Harald Andrés Helfgott: Optimal bounds for sums of arithmetic functions (NTWS 294)

I'm going to talk about a really basic problem, that of summing arithmetic functions. So, of course there's a formal definition, which is that an arithmetic function is just a function on the positive integers to the complex numbers, or equivalently a sequence. That's too general, then we have the definition that we actually like, namely that it's a function of that kind that is interesting to us as number theorists. Well, that's not really a definition, that's a social definition. Something in between the two might be that an arithmetic function is really a sequence a n such that the Dirichlet series A of S has meromorphic continuation. If not to the entire plane, then up to a sensible region. But let's let's say to the to the entire plane to simplify. All right. So, two examples of arithmetic functions, two examples that we could call paradigmatic without being too pretentious, they really serve as examples of subfamilies are a n equal to lambda of n, so the von Mangoldt function, which is basically the characteristic function of the primes times log p. And there the Dirichlet series would just be minus zeta prime over zeta. And then you also have um the Möbius function, which as we know is just minus one if you have an odd number of prime divisors and one if you have an even number of prime divisors. And then um A of S is one over zeta of S. Right. So, the the issue in all of these is where the poles and what are the residues. So, in both cases we know a fair deal because these functions have poles precisely where zeta has zeros. And well, in the first case also where zeta has a pole at one. Um and I have drawn I mean we all hope I have drawn the zeros of zeta in green because green is the color of hope. We hope the Riemann hypothesis is true. But we have two kinds of partial results towards the Riemann hypothesis, namely we can check the Riemann hypothesis rigorously after certain given finite height T. So that I can draw in blue. The zeros are really on the line real part equal to 1/2. Or we can also show that zero free region holds. That you know, you have some sort of narrow ever narrowing region close to real part equal to one. And um I no longer have display problems. Very good. Um And there we are guaranteed not to have zeros. I will really focus here on the first kind of partial results, namely let's say that we have verified the Riemann hypothesis up to height T as we have. What can we deduce then? So I'm having a little computer problem here. Yes. Excuse me, one moment. Yes, hopefully this will work. Very well. So the idea is always how to translate information on the poles of the Dirichlet series to get us estimates on the partial sums of a n. Because many of our problems in analytic number theory reduce to having estimates on these sums. And in this talk I will focus on explicit estimates. Well, as we will see if you really just assume the Riemann hypothesis up to height T, then that's the sort of estimate you're interested in. Um so, on the one hand, you have So, but let us talk about the non-explicit regime first, the regime that most of us are used to working most of the time. Then, um psi of x is just the sum of lambda of n up to x. And then, the fact that psi of x is x plus little o of x is just a prime number theorem. Then, on the other hand, so the Mertens function, big M of x, is the sum of mu of n up to x for n up to x. And there, uh you have the statement that M of x is little o of x. That's morally morally equivalent to a prime number theorem. It's also really equivalent in at least two ways. So, there is an elementary, though non-trivial, proof of equivalence between the two that is almost as old as the prime number theorem. Um and it's also the case, as we know since uh Vinogradov, that both of these statements are really equivalent. Certainly, the fact that this statement on the left is equivalent to zeta of s having no zeros on the line real part equal to one. Uh that's Vinogradov. Now, so since the statement on the zeros of zeta is equivalent to the prime number theorem, and it's also equivalent to M of x being little o of x, the two statements are equivalent to each other. Besides the fact that we also already know that by elementary means. So, uh the world is beautiful at this level, and we are used because of this, we are used to thinking of the problem of estimating psi of x and the problem of estimating M of x as being basically be same. But here comes the interesting part that when you try sit out of you know, you spend your entire life your your education consists in part of being told that they explicit estimates are not interesting that you should know that's engineering or what have you. Well, substitute your favorite kind of snow vicious snow viciousness in there that you really shouldn't look at explicit estimates. But, say out of curiosity or because you need explicit estimates, you look into explicit estimates. And then you realize that there's something really interesting going on. Because the even though these are supposedly equivalent, the estimates that we have in the literature for psi of x and for m of x are of vastly different quality. On the left even even for psi of x the situation with estimates is more complicated than what one might naively expect from you know, what you know from proofs of the prime number theorem that you learn when you first took analytic number theory. But, um nevertheless the situation of of for psi of x is more or less satisfactory thanks to dozens of papers. But, for m of x you also have more than a dozen papers over time well over plenty. But, the situation is very different. So, on the left what you have is that um the kind of proof that we all know and love that this comes from the level of Busan and and Adamar can be adapted with a lot of pain thanks to the work of a lot of people to give good explicit bounds. But, those bounds so true in general pretty much. They start being better than all other bounds only for x rather large, when x is bigger than 10 to the 1,600 or so. Um below that range, you're much better off with another kind of bounds, bounds of the form psi of x minus