am looking for a short proof of the fact that $\zeta(z)\neq 0$ for $\Re z>a$ implies the prime number theorem with an error bound $O(x^{a+\varepsilon})$ for any $\varepsilon>0$, which would be as elementary in spirit as Newman's proof of the PNT.

"Morally", it follows from the Mellin transform formula $$ \psi(x)=-\frac{1}{2\pi i}\int_{\Re s=\sigma}\frac{\zeta'(s)}{\zeta(s)}x^s \frac{ds}{s}=-\frac{x^\sigma}{2\pi}\int_{\mathbb{R}}\frac{\zeta'(\sigma+it)}{\zeta(\sigma+it)}e^{it\log x}\frac{dt}{\sigma+it}, $$

in which one then shifts the line of integration from $\sigma>1$ to $\sigma=a+\varepsilon$, picking up a pole at $s=1$ in the process. However, to rigorously do such shifting one seems to need (at least) estimates of $\zeta'/\zeta$ on the horizontal sides of the rectangle, which I don't see immediately how to obtain just from the non-vanishing of $\zeta$ and its simple properties.

The proofs I have seen in the literature (e. g. in the lecture notes of Elkies or in Edwards book) seem to use the Hadamard product representation of $\zeta$, the related series representation for $\zeta'/\zeta$, and the zero density estimates, and therefore, eventually, the functional equation.

Does an elementary proof of this fact (and, in particular, of the implication between the two forms of Riemann hypothesis) exist? Or is it known to be impossible, in the sense that it genuinely depends on the functional equation? E. g., are there convincing examples where the above "contour shifting" fails?

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    $\begingroup$ I believe that what you are looking for (à la Newman) is Jameson's book "The Prime Number Theorem", Chapter 5, Theorem 5.1.1. In 5.2, it says that by changing the contour in 5.1.1 one could obtain bounds of the form $O(x^{a})$. Section 5.3 seems to be a way to connect the estimates of $\frac{\zeta'}{\zeta}$ with the non-vanishing of $\zeta$ in a elementary way, but I haven't been able to follow the details. It's too difficult for me. $\endgroup$ Commented Aug 10, 2021 at 17:49
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    $\begingroup$ @rgvalenciaalbornoz, thanks for the reference. Unfortunately, the estimates he gives are the standard estimates near the line $\Re s = 1$, which don't really help to push the contour to $\Re s = a+\varepsilon$. $\endgroup$
    – Kostya_I
    Commented Aug 11, 2021 at 12:31

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The "pretentious" Riemann hypothesis (RH) (which is in fact equivalent to the classical formulation of the RH) states that for all $\epsilon>0$, there exists a constant $c_{\epsilon}>0$ such that for every integer $k\geq 1$, we have

$\Big|\Big(\frac{\zeta'(s)}{\zeta(s)}+\frac{1}{s-1}\Big)^{(k)}\Big|\leq c_{\epsilon}k! 2^k (1+t^{\epsilon})$

uniformly for $s=\sigma+it$ with $1\leq\sigma<2$ and $0\leq t\leq e^k$ (see Section 17 of Granville). Using work of Koukoulopoulos, one can show that if the pretentious RH is true, then the error term in the prime number theorem is $O_{\epsilon}(x^{1/2+\epsilon})$ for all fixed $\epsilon>0$. I'm guessing that if $2^k$ is replaced by $\alpha^{-k}$, then the resulting pretentious quasi-RH suffices to obtain $O_{\epsilon}(x^{\alpha+\epsilon})$. Regardless of the value of $\alpha$, the transition from one statement does not appear to touch the functional equation. Whether you find the proof to be "in the spirit of Newman" is highly subjective and dependent on your personal goals; you should read Koukoulopoulos's beautiful paper and determine that for yourself. They are both largely "elementary" (depending on where you stand on the issue of Fourier/Mellin/Laplace inversion), but Newman uses (what I see as) very ad hoc analytic devices whereas Koukoulopoulos uses (what I see as) highly motivated sieve methods.

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  • $\begingroup$ Thanks for the reference. I don't see how to derive the pretentious RH from RH easily. From what Granville writes I surmise that apparently, one estimates the $k$-th derivative by a contour integral, which requires a bound of $|\zeta'(s)/\zeta(s)+\zeta(s)|$ in the critical strip. How does one obtain such a bound without tools like expansions over zeta zeros? $\endgroup$
    – Kostya_I
    Commented Aug 10, 2021 at 12:57
  • $\begingroup$ I don't think that one can. Honestly, without the fact that $\beta+i\gamma$ is a nontrivial zero if and only if $1-\beta+i\gamma$ is too, then I don't see how you could even see that one must have $\alpha\geq 1/2$. For this part of the problem, I think that the zeros / functional equation are pretty crucial. You can "cheat" near the 1-line because there are forces (like nonnegativity of the Dirichlet coefficients of $\zeta(s)$) that conspire to strongly repel zeros from the 1-line. Their impact is greatly diminished far from the 1-line, so you'll need deeper tools (e.g., the FE). $\endgroup$
    – 2734364041
    Commented Aug 10, 2021 at 14:19

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