17
$\begingroup$

The recent post("Long-standing conjectures in analysis ... often turn out to be false") prompted me to think about a question which I have not given much though before: to what extent the Riemann hypothesis (RH) may be regarded as a problem in analysis. It may actually be not as silly as it sounds.

The particular side of it I am curious about is the following. The

Theorem : $\zeta(s)\neq 0$ for $\Re s>1$.

may be considered a very weak consequence of RH. However, this statement is only trivial in the context of number theory. One may ask, is it possible to give it a "purely analytic" proof, without using Euler product and other stuff related to the primes? Apparently, it is not possible to formulate this as a precise mathematical question because all the theories used to formalize analysis contain a good deal of arithmetic, but it should be precise enough for practical purposes. (You know number theory when you see it.)

Most likely such a proof does not exist yet, but some readers may know more about the relevant things then I do. I am honestly curious because while it definitely looks tough, it may be not entirely implausible. If such a proof is found, it might give us a fresh look on the old problem.

Improve this question
$\endgroup$
9
  • 5
    $\begingroup$ Since $\zeta(s)$ is a Dirichlet series for ${\text Re}(s) > 1$ with constant term $1$, we have $\zeta(s)\rightarrow 1$ as the real part of $s$ tends to infinity. Thus it is "trivial" that $\zeta(s) \not= 0$ when the real part of $s$ is sufficiently large. Bringing this fact all the way back to ${\text Re}(s) > 1$ needs some more input about the zeta-function. For example, the finite sums $\sum_{n=1}^{N} 1/n^s$ have zeros with real part greater than $1$ when $N$ is big enough (I think $N > 30$ or $40$ might be enough). $\endgroup$
    – KConrad
    Commented Aug 2, 2018 at 10:23
  • 13
    $\begingroup$ @Gro, it was Barnes who proposed that to the student, Littlewood. $\endgroup$
    – Gerry Myerson
    Commented Aug 2, 2018 at 12:27
  • 7
    $\begingroup$ a dissenting opinion, by J. Brian Conrey: It is my belief that RH is a genuinely arithmetic question that likely will not succumb to methods of analysis. There is a growing body of evidence indicating that one needs to consider families of L-functions in order to make progress on this difficult question. If so, then number theorists are on the right track to an eventual proof of RH, but we are still lacking many of the tools. ams.org/notices/200303/fea-conrey-web.pdf $\endgroup$
    – Carlo Beenakker
    Commented Aug 2, 2018 at 12:36
  • 5
    $\begingroup$ It should be noted that the assignment of RH directly as a thesis problem was in the early 20th century, when its difficulty (particularly when British mathematics was somewhat isolated) was not fully appreciated. Similarly, in the 1930s Hasse assigned to a graduate student the meromorphic continuation and functional equation of the zeta-function of elliptic curves over $\mathbf Q$ as a thesis topic at a time when Weil was not convinced such a result was plausible. See Weil's comments in his collected works (Vol. 2, p. 529) on his paper about Jacobi sums as "Grossencharaktere". $\endgroup$
    – KConrad
    Commented Aug 2, 2018 at 12:44
  • 5
    $\begingroup$ You may wish to look at this recent paper by Banks: arxiv.org/pdf/1709.08551.pdf $\endgroup$
    – Lucia
    Commented Aug 2, 2018 at 21:03

2 Answers 2

Reset to default
10
$\begingroup$

First, let me say that a proof just of $\zeta(s) \not= 0$ for ${\rm Re}(s) > 1$ that does not in any straightforward way also yield a larger nonvanishing half-plane should not be regarded as a "fresh look" at anything as deep as RH. In fact, since the nonvanishing of $\zeta(s)$ on the line ${\rm Re}(s) = 1$ is equivalent to the Prime Number Theorem, it is not realistic that an approach to nonvanishing on ${\rm Re}(s) > 1$ alone is going to get you anything more since you don't just stumble into a new proof of the Prime Number Theorem.

