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So I was reading Hadamard's 1896 paper, here (fr), which is the most celebrated proof of the fact that there are no zeros of the zeta function with real part 1 (and therefore of the prime number theorem). My French is a little shoddy - that may be the problem, but I don't think so. The first four articles are dedicated to proving the lack of zeros, the next six to extending the proof to related functions, and the remainder to using it to demonstrate the PNT and some corollaries. It's article 3 in particular I'm confused by - it seems his argument goes like this:

  • Suppose that $1 + ti$, $t$ real, is a zero of the zeta function.
  • For some $\alpha < \frac{\pi}{2}$, consider the primes for which $t\log p$ is within $\alpha$ of an odd multiple of $\pi$; call them $q$.
  • By some fairly simple algebra (this part I did follow), if $\rho = \limsup_{s\to1}\frac{\sum_q q^s}{\sum_p p^s} < 1$ then $\Re[\log\zeta(1+\epsilon+it)] \ge -(\rho + (1 - \rho)\cos\alpha)\log\zeta(1+\epsilon)$, $\epsilon$ being as always a sufficiently small positive real.

Then comes my problem: Hadamard writes "...ce qui serait en contradiction avec l'hypothèse $\zeta(1+ti) = 0$, ainsi qu'il a été remarqué au numéro précédent," which I understand to mean "...which would be in contradiction to the hypothesis $\zeta(1+ti) = 0$, as was noted in the previous article (paragraph? 'numéro')."

(Article 4 then goes on to show that the limit being 1 would mean a pole at $1+2ti$, and that part was fine.)

Article 2 does mention that if there were a zero there, the approach parallel to the real axis would approach similarly to $\log(s-1)$, and that makes sense, but it seems all that's established here is a constant multiple, and that's allowed, isn't it? What am I missing?

To be clear, I'm not asking for a proof of this fact; I know an easier proof, but due to the place it holds in history, I'm looking to understand this one.

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There seems to be an imprecision in Hadamard's writing. In paragraph 2, he writes that if $\zeta$ had a zero at $1 + ti$, then for $s > 1$, we'd have

$$\operatorname{Re} \log \zeta(s+ti) \sim \log (s-1) \sim -\operatorname{Re} \log \zeta(s),$$

if I interpret it correctly. That is only the case for simple zeros. Generally, if $\zeta$ had a zero of multiplicity $\mu$ at $1 + ti$, writing

$$\zeta(z) = (z - 1 - ti)^{\mu}\cdot g(z)$$

with a holomorphic $g$ that doesn't vanish in a neighbourhood of $1+ti$, we find

$$\operatorname{Re} \log \zeta(s + ti) = \mu \log (s-1) + \log \lvert g(s+ti)\rvert,\tag{$\ast$}$$

where $\log \lvert g(z)\rvert$ is bounded in a neighbourhood of $1 + ti$. So

but it seems all that's established here is a constant multiple

is right. Apart from bounded parts, the behaviour is that of a constant multiple of $\log (s-1)$.

But the constant factor isn't arbitrary, it's a positive integer. Hence it is $\geqslant 1$. And in paragraph 3, Hadamard shows that if $\rho < 1$, then $\operatorname{Re} \log \zeta(s + ti)$ tends to $-\infty$ at most as fast as $\theta\log (s-1)$, where $0 < \theta < 1$. And that contradicts $(\ast)$.

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  • $\begingroup$ So it is impossible for the coefficient to be other than the multiplicity? I thought that might be it, but it seemed so wrong. Now that you've said so explicitly, though, it makes a lot of sense. $\endgroup$ – user361424 Dec 30 '16 at 21:59
  • $\begingroup$ Yes, the coefficient is the multiplicity. By the way, what I didn't think of when I wrote the answer is that the main part, $\log \zeta(z) \approx \sum p^{-z}$ (where $\approx$ means the difference remains bounded as $\operatorname{Re} z \to 1$), shows that a zero on the line $\operatorname{Re} z = 1$ cannot have multiplicity $> 1$, so calling it an imprecision in Hadamard's writing is maybe overly harsh. He just didn't explicitly mention this implication. $\endgroup$ – Daniel Fischer Dec 30 '16 at 22:11

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