x is at most little epsilon times x. Where little epsilon right now, the record before our work, was about 10 to the minus 11. No, and this is true naturally only for x starting at a certain point. But for x up to a certain point, there are computational bounds, be they brute force or computational analytic bounds, which we might talk about at the end. All right, so the situation is already more complicated than we thought, but it's pretty good. Um it's either the bound is either a tiny epsilon times x or um it's this bound that is really little o of x and of roughly the form that you would expect, even if 9.3 is a bit larger than you might have guessed, uh it's still quite decent. All right. But what happens for M of x? Can you hear me clearly? What happens for M of x? Um well, the first bound on M of x in the literature, it's um that it's at most x over 9 plus 8. That the proof is in one page. It's by von Sterneck. It's basically a modification of uh Chebyshev's uh method from before the prime number theorem. That's from the uh 1850s. Um however, uh the problem is that things haven't really progressed all that much since then. Or rather, the best bounds that we have uh are really more and more baroque versions, and I say this for with full respect for lots of good work and for Baroque constructions and so forth. More and more Baroque versions really of um the von Stern neck Chebyshev method. These are not analytic bounds. These are combinatorial methods with coefficients chosen by trial and error basically. And where um actually the strongest bounds which get used all over the place are in papers that are not really self-contained and which really if you read them closely are saying or not saying things such as if you want our coefficients, well, we have them in a floppy disk somewhere. Um and that's not fantastic from any point of view. It's not just this this coefficient 1 over 4,300 is not nearly as good as 10 to the minus 11 and it's not just that these asymmetries bother some, but also that um really things become less self-contained and more complicated. Though this itself here for But it's fairly self-contained, but this itself depends on a long chain of papers. And then um but what what happened to this equivalence this supposed equivalence? Well, if you just try to use this equivalence between this statement and that one, you will get horrible bounds on m of x. But you can use the ideas behind this equivalence to do the following, to combine these bounds on m of x which are not analytic, which are combinatorial with bounds of this form on psi of x minus x to get bounds on m of x which are genuinely of the form little o of x. They will not be fantastic um but they will be useful. They will be much weaker than these ones, but they will be genuinely useful and I was using those bounds uh in uh my work on on on ternary Goldbach uh which has been delayed in part because I you got uh well, you have seen what I you are seeing what I saw, namely I mean you take a trip to the sausage factory and then you don't want to eat sausage anymore unless you make it yourself or what have you. Or or you decide you will eat something else instead. And so you also have bounds a very recent bound on M of X which is stronger but it's really based on this result. It's really a bound of this kind. It's just that you are really using the fact that for X below a certain bound you can use a brute force bound instead. So a bit more complicated than that but that's the spirit. So the real question is why is there this real at some point even if you don't care about quantitative estimates you have to accept that some quantitative difference are so large that they become qualitative. I think it's fair to say that we have a qualitative difference here. Just prima facie. And that's really the case. Analytic people are not are not being silly. There was there were really analytic obstacles. People were not able to um get analytic So on the left you have analytic analytic methods that are more or less like what you learn in a prime in a first analytic number theory course except more complicated because you have to make things explicit and you have to be quite clever if you don't want nasty constants. But um on the right the analytic method gets blocked. Why? Why is there this difference? So there were two good reasons or at least two reasons that looked good. Um so two good analytic reasons and as I will tell you one of them is essential and the other one is not. So one of them is the following. It's not really that Mobius is nasty. It's it's that lambda is unnaturally nice. That um it's uh the Dirichlet series which gets continued to minus zeta prime over zeta is naturally nice. Why? Well, let rho be a zero. Then really what we care about when we work things out, when we shift contours to the left is the residue of minus zeta prime over zeta. But the residue of minus zeta prime over zeta at zero is something beautiful. It's just a multiplicity of the zero. And we count zeros with multiplicity anyhow. So a residue of minus zeta prime over zeta appearing a so-called explicit formula is not a problem at all. You want to you're already taking that into account naturally. So if you count each zero k times where k is its order, which of course it's always one, but we don't know that. Um then you're basically giving each zero a weight of one quite naturally. [clears throat] There's no problem there whatsoever. Whereas the residue of one over zeta is well, goodness knows what. Well, it's not really goodness goodness knows what. It's closely related to the distance of the nearest zero. The problem is that we have no control over that. We believe that zeros cannot just bunch together as closely as they want. But that [clears throat] is felt to be further off than the Riemann hypothesis. We don't even know that there are no double zeros. And just like we don't know there are no double zeros, we don't know that there are no zeros which are almost double, and unspeakably close to double. These two zeros that are extremely extraordinarily close to each other. We don't believe that happens. But we cannot rule that out. We can compute zeros up to height T and show that it doesn't