Non-vanishing of zeta(s), Re(s)=1, without complex analysis? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T18:54:08Z http://mathoverflow.net/feeds/question/39689 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/39689/non-vanishing-of-zetas-res1-without-complex-analysis Non-vanishing of zeta(s), Re(s)=1, without complex analysis? H A Helfgott 2010-09-23T04:53:58Z 2010-11-23T11:20:13Z <p>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 $f$.)</p> <p>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.</p> <p>(Motivation: of course, I am trying to present a proof of the prime number theorem with plenty of analytic ideas but no complex analysis.)</p> http://mathoverflow.net/questions/39689/non-vanishing-of-zetas-res1-without-complex-analysis/39719#39719 Answer by H A Helfgott for Non-vanishing of zeta(s), Re(s)=1, without complex analysis? H A Helfgott 2010-09-23T09:21:34Z 2010-11-17T15:30:19Z <p>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.</p> http://mathoverflow.net/questions/39689/non-vanishing-of-zetas-res1-without-complex-analysis/39728#39728 Answer by David Speyer for Non-vanishing of zeta(s), Re(s)=1, without complex analysis? David Speyer 2010-09-23T11:35:39Z 2010-11-17T17:26:29Z <p>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 <a href="http://en.wikipedia.org/wiki/Taylor%2527s_theorem" rel="nofollow">Taylor's theorem</a>, that will follow if you know that $\zeta(\sigma+i t)$ is a $C^2$ function.</p> <p>Write <code>$$\zeta(s) = \frac{1}{s-1} + \sum_{n=1}^{\infty} \left( \frac{1}{n^s} - \frac{1}{(s-1) n^{s-1}} + \frac{1}{(s-1)(n+1)^{s-1}} \right).$$</code> 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.</p> <p>Differentiating term by term gives you a series which converges uniformly on compact subsets of <code>$\{ \sigma+it : \sigma&gt;0, \ \sigma+it \neq 1 \}$</code> 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 $C^2$.</p> http://mathoverflow.net/questions/39689/non-vanishing-of-zetas-res1-without-complex-analysis/46369#46369 Answer by M Mueger for Non-vanishing of zeta(s), Re(s)=1, without complex analysis? M Mueger 2010-11-17T15:27:27Z 2010-11-17T15:27:27Z <p>You might want to look at the paper "Le théorème des nombres premiers et la transformation de Fourier" by Jean-Bonoit Bost, available at <a href="http://www.math.polytechnique.fr/xups/xups02-01.pdf" rel="nofollow">http://www.math.polytechnique.fr/xups/xups02-01.pdf</a> 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.</p> <p>(I'm sure you know that the brevity of Don Zagier's proof of PNT can't be beaten, provided you accept complex analysis.)</p> http://mathoverflow.net/questions/39689/non-vanishing-of-zetas-res1-without-complex-analysis/46466#46466 Answer by M Mueger for Non-vanishing of zeta(s), Re(s)=1, without complex analysis? M Mueger 2010-11-18T09:27:56Z 2010-11-18T09:27:56Z <p>@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.</p>