# Non-vanishing of zeta(s), Re(s)=1, without complex analysis?

Say you are allowed to use Fourier analysis, complex variables, Euler-Maclaurin, etc., but no complex analysis - no holomorphic continuations, no definition of analytic function, and, in particular, no recourse to the concept that every analytic function $f(s)$ vanishing at $s_0$ behaves like $(s-s_0)$ near $s_0$. (You are still allowed to use that fact for a specific function $f$, if you can prove it for that function $f$.)

How would you prove that $lim_{\sigma->1^+} \zeta(\sigma+it) \ne 0$ ($t$ real and fixed)? All the proofs I know (with or without explicit recourse to $\zeta^3(\sigma) |\zeta(\sigma+it)|^4 |\zeta(\sigma+2it)|\geq 0$ or the like) use the fact that, if the limit were 0, then $\zeta(\sigma+it)\sim (\sigma+it-1)$ for $\sigma$ near 1.

(Motivation: of course, I am trying to present a proof of the prime number theorem with plenty of analytic ideas but no complex analysis.)

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Given your motivation, maybe you just want to present the Erdos-Selberg elementary proof of the Prime Number Theorem. There's an exposition in Tenenbaum and Mendes France, The Prime Numbers and Their Distribution, published by the Amer Math Soc. –  Gerry Myerson Sep 23 '10 at 5:01
Actually, no, I've already decided not to give that. I'm giving a proof via Wiener-Ikehara - it's quite natural, and it introduces many nice ideas from Fourier analysis without using complex analysis. This seems to be the one sticking point. –  H A Helfgott Sep 23 '10 at 5:05

It seems to me that you don't need to know that $\zeta(\sigma + it)$ is proportional to $(\sigma+it-1)^k$ for some $k$, you just need to know that, if it vanishes, then it is $O(\sigma+it-1)$. By Taylor's theorem, that will follow if you know that $\zeta(\sigma+i t)$ is a $C^2$ function.

Write $$\zeta(s) = \frac{1}{s-1} + \sum_{n=1}^{\infty} \left( \frac{1}{n^s} - \frac{1}{(s-1) n^{s-1}} + \frac{1}{(s-1)(n+1)^{s-1}} \right).$$ For $\sigma>1$, this is equal to the standard $\zeta$ by easy manipulations with absolutely convergent series; for $\sigma \in (0,1]$ you can take it as the definition of $\zeta$. Since you don't have analytic continuation, you don't know that this is the "best" extension of $\zeta$ to the critical strip, but that's OK.

Differentiating term by term gives you a series which converges uniformly on compact subsets of $\{ \sigma+it : \sigma>0, \ \sigma+it \neq 1 \}$ Hence $\zeta'$ is represented on this domain by the derivative of this series. The same thing happens when you differentiate again. So $\zeta''$, being the uniformly convergent sum of continuous functions, is continuous and $\zeta$ is $C^2$.

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If nobody has a better idea, I will simply get a (real-variable) Taylor series for $\zeta(\sigma+it)$ up to second-order with remainder. This is just (real) calculus - one can easily get the continuous continuation of $\zeta$, $\zeta'$ and $\zeta''$ up to $Re(s)=1$ by Euler-Maclaurin. Perhaps not ideal, but not horrible either.

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Oh, I guess this is exactly what I wrote in my answer. I didn't see this when I started writing; sorry! –  David Speyer Sep 23 '10 at 14:44
Yes, but it's nice we agree. –  H A Helfgott Sep 24 '10 at 2:43

You might want to look at the paper "Le théorème des nombres premiers et la transformation de Fourier" by Jean-Bonoit Bost, available at http://www.math.polytechnique.fr/xups/xups02-01.pdf It gives a proof of PNT freely using harmonic analysis and some basics of distributions, but as little complex analysis as possible. In particular, section 4 proves a statement that is equivalent to the non-vanishing of zeta(1+it) using only real analysis. Unfortunately, for my taste there still is too much complex reasoning in the preceding sections of the paper.

(I'm sure you know that the brevity of Don Zagier's proof of PNT can't be beaten, provided you accept complex analysis.)

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Typo in author's name –  Yemon Choi Nov 17 '10 at 17:44
Bost, Jean-Benoît –  M Mueger Nov 18 '10 at 9:28

@Marc: Rudin gives a proof of Ingham's theorem, which is a real tauberian theorem and not to be confused with Ikehara-Wiener, which is complex.

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Thanks for the correction! –  Marc Palm Nov 18 '10 at 13:12