1 Introduction↩︎

The study of random multiplicative functions has been very active in recent years, motivated by various connections and applications in number theory, probability, and analysis. Define a Steinhaus random multiplicative function \(f : \mathbb{N}\rightarrow \mathbb{C}\), by taking \((f(p))_{p \; \text{prime}}\) to be independent random variables distributed uniformly on the complex unit circle, and setting \(f(n) := \prod_{p^{a}\| n} f(p)^{a}\) for all natural numbers \(n\) (where \(p^a \| n\) means that \(p^a\) is the highest power of the prime \(p\) that divides \(n\)). These are often considered to be models for number-theoretic functions like Dirichlet characters \(\chi(n)\). A Rademacher random multiplicative function is defined by letting \((f(p))_{p \; \text{prime}}\) be independent, taking values \(\pm 1\) with probability \(\frac{1}{2}\) each, and setting \(f(n) := \prod_{p |n} f(p)\) for all squarefree \(n\), and \(f(n) = 0\) when \(n\) is not squarefree. A fundamental question in the area is to study the statistical properties of partial sums \(\sum_{1\le n \le x}f(n)\), and more generally, weighted partial sums \(\sum_{1\le n \le x} a(n)f(n)\).

One striking result is Harper’s resolution [1] of Helson’s conjecture [2]. Harper proved (in both the Steinhaus and Rademacher settings) that for all large \(x\) \[\label{eqn:32Harper} \mathbb{E}\Big[\Big|\sum_{1\le n \le x} f(n)\Big|\Big] \asymp \frac{\sqrt{x}}{(\log \log x)^{\frac{1}{4}}},\tag{1}\] so that partial sums of random multiplicative functions exhibit “better than square-root cancellation." This implies that the normalized partial sum, with the natural normalization \[\Big(\mathbb{E}\Big[\Big|\sum_{1\le n \le x} f(n)\Big|^2\Big]\Big)^{-\frac{1}{2}} \asymp \frac{1}{\sqrt{x}},\] has a trivial limiting distribution. In earlier work [3], it had been proved (in the Rademacher case) that \[\Big(\mathbb{E}\Big[\Big|\sum_{1\le n \le x} f(n)\Big|^2\Big] \Big)^{-\frac{1}{2}} \sum_{n \leq x} f(n)\] could not converge in distribution to a standard Gaussian \(N(0,1)\) (though this did not exclude the more subtle possibility that it might converge to some Gaussian with variance different from \(1\)).

A central limit theorem may emerge in cases with general weights \(a(n)\) that somewhat disrupt the multiplicative structure. One particular example that has attracted considerable attention is when \(a(n)\) is an indicator function of a short interval, namely we study the limiting distribution of \(\sum_{x\le n \le x+y}f(n)\) where \(y \rightarrow \infty\), but \(y=o(x)\). It was first proved by Chatterjee and Soundararajan in [4] (for Rademacher \(f(n)\)) that the limiting distribution, with usual normalization \(1/\sqrt {y}\), is Gaussian as long as \(Cx^{1/5} \log x \leq y = o(x/\log x)\). Note that the lower bound condition on \(y\) is, in a sense, much less interesting here; it was imposed so that one knows unconditionally that all intervals \([x, x+y]\) contain roughly the expected quantity of squarefree numbers. This result was improved by Soundararajan and Xu in [5] to the wider range \(Cx^{1/5} \log x \leq y\le x/(\log x)^{\log4-1 + \epsilon}\), and also extended to allow Steinhaus \(f(n)\) whenever \(y \rightarrow \infty\) and \(y\le x/(\log x)^{\log4-1 + \epsilon}\). On the other hand, by using the triangle inequality and Harper’s theorem 1 , one can see that when \(y\) is sufficiently close to \(x\), \(\frac{1}{\sqrt{y}} \sum_{x\le n \le x+y}f(n)\) has a trivial limiting distribution. In a recent work of Caich [6], it is proved that such trivial limiting distribution holds as long as \(y\ge \frac{x}{\exp((\log \log x)^{1/2-\epsilon})}\). It has been an intriguing question to properly understand all these transitions in behavior. In particular, one would like to know whether \(\frac{1}{\sqrt{y}} \sum_{x\le n \le x+y}f(n)\) actually has Gaussian behavior for \(y\) larger than \(x/(\log x)^{\log4-1 + \epsilon}\), and what can be said about the distribution for those very large \(y\) where the limit becomes trivial.

Perhaps surprisingly, we prove a Gaussian limit theorem for partial sums \(\sum_{x\le n \le x+y}f(n)\) of a Steinhaus random multiplicative function whenever \(y \rightarrow \infty\) with \(y=o(x)\). This completely determines the distribution over all short intervals, thereby completing the previous results in [4]–[6].

The value of \(V(x,y)\) that emerges naturally from our arguments is expressed in terms of a quantity \(\mu(x,y)\), which is (up to scaling) a certain random walk probability. See Proposition 6 and 48 below, and the discussion in sections 1.1 and 1.2. However, we are free to replace \(V(x,y)\) in Theorem 1 by any quantity to which it is asymptotic (as \(y\to \infty\) with \(y=o(x)\)), and our estimation of \(\mu(x,y)\) (again see Proposition 6) shows that 2 is such a choice.

This is the first time that partial sums of random multiplicative functions with non-obvious normalization have been found to have a Gaussian limiting distribution. There are a few other number-theoretic situations where somewhat similar (and very interesting) phenomena arise. For example, Montgomery and Soundararajan [7] show, assuming various strong but plausible hypotheses, that the distribution of the von Mangoldt function in short intervals \(\sum_{x \leq n \leq x+y} \Lambda(n)\) as \(x \leq N\) varies and with \(y\) in the range \(N^{\delta} \leq y \leq N^{1-\delta}\), should be roughly Gaussian, but with variance differing from the obvious guess by a constant factor. Gorodetsky, Mangerel and Rodgers [8] recently gave a beautiful proof that the analogous count of squarefree numbers in short intervals, where \(x \leq N\) varies and \(y \rightarrow \infty\) with \(y = N^{o(1)}\), is roughly Gaussian with variance differing from the obvious guess by a power. In the squarefree case, the variance drop reflects the highly structured nature of the sequence of squarefree numbers. For the von Mangoldt function (equivalently primes) the situation is closer to our case here, but again the variance drop reflects a structural property of the sequence, namely its connection with the zeros of the Riemann zeta function. We are not aware of a similar phenomenon having been observed before in any random or deterministic context quite analogous to the present one (e.g. involving character sums). We also note that in both the primes and the squarefree examples, one sees the variance drop simply by computing the variance of the sequence itself (and the results on limiting distributions are obtained by a moment method). In contrast, the variance of our random sums has the obvious size one would expect (from orthogonality), it is only in the limiting distribution that one sees a change of scaling.

Theorem 1 contrasts significantly with the distributional behaviour of the long sums \(\sum_{n \leq x} f(n)\), or \(\sum_{x\le n \le x+y}f(n)\) where \(y \asymp x\). In view of 1 and Theorem 1, one might initially suppose that if one rescaled these sums by something like \(\frac{\sqrt{x}}{(\log \log x)^{1/4}}\), then a Gaussian limiting distribution should appear. However, the ideas in the proofs, and specifically the connection with so-called multiplicative chaos (on which we elaborate below), suggest otherwise. Harper [1] proposed, in both the Steinhaus and Rademacher cases, that \(\frac{(\log \log x)^{1/4}}{\sqrt{x}} \sum_{n \leq x} f(n)\) should have limiting distributions related to the total mass of critical multiplicative chaos. Later, Gorodetsky and Wong made an explicit conjecture of the limiting distribution in [9]. In particular, this would imply that the tail probabilities \(\mathbb{P}\left(|\sum_{n \leq x} f(n)| \geq \lambda \frac{\sqrt{x}}{(\log\log x)^{1/4}} \right)\) should be of size \(\asymp 1/\lambda^2\) for fixed large \(\lambda\), much heavier than a Gaussian. Recent breakthrough work of Gorodetsky and Wong [10] determines, in the Steinhaus case (although not yet the Rademacher case), that the limiting distribution is indeed of this form. Their arguments also apply to \(\sum_{x\le n \le x+y}f(n)\) when \(y \asymp x\).

Exploiting the known blow-up of the low moments \((\frac{(\log\log x)^{1/4}}{\sqrt{x}})^{2q} \mathbb{E}[|\sum_{n \leq x} f(n)|^{2q}]\) as \(q\) approaches \(1\), we establish the following heavy tail bound.

Proposition 2 will be quickly proved in Section 5, essentially independently of the rest of the proofs in this paper. As we describe explicitly there, this implies that there is no choice of normalizing factor \(V(x)\) for which \(\frac{1}{\sqrt{V(x)}} \sum_{n \leq x} f(n)\) can converge in distribution to a non-degenerate Gaussian.

We shall comment more on the work of Gorodetsky and Wong [10] (and also S. Hardy [11]) on the full-sum limiting distribution in Section 1.4. Here the limiting distribution is non-Gaussian, or more precisely it may be thought of as a Gaussian whose variance is itself a genuine, heavy-tailed random variable. Perhaps it is then surprising that in our short interval case, there exists a deterministic normalization which makes the random sum behave like a Gaussian. This new feature is due to a concentration phenomenon of the conditional variance, which we will try to explain below (see Section 1.2).

1.1 Outline proof strategy for Theorem 1↩︎

The usual framework for proving a limiting distribution for partial sums of random multiplicative functions (originating in Harper’s work [3] on sums with few prime factors) is by applying a suitable version of the martingale central limit theorem. This is the same framework we will use but with several new additional features, both conceptual and technical. Essentially, we require a fusion of martingale technique with some quite delicate barrier arguments, developing the previous work of e.g. Caich [6], Harper [1], [12] and Xu [13].

We apply a version of McLeish’s central limit theorem [14] developed in [5]. A direct application would require showing that the fourth moment of the random sum \(\sum_{x\le n \le x+y}f(n)\) (after a suitable pruning of the terms) is dominated by the diagonal contribution. However, in fact the fourth moment blows up once \(y\) is large enough. This forces the applicable range of \(y\) to be \(\ll x/(\log x)^{c}\) for certain \(c>0\), exactly the limitation of the results in [4] and [5].

To overcome the limitation, our strategy is to apply the martingale central limit theorem only after conditioning on small primes \(f(p)\) for \(p\le z\) (for a parameter \(z\) that needs to be carefully chosen), instead of applying it to the full sum from the beginning. For our choice of \(z\), the set of integers in \([x,x+y]\) with all prime factors below \(z\) has negligible size, and may be ignored. By using the multiplicative property of \(f\), the sum over the remaining integers may be decomposed as \[\sum_{\substack{x \le n \le x+y \\ P(n) > z } } f(n) = \sum_{ \substack{ 1<m\le x+y \\ p|m \implies p>z}} f(m) \sum_{\substack{x/m\le n \le (x+y)/m \\ p|n \implies p\le z } } f(n),\] where henceforth we write \(P(n)\) for the largest prime factor of \(n\). After the conditioning, the inner sum is fixed and can be viewed as a constant coefficient \(a(m)\) indexed by \(m\). Then one can apply the version of the martingale central limit theorem [5] for general sums \(\sum_{m}a(m)f(m)\) aiming to get a conditional central limit theorem. As noted, to make this work we need to choose \(z\) carefully. Suppose such a choice of \(z\) exists and the theorem in [5] can be successfully applied. Then this implies that given a fixed choice of \(f(p)\) for \(p\le z\), the random sum is approximately Gaussian, but with a conditional variance \(V_f(x, y) = \sum_{m}|a(m)|^2\) which depends on the values of \(f(p)\) for \(p\le z\).

In order to get a limiting distribution for the original random sum, we need to “reveal” the conditioned choices of \(f(p)\) with \(p\le z\) and understand the behavior of the conditional random variance \(V_f(x, y)\). It is not clear a priori whether \(V_f(x, y)\) has genuinely nontrivial distribution (as the analogous object does for the long sum \(\sum_{n \leq x} f(n)\)) or typically behaves like a deterministic quantity, i.e., for asymptotically almost all realizations of \(f(p)\) with \(p\le z\), it is asymptotic to some deterministic quantity \(V(x, y)\) (which doesn’t depend on \(f\), only on \(x, y\)). One important feature of our proof is a concentration phenomenon which happens with the choice \(z = x^{\frac{1}{\log \log \log x}}\) (say): namely, for a deterministic quantity \(V(x,y)\), one has typically \(V_f(x, y)\sim V(x, y)\). Moreover, at the same time, this choice of \(z\) is eligible for passing through the conditions of the martingale central limit theorem in [5].

In the next couple of subsections, we will try to explain a little more of what is involved in the conditioned application of the martingale central limit theorem (which amounts to needing to understand something like the “off-diagonal” contribution to a conditional fourth moment), and in showing the crucial concentration of the conditional variance.

In the process of proving the concentration result, we see the phase transition in \(V(x, y)\) when \(y\) gets close to \(x\). As found in previous work [4], [5], \(\frac{1}{\sqrt{y}} \sum_{x\le n \le x+y}f(n)\) has the expected Gaussian limiting distribution when \(y\ll x/(\log x)^{c}\) for suitable \(c\), so our choice of \(V(x, y)\) should satisfy \(V(x, y)\sim y\) (the obvious value, namely the variance of \(\sum_{x\le n \le x+y}f(n)\)) for such \(y\). However, as \(y\) approaches \(x\), a “better than square-root cancellation" phenomenon shows up, produced by multiplicative chaos effects. Indeed, the transition point of \(y\) is closely connected to a ballot-type problem in the theory of random walks. The work of Caich [6] gives that \(y\approx x/\exp((\log \log x)^{1/2})\) is a transition point for”better than square-root cancellation" (working at the level of order of magnitude of low moments), which is also the transition point for the shape of \(V(x, y)\) changing from \(y\) to \(\asymp y \cdot \frac{\log(x/y)}{\sqrt{\log \log x}}\). We also refer readers to [13], [15] for related proofs exploring such transitions (again at the order of magnitude level) in other settings.

At a very high level, one might have in mind that a standard way of proving a central limit theorem (although generally not easy to apply to random multiplicative function problems) is the method of moments, computing all moments of the random variables of interest and showing they converge to Gaussian moments. More sophisticated methods, such as martingale techniques, can essentially reduce from needing to compute all moments to just the fourth moment. Chatterjee and Soundararajan’s work [4] on short interval sums, using Stein’s method, operates on this level. To obtain their improvement in the range of \(y\), Soundararajan and Xu [5] continue to work on the fourth moment level (now using martingales), but first remove a sparse subset of integers from the short interval sum so that the fourth moment behaves well on a wider range. In proving Theorem 1 on the full range of \(y\), we work in a regime where not only the fourth moment, but even the second moment, does not behave well and reflect the genuine distributional behaviour. As we shall try to explain, to address this we make use of “barriers” editing our random variables (after conditioning) on the “Fourier side” of random Euler products, rather than removing terms from the original sums. The argument thus becomes somewhat indirect, but ultimately explains clearly the cause of the transition in \(V(x,y)\), and the difference between short interval sums (where \(y=o(x)\)) and long sums. We are not aware of any way to understand this transition operating simply with \(\sum_{x\le n \le x+y}f(n)\) in “physical space”.

1.2 Concentration of conditional variance↩︎

As a first step, one can see that the definition \(\sum_{m}|a(m)|^2\) of \(V_f(x, y)\) is a discrete mean square of certain random sums (see 5 , below). As usual in this area, we apply Parseval’s identity to transform the sum to its Fourier side, i.e. to some continuous second moment \(\int |F_z(s)|^{2}\,ds\) of the random Euler product \(F_z(s) := \prod_{p\le z}(1-\frac{f(p)}{p^{s}})^{-1}\). This step is standard but more demanding than in many previous works (compare with e.g. Caich’s paper [6]), since here we need an exact asymptotic expression for \(V_f(x, y)\). Even the loss of a multiplicative constant factor would break the proof of concentration and of a limiting distribution. This generates some technical issues, but these are manageable and we find (with high probability over the \((f(p))_{p \leq z}\)) that \[V_f(x, y) \sim \frac{e^{-\gamma}}{4\pi} \frac{y}{\log z} \frac{1}{T} \int_{-T(\log T)^{100}}^{T(\log T)^{100} } |F_z(\tfrac{1}{2}+it)|^{2}|K_T(t)|^{2} dt,\] where \(K_T(t)\) is a certain real-valued kernel which decays rapidly for large \(t\) and thus allows the integral to be truncated at \(|t|\approx T\) with \(T \approx 2x/y\). Notice that going from an interval \([x,x+y]\) in physical space, to an integral of effective length \(\asymp x/y\) on the Fourier side, is consistent with the usual numerology in multiplicative number theory, in particular shorter intervals require more frequencies \(t\) on the Fourier side. This reformulating of \(V_f(x, y)\) forms the bulk of Section 2.

We must proceed to analyze \(V_f(x, y)\) further, needing to understand this not just in (typical) order of magnitude, but seeking a more delicate concentration result. One might hope that an estimate like \[\label{dream} \mathbb{E}[|V_f(x, y) - V(x, y)|^{2}] = o(V(x, y)^{2})\tag{4}\] holds for our candidate \(V(x, y)\), which if true would establish that typically \(V_f(x, y)\sim V(x, y)\). However, when \(y\) is large this second moment type computation (which, since \(V_f(x, y)\) is itself a mean square, is really like a fourth moment) is dominated by certain relatively rare events, and (as expected given the failure of a direct application of McLeish’s theorem) we cannot achieve such concentration directly. To overcome this, the crucial idea is to impose barrier events \(\mathcal{G}(t)\) on the growth of the partial Euler products \(F_z(\frac{1}{2}+it)\) at all different “scales”. We hope that \(\mathcal{G}(t)\) holds, simultaneously at all relevant points \(t\), with high probability, and then it would suffice to show that 4 holds with the restriction \(\mathcal{G}(t)\) inside our integral approximation to \(V_f(x, y)\). The purpose of \(\mathcal{G}(t)\) is to suppress any blow-up of 4 created by rare extreme events. Setting barrier events is a well developed tool in the probabilistic study of so-called multiplicative chaos, and it first shows up in [1] for the study of random multiplicative functions.

However, here it turns out that inserting a high probability event \(\mathcal{G}(t)\) is still not enough to establish the concentration estimate. We solve this by inserting a further refined constraint \(\mathcal{H}(t)\) and show that \(\mathbb{E}[\mathbb{1}_{\mathcal{G}(t)}\mathbb{1}_{\mathcal{H}(t)~\text{fails}}|F_z(\frac{1}{2}+it)|^{2} ]\) is very small. We emphasize that the combination of \(\mathcal{G}(t)\) with \(\mathcal{H}(t)\), and of analyzing what happens in high probability along with what happens in mean square, is crucial here. We cannot show that the stronger barrier event \(\mathcal{H}(t)\) holds at all points \(t\) with high probability, and neither is it true that \(\mathbb{E}[\mathbb{1}_{\mathcal{H}(t)~\text{fails}}|F_z(1/2+it)|^{2} ]\) (without the initial barrier \(\mathbb{1}_{\mathcal{G}(t)}\)) is small when \(y\) is large (in fact it would be \(\sim \mathbb{E}[|F_z(\frac{1}{2}+it)|^{2} ]\)). The combination of the two barriers allows us further to assume \(\mathcal{H}(t)\) holds, and then we aim for a concentration result like \[\mathbb{E}\Big[\Big|\int_{-T(\log T)^{100}}^{T(\log T)^{100}}|F_z(1/2+it)|^{2} K_T(t)^{2} \mathbb{1}_{\mathcal{H}(t)} dt - \text{Main Term}\Big|^{2}\Big] = o((\text{Main Term})^{2}).\] This is accomplished in Section 3. Here the Main Term is simply \[\mathbb{E}\Big[ \int_{-T(\log T)^{100}}^{T(\log T)^{100}}|F_z(1/2+it)|^{2} K_T(t)^{2} \mathbb{1}_{\mathcal{H}(t)} dt\Big],\] and thus we take \(V(x, y)\) to be a scaled version of this. The transition 3 in \(V(x,y)\), as \(y\) varies with \(x\), precisely reflects the effect of \(\mathbb{1}_{\mathcal{H}(t)}\) (which translates into multiplication by a certain random walk probability) inside this expectation.

This is the point in the argument where we see the conceptual origin of \(V(x, y)\), and we shall try to expand on what is happening. It turns out that the Euler products \(F_z(\frac{1}{2}+it)\) behave more or less independently on points \(t\) that are at least 1 apart, say. Thus our integral, which is (recalling that \(K_T(t)\) basically serves to truncate things at \(|t|\approx T\)) \[\approx \sum_{|n| \leq T(\log T)^{100}} \int_{n-\frac{1}{2}}^{n+\frac{1}{2}}|F_z(\tfrac 12+it)|^{2} K_T(t)^{2} \mathbb{1}_{\mathcal{H}(t)} dt \approx \sum_{|n| \leq T} \int_{n-\frac{1}{2}}^{n+\frac{1}{2}}|F_z(\frac{1}{2}+it)|^{2} \mathbb{1}_{\mathcal{H}(t)} dt,\] has the shape of a sum of \(\approx T\) roughly independent terms. Provided that \(T \rightarrow \infty\) (which exactly means \(y=o(x)\)), results like the Law of Large Numbers lead one to expect this to concentrate around its mean. We find this is indeed the case, but since each piece \(\int_{n-\frac{1}{2}}^{n+\frac{1}{2}}|F_z(\frac{1}{2}+it)|^{2} \mathbb{1}_{\mathcal{H}(t)} dt\) is heavy-tailed it is a delicate matter to establish this.

In fact, when we expand the square we need to deal with the expectation of a product of two factors \(\mathbb{E}[|F_{z}(\frac{1}{2}+it_1)|^{2}\mathbb{1}_{\mathcal{H}(t_1)} |F_{z}(\frac{1}{2}+it_2)|^{2} \mathbb{1}_{\mathcal{H}(t_2)}]\). When \(|t_1 - t_2| \rightarrow \infty\) we can show almost perfect decorrelation; this requires a slightly different analysis than in previous works (see Lemma 2, and its application inside the proof of Proposition 8 below). We can also prove a decorrelation result when \(|t_1-t_2|\) is smaller (see Lemma 4), approximating the expectation by the product of two expectations (each with the corresponding barrier events involved), but now this can only be done for the parts of the Euler products involving sufficiently large primes in terms of \(|t_1-t_2|\). The barrier conditions \(\mathcal{H}(t_1), \mathcal{H}(t_2)\) are invoked to control the small prime contributions. More precisely, to succeed we need the barriers to restrict the small prime contribution to be (up to some normalization) a bit less than \(\sqrt{T}\), so when squared inside \(|F_{z}(1/2+it_1)|^{2}\) we get a factor a bit smaller than the total integral length \(T\). The high probability barrier \(\mathcal{G}(t_1)\) can only impose a restriction a bit larger than \(\sqrt{T}\), but the refined barrier \(\mathcal{H}(t_1)\) enables us to go a bit below this, and so complete the concentration argument. The factor \(\min \{1, \frac{\log (x/y)}{\sqrt{\log \log x}}\} \sim \min \{1, \frac{\log T}{\sqrt{\log \log x}}\}\) that arises in \(V(x, y)\) reflects this size \(\approx \sqrt{T}\), on taking a logarithm we get a restriction \(\approx \frac{1}{2}\log T\) (up to some recentering) on the logarithms of the partial Euler products (which are random sums over primes), the denominator \(\sqrt{\log \log x}\) reflects the “number of scales” in the sums.

In summary, one should think there is a sort of competition between the effect of the heavy-tailedness of each piece \(\int_{n-\frac{1}{2}}^{n+\frac{1}{2}}|F_z(\frac{1}{2}+it)|^{2} \mathbb{1}_{\mathcal{H}(t)} dt\) (kept somewhat under control by the barrier \(\mathcal{H}(t)\)), and the concentration effect of summing \(\approx T\) pieces. Provided \(T \rightarrow \infty\), the concentration effect wins out, but with the “rate” of concentration (i.e. the exceptional probability that \(V_f(x, y)\) is not close to \(V(x, y)\)) becoming worse when \(T\) grows more slowly. Once \(y \asymp x\), and so \(T \asymp 1\), the heavy-tailedness dominates and one no longer gets concentration.

1.3 Asymptotic of the fourth moment↩︎

In the application of the (conditioned) martingale central limit theorem, the main issue is a certain conditional fourth moment computation. We need to demonstrate that all “off-diagonal” contributions are negligible. To illustrate the problem, one major task is to show that, with high probability over the \((f(p))_{p \leq z}\) and as \(x\to \infty\), \[\Big| \sum_{\substack{ m_1, m_2, m_3, m_4 \in \mathcal{A}\\ m_1 m_2= m_3 m_4 \\ m_1 \neq m_3, m_2\neq m_4 \\ P(m_1)=P(m_3) \\ P(m_2)=P(m_4)}} \;\;\; \sum_{\substack{n_1, n_2, n_3, n_4 \\ \frac{x}{m_j} \le n_j \le \frac{x+y}{m_j} \\ P(n_j)\le z}} f(n_1)f(n_2)\overline{f(n_3) f(n_4)} \Big| =o(V_f(x, y)^2),\] where \(\mathcal{A}\) is the set of integers in \((1,x+y]\) with all their prime factors \(> z\). Our approach is to write each inner sum over \(n_j\) as an integral involving \(F_z(s)\) by a suitable smoothed Perron formula: \[\sum_{\substack{n \in [\frac{x}{m}, \frac{x+y}{m}]\\P(n)\le z}} f(n) = \frac{1}{2\pi }\int_{\infty}^{\infty} F_z(\tfrac 12 +it) \Big(\frac{x}{m}\Big)^{\frac{1}{2}+it} W(t)dt,\] where the smooth weight \(W(t)\) effectively restricts the integral to \(|t|\le (\log x)^{100}\). Then the quantity we need to estimate is a quadruple integral involving the product of four copies of random Euler products \(F_z(\frac{1}{2}+ it_j)\) (\(1\le j \le 4\)), and importantly a sort of harmonic sum that comes from the \(1/m^{s}\) factors, namely \[G(t_1, t_2, t_3, t_4):= \sum_{\substack{m_1, m_2, m_3, m_4 >1\\ p|m_j \implies p\in (z,x+y] \\ m_1m_2=m_3m_4\\ m_1\neq m_3, m_2\neq m_4 \\ P(m_1)= P(m_3) \\ P(m_2) =P(m_4) }} \frac{1}{m_1^{\frac{1}{2} + it_1}m_2^{\frac{1}{2} + it_2}} \cdot \frac{1}{m_3^{\frac{1}{2} - it_3}m_4^{\frac{1}{2} - it_4}} .\] One can show (with some technical work, see Lemmas 6 and 7 below) that the \(G(t_1, t_2, t_3, t_4)\) factor provides a good saving unless \(t_1, t_2, t_3, t_4\) are all very close to each other. Thus we may restrict our attention to the case that all four \(t_j\) are close, and the resulting quadruple product would behave like the product of two squares \(|F_z(1/2+it_1)|^{2}\) and \(|F_z(1/2+it_2)|^{2}\) with \(t_1\) and \(t_2\) close. Indeed, we are left to deal with a quantity like (with a suitable weight \(W\)) \[\int_{\substack{t_1\approx t_2 \\ |t_1|, |t_2|\le (\log x)^{100}}} |F_z(1/2+it_1)|^{2}|F_z(1/2+it_2)|^{2} |W(t_1)|^2 |W(t_2)|^2 dt_1 dt_2.\]

It would suffice to show that the expectation of this integral is “small” (essentially compared with \(V(x,y)^2\), up to some scaling factors that we have suppressed). But as with 4 , or a direct computation of the fourth moment, the expectation would in fact blow up. Instead, we use the method of barrier events again, and just like before, we need to put some barriers \(\mathcal{G}^{*}(t)\) and \(\mathcal{H}^{*}(t)\) on the growth of the Euler products. With \(\mathcal{G}^{*}(t)\) and \(\mathcal{H}^{*}(t)\) set appropriately, this final analysis becomes relatively straightforward, see Propositions 11, 12 and the surrounding calculations below. We note, however, that this is a point in the proof where the choice of \(z\) is crucial. We lose factors of the shape \((\frac{\log x}{\log z})^{O(1)}\) and need the saving coming from \(\mathcal{H}^{*}(t)\), which is very limited, to (more than) compensate for this. Thus although \(z\) cannot be too close to \(x\) (in order for earlier steps of the argument to work), it also cannot be too far from \(x\).

One small difference between the barrier events here and the previous \(\mathcal{G}(t)\) and \(\mathcal{H}(t)\), is that our integral here has long length \((\log x)^{100}\). Thus our barrier events must take the size of \(t\) into account, in fact when \(|t| \geq T\) we need the barrier to increase a bit with \(t\) (becoming weaker) so that the exceptional probability of exceeding the barrier decreases proportional to \(|t|\), otherwise a union bound on the exceptional probabilities would blow up. But the decay factors involving \(W(t_1)\) and \(W(t_2)\) compensate for this weakening.

1.4 Further remarks↩︎

We already mentioned that, for the appropriately normalized long sum \(\frac{(\log \log x)^{\frac{1}{4}}}{\sqrt{x}} \sum_{n \leq x} f(n)\) of a Steinhaus random multiplicative function, a recent breakthrough of Gorodetsky and Wong [10] establishes convergence in distribution to a non-Gaussian, heavy tailed limit. A little earlier, S. Hardy [11] established the analogous result for the sub-sum \(\frac{(\log \log x)^{\frac{1}{4}}}{\sqrt{x}} \sum_{n \leq x, P(n) > \sqrt{x}} f(n)\) over integers having a large prime factor. The simplifying condition \(P(n) > \sqrt{x}\) means that Hardy does not require any martingale techniques (after conditioning on \((f(p))_{p \leq \sqrt{x}}\), he has a classical weighted sum of independent random variables to work with). The bulk of his task is understanding the conditional variance, which he finds to converge to a heavy-tailed distribution closely related to the total mass of critical multiplicative chaos (unlike the concentration that we find in the short interval setting here). Hardy’s proof of convergence ultimately relies on some existing results from the theory of multiplicative chaos, but to make these applicable he must bring the conditional variance into an appropriate form. This uses a barrier-weighted mean square argument, sharing some features with our Proposition 8 below.

Gorodetsky and Wong [10], handling the full sum, require martingale theory. A major task for them is the analysis of the arising “bracket process”, which is the same type of object as the conditional variance investigated by Hardy [11] and by us. Gorodetsky and Wong’s beautiful proof has many features in common, and also many differences from, the work of Hardy [11] and our work here, and we only mention a few points. Firstly, to bring the conditional variance into a nice form we work mostly on the “Fourier analytic” side, and Hardy does likewise. Gorodetsky and Wong’s bracket process has a bit more complicated structure, and they require a non-trivial “truncation” procedure on the physical side to produce an object they can work with. Secondly, Gorodetsky and Wong eschew the use of barriers, which are crucial to our work and which Hardy also makes some use of. Instead, they develop and use a new penalized second moment method, see their paper [10] for a detailed discussion of this. (Although behind this lie some of the same tools, like an approximate Girsanov theorem, that lie behind our barrier calculations.) Thirdly, note that about a third of our work here (almost all of section 4) is the analysis of a conditional fourth moment contribution, to make the martingale central limit theorem applicable. Because of the form of martingale central limit theorem that they use, Gorodetsky and Wong do not require this type of calculation, but instead they require (and establish) demanding information about the nature of the convergence of their bracket process.

Inspecting our proofs (see Theorem 4 below), the reader will see that we show \(V_f(x, y)\sim V(x, y)\) holds with probability \(1- O((\log T)^{-\frac{1}{5}})\). Recall that \(T \approx 2x/y\). On first glance this rate estimate may look weak, but in fact it probably is not far from the truth, apart from the precise exponent of \(\log T\). For examining our integral expression for \(V_f(x, y)\), we see that if any of the integrals \(\int_{n-\frac{1}{2}}^{n+\frac{1}{2}}|F_z(\frac{1}{2}+it)|^{2} dt\), for \(|n| \leq T\), exceeds a large multiple of \(T \log T \frac{\log z}{\sqrt{\log \log z}}\) then \(V_f(x, y)\) will be significantly larger than \(V(x, y)\). If \(T\) isn’t too big compared with \(z\) (i.e. if \(y\) is reasonably close to \(x\)), the probability of this for any given \(n\) should be \(\approx \frac{1}{T\log T}\), see e.g. the discussion of tail probabilities in the introduction of Harper’s paper [1]. And since these integrals should behave more or less independently for different \(n\), the probability of it happening for some \(|n| \leq T\) will be \(\approx \frac{T}{T\log T} = \frac{1}{\log T}\). This further suggests that our arguments, although involved, are capturing the true behavior of our objects of study.

Theorem 1 solves the short interval story completely for Steinhaus random multiplicative functions, but not the Rademacher case. Most steps in our proof should transfer to the Rademacher case, either immediately or with quite obvious modifications, and we certainly expect the analogous theorem to hold (now with a real Gaussian limit, and ignoring any issues with counting square-free numbers when \(y\) is very small). However, there is at least one place where some less trivial work would be needed. In the proof of Proposition 8, after applying a decorrelation estimate to the “large” prime contributions, we factor these out exploiting the fact that the distribution of \(\mathbb{1}_{\mathcal{H}(t)} \prod_{ z^{e^{-\tau}}<p\le z} |1-\frac{f(p)}{p^{\frac{1}{2}+it}}|^{-2}\) is exactly the same for all \(t \in \mathbb{R}\) (because the joint distribution of the sequence \((f(p)p^{-it})_{p}\) is exactly the same for all shifts \(t \in \mathbb{R}\)). This “translation invariance in law” does not hold in the Rademacher case. It should be possible to address this (e.g. although translation invariance in law does not hold, it should “almost” hold provided \(t\) isn’t too small); we hope that an interested reader will take up this task.

1.5 Organization and notations↩︎

It will suffice to prove Theorem 1 in the range \(\frac{x}{\sqrt{\log x}} \leq y = o(x)\), since smaller \(y\) are already handled by the work of Soundararajan and Xu [5]. Although not essential, restricting to this range for \(y\) will streamline the writing of several of our arguments.

We write the conditional variance in a more useful asymptotic form, in terms of the random Euler product \(F_z(s)\), in the fairly short Section 2. The concentration of conditional variance is established in Section 3 and the fourth moment computation together with other verifications of conditions in the martingale central limit theorem is done in Section 4. Thus, Theorem 1 is proved by combining results in Section 2, Section 3 and Section 4. Finally, we prove Proposition 2 in Section 5.

We use standard notations of analytic number theory. Write \(f(x)\ll g(x)\) and \(f(x) =O(g(x))\) to denote that there exists a constant \(C>0\) such that \(|f(x)|\le Cg(x)\) for all \(x\). Lastly \(f(x)\asymp g(x)\) means that \(g(x)\ll f(x)\ll g(x)\).

Acknowledgements↩︎

A.J.H. was supported in part by the Engineering and Physical Sciences Research Council of the United Kingdom [grant EP/V055755/1]. Some of the research for this paper was conducted when A.J.H. visited K.S. and M.W.X. in April 2023, and he would like to thank Stanford University for their hospitality during this visit. Some of the final writing up was done with support from the Simons Foundation and the Centre de Recherches Mathématiques, Montréal, while A.J.H. was in residence as Aisenstadt Chair during the 2026 Universal Statistics in Number Theory thematic semester. K.S. is partially supported through a grant from the National Science Foundation; some of this work was also done while he was supported by a Simons Investigator grant from the Simons Foundation. M.W.X. was supported by a Simons Junior Fellowship from the Simons Foundation. He appreciates the hospitality of Warwick Mathematics Institute, and he thanks Ye Tian for hosting him at the Morningside Center of Mathematics, where some of the writing up was done.

We would like to thank Ofir Gorodetsky, Seth Hardy, Youness Lamzouri and Mo Dick Wong for their helpful comments on a draft of this paper. We especially thank Andrew Granville, whose comments and questions led us to determine an explicit form for the normalization factor \(V(x,y)\).

For the purpose of open access, the authors have applied a Creative Commons Attribution (CC-BY) licence to any Author Accepted Manuscript version arising from this submission.

2 Conditional variance↩︎

Recall that \(z = x^{\frac{1}{\log \log \log x}}\), and that we say a number \(n\) is \(z\)-smooth if its largest prime factor \(P(n)\) is \(\leq z\). As is usual, we write \(\Psi(x,z)\) to mean the count of \(z\)-smooth numbers \(\leq x\) (although we will not need this notation very much).

Since the \(f(n)\) are orthogonal, \[\mathbb{E}\Big| \sum_{\substack{x \le n \le x+y \\ P(n)\le z } } f(n) \Big|^{2} = \sum_{\substack{x \le n \le x+y \\ P(n)\le z } } 1 = \Psi(x+y, z) - \Psi(x, z) + O(1) \ll \frac{y}{(\log\log x)^{100}} ,\] where the last inequality follows upon using the sub-additivity \(\Psi(x+y, z) - \Psi(x, z) \le \Psi(y, z)\) proved in [16] (for \(x,y \geq z\)) and a classical estimate on \(\Psi(y, z)\), see for example [17]. Since this is negligible compared with \(V(x,y)\) in Theorem 1, we see that we may discard the sub-sum over \(z\)-smooth numbers, and it will suffice to prove the theorem for \(\sum_{\substack{x \le n \le x+y \\ P(n) > z } } f(n)\). Using the multiplicativity of \(f\), this sum may be rewritten as \[\sum_{\substack{x \le n \le x+y \\ P(n) > z } } f(n) = \sum_{ \substack{ 1<m\le x+y \\ p|m \implies p>z}} f(m) \sum_{\substack{x/m\le n \le (x+y)/m \\ p|n \implies p\le z } } f(n) ,\] which we shall analyze by first conditioning on the behavior of \((f(p))_{p \leq z}\).

In this section, we establish an asymptotic expression for the conditional variance \(V_f(x, y)\), which holds with asymptotic probability 1 (over realizations of the \((f(p))_{p \leq z}\)). By orthogonality, note that the conditional variance satisfies \[\label{eqn:32Vf} V_f(x, y) = \sum_{ \substack{ 1<m\le x+y \\ p|m \implies p>z}} \Big| \sum_{\substack{x/m\le n \le (x+y)/m \\ p|n \implies p\le z } } f(n) \Big|^{2}.\tag{5}\]

The specific value \(T=2/\log(1+y/x)\) simply arises from a change of variables in the course of the proof; note that \(T=2x/y + O(1)\). In section 3, we will prove that with probability \(1-O((\log T)^{-\frac{1}{5}})\), \[V_f(x, y) \gg \frac{y}{\sqrt{\log \log x}},\] so that the remainder term in ?? is negligible with high probability.

3 Concentration of the conditional variance↩︎

In this section, we prove that with asymptotic probability 1 over realizations of the \((f(p))_{p \leq z}\), the conditional variance \(V_f(x, y)\) is concentrated around a deterministic quantity \(V(x,y)\) with the properties claimed in 3 and 2 . In the next section, this will allow us to establish an unconditional central limit theorem for \(\sum_{\substack{x \le n \le x+y \\ P(n) > z } } f(n)\) (and thus for \(\sum_{x\le n \le x+y}f(n)\)).

In view of Proposition 3, to prove Theorem 4 it will suffice to show that with probability \(1- O((\log T)^{-\frac{1}{5}})\) we have (with \(\widetilde{T}=T(\log T)^{100}\) as before) \[\frac{1}{4\pi T} \int_{-\widetilde{T}}^{\widetilde{T}} |F_z(\tfrac{1}{2}+it)|^{2} K_T(t)^{2} dt = \mu \Big(1 + O\Big(\frac{1}{(\log T)^{\frac{1}{5}}}\Big)\Big),\] where \(\mu\) has the properties claimed in the theorem. The integral only depends on \(x\) and \(y\) via \(z = x^{\frac{1}{\log \log \log x}}\) and \(T=2/\log(1+y/x) = 2x/y + O(1)\). Likewise, \(\mu\) will only depend on \(x\), \(y\) via \(z\) and \(T\), see Proposition 6 below.

Overall strategy for establishing Theorem 4↩︎

Our proof will proceed in three steps:

1. We set up an initial barrier event \(\mathcal{G}(t)\) holding with high probability for all \(|t| \leq \widetilde{T}\), such that with high probability \[\int_{-\widetilde{T}}^{\widetilde{T}} |F_z(\tfrac{1}{2}+it)|^{2} K_T(t)^{2} dt = \int_{-\widetilde{T}}^{\widetilde{T}} |F_z(\tfrac{1}{2}+it)|^{2} K_T(t)^{2} \mathbb{1}_{\mathcal{G}(t)} dt.\]

2. With high probability, we replace \(\mathcal{G}(t)\) by a stronger (more restrictive) barrier event \(\mathcal{H}(t)\) with little loss. That is, with high probability one has \[\int_{-\widetilde{T}}^{\widetilde{T}} |F_z(\tfrac{1}{2}+it)|^{2} K_T(t)^{2} \mathbb{1}_{\mathcal{G}(t)} dt \approx \int_{-\widetilde{T}}^{\widetilde{T}} |F_z(\tfrac{1}{2}+it)|^{2} K_T(t)^{2} \mathbb{1}_{\mathcal{H}(t)} dt.\] Unlike the event \(\mathcal{G}(t)\) which holds with high probability for all \(|t|\le T(\log T)^{100}\), the event \(\mathcal{H}(t)\) is not required to hold for all \(t\) with high probability. Instead, the total contribution from points \(t\) where it fails (but \(\mathcal{G}(t)\) holds) is adequately small with high probability.

3. We establish the desired concentration estimate (in fact a much stronger one) for \(\int_{-\widetilde{T}}^{\widetilde{T}} |F_z(\frac{1}{2}+it) |^{2} K_T(t)^{2} \mathbb{1}_{\mathcal{H}(t)} dt\), via a variance calculation heavily relying on the presence of \(\mathbb{1}_{\mathcal{H}(t)}\).

Before carrying out this strategy, we first prove a much simpler concentration result (a variance type calculation without any barriers) that performs well when only rather small primes (compared with \(T\)) are involved.

Note that \[\sum_{\substack{n\ge 1\\ p|n \implies p\le w}} \frac{1}{n} \int_{-H}^{H} K_T(t)^{2} dt = \prod_{p \leq w} \Big(1 - \frac{1}{p}\Big)^{-1} \int_{-H}^{H} K_T(t)^{2} dt \asymp T\log w,\] so the lemma provides genuine concentration if the right hand side is smaller than \((T\log w)^2\); that is, if \(\log w\) is smaller than \(\sqrt{T}\). This lemma will be useful in the proof of Proposition 8, for an appropriate \(w\) chosen in terms of \(T\), to handle the contribution from the small primes while a different argument takes care of larger primes.

We now begin work on executing the proof strategy described earlier. Let \(\tau\) denote the smallest natural number such that \[z^{e^{-\tau}} \le \exp( \exp( (\log T)^{1/200})),\] so that (by our choice of \(z\) and as \(T \ll \sqrt{\log x}\)) \[\tau = \log \log z - (\log T)^{\frac{1}{200}} +O(1) \sim \log \log z \sim \log \log x.\] Let \(B\) be a fixed large constant, chosen sufficiently large to ensure that the probabilistic results invoked below hold.

For each integer \(0\le j\le \tau\), it will be convenient to define the partial random Euler product \[\label{3463} {\mathcal{F}}_j(s) = \prod_{z^{e^{-\tau}} < p \le z^{e^{-j}}} \Big( 1- \frac{f(p)}{p^s}\Big)^{-1};\tag{15}\] note that the final Euler product \({\mathcal{F}}_\tau\) is empty (and thus equals \(1\)). Note that for any \(t\in \mathbb{R}\) (and large \(T\)) \[\label{3464} \mathbb{E}[ |{\mathcal{F}}_j(1/2+it)|^2 ] = \prod_{z^{e^{-\tau}} < p \le z^{e^{-j}}} \Big( 1- \frac{1}{p}\Big)^{-1} \sim e^{\tau-j}.\tag{16}\]

For each \(t \in \mathbb{R}\), let \(\mathcal{G}(t)\) denote the event that for all \(0\le j\le \tau\) \[\label{eqn:32g40t41} (\sqrt{T} e^{\tau-j})^{-B} \le |{\mathcal{F}}_j(\tfrac 1 2 +it)| \le \sqrt{T} e^{\tau-j}\exp((\log T)^{\frac{1}{100}}) (\log \log z - j).\tag{17}\] The stronger barrier event \(\mathcal{H}(t)\) demands that the above holds and in addition (for all \(0\le j\le \tau\)), \[\label{eqn:32AJH32h40t41} |{\mathcal{F}}_j(\tfrac 12+it)| \le \sqrt{T} e^{\tau-j} \frac{\exp(-(\log T)^{\frac{1}{100}})}{ (\log \log z - j)^{5}}.\tag{18}\] Note that when \(j=\tau\) the Euler product \({\mathcal{F}}_j\) is empty, and the conditions 17 and 18 hold automatically.

The lower bound in the definition of \(\mathcal{G}(t)\) is purely technical, the upper bound constraint is the important part. Then \(\mathcal{H}(t)\) will give us a small, but crucial, extra saving in terms of both \(T\) and \((\log\log z - j)\) (which, as the reader may check, cannot be incorporated directly into our proof that \(\mathcal{G}(t)\) holds with high probability).

More specifically, the factor \((\log \log z - j)^{-5}\) ultimately leads to the uniform boundedness of an integral appearing in the proof of Proposition 8, below. The factor \(\exp(-(\log T)^{\frac{1}{100}})\) is needed both to overcome losses because our barrier conditions only start at \(z^{e^{-\tau}}\), and in producing the final saving in Proposition 8 (we note in passing that we could afford to save much less there and would still obtain Theorem 4, the bounds flowing from Proposition 7 below are anyway weaker). There is quite a lot of flexibility in the choice of the factor \(\exp(-(\log T)^{\frac{1}{100}})\) and of the start point \(z^{e^{-\tau}}\), but this must grow with \(T\) at a certain rate so that the probabilistic results we shall invoke are valid (this ultimately corresponds to having good enough error terms for prime number sums of length \(z^{e^{-\tau}}\)).

We now establish our first step, showing that the barrier event \({\mathcal{G}}(t)\) holds for all \(|t|\le \widetilde{T}\) with high probability. The proof combines second moment calculations together with the union bound and a discretization modeled after Harper [1] and (especially) Soundararajan and Zaman [20].

Our next proposition will compute the expectation of \(|F_z(\frac{1}{2} + it)|^{2}\) restricted to the barrier event \(\mathcal{H}(t)\). Here we see the appearance and properties of the quantity \(\mu\), which is related to the value to which our variance will concentrate. In order to obtain sufficiently strong information about \(\mu\) to deduce 2 , we will need the following Gaussian random walk lemma, which (although not hard to prove) is more precise than usually required in random multiplicative function problems.

We next show that the expected contribution of \(|F_z(\tfrac 12+it)|^2\) when \(\mathcal{G}(t)\) holds but \(\mathcal{H}(t)\) fails is small when compared with \(\mu\). In combination with Markov’s inequality, this will later allow us to replace \(\mathcal{G}(t)\) with \(\mathcal{H}(t)\), as per Step 2 of the strategy outlined above. Note that since \({\mathcal{H}}(t)\) imposes stronger restrictions than \({\mathcal{G}}(t)\), the expression \({\mathbb{1}}_{{\mathcal{G}}(t)}- {\mathbb{1}}_{{\mathcal{H}}(t)}\) is the indicator function of the event that \({\mathcal{G}}(t)\) holds but \({\mathcal{H}}(t)\) fails.

The final (vital) ingredient in proving Theorem 4 will be the following “barrier-modified" variance estimate, crucially exploiting the presence of \(\mathcal{H}(t)\) (to control the near diagonal contribution when everything is expanded out), whose proof we postpone for a moment.

From the definition of \(K_T(t)\) (see ?? ) note that \[K_T(t)^2 = \frac{T^2 (e^{1/T} +e^{-1/T}-2)}{1/4+t^2} + 4 \frac{T^2 (\sin (t/T))^2}{1/4+t^2},\] from which it follows that \[\begin{align} \label{34621} \int_{-\widetilde{T}}^{\widetilde{T}} K_T(t)^2 dt &= \int_{-\infty}^{\infty} K_T(t)^2 dt +O\Big(\int_{\widetilde{T}}^{\infty} \frac{T^2}{t^2}dt\Big) = \int_{-\infty}^{\infty} \frac{4T^2 (\sin (t/T))^2}{1/4+t^2} dt +O\Big(\frac{T}{(\log T)^{100}}\Big) \nonumber \\ &= T \int_{-\infty}^{\infty} \frac{4 (\sin u)^2}{1/(4T^2) +u^2} du + O\Big(\frac{T}{(\log T)^{100}}\Big) = 4\pi T + O\Big(\frac{T}{(\log T)^{100}}\Big). \end{align}\tag{30}\] Thus the bound in Proposition 8 shows that the variance computed there is indeed small, compared with \(( \mu \int_{-\widetilde{T}}^{\widetilde{T}}K_T(t)^{2} dt )^2\). We postpone the proof of this proposition, showing first how Theorem 4 may be derived from it.

We now begin the proof of Proposition 8, which will take up the rest of this section.

4 Conditional Gaussian limiting distribution↩︎

In this section, we show a conditional central limit theorem (conditioning on all the values \((f(p))_{p\le z}\)) for the sum \(\sum_{\substack{x\le n \le x+y \\ P(n)>z} } f(n)\). More precisely, we will show that with high probability over all realizations of the \((f(p))_{p\le z}\), the conditional characteristic function (and therefore the conditional distribution) of \(\sum_{\substack{x\le n \le x+y \\ P(n)>z} } f(n)\) is close to complex Gaussian.

Recall the decomposition \[\sum_{\substack{x\le n \le x+y \\ P(n)>z} } f(n) = \sum_{\substack{1<m\le x+y\\ p|m \implies p>z}} f(m) \sum_{\substack{\frac{x}{m} \le n \le \frac{x+y}{m} \\ P(n)\le z}} f(n).\] By using the orthogonality of \(f(n)\), the conditional variance is \[V_f(x, y) = \sum_{\substack{1<m\le x+y\\ p|m \implies p>z}} \Big| \sum_{\substack{\frac{x}{m} \le n \le \frac{x+y}{m} \\ P(n)\le z}} f(n) \Big|^{2}.\] The goal of this section is to show the following.

We state this a little loosely (e.g. we do not bother to specify what constitutes a “nice region”), as noted above we will actually prove a quantitative statement about the proximity of probabilistic characteristic functions, which would imply a quantitative version of Theorem 9. To establish this, we shall apply the following complex-valued version of the martingale central limit theorem [14], established in [5].

Let \(\mathcal{A}:=\{1<m \le x+y: p|m \implies p>z\}\). We apply the above criteria to our setting with the following: \[M=x+y, \quad a_m = \mathbb{1}_{\mathcal{A}}(m) \cdot \sum_{\substack{\frac{x}{m} \le n \le \frac{x+y}{m} \\ P(n)\le z}} f(n), \quad V= V_f(x, y).\] Note that once we condition on the \((f(p))_{p\le z}\), the \(a_m\) and \(V\) become fixed complex numbers.

The goal is to show that with probability \(1- O((\log T)^{-\frac{1}{5}}) -O((\log \log z)^{-\frac{1}{10}})\) over realizations of \((f(p))_{p\le z}\), the following conditions hold: \[\label{eqn:32cross} \Big| \sum_{\substack{ m_1, m_2, m_3, m_4 \in \mathcal{A}\\ m_1 m_2= m_3 m_4 \\ m_1 \neq m_3, m_2\neq m_4 \\ P(m_1)=P(m_3) \\ P(m_2)=P(m_4)}} \sum_{\substack{n_1, n_2, n_3, n_4 \\ \frac{x}{m_j} \le n_j \le \frac{x+y}{m_j} \\ P(n_j)\le z}} f(n_1)f(n_2)\overline{f(n_3) f(n_4)} \Big| \ll \frac{V_f(x, y)^2}{(\log\log z)^{\frac{1}{5}}} ,\tag{46}\] and \[\label{eqn:32linden} \Big| \sum_{\substack{ m_1, m_2, m_3, m_4 \in \mathcal{A}\\ m_1 m_2= m_3 m_4 \\ P(m_1)=P(m_2) =P(m_3)=P(m_4)}} \sum_{\substack{n_1, n_2, n_3, n_4 \\ \frac{x}{m_j} \le n_j \le \frac{x+y}{m_j} \\ P(n_j)\le z}} f(n_1)f(n_2)\overline{f(n_3) f(n_4)} \Big| \ll \frac{V_f(x, y)^2}{\sqrt{z}} .\tag{47}\] Verifying these conditions, especially the difficult 46 , will occupy most of this section. Before establishing them, we first show how 46 and 47 may be used with Theorem 10 to obtain our main result Theorem 1.

Thus it remains only to establish the estimates in 46 and 47 with suitably high probability. We begin with a lemma which will quickly lead to a proof of 47 .

It remains to verify the more demanding condition 46 . The argument is a bit lengthy, but breaks into a few distinct phases. We shall bound the left hand side of 46 by a (smoothed) quadruple integral of our random Euler products. We will then show that (with high probability) the ranges of integration may be significantly truncated to a very near diagonal portion. Finally, a barrier analysis (similar as in section 3, but less delicate) will give an acceptable bound for this portion of the quadruple integral.

Put \[\label{4464} \delta= \log (1 +y/x) \asymp y/x, \qquad \text{ and } \qquad \delta_1 = (\log x)^{-40}.\tag{49}\] For positive real numbers \(u\), \(v\) define two functions \(a(u;v)\) and \(b(u;v)\) by setting them both to be \(0\) if \(u<ve^{-\delta_1}\) or if \(u> ve^{\delta}\). In the range \(ve^{-\delta_1} \le u \le v e^{\delta}\) define \[\label{4465} a(u;v) = \begin{cases} \delta_1^{-1} \log(ue^{\delta_1}/v) &\text{ if } ve^{-\delta_1} \le u \le v \\ 1 & \text{ if } v \le u \le ve^{\delta-\delta_1}\\ \delta_1^{-1} \log (e^{\delta}v/u) &\text{ if } ve^{\delta-\delta_1} \le u\le ve^{\delta} .\\ \end{cases}\tag{50}\] The function \(b(u;v)\) is defined by setting \[\label{4466} b(u;v) = \begin{cases} 1- a(u;v) &\text{ if } v\le u \le e^{\delta}v\\ -a(u;v) &\text{ if } ve^{-\delta_1} \le u\le v .\\ \end{cases}\tag{51}\] Thus \(a(u;v)+b(u;v)\) is the indicator function of the condition \(u \in [v,ve^{\delta}]\); the function \(a(u;v)\) is a smoothed approximation to this indicator function, and \(b(u;v)\) is the error incurred in the smoothing.

Examining the left side of 46 , we use the above notation to write the inner sum over \(n_1\) as \[\sum_{P(n_1) \le z} a(n_1;x/m_1) f(n_1) + \sum_{P(n_1) \le z} b(n_1; x/m_1) f(n_1).\] Similar expressions hold for the sums over \(n_2\), \(n_3\), \(n_4\), taking care to replace \(f\) by its conjugate in the cases of \(n_3\) and \(n_4\). Thus the left side of 46 may be bounded in terms of \(16\) sums, depending on which combination of \(a(n_j;x/m_j)\) or \(b(n_j; x/m_j)\) arises. Of these \(16\) terms, we now show that only the term with \(a(n_j;x/m_j)\) appearing for all \(1\le j\le 4\) is significant, and the remaining \(15\) terms may be bounded easily.

Consider one of the \(15\) terms where \(b(n_j;x/m_j)\) appears at least once. Suppose \(b(n_1;x/m_1)\) appears, and for \(j=2\), \(3\), \(4\) we are indifferent to whether \(a(n_j;x/m_j)\) or \(b(n_j;x/m_j)\) occurs, and denote by \(c(n_j;x/m_j)\) either of these possibilities. We now bound the expected value of the contribution of such terms to 46 : namely, \[\sum_{\substack{ m_1, m_2, m_3, m_4 \in \mathcal{A}\\ m_1 m_2= m_3 m_4 \\ m_1 \neq m_3, m_2\neq m_4 \\ P(m_1)=P(m_3) \\ P(m_2)=P(m_4)}} \mathbb{E}\Big[ \Big| \sum_{\substack{ n_1, n_2, n_3, n_4 \\ P(n_j) \le z}} b(n_1;x/m_1) f(n_1) c(n_2;x/m_2) f(n_2) \prod_{j=3}^{4} c(n_j; x/m_j) \overline{f(n_j)}\Big|\Big].\] By applications of Hölder’s inequality followed by Lemma 5 we may bound the inner expectation by \[\begin{align} &\Big( \mathbb{E}\Big[ \Big| \sum_{P(n_1)\le z} b(n_1;x/m_1) f(n_1)\Big|^4\Big]\Big)^{\frac{1}{4}} \prod_{j=2}^{4} \Big(\mathbb{E}\Big[ \Big| \sum_{P(n_j) \le z} c(n_j;x/m_j) f(n_j)\Big|^4 \Big] \Big)^{\frac{1}{4}} \\ &\ll \Big( \mathbb{E}\Big[ \Big| \sum_{P(n_1)\le z} b(n_1;x/m_1) f(n_1)\Big|^4\Big]\Big)^{\frac{1}{4}} \prod_{j=2}^{4} \Big( \frac{x^2}{m_j^2} (\log x)^3 \Big)^{\frac{1}{4}} \end{align}\] We sum this over \(m_1\), \(m_2\), \(m_3\), \(m_4\), keeping only the constraints that these variables are \(\le x+y\) and satisfy \(m_1 m_2 = m_3 m_4\). Given \(m_1\), \(m_2\), there are at most \(d(m_1 m_2) \le d(m_1) d(m_2)\) choices for \(m_3\) and \(m_4\). Thus the total contribution to 46 from this case is \[\begin{align} &\ll \sum_{m_1, m_2 \le x+y} \Big( \mathbb{E}\Big[ \Big| \sum_{P(n_1)\le z} b(n_1;x/m_1) f(n_1)\Big|^4\Big]\Big)^{\frac{1}{4}} x^{\frac{3}{2}} (\log x)^{\frac{9}{4}} \frac{d(m_1)d(m_2)}{\sqrt{m_1} m_2} \\ &\ll x^{\frac{3}{2}} (\log x)^{\frac{17}{4}} \sum_{m_1 \le x+y} \frac{d(m_1)}{\sqrt{m_1}} \Big( \mathbb{E}\Big[ \Big| \sum_{P(n_1)\le z} b(n_1;x/m_1) f(n_1)\Big|^4\Big]\Big)^{\frac{1}{4}}. \end{align}\] To estimate this sum, we distinguish two cases: (i) when \(m_1 \le x/(\log x)^{40}\) and (ii) when \(x/(\log x)^{40} < m_1 \le x+y\). In the first case, note that (when \(b(n_1;x/m_1)\) is non-zero) \(n_1\) lies in two intervals near \(x/m_1\) and \((x+y)/m_1\) of length \(\asymp \delta_1 (x/m_1) = (\log x)^{-40}(x/m_1)\). Therefore using Lemma 5 \[\mathbb{E}\Big[ \Big| \sum_{P(n_1)\le z} b(n_1;x/m_1) f(n_1)\Big|^4\Big] \ll \frac{x}{m_1} \frac{x}{m_1(\log x)^{40} }(\log x)^3 \ll \frac{x^2}{m_1^2 (\log x)^{37}}.\] Thus the contribution of this range of \(m_1\) to the expectation is \[\ll x^2 (\log x)^{-5} \sum_{m_1 \le x/(\log x)^{40}} \frac{d(m_1)}{m_1} \ll x^2 (\log x)^{-3}.\] In the second case, note that the sum over \(n_1\) again runs over integers in two intervals around \(x/m_1\) and \((x+y)/m_1\) of length \(\asymp \delta_1 x/m_1\), and now there are at most a bounded number of integers in these intervals. Thus the expectation of the sum over \(n_1\) is bounded. Moreover, usually there are no integers \(n_1\) so close to \(x/m_1\) or \((x+y)/m_1\), and the sum is non-zero only if for some integer \(k\le 2(\log x)^{40}\) one has \(m_1\) lying in an interval of length \(O(x(\log x)^{-40}/k)\) around \(x/k\) or \((x+y)/k\). Thus the contribution from the second case is \[\begin{align} & \ll x^{\frac{3}{2}} (\log x)^{\frac{17}{4}} \sum_{k\le 2(\log x)^{40}} \Big( \sum_{|m_1 -x/k| \ll x(\log x)^{-40}/k} \frac{d(m_1)}{\sqrt{m_1}} + \sum_{|m_1-(x+y)/k| \ll x(\log x)^{-40}/k} \frac{d(m_1)}{\sqrt{m_1}} \Big) \\ &\ll x^{\frac{3}{2}} (\log x)^{\frac{17}{4}} \sum_{k\le 2(\log x)^{40}} \frac{\sqrt{x}}{\sqrt{k}} (\log x)^{-39} \ll \frac{x^2}{(\log x)^{10}}. \end{align}\] We conclude that the expected value of the contribution of the \(15\) remainder terms to 46 is \(\ll x^2/(\log x)^3 \ll y^2/(\log x)^2\), since \(y \ge x/(\log x)^{\frac{1}{2}}\). By Markov’s inequality we conclude that with probability \(1- O((\log x)^{-1})\) the contribution of these terms to 46 is \(\ll y^2/(\log x)\), which is more than satisfactory.

Thus our goal now is to understand \[\label{4467} \sum_{\substack{ m_1, m_2, m_3, m_4 \in \mathcal{A}\\ m_1 m_2= m_3 m_4 \\ m_1 \neq m_3, m_2\neq m_4 \\ P(m_1)=P(m_3) \\ P(m_2)=P(m_4)}} \sum_{\substack{ n_1, n_2, n_3, n_4 \\ P(n_j) \le z}} a(n_1;x/m_1) f(n_1) a(n_2;x/m_2) f(n_2) a(n_3; x/m_3) \overline{f(n_3)} a(n_4;x/m_4) \overline{f(n_4)}.\tag{52}\] The function \(a(u;v)\) may be expressed as a contour integral \[a(u;v) = \frac{1}{2\pi i} \int_{(c)} \Big(\frac{v}{u}\Big)^s \Big( \frac{e^{\delta s}-1}{s}\Big) \Big( \frac{1-e^{-\delta_1 s}}{\delta_1 s} \Big) ds,\] where the integral is taken over the line Re\((s)=c >0\). We take the line of integration to be \(c=\tfrac 12\), and writing \(s=\tfrac 12+it\) express the above as \[a(u;v) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \Big(\frac{v}{u}\Big)^{\frac{1}{2}+it} W(t) dt,\] where the kernel \(W(t)=W(t;\delta,\delta_1)\) is defined by \[\label{4468} W(t)= \Big(\frac{e^{\delta(\frac{1}{2}+it)} -1}{\frac{1}{2}+it}\Big) \Big(\frac{1-e^{-\delta_1 (\frac{1}{2}+it)}}{\delta_1 (\frac{1}{2}+it)}\Big).\tag{53}\] We may readily check that \(W(t)\) satisfies the bound \[\label{4469} W(t) \ll \min\Big( \delta, \frac{1}{1+|t|}\Big) \min\Big( 1, \frac{1}{\delta_1 (1+|t|)}\Big) = \min \Big( \delta, \frac{1}{1+|t|}, \frac{1}{\delta_1 (1+|t|)^2}\Big).\tag{54}\] With this notation, it follows that \[\sum_{P(n_1)\le z} a(n_1;x/m_1) f(n_1) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F_z(\tfrac 12+it_1) \Big(\frac{x}{m_1}\Big)^{\frac{1}{2}+it_1} W(t_1) dt_1.\] An identical expression holds for the sum over \(n_2\), while for \(n_3\) and \(n_4\) we have closely analogous expressions after taking into account complex conjugation: for instance, \[\sum_{P(n_3) \le z} a(n_3;x/m_3) \overline{f(n_3) }= \frac{1}{2\pi} \int_{-\infty}^{\infty} \overline{F_z(\tfrac 12+it_3)} \Big(\frac{x}{m_3}\Big)^{\frac{1}{2}-it_3} \overline{W(t_3)} dt_3.\]

We use these expressions for the sums over \(n_j\) in 52 , and then bring in the outer sums over the \(m_j\), replacing the condition \(m_j \in {\mathcal{A}}\) by just requiring \(m_j >1\) and having prime factors in the range \((z,x+y]\) (since the terms with \(m_j>x+y\) have \(a(n_j;x/m_j) = 0\)). For ease of notation, define \[\label{44610} G(t_1, t_2, t_3, t_4) = \sideset{_{} ^{}} {_{} ^{\star}}{\sum }_{m_1, m_2, m_3, m_4 > 1} \frac{1}{m_1^{\frac{1}{2}+it_1}} \frac{1}{m_2^{\frac{1}{2}+it_2}} \frac{1}{m_3^{\frac{1}{2}-it_3}}\frac{1}{m_4^{\frac{1}{2}-it_4}},\tag{55}\] where the \(\star\) expresses the constraints \[\begin{align} \label{44611} p|m_j \implies p\in (z,x+y]; \;\;&m_1 m_2 =m_3m_4; \;\;m_1 \neq m_3, m_2 \neq m_4; \nonumber\\ &\;\;P(m_1) = P(m_3), P(m_2)=P(m_4). \end{align}\tag{56}\] Gathering our argument so far, we conclude that the desired quantity in 52 may be bounded by \[\label{44612} \ll x^2 \int_{t_1, t_2, t_3, t_4 \in {\mathbb{R}}} \prod_{j=1}^{4} |F_z(\tfrac 12+it_j) W(t_j)| |G(t_1,t_2,t_3,t_4)| dt_1 dt_2 dt_3 dt_4.\tag{57}\]

Our next step is to show that in 57 we may restrict attention to the range when all \(|t_j|\) are at most \((\log x)^{100}\). To see this, we first record a simple bound on \(G(t_1,t_2,t_3,t_4)\). Using the triangle inequality, and keeping only the conditions that \(m_1m_2 =m_3 m_4\) (so that given \(m_1\), \(m_2\) there are at most \(d(m_1 m_2) \le d(m_1)d(m_2)\) choices for \(m_3\) and \(m_4\)) and that all prime factors of \(m_j\) are in \((z,x+y]\), we see that \[\begin{align} |G(t_1, t_3, t_3, t_4)| &\le \sum_{\substack{m_1, m_2 \\ p|m_j \implies p\in (z,x+y]}} \frac{d(m_1 m_2)}{m_1 m_2} \le \Big( \sum_{p|m_1 \implies p\in (z,x+y]} \frac{d(m)}{m}\Big)^2 \\ &= \prod_{z < p \le x+y} \Big(1-\frac{1}{p}\Big)^{-4} \ll \Big(\frac{\log x}{\log z} \Big)^4. \end{align}\] Further, note that \[\mathbb{E}\Big[ \prod_{j=1}^{4} |F_z(\tfrac 12+it_j)|\Big] \le \frac{1}{4} \sum_{j=1}^{4} \mathbb{E}[ |F_z(\tfrac 12+it_j)|^4] = \mathbb{E}[ |F_z(\tfrac 12)|^4] \ll (\log z)^4,\] where the third estimate follows by translation invariance, and the fourth moment bound follows by a straightforward calculation (or see [12]). Thus the expected value of the terms where one of the \(|t_j|\) exceeds \((\log x)^{100}\) is \[\ll x^2 \Big(\frac{\log x}{\log z}\Big)^4 (\log z)^4 \Big( \int_{|t| \ge (\log x)^{100}} |W(t)| dt \Big) \Big( \int_{-\infty}^{\infty} |W(t)| dt \Big)^3 \ll \frac{x^2}{(\log x)^{50}},\] upon using 54 . Thus by Markov’s inequality, with probability \(1-O((\log x)^{-1})\) the contribution of terms to 57 with one of the variables \(|t_j| > (\log x)^{100}\) is \(\le y^2/(\log x)^{40}\) which is satisfactory.

Henceforth we restrict attention to the case where all \(|t_j|\) are below \((\log x)^{100}\). Here a key feature will be that \(|G(t_1, t_2, t_3, t_4)|\) provides a saving unless the variables \(t_j\) are all close to each other. For prime number theoretic reasons, we will need such a restriction on the maximum size of the \(t_j\) to show the bound we want for \(|G(t_1, t_2, t_3, t_4)|\). Indeed, it follows easily from the prime number theorem and partial summation (or consult [12]) that for any \(z \le u \le x+y\), and \(|t| \le 2(\log x)^{100}\) one has \[\label{44613} \Big| \sum_{z < p\le u} \frac{1}{p^{1+it}} \Big| \le \min \Big( \log \frac{\log u}{\log z} + O(1), O\Big( \frac{1}{|t| \log z} \Big) \Big) .\tag{58}\] With this estimate in hand, we proceed to bounding \(|G(t_1,t_2,t_3,t_4)|\).

As discussed earlier, the key feature of the above bound is that unless all four points \(t_1, t_2, t_3, t_4\) are close to each other (on a scale of \(\approx 1/\log x\)), we must have a saving. When we apply the bound, it will be very important that the extra factor \((\log x/\log z)^4\) is not too large (a very small power of \(\log\log x\) at most), and here it is crucial that \(z\) was chosen fairly close to \(x\) (on a logarithmic scale).

We return to the problem of estimating 57 with the \(t_j\) restricted to \(|t_j| \le (\log x)^{100}\). Lemma 6 will allow us to make a further important truncation, to the situation where the four variables \(t_j\) are all very nearly equal. Define \[\mathcal{R}: = \Big\{(t_1, t_2, t_3, t_4): |t_j| \le (\log x)^{100}; \;\; |t_1-t_3|, |t_2-t_4|\le \frac{(\log \log z)^{3}}{\log z}; \;\; |t_1-t_4|\le \frac{(\log \log z)^{9}}{\log z} \Big\},\] and let \({\mathcal{R}}^{c}\) denote the complement of \({\mathcal{R}}\) in \([-(\log x)^{100}, (\log x)^{100}]^4\).

In view of Lemma 7, it remains now to estimate the integral in 57 when restricted to the region \({\mathcal{R}}\), where in particular all four variables \(t_1\), \(t_2\), \(t_3\), \(t_4\) are near each other (precisely, within \(2(\log \log z)^9/\log z\) of each other). We use the bound \(|G(t_1,t_2,t_3,t_4)| \ll (\log x/\log z)^4 \gamma(t_1 -t_3) \gamma(t_2-t_4)\) and note that \[\prod_{j=1}^{4} |F_z(\tfrac 12+it_j)| \le |F_z(\tfrac 12+it_1)F_z(\tfrac 12+it_2)|^2 + |F_z(\tfrac 12+it_3) F_z(\tfrac 12+it_4)|^2.\] Thus the integral over the region \({\mathcal{R}}\) in 57 may be bounded by \[\begin{align} &\ll x^2 \Big( \frac{\log x}{\log z}\Big)^4 \int_{\substack{|t_1|, |t_2| \le (\log x)^{100}\\ |t_1 -t_2| \le 2 (\log \log z)^9/\log z}} |F_z(\tfrac 12+it_1)F_z(\tfrac 12+it_2)|^2 |W(t_1) W(t_2)| \\ &\times \int_{|t_1-t_3| \le 2(\log \log z)^9/\log z} \gamma(t_1-t_3) |W(t_3)| dt_3 \int_{|t_2-t_4|\le 2 (\log \log z)^9/\log z} \gamma(t_2-t_4) |W(t_4)| dt_4 dt_1 dt_2. \end{align}\] Here we have omitted the term corresponding to \(|F_z(\tfrac 12+it_3)F_z(\tfrac 12+it_4)|^2\), which makes an identical contribution.

Now \[\begin{align} \int_{|t_1-t_3| \le 2(\log \log z)^9/\log z} \gamma(t_1-t_3) |W(t_3)| dt_3 &\ll \min\Big(\delta, \frac{1}{1+|t_1|}\Big) \frac{\log \log \log x}{\log x}\\ & \ll \frac{1}{\log z} \min \Big(\delta, \frac{1}{1+|t_1|}\Big), \end{align}\] and a similar bound holds for the integral over \(t_4\). Thus we are left with the task of bounding \[\label{44619} \frac{x^2}{(\log z)^2} \Big(\frac{\log x}{\log z}\Big)^4 \int_{\substack{|t_1|, |t_2| \le (\log x)^{100}\\ |t_1 -t_2| \le 2 (\log \log z)^9/\log z}} \prod_{j=1}^{2} |F_z(\tfrac 12+it_j)|^2 \min\Big(\delta^2,\frac{1}{1+|t_j|^2}\Big) dt_1dt_2.\tag{64}\]

We use the barrier events method again to analyze the above double integral. The situation is similar to our work in section 3; while what we require now is less delicate than our work in that section, there is one further complication that the range of the \(t_j\) here can be quite large compared with \(T\). Thus let \(\tau^{*}\) be the smallest natural number for which \(z^{e^{-\tau^*}} \leq e^{(\log\log z)^3}\), and \(B\) be a suitably large fixed number. For \(|t| \leq (\log x)^{100}\), let \(\mathcal{G}^{*}(t)\) be the event that for all \(0\le j\le \tau^* -1\) \[\label{44620} (\sqrt{T+|t|} e^{\tau^*-j} \log\log z)^{-B} \le \prod_{z^{e^{-\tau^*}} <p \le z^{e^{-j}} } \Big|1-\frac{f(p)}{p^{\frac{1}{2}+ it}}\Big|^{-1} \le \sqrt{T+|t|} e^{\tau^*-j} (\log \log z)^5 .\tag{65}\] Let \(\mathcal{H}^{*}(t)\) denote the event that \(\mathcal{G}^{*}(t)\) holds and in addition \[\label{44621} \prod_{z^{e^{-\tau^*}} <p \le z } \Big|1-\frac{f(p)}{p^{\frac{1}{2}+ it}}\Big|^{-1} \le \sqrt{T+|t|} \frac{e^{\tau^*}}{ (\log \log z )^{50}}.\tag{66}\] Note that the barriers here weaken when \(|t| \geq T\), which is needed to show that \(\mathcal{G}^{*}(t)\) holds for all \(|t|\le (\log x)^{100}\) with high probability. This weakening will be compensated by the decay of \(|W(t)|\) for large \(t\). Another feature is that, compared to \(\mathcal{H}(t)\) where there is an extra restriction at all scales \(j\), the event \(\mathcal{H}^{*}\) is simpler and only imposes an extra restriction on the full Euler product up to \(z\). The very close proximity of \(t_1\) and \(t_2\) in 64 means that we will only use the barrier \({\mathcal{H}}^*(t)\) to factor out and bound a copy of the full Euler product. The extra saving \((\log\log z)^{50}\) in 66 , showing that (when \({\mathcal{H}}^*(t)\) holds) the Euler products cannot be too large, will then lead to an acceptable bound for 64 .

We first modify the argument of Proposition 5 to show that the event \({\mathcal{G}}^{*}(t)\) holds for all \(|t|\le (\log x)^{100}\) with high probability.

We are now ready to resume the task of bounding 64 . In view of Proposition 11, with high probability we may assume that in 64 both \({\mathcal{G}}^*(t_1)\) and \({\mathcal{G}}^*(t_2)\) hold: thus, we seek now to bound \[\label{eqn:32rest} \frac{x^2}{(\log z)^{2}} \Big(\frac{\log x}{\log z}\Big)^4 \int_{\substack{|t_1|, |t_2| \le (\log x)^{100}\\ |t_1 -t_2| \le 2 (\log \log z)^9/\log z}} \prod_{j=1}^{2} {\mathbb{1}_{\mathcal{G}^{*}(t_j)}} |F_z(\tfrac 12+it_j)|^2 \min\Big(\delta^2,\frac{1}{1+|t_j|^2}\Big) dt_1dt_2.\tag{67}\]

To bound 67 we first show that, up to acceptable errors, one can replace \(\mathbb{1}_{\mathcal{G}^{*}(t_2)}\) by the stronger condition \(\mathbb{1}_{\mathcal{H}^{*}(t_2)}\). Define \[I_1: = \frac{1}{\log z}\int_{\substack{|t| \le (\log x)^{100} }} |F_z(\tfrac 12+it)|^{2} \mathbb{1}_{\mathcal{G}^{*}(t)} \min\Big(\delta^2, \frac{1}{1+t^2} \Big)dt ,\] and \[I_2: = \frac{1}{\log z} \int_{\substack{|t| \le (\log x)^{100} }} |F_z(\tfrac 12+it)|^{2} (\mathbb{1}_{\mathcal{G}^{*}(t)} - \mathbb{1}_{\mathcal{H}^{*}(t)}) \min\Big(\delta^2, \frac{1}{1+t^2} \Big) dt.\] The first part of Proposition 12 shows (recalling \(\delta \asymp y/x \asymp 1/T\)) that \[\begin{align} \mathbb{E}[I_1] &\ll \int_{\substack{|t| \le (\log x)^{100} }} \min\Big(1, \frac{\log (T+|t|)+\log \log \log z}{\sqrt{\log \log z}}\Big) \min\Big(\delta^2, \frac{1}{1+t^2}\Big) dt \\ &\ll \delta \min\Big(1, \frac{\log T +\log \log \log z}{\sqrt{\log \log z}}\Big). \end{align}\] By Markov’s inequality, it follows that with probability at least \(1-O((\log \log z)^{-\frac{1}{10}})\), we have \[\label{44625} I_1 \ll (\log \log z)^{\frac{1}{9}} \delta \min\Big(1, \frac{\log T}{\sqrt{\log \log z}}\Big).\tag{68}\] The second part of Proposition 12 shows that \[\begin{align} \mathbb{E}[I_2] &\ll \int_{\substack{|t| \le (\log x)^{100} }} \min\Big(1, \frac{\log (T+|t|)}{\sqrt{\log \log z}}\Big) \frac{(\log \log \log z)^3}{\log \log z} \min\Big(\delta^2, \frac{1}{1+t^2}\Big) dt \\ & \ll \delta \min\Big(1, \frac{\log T}{\sqrt{\log \log z}}\Big) \frac{(\log \log \log z)^3}{\log \log z}, \end{align}\] so that with probability \(1-O((\log \log z)^{-\frac{1}{10}})\) we have \[\label{44626} I_2 \ll \delta \min\Big(1, \frac{\log T }{\sqrt{\log \log z}} \Big) (\log \log z)^{-\frac{8}{9}}.\tag{69}\] Thus with probability \(1-O((\log \log z)^{-\frac{1}{10}})\) both 68 and 69 hold, and the contribution to 67 from the portion of the integral where \({\mathcal{H}}^*(t_2)\) fails is at most (temporarily dropping the condition that \(t_1\) and \(t_2\) are to be close) \[\ll x^2\Big(\frac{\log x}{\log z}\Big)^4 I_1 \cdot I_2 \ll \frac{y^{2}}{(\log \log z)^{\frac{3}{4}}} \min\Big( 1, \frac{(\log T)^2}{\log \log z}\Big) \ll \frac{V(x,y)^2}{(\log \log z)^{\frac{3}{4}}},\] which is satisfactory. We remark that the small size of \((\log x)/\log z\) is very important at this step.

It remains lastly to consider the contribution to 67 where \(\mathcal{H}^{*}(t_2)\) holds. Note that (using 66 at \(t_2\) and dropping the condition \({\mathcal{G}}^*(t_1)\)) \[\begin{align} \mathbb{E}\Big[ {\mathbb{1}}_{{\mathcal{G}}^*(t_1)} |F_z(\tfrac 12+it_1)|^2 {\mathbb{1}}_{{\mathcal{H}}^*(t_2)}& |F_z(\tfrac 12+it_2)|^2] \\ &\ll \frac{e^{2\tau^*}(T+|t_2|)}{(\log \log z)^{100}} \mathbb{E}\Big[ |F_z(\tfrac 12+it_1)|^2 \prod_{p\le z^{e^{-\tau^*}}} \Big| 1- \frac{f(p)}{p^{\frac{1}{2}+it_2}}\Big|^{-2}\Big]. \end{align}\] Using the independence of \(f\) on distinct primes, and Euler Product Result 1 in [12], we obtain \[\begin{align} \mathbb{E}\Big[ &|F_z(\tfrac 12+it_1)|^2 \prod_{p\le z^{e^{-\tau^*}}} \Big| 1- \frac{f(p)}{p^{\frac{1}{2}+it_2}}\Big|^{-2}\Big] \\ &=\mathbb{E}\Big[ \prod_{p\le z^{e^{-\tau^{*}}}} |1- \frac{f(p)}{p^{\frac{1}{2}+it_1}}|^{-2} \prod_{p\le z^{e^{-\tau^{*}}}} \Big|1- \frac{f(p)}{p^{\frac{1}{2}+it_2}}\Big|^{-2} \Big] \prod_{z^{e^{-\tau^*}} < p\le z}\Big(1-\frac{1}{p}\Big)^{-1} \ll (e^{-\tau^*} \log z)^4 e^{\tau^*}.\\ \end{align}\] Thus the expected contribution to 67 from the terms where \(\mathcal{H}^{*}(t_2)\) holds is (recall that \(e^{-\tau^*} \log z\) is about \((\log \log z)^{3}\) by definition) \[\begin{align} &\ll x^2 \frac{e^{-\tau^*} (\log z)^2}{(\log \log z)^{100}} \Big(\frac{\log x}{\log z}\Big)^4 \int_{\substack{ |t_1|, |t_2| \le (\log x)^{100} \\ |t_1-t_2| \le 2 (\log \log z)^9/\log z}} (T+|t_2|) \min\Big(\delta^2,\frac{1}{1+t_2^2}\Big)^2 dt_1 dt_2 \\ &\ll \frac{x^2}{(\log \log z)^{87}} \int_{|t_2|\le (\log x)^{100}} (T+|t_2|) \min\Big(\delta^2,\frac{1}{1+t_2^2}\Big)^2 dt_2 \ll \frac{y^2}{(\log \log z)^{87}}. \end{align}\] At this step, we crucially retained and used the restriction that \(|t_1-t_2| \le 2 (\log \log z)^9/\log z\). By Markov’s inequality it follows that with probability \(1-O((\log \log x)^{-1})\) this contribution to 67 is \(\ll y^2 (\log \log z)^{-86}\), which is better than needed.

Tracking back through all our reductions, and keeping in mind 48 , at last we conclude that 46 holds with the desired high probability. 0◻

5 Long sums↩︎

5.1 Proof of Proposition 2↩︎

The main theorems of Harper [1] imply the existence of absolute constants \(c\), \(C\) with \(0 < c \leq C\), such that uniformly for all large \(x\) and \(0 \leq q \leq 1\) we have \[\label{eqn:32AJHmomcC} c \left( \frac{x}{1 + (1-q)\sqrt{\log\log x}} \right)^q \leq \mathbb{E}[|\sum_{n \leq x} f(n)|^{2q}] \leq C \left( \frac{x}{1 + (1-q)\sqrt{\log\log x}} \right)^q .\tag{70}\] For ease of notation put temporarily \(g= (\log \log x)^{\frac{1}{4}} x^{-\frac{1}{2}} \sum_{n\le x} f(n)\), so that uniformly in \(q\le 1- 1/\sqrt{\log \log x}\) we find from 70 \[\label{5462} \frac{c}{3(1-q)} \le \mathbb{E}[|g|^{2q}] \le \frac{C}{(1-q)}.\tag{71}\] Here we used that \((2(1-q))^{-q} \ge 1/(3(1-q))\) and that \((1-q)^{-q} \le(1-q)^{-1}\) for all \(0\le q<1\).

Let \(\lambda\) be large, and take \[p = 1-\frac{1}{\log \lambda}, \qquad q = 1- \frac{c}{50 C \log \lambda}, \qquad r= 1- \frac{c}{100 C \log \lambda},\] so that \(0< p < q < r \le 1- 1/\sqrt{\log \log x}\) for \(x\) sufficiently large (in terms of \(\lambda\)). We will apply 71 to analyze the \(2p\), \(2q\) and \(2r\)-th moments of \(g\). The increased size of the \(2q\)-th moment compared with the \(2p\)-th moment will allow us to conclude that a large portion of the \(2q\)-th moment must be produced by values \(|g| > \lambda\). On the other hand, the fact that the \(2r\)-th moment isn’t too much bigger than the \(2q\)-th will imply that this size isn’t all produced by values much larger than \(\lambda\), and so \(|g|\) must exceed \(\lambda\) with reasonable probability.

Note that \[\begin{align} \mathbb{E}[ |g|^{2q} {\mathbb{1}}_{|g|>\lambda}] &= \mathbb{E}[|g|^{2q}] - \mathbb{E}[|g|^{2q} {\mathbb{1}}_{|g| \le \lambda}] \ge \mathbb{E}[|g|^{2q}] - \lambda^{2(q-p)} \mathbb{E}[ |g|^{2p}] \\ &\ge \frac{c}{3(1-q)} - e^2 \frac{C}{(1-p)} \ge 8C \log \lambda. \end{align}\] On the other hand, by Hölder’s inequality \[\begin{align} \mathbb{E}[ |g|^{2q} {\mathbb{1}}_{|g|>\lambda}] &\le (\mathbb{P}[|g| >\lambda])^{1- \frac{q}{r}} (\mathbb{E}[|g|^{2r}])^{\frac{q}{r}} \le (\mathbb{P}[|g| >\lambda])^{1- \frac{q}{r}} (100C^2 c^{-1} \log \lambda)^{\frac{q}{r}} \\ &\le (\mathbb{P}[|g| >\lambda])^{1- \frac{q}{r}} (100C^2 c^{-1} \log \lambda) . \end{align}\] We conclude that \[\mathbb{P}[|g| >\lambda] \ge \Big( \frac{2c}{25C}\Big)^{r/(r-q)} \ge \Big( \frac{2c}{25C}\Big)^{1/(r-q)} = \lambda^{-A}\] with \(A = 100(C/c) \log (25C/2c)\), which proves the desired bound.

Similarly for \(\sum_{x \leq n \leq (1+\delta)x} f(n)\), there exist constants \(0 < c(\delta) \leq C(\delta)\) such that \[c(\delta) \Big( \frac{\delta x}{1 + (1-q)\sqrt{\log\log x}} \Big)^q \leq \mathbb{E}[\Big|\sum_{x \leq n \leq (1+\delta)x} f(n) \Big|^{2q}] \leq C(\delta) \Big( \frac{\delta x}{1 + (1-q)\sqrt{\log\log x}} \Big)^q\] for all large enough \(x\) (depending on \(\delta\)) and \(0 \leq q \leq 1\). The existence of \(C(\delta)\) follows immediately from 70 and the triangle inequality, or from Theorem 1.1 of Caich [6] (which would yield a much better dependence on \(\delta\), in fact with \(C(\delta)\) uniformly bounded if \(q\) is close enough to 1 depending on \(\delta\)). The existence of \(c(\delta)\) follows by slightly adapting the arguments of Harper [1] along the lines of section 2 here. For example one could show that (omitting smaller order terms below) \[\begin{align} \mathbb{E}[\Big|\sum_{x \leq n \leq (1+\delta)x} f(n) \Big|^{2q}] & \gg & (\delta x)^q \mathbb{E}\Big[\Big(\frac{\delta}{\log x} \int_{-1/(100\delta)}^{1/(100\delta)} \Big|F_{x^{3/4}}(\tfrac{1}{2} + \tfrac{B\log(1/\delta)}{\log x} +it)\Big|^{2} dt \Big)^{q}\Big] \nonumber \\ & \geq & (\delta x)^q \mathbb{E}\Big[\Big(\frac{\delta}{\log x} \int_{-1/2}^{1/2} \Big|F_{x^{3/4}}(\tfrac{1}{2} + \tfrac{B\log(1/\delta)}{\log x} +it)\Big|^{2} dt \Big)^{q}\Big] \nonumber \end{align}\] for a suitable large constant \(B\), and then the results in section 5 of Harper [1] show this is indeed \(\geq c(\delta) \left( \frac{\delta x}{1 + (1-q)\sqrt{\log\log x}} \right)^q\). The claimed lower bound in Proposition 2 can then be obtained by exactly the same argument as for \(\sum_{n \leq x} f(n)\). 0◻

5.2 Long sums cannot converge to a non-degenerate Gaussian↩︎

As mentioned in the Introduction, a particular consequence of Proposition 2 (and other known results) is that there is no normalizing factor \(V(x)\) for which \(\frac{1}{\sqrt{V(x)}} \sum_{n \leq x} f(n)\) can converge in distribution to a non-degenerate Gaussian. Indeed 1 gives \[\frac{1}{\sqrt{V(x)}} \mathbb{E}\Big[\Big|\sum_{n \leq x} f(n)\Big|\Big] \ll \frac{1}{\sqrt{V(x)}} \frac{\sqrt{x}}{(\log\log x)^{\frac{1}{4}}},\] so that \(V(x)\) must necessarily be bounded by \(L {x}/(\log \log x)^{\frac{1}{2}}\) for some constant \(L\). But in this case, Proposition 2 yields that \[\mathbb{P}\Big(\frac{1}{\sqrt{V(x)}} \Big|\sum_{n \leq x} f(n)\Big| \geq \lambda \Big) \geq \mathbb{P}\Big(\Big|\sum_{n \leq x} f(n)\Big| \geq \lambda \sqrt{L} \frac{\sqrt{x}}{(\log\log x)^{1/4}} \Big) \geq \frac{1}{L^{A/2} \lambda^A} .\] For sufficiently large fixed \(\lambda\), this greatly exceeds the Gaussian tail, preventing convergence in distribution.

In fact, there is an even softer argument for showing that \(\frac{(\log \log x)^{1/4}}{\sqrt{x}} \sum_{n \leq x} f(n)\) cannot converge in distribution to a Gaussian (although the probability lower bound in Proposition 2 seems of independent interest). For any fixed \(q_0 < 1\), the moment bounds 70 imply that \((\frac{(\log\log x)^{1/4}}{\sqrt{x}})^{2q_0} \mathbb{E}[|\sum_{n \leq x} f(n)|^{2q_0}]\) is uniformly bounded as \(x \rightarrow \infty\). By general probability theory (see e.g. Theorem 4.2 of Gut [23]), this means that for any fixed \(q < q_0\) (so for any fixed \(q < 1\)), the sequence of random variables \((\frac{(\log\log x)^{1/4}}{\sqrt{x}})^{2q} |\sum_{n \leq x} f(n)|^{2q}\) is uniformly integrable. Again by general theory (see e.g. Theorem 5.9 of Gut [23]), if \(\frac{(\log \log x)^{1/4}}{\sqrt{x}} \sum_{n \leq x} f(n)\) were converging to a Gaussian, we would then get that all of the moments \((\frac{(\log\log x)^{1/4}}{\sqrt{x}})^{2q} \mathbb{E}[|\sum_{n \leq x} f(n)|^{2q}]\) with \(q < 1\) would converge to the moments of that Gaussian limit. But the \(2q\)-th moments of any fixed Gaussian are uniformly bounded for all \(q \leq 1\), whereas the lower bound part of 70 shows the moments \((\frac{(\log\log x)^{1/4}}{\sqrt{x}})^{2q} \mathbb{E}[|\sum_{n \leq x} f(n)|^{2q}]\) can be made arbitrarily large by taking \(q\) close enough to 1.

For any fixed small \(\delta > 0\), exactly the same considerations apply to \(\sum_{x \leq n \leq (1+\delta)x} f(n)\).

References↩︎