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Timeline for Scaled Riemann zeta function with no zero in the critical strip

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Apr 29, 2021 at 6:04 comment added Vincent Granville @Wojowu: Assume $x, s$ are real. The product converges if $\Re(s)> 1$. Multiply it by $\prod_k(1+x/k^s)$. Now it converges if $\Re(s)>1/2$, it is an analytic continuation, and sure I added a bunch of new roots which can be proved to NOT being roots of the original $f(x,s)$. But in essence, I preserved the roots of the original function $f$. Now think about $f(s)=\prod_k(1-ix/k^s)$. Multiplying by $\prod(1+ix/k^s)$ to extend analyticity, I would kill all the roots of the original function in the process. That is why we need a theorem about root-preserving when making such manipulations.
Apr 29, 2021 at 6:03 comment added Vincent Granville @Wojowu: The example $\tau_k(s)=|1-p_k^s|$ makes the whole argumentation fails as you pointed out, and I agree. Even if that was the case for just one $k$ alone (say $k=3$), it would make my argumentation to fail. The more I think about it, including such a factor is similar to potentially introducing a $\frac{0}{0}$ in the product. You need to first test the mechanics on a product that does actually have zeroes, like $f(x,s)=\prod_k (1-x/k^s)$ (then $x=k^s$ is a root).
Apr 29, 2021 at 3:23 comment added Vincent Granville @Conrad: Regarding analytic continuation, I will say this. It two functions $a(s), b(s)$ have analytic continuation, does $a^2(s)+b^2(s)$ have analytic continuation? If there is a theorem saying so, then analytic continuation is no longer a problem in my case. But the choice of $\tau(s)$ still remains an issue.
Apr 29, 2021 at 2:57 comment added Vincent Granville @Conrad: I have very little leeway in my choice for $\tau_k(s)$. Essentially, I am trying to extend $\eta^*(s)$ from $\Re(s)>1$ to $\Re(s)>\frac{1}{2}$. Essentially, it is almost like trying to make $\prod(1-1/k^\sigma)$ extend to convergence by multiplying it by $\prod(1+1/k^\sigma)$. Both considered separately are divergent if $\sigma>1$, but when blended together in the right way, it converges if $\sigma>1/2$. This is just an analogy. Replace $\prod(1+1/k^\sigma)$ by $\prod(1+ (1+2^{-1000000})/k^\sigma)$, and convergence fails. This is how much lille leeway I have.
Apr 29, 2021 at 2:39 comment added Vincent Granville @Conrad: Thank you for your useful comments. I have no doubt I am very far away from proving RH, and at the same times I am trying to bring a new approach (products rather than sums or integrals), but as you said, in the end, if my function $\eta^2(s)$ is not indefinitely differentiable, (and proving it may be as hard as proving RH), then the "analytic continuation" argument fails.
Apr 29, 2021 at 1:33 comment added Conrad it is unclear what smoothness does the product have - the factors are not (complex) analytic so the usual normal convergence theorem for holomorphic products doesn't apply; the product is continuous and I could buy differentiable infinitely many times (though would need proof as infinite products are tricky in general) but not sure about any kind of analyticity
Apr 29, 2021 at 0:56 comment added Vincent Granville @Conrad: What about $1/(\eta^*)^2$? At least that one has all the square roots gone.
Apr 29, 2021 at 0:23 comment added Conrad What puzzles me is in what sense you talk about $\eta^*$ being analytic continuation because that function is not analytic
Apr 28, 2021 at 23:58 comment added Vincent Granville @Wojowu: I am working (with anyone who wants to help) on a theorem about "root-preserving scaled analytic continuations of infinite products". The example you mentioned, $\tau_k(s)=|1-p_k^{-s}|$ would not meet the requirement of that theorem. Indeed if just one of the $\tau_k(s)$ (say for $k=3$) is equal to $|1-p_k^{-s}|$, it would not meet the conditions of that theorem.
Apr 28, 2021 at 23:38 comment added Vincent Granville @Conrad: The function $\eta^*(s)$ is defined on $\mathbb{C}$ but takes only real values. For instance, if $\tau_k(s)=1$, then $\eta^*(s)=|\zeta(s)|$.
Apr 28, 2021 at 23:32 comment added Vincent Granville @Conrad: I think if any progress has to be made, it needs to be shown on a very simple function with known roots in the critical strip, that the re-scaling / analytic continuation works. In essence, you need a theorem about "root-preserving scaled analytic continuations of infinite products", showing how it works on a more basic function before applying it to RH. I will ask a question about that. Examples will be so simple that actually it will be about an analytic continuation on the real line. from $]1,\infty]$ to $\frac{1}{2},\infty]$.
Apr 28, 2021 at 4:47 vote accept Vincent Granville
Apr 27, 2021 at 22:05 history edited Vincent Granville CC BY-SA 4.0
deleted 110 characters in body
Apr 27, 2021 at 21:50 comment added Vincent Granville Yes $1-2^{-s}$ should be removed, but I've made already many edits. I will remove it later for sure.