With that in mind, consider the following integral representation of $\zeta(s)$ for ${\rm Re}(s) > 1$, which follows from partial summation on the Dirichlet series for $\zeta(s)$ (it is in Serre's Course in Arithmetic, for example): $$ \zeta(s) = s\int_1^\infty \frac{[x]}{x^{s+1}}\,dx = \frac{s}{s-1} + s\int_1^\infty \frac{\{x\}}{x^{s+1}}\,dx. $$

If $\zeta(s) = 0$ at an $s$ with ${\rm Re}(s) > 1$, then by the above formula $$ \frac{1}{1-s} = \int_1^\infty \frac{\{x\}}{x^{s+1}}\,dx. $$ Taking absolute values of both sides and using $|\{x\}|<1$ and writing $s=\sigma + it$, $$ \frac{1}{|1-s|} < \frac{1}{\sigma}, $$ so $\sigma < |1-s|$. Squaring both sides, $\sigma^2 < (1-\sigma)^2 + t^2$, which is the same as $\sigma < (1+t^2)/2$. Combining this with the restriction $1 < \sigma$ puts all $s$ with ${\rm Re}(s) > 1$ and $\zeta(s) = 0$ in a restricted range in the half-plane ${\rm Re}(s) > 1$, but not into the empty set. Perhaps by not using something as simple as $|\{x\}| < 1$ in the estimate on the integral someone can restrict the range of possible $s$ even further.

Improve this answer
$\endgroup$
4
  • $\begingroup$ I think we can prove the result for the whole half plane $\Re(s) > 1$ by noting that $\int_1^{\infty} \frac{\{x \}}{x^{s+1}} {\rm d} x < \frac{1}{2} \cdot \int_1^{\infty} \frac{1}{x^{s+1}} {\rm d} x$. To see this, write $\int_n^{n+1} \frac{\{ x\}}{x^{s+1}} {\rm d} x = \int_{0}^{1/2}\Big( \frac{u}{(n+u)^{s+1}} + \frac{1-u}{(n+1-u)^{s+1}} \Big) {\rm d} u \\= \int_{0}^{1/2}\Big( \frac{1}{(n+1-u)^{s+1}} + u \cdot \Big( \frac{1}{(n+u)^{s+1}} - \frac{1}{(n+1-u)^{s+1}} \Big)\Big) {\rm d} u < \frac{1}{2}\int_{0}^{1/2}\Big(\frac{1}{(n+1-u)^{s+1}} + \frac{1}{(n+u)^{s+1}} \Big) {\rm d} u$. $\endgroup$
    – Noah Stephens-Davidowitz
    Commented Aug 2, 2018 at 23:37
  • 3
    $\begingroup$ When $s$ is complex it does not make sense to have an inequality of such integrals. $\endgroup$
    – KConrad
    Commented Aug 3, 2018 at 4:15
  • $\begingroup$ Oh, of course. Sorry for the stupidity. $\endgroup$
    – Noah Stephens-Davidowitz
    Commented Aug 3, 2018 at 5:23
  • 2
    $\begingroup$ Replacing the bound $|\{ x \}|<1$ with $|\{ x \} - 1/2 | < 1/2$ pushes the zero free region out further, but still doesn't do it. Does anyone want to try some higher Bernoulli polynomials and see what happens? $\endgroup$
    – David E Speyer
    Commented Aug 8, 2018 at 16:49
10
$\begingroup$

Andrew Booker is of the opinion that it is the nonvanishing of zeta (or an L-function) to the right of the critical strip which is more fundamental than the Euler product. See Slide 10 of his recent talk https://heilbronn.ac.uk/wp-content/uploads/2018/07/Booker-talk.pdf

Improve this answer
$\endgroup$
7
  • 3
    $\begingroup$ I did enjoy the uses of xkcd 927 and xkcd 483, but (being a non-specialist, and hence rather lost towards the end) I half expected to see xkcd 349 ... $\endgroup$
    – Yemon Choi
    Commented Aug 8, 2018 at 16:31
  • $\begingroup$ This $L$-datum business seems like an impossible way to define anything, even the Riemann zeta-function, since the third ingredient $m(z)$ in an $L$-datum can't be described in any direct way. $\endgroup$
    – KConrad
    Commented Aug 8, 2018 at 23:01
  • 2
    $\begingroup$ For reference, the conference page is heilbronn.ac.uk/2017/08/08/…, where apparently videos will be uploaded at some point. $\endgroup$
    – David Roberts ♦
    Commented Aug 9, 2018 at 0:59
  • $\begingroup$ @KConrad and it looks as if $m$ can be an arbitrary function? Certainly it must be discontinuous. $\endgroup$
    – David Roberts ♦
    Commented Aug 9, 2018 at 1:00
  • $\begingroup$ This may be interesting (and other talks too). Waiting for videos. $\endgroup$
    – Alex Gavrilov
    Commented Aug 9, 2018 at 10:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .