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

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

2010-09-23T04:53:58Z

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.)

Answer by H A Helfgott for Non-vanishing of zeta(s), Re(s)=1, without complex analysis?

2010-09-23T09:21:34Z

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.

Answer by David Speyer for Non-vanishing of zeta(s), Re(s)=1, without complex analysis?

2010-09-23T11:35:39Z

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 $C^2$.

Answer by M Mueger for Non-vanishing of zeta(s), Re(s)=1, without complex analysis?

2010-11-17T15:27:27Z

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.)

Answer by M Mueger for Non-vanishing of zeta(s), Re(s)=1, without complex analysis?

2010-11-18T09:27:56Z

@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.