You are currently browsing the monthly archive for February 2015.

In analytic number theory, it is a well-known phenomenon that for many arithmetic functions ${f: {\bf N} \rightarrow {\bf C}}$ of interest in number theory, it is significantly easier to estimate logarithmic sums such as

$\displaystyle \sum_{n \leq x} \frac{f(n)}{n}$

than it is to estimate summatory functions such as

$\displaystyle \sum_{n \leq x} f(n).$

(Here we are normalising ${f}$ to be roughly constant in size, e.g. ${f(n) = O( n^{o(1)} )}$ as ${n \rightarrow \infty}$.) For instance, when ${f}$ is the von Mangoldt function ${\Lambda}$, the logarithmic sums ${\sum_{n \leq x} \frac{\Lambda(n)}{n}}$ can be adequately estimated by Mertens’ theorem, which can be easily proven by elementary means (see Notes 1); but a satisfactory estimate on the summatory function ${\sum_{n \leq x} \Lambda(n)}$ requires the prime number theorem, which is substantially harder to prove (see Notes 2). (From a complex-analytic or Fourier-analytic viewpoint, the problem is that the logarithmic sums ${\sum_{n \leq x} \frac{f(n)}{n}}$ can usually be controlled just from knowledge of the Dirichlet series ${\sum_n \frac{f(n)}{n^s}}$ for ${s}$ near ${1}$; but the summatory functions require control of the Dirichlet series ${\sum_n \frac{f(n)}{n^s}}$ for ${s}$ on or near a large portion of the line ${\{ 1+it: t \in {\bf R} \}}$. See Notes 2 for further discussion.)

Viewed conversely, whenever one has a difficult estimate on a summatory function such as ${\sum_{n \leq x} f(n)}$, one can look to see if there is a “cheaper” version of that estimate that only controls the logarithmic sums ${\sum_{n \leq x} \frac{f(n)}{n}}$, which is easier to prove than the original, more “expensive” estimate. In this post, we shall do this for two theorems, a classical theorem of Halasz on mean values of multiplicative functions on long intervals, and a much more recent result of Matomaki and Radziwiłł on mean values of multiplicative functions in short intervals. The two are related; the former theorem is an ingredient in the latter (though in the special case of the Matomaki-Radziwiłł theorem considered here, we will not need Halasz’s theorem directly, instead using a key tool in the proof of that theorem).

We begin with Halasz’s theorem. Here is a version of this theorem, due to Montgomery and to Tenenbaum:

Theorem 1 (Halasz-Montgomery-Tenenbaum) Let ${f: {\bf N} \rightarrow {\bf C}}$ be a multiplicative function with ${|f(n)| \leq 1}$ for all ${n}$. Let ${x \geq 3}$ and ${T \geq 1}$, and set

$\displaystyle M := \min_{|t| \leq T} \sum_{p \leq x} \frac{1 - \hbox{Re}( f(p) p^{-it} )}{p}.$

Then one has

$\displaystyle \frac{1}{x} \sum_{n \leq x} f(n) \ll (1+M) e^{-M} + \frac{1}{\sqrt{T}}.$

Informally, this theorem asserts that ${\sum_{n \leq x} f(n)}$ is small compared with ${x}$, unless ${f}$ “pretends” to be like the character ${p \mapsto p^{it}}$ on primes for some small ${y}$. (This is the starting point of the “pretentious” approach of Granville and Soundararajan to analytic number theory, as developed for instance here.) We now give a “cheap” version of this theorem which is significantly weaker (both because it settles for controlling logarithmic sums rather than summatory functions, it requires ${f}$ to be completely multiplicative instead of multiplicative, it requires a strong bound on the analogue of the quantity ${M}$, and because it only gives qualitative decay rather than quantitative estimates), but easier to prove:

Theorem 2 (Cheap Halasz) Let ${x}$ be an asymptotic parameter goingto infinity. Let ${f: {\bf N} \rightarrow {\bf C}}$ be a completely multiplicative function (possibly depending on ${x}$) such that ${|f(n)| \leq 1}$ for all ${n}$, such that

$\displaystyle \sum_{p \leq x} \frac{1 - \hbox{Re}( f(p) )}{p} \gg \log\log x. \ \ \ \ \ (1)$

Then

$\displaystyle \frac{1}{\log x} \sum_{n \leq x} \frac{f(n)}{n} = o(1). \ \ \ \ \ (2)$

Note that now that we are content with estimating exponential sums, we no longer need to preclude the possibility that ${f(p)}$ pretends to be like ${p^{it}}$; see Exercise 11 of Notes 1 for a related observation.