happen there. But we cannot say that in general it doesn't We don't know how to prove that. Even under our age, we don't know how to prove that. So, because of that, we cannot just shift the contour to the left all the way to the left when we are estimating m of x. We could instead shift the contour within a zero-free region. Um and that's what we would do when we are proving uh non-explicit estimates when teaching analytic number theory. But that actually gives pretty ugly explicit bounds. Um that's not how you prove, by the way, explicit bounds for psi of x. You really do want to shift the contour all the way to the left. Um and if you try to do that but for uh for m of x, in principle, that's doable, but that will give you pretty horrible bounds. Uh and nobody has done that correctly in the literature. So, there's an incorrect result on archive. Um but um somebody should really do that uh just for comparison, but the result will be very ugly and fairly useless. That's the truth. Um simply because um our bounds for 1 over zeta within a zero-free region are so bad. You We get them by exponentiating really a bound that is already not fantastic. Still, somebody should do that and do that well, but it's it's actually quite tricky. I need All right. So, that's already a serious difficulty. Um then there's another thing. I told you that Well, if we could use We can check that the residues are We can check the computer residues using integral arithmetic. So, up to high standards, we can really prove that they are bounded by something uh for all the zeros of height up to T. That is of imaginary part up to T. But then you have another difficulty, uh which is really more of an psychological block more than an block, but it looks like an analytic block. So, um how do we bound uh a partial sum? We are modern people. So, you know, I think Yeah, everybody who is an organizer is more or less in my generation, maybe a bit younger, maybe a bit older. So, we were told um to always smooth uh in part by some people who may be in the audience. I haven't checked. Um So, our advisors told us to always smooth. Or they also told us smoothing is given by nature. I'm referring to somebody in particular. Is he online or not? Which is actually a deeper statement, but in this case it's not given by nature. We have to choose a smoothing. So, we approximate, being modern people, um our sum which has a uh brutal truncation, as people say in the in in French, a sharp truncation, by a sum with a smoothing. And then you have something nicer and more civilized than the ones formula. You have an inverse Mellin transform, and in here you have the Mellin transform of your smoothing, your continuous weight. Um and then uh if I were, you know, this would come with a warning, because if say you were uh I were giving this talk to young and sensitive um first year grad students, say, then somebody might one of them might stop listening and say, "Aha, I have a brilliant idea. I'm going to revolutionize everything. I'm going to choose an eta such that um the Mellin transform has compact support, or at least its restriction to uh uh the line the restriction to the line imaginary real part equal to one has compact support. But in my stylus though, there's an I missing here. Okay. Uh why would you want to do that? Because, um then, uh we could shift the contour to real part equal to minus infinity, and the only residues uh that we would take be taking into into account would be uh the residues with imaginary part at most T, and we would be done. Now, of course, the problem is that's completely impossible. You cannot have a million trans- The million transform is holomorphic within a band. Um uh within a band. And the holomorphic function just can- cannot have compact support. Its restriction to a line cannot have compact support. Even even the continuous even if the uh you are holomorphic within a band uh we within a band, and you continue your function continuously, your holomorphic function just continuously to the to the border of the band, uh the function cannot be compactly supported there on the border. So, it's a little bit subtler. Uh and so, you know, the graduate student just the imaginary first-year graduate student goes home and cries. Dreams are dashed. You fall off a purple cloud. Um yeah. Okay. Uh sad story. Uh imaginary story. Um however, I claim that you can you can do something that is as good as that. We can manage to use only the residues of, say, one over zeta with imaginary part with absolute value at most T, and we can do that optimally. And the zeros of zeta with imaginary part greater than T simply do not appear at all. And so, we have two main theorems here. Um if you find that it is a bit full, you can decide to use to look only at half of the screen, you are free to do that. Yes, Henrik, I meant you. And I I I I did mean that you So, I should credit the Sattler version of the two statements to Henrik. Okay? Namely, that smoothing is given by nation. >> Yes. >> Sorry? >> I expected you meant me. >> Yes. Yes. Not The second one was you. Uh yeah. >> At any rate, moves moves every time. >> Excellent. >> Okay. Bye. >> Of course, as we will see, one should not be too picky. Often, um you don't want a smooth smoothing as in C infinity. Sometimes, C1 is optimal, in fact, or even less than C1. We We shall see. >> And be more right. >> Excellent. Very good. So, if this is too confusing, just look at your favorite half of the screen. So, if you care about bounded functions, such as mu of n, look at the left half of the screen. If you care about non-negative functions, such as lambda of n, look at the right half of the screen. Why do you have to assume one or the other? Well, you don't really have to, but having one one assumption or the other really simplifies things vastly, and having one of these two assumptions allows us to give a fully general statement. So, mu of n and lambda of n become really special cases. So, let I because the formula is a bit nicer for mu of n, let me focus on mu of n. So, for mu of n, as for for mu of n, you generally get a an explicit formula that involves only uh, the zeros with imaginary part up to T. Because here, so this is the formula. So, what