Apr 27, 2021 at 21:30 comment added Conrad in this last form, not sure why you want the $1-2^{1-s}$ term as that changes nothing in terms of convergence etc - just adds some zeroes on $\Re s =1$; second with the choice of $\theta_k(s)=\arg(1-p_k^{-s})$ it looks like the product is one of positive numbers $\frac{\tau_k(s)}{|1-p_k^{-s}|}$, so not sure what has that to do with RZ where a lot of the subtlety is in the argument
Apr 27, 2021 at 20:59 comment added Vincent Granville Thanks, I haven't thought about that, you are right. Not sure if there are other counter-examples. I will continue to look into this. And anyway, as I said, I can't imagine myself being able to prove RH, not even in a small sub-strip for the most basic $L$-function (other than finite products, which are irrelevant)
Apr 27, 2021 at 20:44 history edited Vincent Granville CC BY-SA 4.0
Some little fixes were needed after introducing the complex argument in the definition of $\eta^*$, but it does not change the final result.
Apr 27, 2021 at 20:44 comment added Wojowu Right now you can take $\tau_k(s)$ to be the absolute value of $1-p_k^{-s}$, which would render all factors equal to $1$. The resulting product indeed has no zeros for $s\in S$, but that is not known to imply RH.
Apr 27, 2021 at 20:18 history edited Vincent Granville CC BY-SA 4.0
I modified the definition of $\eta^*(s)$ to (hopefully) address the convergence issue, as well as the analytic continuation
Apr 27, 2021 at 18:19 comment added Vincent Granville @Wojowu: my question in its current form is beyond repair. I may try one last thing, but I hesitate between deleting this question / writing a new one, or updating the question. What would you suggest?
Apr 27, 2021 at 14:12 history edited Vincent Granville CC BY-SA 4.0
edited body
Apr 27, 2021 at 14:10 comment added Vincent Granville @Conrad: I am working on changing $\zeta^*$. Right now it has become $\eta^*$ but more changes are needed.
Apr 27, 2021 at 14:03 comment added Conrad regarding the conclusion, in what sense is $\zeta^*$ analytic? definitely not complex analytic even for $\Re s >1$ as the ratio of two analytic functions cannot be real nonconstant;
Apr 27, 2021 at 14:00 history edited Vincent Granville CC BY-SA 4.0
I replaced $\zeta^*$ (based on $\zeta$) by $\eta^*$ (based on $\eta$). Further changes will be required, but this is needed first to go in the right direction.
Apr 27, 2021 at 11:59 answer added Wojowu timeline score: 2
Apr 27, 2021 at 11:50 comment added Vincent Granville Sorry my previous reply was wrong, I deleted it. Actually, the larger $\sigma>1$, the faster the product for $\tau(s)$ will converge. If $\sigma\leq 1$ that product diverges. One problem I am interested in is computing $\zeta^*(1)$. We have $\eta(1)=\log 2$, and I am hoping I can get an exact known math constant for $\zeta^*(1)$ as well. I also need things to work out for $\sigma>1$ in order to be able to claim (scaled) "analytic continuation".
Apr 27, 2021 at 11:31 comment added Wojowu You only demand that $\tau_k(s)$ be defined if $s\in S$. If $\sigma>1$ then $s\not\in S$, so $\tau(s)$ won't converge, nor even need $\tau_k(s)$ be defined
Apr 27, 2021 at 11:15 comment added Vincent Granville @Wojowu: you need $\sigma>1$ for $\tau(s)$ to converge. If $\sigma\leq 1$, each $\tau_k(s)$ must be attached to its "sister" factor $(1-p_k^{-s})^{-1}$ to make $\zeta^*(s)$ to converge. You can't separate them if $\sigma\leq 1$. This is the same as saying you can separate $1-2^{1-s}$ from $1^{-s}+2^{-s}+3^{-s}+\cdots$ in the $\eta$ function. You can do it if $\sigma>1$, not if $\sigma\leq 1$.
Apr 27, 2021 at 10:48 comment added Wojowu From what I can see, your conditions on $\tau_k$ do not imply that it $\tau(s)$ converges anywhere, nor that it is even defined for $\Re(s)>1$ (you only ask for $\tau_k$ to be defined on $S$). Even if defined, $\tau$ cannot be analytic unless it is constant, given your conditions imply it is always real.
Apr 27, 2021 at 10:33 comment added Vincent Granville @Conrad: I added a conclusion to my question, hope it helps.
Apr 27, 2021 at 10:21 history edited Vincent Granville CC BY-SA 4.0
I added a conclusion section at the bottom
Apr 27, 2021 at 9:59 history edited Vincent Granville CC BY-SA 4.0
added 178 characters in body
Apr 27, 2021 at 7:27 comment added Vincent Granville @Conrad: I am not sure myself either, thus my question. I will update this post, maybe it will become more clear what the relation is between $\zeta$ and $\zeta^*$.
Apr 27, 2021 at 2:35 comment added Conrad not sure why you think that the existence of $\zeta^*$ has anything to do with the behavior of $\zeta$ in the strip $1/2 < \sigma \le 1$, its zeroes, RH etc;
Apr 26, 2021 at 19:31 comment added Vincent Granville Of course, I know $\zeta$ has infinitely many zeroes on the critical line. My point is that the above product for $\zeta^*(s)$ is not defined for $\Re(s)=\frac{1}{2}$.
Apr 26, 2021 at 19:13 history asked Vincent Granville CC BY-SA 4.0
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