does this mean? So, uh, for me epsilon means something extremely In fact, epsilon is tiny both in theory and practice. Epsilon is little o of one, and it's also really, really tiny. Uh, so it it's less than one over T squared. So, for where T you should think of the height up to which you verify things, you should think of it as being at least 10 to the 9 even on your desktop. In fact, you ask a a very good programmer, in this case David Platt, and he will carry out the computation much higher than that. Um, so epsilon is really tiny. For me, iota is something bigger than epsilon, but still quite small. That's the way in which normal people use iota. Um, and what is iota here? Iota is some quantity smaller than uh, hyperbolic tangent of pi over 2T, which is of course almost pi over 2T. Um, this we will see, this is really interesting This is really there. This is optimal in in two strong senses, we shall see. So, you have you cannot avoid having this little term that is smaller than pi over 2T. And more precisely, it's bounded by hyperbolic tangent of pi over 2T. And there here you have what you expect in an explicit formula. It's a sum over zeros, but you what you don't over poles, but what you don't expect it say that it's a sum only over those poles that have imaginary part at most T. And then what you do expect or almost is what you have here, so a residue. The residue at rho of I of S times X to the S minus 1 times a weight. And the weight is the optimal weight, which turns out to be beautiful as optimal things sometimes are. At least I can it beautiful because it's basically just hyperbolic cotangent minus minus a constant which is hyperbolic tangent of pi over 2t. So, the weight is just hyperbolic cotangent which is because s is something vertical, it's it's really basically cotangent. Right. And and that's really it. There's not much and much else to it. The formula for a n non-negative is a bit more complicated and you do have some other o star terms uh which don't dominate but you still have to take care of things. It's a bit more painful and we will later see why. O star just means something with implicit constant at most one. But again, you have um that yeah, you have uh trigonometric functions lurking in. All right. Before we see where all of these functions come from, let us see what they give us. And you might think that these results are ugly because, you know, though they are one line in this case, this one is one line, you need to explain what the terms are. Well, let's let's see. So, first let let me just I've written it down to to explain that I'm not lying. That here the iotas here all of these terms pi over t cotan No, actually this term this term plays a role of one. This is very funny. That you will see this phenomenon later but the iota here that the role of iota here is played by pi over t. And whereas here is tan hyperbolic tangent of pi over t. And these these intrinsic tiny but constant error terms are really optimal. In what sense? That we can actually construct Dirichlet series that don't have poles you we can construct sequences such that the Dirichlet series don't have poles with imaginary part bounded by t at all. They have no poles there. No poles whatsoever. The sum over poles is empty, but they still oscillate. They oscillate and by how much? Exactly by in this case for A and bounded by hyperbolic tangent of pi over 2t. Or it would be pi over t in the case of non-negative functions. So we can actually have construct bounded A and such that the the sums don't obey the prime number theorem. They miss it just by this tiny term hyperbolic tangent. They oscillate by that. Now, I'm claiming that these results are optimal. I'm also claiming that they're optimal in general. And I but for mu one can do better. So what how can you do better than optimal result? Well, mu has um uh well, it's not just that it's optimal given a verification up to T. I claim that you you can do better just with some verification of zeros up to T. Simply because mu of n is not has support on the square-free integers. So you should be able to win that to gain that factor of 6 over pi squared and and we are able to. So in the end um we have pretty clean bounds. So we get for M of X um using a rigorous computation of the residues of one over zeta for imaginary part of the 10 to the 10, we get that M of X is bounded by 3 over pi times X over 10 to the 10 plus something in this case a constant times square root of X. You know, in general, you know, we this is not guaranteed to be a constant. It's a sum over zeros, but it turns out to grow very reasonably. There's some cancellation here. No, actually this is without cancellation. We will later talk about cancellation, but without even without cancellation, this is something reasonable. So you get this bound which you should compare to what we had before. Awesome. This anonymous bound, valid for all x greater than equal to one. And this is the 10 to the 10 is really the level up to which we have checked t and 3 over pi is precisely zeta of 2 * pi over 2. Now, in fact, we also improve on existing bounds for lambda of n. I mean, we improve them by less because the results that existed were better. And in fact, in some sense, the the optimal bounds that we get for non-negative functions are worse by a factor of two. In this case, they are they are better simply because the Riemann hypothesis has already been checked up to height 3 * 10 to the 12 by Platt and Trudgian. And you don't need to compute residues here. But you get something very funny that I did not expect. Maybe some people in the audience knew this, but I never knew that I was we were going to get an error interval for psi of x that is not centered on x. It's a little bit off-centered. Of course, the error interval contains x, otherwise you know, the world would collapse. But still, it's something very funny that I did not expect. And then you have this iota, which just to simplify the iota plus the epsilon is bounded by pi over t minus one times x. So, we can give this general result that if RH holds for all imaginary for all zeros which imaginary part up to t, this should be an absolute value, obviously. Then, you have this formula. And this is optimal, as I said, in this generality, you know, when you want a result valid for non-negative functions. You are not using a zero-free region at all here. This coefficient for square root of x is basically what you could expect, except that we actually get for free a little bit a small gain on what people had before, which was these, just because the weight function decreases, no? So, unconditionally we get the following, that for all x greater than or equal to 1 this holds. Now, this constant is a little bit uglier for it's an interesting phenomenon, but this works quite nicely. All right. Now that we are happy that we have results, we have, I guess, 20 minutes for the proof? 25 minutes for the proof? Very good. So, what is the proof strategy? So, again, smooth smooth smooth or We will approximate one not by an arbitrary eta and then work with Mellin transform of eta. We already know know that that could lead us into trouble. We will approximate it by a Fourier transform, phi hat. Okay? And then we don't use the inverse Mellin transform. We just use Basically, we don't even have to hurt the the Fourier inversions here. We just use the definition of the Fourier transform and we get this, which is roughly analogous to the Mellin inversion formula, but in fact it you can prove it in a paragraph. And this is not new. Statements like this exist um uh in some form, not sometimes we say hard switched, but this statement exists in proof of the Wiener-Ikehara theorem. Um and uh you even have there in in proofs via Wiener-Ikehara of the prime number theorem the idea that you can choose phi to So, here here you don't you're not phi just needs to be defined on the real line. So, you are you don't have to assume you you are not hurting holomorphic functions at all. And phi particular does not have phi I be of compact support. That's completely fair. Uh phi yes has to be in L1. Its Fourier transform should be in L1, but that's completely compatible with being of compact support. You can have for instance trian- the triangle, or you could have the truncated cosine. You can have whatever you want. It's It's a very you know soft condition that phi be in L1, phi hat be in L1, and you let's say that you also impose a condition that phi be of compact support. Support in minus one comma one, and then this um improper integral becomes proper. It just becomes an integral from one minus IT to one plus IT. Again, nothing new here because the proof of inner product had a status this way. Now, the following step is already new. So, um surprisingly so uh we want to know since we we're going to do this, okay? But we we said that we are going to approximate our brutal truncation by uh uh trun- by a sum that is weighted by phi hat. So, what is the difference between the sum that we want and the sum we actually estimate? So, um you know, this is the sum we will actually estimate, and the sum we want is the sum with a brutal truncation. So, we want to because AN is bounded, we are doing that version of the proof, we want to bound this quantity here. Now, just by some change of variables um we end up having this sum over here. That's nothing. And then the following step is surprisingly already new. So, this looks like an L1 bound, and it is generally basically an L1 bound. You can You can bound this sum here by 2 pi over T times the difference in L1 norm between phi hat and I over in the real line plus some tiny tiny tiny error terms, which um depend on being given that you know, you're given the input that phi hat minus i t case quadratically. Where in fact I you barely need it here. I here um here and here I is not um uh the characteristic function of an interval. Here by I I mean a truncated exponential. Okay. So uh everything this is bounded by 2 pi over t times the difference in L1 norm, the distance in L1 norm between phi hat and the truncated exponential and you have some very terms coming from some tail bounds on phi hat basically. Uh here, you know, the this is important. If you just try to use the the tail bounds and work only with that, you're going to get something very suboptimal. You should really see this as an L1 bound. Aided a bit by a by a tail bound. Okay. Now comes an interesting step. Uh how many of you have played Mafia or Werewolf? Okay. So um we uh have this integral here, this integral over the finite segment 1 minus i t 1 plus i t. Now everybody goes to sleep. Now, close your eyes. A benevolent thief, a philanthropic thief, comes in the night and replaces little phi, which is just defined on the real line, by big phi. Big phi is meromorphic. And not only is it meromorphic, but uh it's identical to phi not not on the whole real line. That would be impossible. Just on the segment minus 1 {comma} 1. That's the case. Now everybody wake up. So the value of our integral has not changed because we have just replaced little phi, the thief replaced little phi by big phi, uh Uh, and you know, and the argument goes from minus one to one, so there's no difference here. You still have this ugly image over T, but you're on a line, so this is just S minus one over IT. So, you now have a beautiful very a beautiful complex integral. And now, you can shift the contour over to the left. You you get these skid marks, horizontal skid marks, but they are no big deal. Certainly for mu of n, they contribute something very very tiny. Uh, that's these contour integrals, so you're horizontal contour integrals, and then you just are capturing the residues given by the zeros of height up to T. So, by the way, the closest anybody has got to these, and this was the main inspiration, you will find these in Ramana not these, but you will find something a bit like these in Ramana Ramare because um at when they So, Ramare really made me aware or made many people aware a long time ago how frustrating this lack of parallel between mu of x and psi of x was. But, interestingly, I think the closest that uh that he came to these was in a paper on a different problem, a paper by Ramanujan and Ramare. And Ramana had also worked on um variations of Ingham-Jessen where they were using they were not looking at this problem in full generality, and they did not have the previous bound on the one norms, but they were working with piecewise polynomial functions. Uh, a piecewise polynomial phi indeed. And and of course the issue with a piecewise polynomial phi is well, a polynomial isn't entire. So, then you can shift. Yeah. So, that's the the I think that it's important to give them credit for that. But, you really should look at this problem in full generality. And in full generality, what we you want is a thief in the night. Whenever you you have little phi such that you have a thief in the night, you can in fact shift the contour. Yeah, somebody say something. No, not yet. Um All right. So, uh Very good. Very good. There's a finite number of these poles, and you can just compute the residues or really ask David Glad to compute them. Um and uh this fantastic. Of course, I was telling you that I was giving you the version for not the bounded version. What about the non-negative version? For the non-negative version, you don't even need this. Uh number two. However, for the non-negative version, um you you will see that we get a bit lucky with phi that you will find that our for our optimal phi, there really is a big phi. Uh for the optimal phi for A and non-negative, uh you have to paste two phis, two big phis. So, um uh and you have to not be you should think of them as them as being sewed together, and you you Now, you have a little suture here, and that gets a little bit messy, but it's not too bad. These are the suture lines, so to speak. Okay. Let's proceed. So, the problem is to find phi such that phi hat minus I, I being a truncated exponential, is minimal in L1 in L1 norm. But, that's uh an optimization problem of known type. And in fact, it it's has been solved uh over the course of time for by different sets of people for different uh uh different things. Uh certainly for the for the functions we we are considering here, bounded functions, non-negative functions, it has been solved. So, in the special case that would come if we were estimating as we can sums of the form lambda n over n, that was solved by Bohr and Selberg. They never published or Bohr never published, and his student Riedel date, and he didn't actually publish in the paper. Well, never mind. Uh they had their own publication. It's difficult to give a precise date, but this is known to people since the '70s. Uh and then there's um but we really need a more general thing. So, Beurling-Selberg has been used a lot in analytic number theory, so it's a bit surprising that this hasn't been done. But we really need a more general thing, but these things that are known to number theorists as well. So, Graham and Vaaler that's uh they solved the problem, the right problem for uh sums of lambda of n. And for uh sums of mu of n, really the the solution is due to Vaaler and to Carneiro-Litman. Because what are the problems here? It depends on the constraints. So, um if you have bounded functions, it's really just that. You want the phi which supports on minus one comma one such that uh the difference between phi hat and I is minimal in L1 norm. Uh no other constraints. Yes, that the support of phi is um is in minus one comma one, so it's it's it's continuous at minus one, right? It's continuous everywhere, that's all. And then then it's Yeah, it's known what what uh what the problem is. So, um it's this [snorts] thing on the left. So, um it is approximate what people call approximate. So, this is what you get from mu of n over n. This is closest to Beurling-Selberg. Uh that was uh Graham uh no, Vaaler. That was Vaaler. No, that was Graham-Vaaler. And then Vaaler um no, Vaaler was this, Graham-Vaaler was this. Uh no, I'm getting this completely backwards. So, this was Vaaler and this was Carneiro-Litman. Right. So, these are the best approximates. We can plot them quite nicely. Uh I mean, they are special functions, but it has um the approximates, but we will see that they are Fourier transforms of very nice things, not really special functions. And um whereas uh when you also have the added condition which you need for when the functions are not bounded but non-negative that the function your approximate might are majorants or minorants of the functions you're approximating. There you have the in fact more familiar Bolling Selberg problem. People have seen this plot before probably. And here is if you have the truncated exponential. Notice how these functions are equal to the function they're approximating at the integers except for zero in this case. And at all half integers in this case. That's not a coincidence. There are several ways to do this. You can see this by calculus of variations. I don't see it standard way in the literature. It seems as an interesting heuristic. That's how we first saw it before we realized that it was not just inspired by the literature but contained in the literature. Or you can do it in other ways. But this is really not not chance. You end up using an interpolation formula. And something that is really not clear is stated in the literature because people tend to phrase people tend to work most of these papers work out phi hat not phi itself. Most of them is that phi is actually much nicer than phi hat. Phi hat is already nice enough. It's a sum. You can write it in terms of the digamma function sometimes or the large function. But phi is something beautiful. Phi is something you can write in terms of hyper- hyperbolic trigonometric functions. Well, really just trigonometric plane or trigonometric functions because you get that as um Yeah, you you you get that by you know you get a Q series from interpolation formula and every and precisely because it is a trigon- a trigonometric function but basically a trigon- a trigonometric function shifted a bit in the complex plane. Um yeah, you get that because trigonometric functions are meromorphic. So, phi the optimal phi is a trigonometric function restricted to minus one comma one, so big phi always exists. Or for majorants and minorants, you really get something of this kind, you know, a phi little circle plus sine of t and as I said, you have two functions summed together close to the x-axis. And this is the greatest contribution of AI to this paper, namely it um I mean So I I I have to credit, I think ChatGPT with This is just a center. Uh so this is a proof that this half Fourier, this is supposed to be Fourier and half analytic. So you are really um if any you remember anything from uh the proof, it should be this. Right? So you start as in a free analytic proof of the prime number theorem, and then you shift gears entirely and it becomes a complex analytic proof. Okay, what now? So uh yeah, so this is optimal. You should So you can and you should redo large parts of explicit analytic number theory now. This sounds insufferable, but it's true because much reduces to bounds on psi of x and mu of x, and the same techniques apply to many other L-functions. Uh it's you know, I don't I you have to I have to be a bit careful, but I would say morally speaking that this applies to a class of functions broader than Selberg class because you don't really need a functional equation, even though it's helpful if you really want to shift to minus infinity. And there's really no point in waiting since bounds are essentially optimal as a function of t. Certainly the leading terms are. Um and in fact, there's now a team of people um of people and not people who are uh and AI really does come into play there for working at formalizing this in Lean. Uh there is an entire project to to do I mean, it's really important to, you know, clean up the social factory. And this has become part of this. So, uh formalizing this paper these two papers has become a priority. I think both should be a priority, but people have started with fire effects. Uh I've had a small role in that formalization effort. Um I'm not really fluent in Lean by any stretch of the imagination, but there were all sorts of surprises as to what and how they had not been formalized. Um okay, people people have told me not MathLib, people have told me not to say this, but I will say this. As far as number theories are concerned, MathLib has all sorts of sophisticated and beautiful things, but it did not have integration by parts in the generality that we wanted because it had integration by parts when for two functions that are in C1, but of course that's not enough. We want integration by parts when one function is C1 and the other function is of bounded And that was not in MathLib and could not be in MathLib because um the theory of uh Lebesgue-Stieltjes integration was not ready yet and it will not be ready yet for a couple of years, I think. Um and uh people in MathLib thought that Riemann integration was beneath them, I think. They had something more general called box integral, which is helpful, but it was not, you know, it only the basic theory was worked out. So, Riemann-Stieltjes integration was not there. So, something that I did, everybody wants to hear about AI nowadays. So, uh something I did in January was I was playing with AI. Uh and AI can already form help in formalizing things badly. It produces ugly code often, but it's very good as a um proof of concept. So, I found that um there there is uh Aristotle. It's uh yes, it's it's an AI tool, but my my much of the time you can also just use by now really powerful general LLMs uh that um if you choose your textbook well, it can sometimes chew on a textbook for a couple of the few pages of a textbook and spit out formalized uh code. And so for instance, I found that um textbooks that I like were also liked by Aristotle the AI. So uh I I think um uh Zigmond is fantastic and Aristotle felt the same way. Uh and um of course the code that uh Aristotle produced was much uglier than Zigmond, but this uh but I by mid-February I had that you could use Zigmond and also appendix A in um Montgomery Vol. 1 to get uh to formalize enough Riemann-Stieltjes integration for the purposes of analytic number theory for many purposes. In what we really care is that we are able to work um to have uh that F hat decays quadratically when F is of bounded variation, right? And that comes from integration by parts of the generality that we need and that was not there in MathLib. Or also we need there was a statement in in MathLib. MathLib people, please take this as friendly criticism that said literally, "This is the most general statement of the Poisson summation formula." And it was not a statement of the generality we need in analytic number theory because it assumed uh you know uh that F was nicer than many It was It was assuming that F was C1. And we want to we don't we we we often want Poisson sum in the Poisson summation formula for functions that are not necessarily C1. Okay. So what what has really happened there is that people were delighted and grossed out by how ugly the Lean code was. And uh people got together including me, but I you know, as he said, I'm I'm bad at lean still, uh in May and a team of people at I sum, I should really have written down all their names, but you can look them up in I sum and I think this will be uploaded and you will see everybody including young people that should be given lots of credit. In a week people managed to redo at least a basic Riemann-Stieltjes integration in appendix A of Montgomery von, and now it's really really clean and done by humans, uh and people will be able to use that and build on that to to be able to to have what we need in analytic number theory. In particular, now we have integration by parts finally in the generality we need. Somebody's in the chat. Uh yes, Henrik, later. One moment. Do you have a mythological animal that is Ah, yes. Yeah, but it's the wrong half of the elephant. Well, I do not know. We we shall talk about that later. At any rate, um yeah. Okay. Moreover, there are some other things in the next 4 minutes that you get many other things. Let's backtrack from formalization, let's backtrack from AI and so forth. Let's get back to human classical things. So, first of all, just we weren't intending to but we get a self-contained proof, yet another self-contained proof of PNT early on for free. As in, you know, in in section two we're just building up foundations and then in section 3.1 of both papers we work out a self-contained proof of PNT because in section 3.1 we get Yeah, I should really say that what we get is that um the statement that is equivalent to PNT elementarily because we get very very easily. So, it's it's really a variant of Vinogradov but it's really very clean and and short. I've been frustrated trying to give frustrated trying to give Vinogradov in class. And so, you get that a very, very clean Vinogradov Vinogradov type proof just in a paragraph of the statement that m of x is little o of x just given the basic theory that we had to develop anyhow. in particular Not not not three or four, just just these bounds. So, just using one and two, which we need anyhow, we get basically for free Vinogradov for m of x equal to little o of x. Now, of course, I said self-contained and you could say, "Vinogradov, you are not self-contained. You still have to use the the ancient trigonometric trick or something which Cauchy-Schwarz to get that you have no zeros with real part of one." But then, when I was giving this talk on an earlier version of this talk in Nancy, and then our tone pointed out, "Harald, you may want to look at something that Jean-Pierre Kahane did a proof of the prime number theorem." And indeed, you can give what is basically a variant of the first half of Kahane's proof to show in a way that uses again the foundation section of the psi of x the non-negative paper that zeta of 1 + it does not vanish. Yeah, but zeta does not vanish in the real part of one. Now, I I I like to make it to do it differently. Again, it's two paragraphs. Yes, use a bit of Diophantine approximation. Maybe it would be it's maybe Kahane would feel that it's a bit impure. He's not now. He's unfortunately no longer with us. But it's turns out quite nicely. Okay, but that's cute and all. Now, let's uh go for things that we intend to do, but people are welcome to try their hand at it because we are already sketching them in final remarks. They will be a pain to do, but um so, what about bounds of the form m of x is little o of x, but explicit bounds? As I said, all such bounds in literature are based on bounds of the form m of x uh bounded by epsilon x. So, we could redo the old papers to get bounds of the form m of x little o of x, but in but explicit bounds. But, in fact, it's much better to use the idea there and redo everything, do everything from scratch. Because the basic idea is to use a combinat- combinatorial identity. Uh and by the way, something that came So, did somebody Yeah, I think this also came out came from uh I think Balazard pointed out to me that I might in these combinatorial identities, No, it is some book in a book by Balazard. There's these combinatorial identities that people use that are related to Scherk-Selberg. I mean, all communi- lots of combinatorial identities in number theory are related to each other. Um one of them is very funny. Um You you could see this as being related to elementary proofs of the prime number theorem. Uh in fact, in particular, I say that think this is closest. This was pointed out by Balazard to um the the combinatorial identities that people use to prove m of x is little o of x. This is closest to um Kaletsky's variant of the version of two Soviet mathematicians, I've forgotten, sorry, of uh Selberg's uh Selberg or Selberg-Erdos, but let's say Selberg's proof, elementary proof. Um and by Kaletsky, I do mean the economist. That's a long story. Um uh and here um Yeah, but so, that's the idea that existed, but you really want to use those identities and do everything from scratch. Why? Because I wrote these explicit formulas not just out of say this and I mean the bounds may look clean and all, but it's better to have an explicit formula because you can plug in an explicit formula into something else and you can play with it. You can do all sorts of things with this explicit formula. Um and you can get cancellation. Even if, you know, you don't assume anything about the residues, you can get cancellation. And that's precisely what happens. And you know, this is only sketched in the in the paper at the end, but you really do get even better balance than what we would get. Because just by by using these bounds, not just using these bounds, should correct myself, but really using the explicit formula. And the other thing is that I talked briefly about analytic computational bounds in that it's not just that you have brute force bounds, then epsilon X bounds and then little o of X bounds. There's this twilight zone here between brute force bounds and analytic bounds of the type epsilon X, analytic computational bounds. So there's a paper there due to Buthe and basic idea is implicit in Odlyzko. Um but Buthe is really using a function that was optimized that is in Odlyzko and um somebody else, but uh that is really optimized for a different purpose and you should not use that. Um you should use other functions uh but moreover, uh what you but the basic idea is good. The basic idea is that you want to use interpolation formulas and the fast Fourier transform. But you cannot do the fast Fourier transform on 10 to the 12 zeros and 10 to the 10 zeros. You chop things into pieces, but here comes the how you can improve things vastly, we believe. We shall see how big the improvement is at the end. You shouldn't chop things brutally. Always smooth, remember. Um you should or always move in a continuous way at least. Instead of chopping chopping the zeros, think of it as being the the line real part equal to 1/2. Instead of chopping the this this line parallel to the Y axis brutally into segments, you should chop them with um shearing shearing pinks. No, pinking shears, pinking shears, right? Uh those crocodile type scissors. So, you get some these nice beautiful triangular functions. Two of them added together give you a constant function. And then what happens is Fejer happens that you end up having an interpolation formula where the coefficients decay quadratically. So, your interpolation formula converges absolutely. So, everything works out much more nicely computationally and otherwise. And I think this is new. If this is not new, people should tell me. And they should have other applications as well. Okay. My time is up. I have um another slide that is not part of the talk, but just as an answer to a certain kind of question. Very good. Thank you.