To prove this theorem, we first need a special case of the Turan-Kubilius inequality.

Lemma 3 (Turan-Kubilius) Let ${x}$ be a parameter going to infinity, and let ${1 < P < x}$ be a quantity depending on ${x}$ such that ${P = x^{o(1)}}$ and ${P \rightarrow \infty}$ as ${x \rightarrow \infty}$. Then

$\displaystyle \sum_{n \leq x} \frac{ | \frac{1}{\log \log P} \sum_{p \leq P: p|n} 1 - 1 |}{n} = o( \log x ).$

Informally, this lemma is asserting that

$\displaystyle \sum_{p \leq P: p|n} 1 \approx \log \log P$

for most large numbers ${n}$. Another way of writing this heuristically is in terms of Dirichlet convolutions:

$\displaystyle 1 \approx 1 * \frac{1}{\log\log P} 1_{{\mathcal P} \cap [1,P]}.$

This type of estimate was previously discussed as a tool to establish a criterion of Katai and Bourgain-Sarnak-Ziegler for Möbius orthogonality estimates in this previous blog post. See also Section 5 of Notes 1 for some similar computations.

Proof: By Cauchy-Schwarz it suffices to show that

$\displaystyle \sum_{n \leq x} \frac{ | \frac{1}{\log \log P} \sum_{p \leq P: p|n} 1 - 1 |^2}{n} = o( \log x ).$

Expanding out the square, it suffices to show that

$\displaystyle \sum_{n \leq x} \frac{ (\frac{1}{\log \log P} \sum_{p \leq P: p|n} 1)^j}{n} = \log x + o( \log x )$

for ${j=0,1,2}$.

We just show the ${j=2}$ case, as the ${j=0,1}$ cases are similar (and easier). We rearrange the left-hand side as

$\displaystyle \frac{1}{(\log\log P)^2} \sum_{p_1, p_2 \leq P} \sum_{n \leq x: p_1,p_2|n} \frac{1}{n}.$

We can estimate the inner sum as ${(1+o(1)) \frac{1}{[p_1,p_2]} \log x}$. But a routine application of Mertens’ theorem (handling the diagonal case when ${p_1=p_2}$ separately) shows that

$\displaystyle \sum_{p_1, p_2 \leq P} \frac{1}{[p_1,p_2]} = (1+o(1)) (\log\log P)^2$

and the claim follows. $\Box$

Remark 4 As an alternative to the Turan-Kubilius inequality, one can use the Ramaré identity

$\displaystyle \sum_{p \leq P: p|n} \frac{1}{\# \{ p' \leq P: p'|n\} + 1} - 1 = 1_{(p,n)=1 \hbox{ for all } p \leq P}$

(see e.g. Section 17.3 of Friedlander-Iwaniec). This identity turns out to give superior quantitative results than the Turan-Kubilius inequality in applications; see the paper of Matomaki and Radziwiłł for an instance of this.

We now prove Theorem 2. Let ${Q}$ denote the left-hand side of (2); by the triangle inequality we have ${Q=O(1)}$. By Lemma 3 (for some ${P = x^{o(1)}}$ to be chosen later) and the triangle inequality we have

$\displaystyle \sum_{n \leq x} \frac{\frac{1}{\log \log P} \sum_{p \leq P: p|n} f(n)}{n} = Q \log x + o( \log x ).$

We rearrange the left-hand side as

$\displaystyle \frac{1}{\log\log P} \sum_{p \leq P} \frac{f(p)}{p} \sum_{m \leq x/p} \frac{f(m)}{m}.$

We now replace the constraint ${m \leq x/p}$ by ${m \leq x}$. The error incurred in doing so is

$\displaystyle O( \frac{1}{\log\log P} \sum_{p \leq P} \frac{1}{p} \sum_{x/P \leq m \leq x} \frac{1}{m} )$

which by Mertens’ theorem is ${O(\log P) = o( \log x )}$. Thus we have

$\displaystyle \frac{1}{\log\log P} \sum_{p \leq P} \frac{f(p)}{p} \sum_{m \leq x} \frac{f(m)}{m} = Q \log x + o( \log x ).$

But by definition of ${Q}$, we have ${\sum_{m \leq x} \frac{f(m)}{m} = Q \log x}$, thus

$\displaystyle [1 - \frac{1}{\log\log P} \sum_{p \leq P} \frac{f(p)}{p}] Q = o(1). \ \ \ \ \ (3)$

From Mertens’ theorem, the expression in brackets can be rewritten as

$\displaystyle \frac{1}{\log\log P} \sum_{p \leq P} \frac{1 - f(p)}{p} + o(1)$

and so the real part of this expression is

$\displaystyle \frac{1}{\log\log P} \sum_{p \leq P} \frac{1 - \hbox{Re} f(p)}{p} + o(1).$

By (1), Mertens’ theorem and the hypothesis on ${f}$ we have

$\displaystyle \sum_{p \leq x^\varepsilon} \frac{(1 - \hbox{Re} f(p)) \log p}{p} \gg \log\log x^\varepsilon - O_\varepsilon(1)$

for any ${\varepsilon > 0}$. This implies that we can find ${P = x^{o(1)}}$ going to infinity such that

$\displaystyle \sum_{p \leq P} \frac{(1 - \hbox{Re} f(p)) \log p}{p} \gg (1-o(1))\log\log P$

and thus the expression in brackets has real part ${\gg 1-o(1)}$. The claim follows.

The Turan-Kubilius argument is certainly not the most efficient way to estimate sums such as ${\frac{1}{n} \sum_{n \leq x} f(n)}$. In the exercise below we give a significantly more accurate estimate that works when ${f}$ is non-negative.

Exercise 5 (Granville-Koukoulopoulos-Matomaki)

• (i) If ${g}$ is a completely multiplicative function with ${g(p) \in \{0,1\}}$ for all primes ${p}$, show that

$\displaystyle (e^{-\gamma}-o(1)) \prod_{p \leq x} (1 - \frac{g(p)}{p})^{-1} \leq \sum_{n \leq x} \frac{g(n)}{n} \leq \prod_{p \leq x} (1 - \frac{g(p)}{p})^{-1}.$

as ${x \rightarrow \infty}$. (Hint: for the upper bound, expand out the Euler product. For the lower bound, show that ${\sum_{n \leq x} \frac{g(n)}{n} \times \sum_{n \leq x} \frac{h(n)}{n} \ge \sum_{n \leq x} \frac{1}{n}}$, where ${h}$ is the completely multiplicative function with ${h(p) = 1-g(p)}$ for all primes ${p}$.)

• (ii) If ${g}$ is multiplicative and takes values in ${[0,1]}$, show that

$\displaystyle \sum_{n \leq x} \frac{g(n)}{n} \asymp \prod_{p \leq x} (1 - \frac{g(p)}{p})^{-1}$

$\displaystyle \asymp \exp( \sum_{p \leq x} \frac{g(p)}{p} )$

for all ${x \geq 1}$.

Now we turn to a very recent result of Matomaki and Radziwiłł on mean values of multiplicative functions in short intervals. For sake of illustration we specialise their results to the simpler case of the Liouville function ${\lambda}$, although their arguments actually work (with some additional effort) for arbitrary multiplicative functions of magnitude at most ${1}$ that are real-valued (or more generally, stay far from complex characters ${p \mapsto p^{it}}$). Furthermore, we give a qualitative form of their estimates rather than a quantitative one:

Theorem 6 (Matomaki-Radziwiłł, special case) Let ${X}$ be a parameter going to infinity, and let ${2 \leq h \leq X}$ be a quantity going to infinity as ${X \rightarrow \infty}$. Then for all but ${o(X)}$ of the integers ${x \in [X,2X]}$, one has

$\displaystyle \sum_{x \leq n \leq x+h} \lambda(n) = o( h ).$

Equivalently, one has

$\displaystyle \sum_{X \leq x \leq 2X} |\sum_{x \leq n \leq x+h} \lambda(n)|^2 = o( h^2 X ). \ \ \ \ \ (4)$

A simple sieving argument (see Exercise 18 of Supplement 4) shows that one can replace ${\lambda}$ by the Möbius function ${\mu}$ and obtain the same conclusion. See this recent note of Matomaki and Radziwiłł for a simple proof of their (quantitative) main theorem in this special case.

Of course, (4) improves upon the trivial bound of ${O( h^2 X )}$. Prior to this paper, such estimates were only known (using arguments similar to those in Section 3 of Notes 6) for ${h \geq X^{1/6+\varepsilon}}$ unconditionally, or for ${h \geq \log^A X}$ for some sufficiently large ${A}$ if one assumed the Riemann hypothesis. This theorem also represents some progress towards Chowla’s conjecture (discussed in Supplement 4) that

$\displaystyle \sum_{n \leq x} \lambda(n+h_1) \dots \lambda(n+h_k) = o( x )$

as ${x \rightarrow \infty}$ for any fixed distinct ${h_1,\dots,h_k}$; indeed, it implies that this conjecture holds if one performs a small amount of averaging in the ${h_1,\dots,h_k}$.

Below the fold, we give a “cheap” version of the Matomaki-Radziwiłł argument. More precisely, we establish

Theorem 7 (Cheap Matomaki-Radziwiłł) Let ${X}$ be a parameter going to infinity, and let ${1 \leq T \leq X}$. Then

$\displaystyle \int_X^{X^A} \left|\sum_{x \leq n \leq e^{1/T} x} \frac{\lambda(n)}{n}\right|^2\frac{dx}{x} = o\left( \frac{\log X}{T^2} \right), \ \ \ \ \ (5)$

for any fixed ${A>1}$.

Note that (5) improves upon the trivial bound of ${O( \frac{\log X}{T^2} )}$. Again, one can replace ${\lambda}$ with ${\mu}$ if desired. Due to the cheapness of Theorem 7, the proof will require few ingredients; the deepest input is the improved zero-free region for the Riemann zeta function due to Vinogradov and Korobov. Other than that, the main tools are the Turan-Kubilius result established above, and some Fourier (or complex) analysis.

In the previous set of notes, we saw how zero-density theorems for the Riemann zeta function, when combined with the zero-free region of Vinogradov and Korobov, could be used to obtain prime number theorems in short intervals. It turns out that a more sophisticated version of this type of argument also works to obtain prime number theorems in arithmetic progressions, in particular establishing the celebrated theorem of Linnik:

Theorem 1 (Linnik’s theorem) Let ${a\ (q)}$ be a primitive residue class. Then ${a\ (q)}$ contains a prime ${p}$ with ${p \ll q^{O(1)}}$.

In fact it is known that one can find a prime ${p}$ with ${p \ll q^{5}}$, a result of Xylouris. For sake of comparison, recall from Exercise 65 of Notes 2 that the Siegel-Walfisz theorem gives this theorem with a bound of ${p \ll \exp( q^{o(1)} )}$, and from Exercise 48 of Notes 2 one can obtain a bound of the form ${p \ll \phi(q)^2 \log^2 q}$ if one assumes the generalised Riemann hypothesis. The probabilistic random models from Supplement 4 suggest that one should in fact be able to take ${p \ll q^{1+o(1)}}$.

We will not aim to obtain the optimal exponents for Linnik’s theorem here, and follow the treatment in Chapter 18 of Iwaniec and Kowalski. We will in fact establish the following more quantitative result (a special case of a more powerful theorem of Gallagher), which splits into two cases, depending on whether there is an exceptional zero or not:

Theorem 2 (Quantitative Linnik theorem) Let ${a\ (q)}$ be a primitive residue class for some ${q \geq 2}$. For any ${x > 1}$, let ${\psi(x;q,a)}$ denote the quantity

$\displaystyle \psi(x;q,a) := \sum_{n \leq x: n=a\ (q)} \Lambda(n).$

Assume that ${x \geq q^C}$ for some sufficiently large ${C}$.

• (i) (No exceptional zero) If all the real zeroes ${\beta}$ of ${L}$-functions ${L(\cdot,\chi)}$ of real characters ${\chi}$ of modulus ${q}$ are such that ${1-\beta \gg \frac{1}{\log q}}$, then

$\displaystyle \psi(x;q,a) = \frac{x}{\phi(q)} ( 1 + O( \exp( - c \frac{\log x}{\log q} ) ) + O( \frac{\log^2 q}{q} ) )$

for all ${x \geq 1}$ and some absolute constant ${c>0}$.

• (ii) (Exceptional zero) If there is a zero ${\beta}$ of an ${L}$-function ${L(\cdot,\chi_1)}$ of a real character ${\chi_1}$ of modulus ${q}$ with ${\beta = 1 - \frac{\varepsilon}{\log q}}$ for some sufficiently small ${\varepsilon>0}$, then

$\displaystyle \psi(x;q,a) = \frac{x}{\phi(q)} ( 1 - \chi_1(a) \frac{x^{\beta-1}}{\beta} \ \ \ \ \ (1)$

$\displaystyle + O( \exp( - c \frac{\log x}{\log q} \log \frac{1}{\varepsilon} ) )$

$\displaystyle + O( \frac{\log^2 q}{q} ) )$

for all ${x \geq 1}$ and some absolute constant ${c>0}$.

The implied constants here are effective.

Note from the Landau-Page theorem (Exercise 54 from Notes 2) that at most one exceptional zero exists (if ${\varepsilon}$ is small enough). A key point here is that the error term ${O( \exp( - c \frac{\log x}{\log q} \log \frac{1}{\varepsilon} ) )}$ in the exceptional zero case is an improvement over the error term when no exceptional zero is present; this compensates for the potential reduction in the main term coming from the ${\chi_1(a) \frac{x^{\beta-1}}{\beta}}$ term. The splitting into cases depending on whether an exceptional zero exists or not turns out to be an essential technique in many advanced results in analytic number theory (though presumably such a splitting will one day become unnecessary, once the possibility of exceptional zeroes are finally eliminated for good).

Exercise 3 Assuming Theorem 2, and assuming ${x \geq q^C}$ for some sufficiently large absolute constant ${C}$, establish the lower bound

$\displaystyle \psi(x;a,q) \gg \frac{x}{\phi(q)}$

when there is no exceptional zero, and

$\displaystyle \psi(x;a,q) \gg \varepsilon \frac{x}{\phi(q)}$

when there is an exceptional zero ${\beta = 1 - \frac{\varepsilon}{\log q}}$. Conclude that Theorem 2 implies Theorem 1, regardless of whether an exceptional zero exists or not.

Remark 4 The Brun-Titchmarsh theorem (Exercise 33 from Notes 4), in the sharp form of Montgomery and Vaughan, gives that

$\displaystyle \pi(x; q, a) \leq 2 \frac{x}{\phi(q) \log (x/q)}$

for any primitive residue class ${a\ (q)}$ and any ${x \geq q}$. This is (barely) consistent with the estimate (1). Any lowering of the coefficient ${2}$ in the Brun-Titchmarsh inequality (with reasonable error terms), in the regime when ${x}$ is a large power of ${q}$, would then lead to at least some elimination of the exceptional zero case. However, this has not led to any progress on the Landau-Siegel zero problem (and may well be just a reformulation of that problem). (When ${x}$ is a relatively small power of ${q}$, some improvements to Brun-Titchmarsh are possible that are not in contradiction with the presence of an exceptional zero; see this paper of Maynard for more discussion.)

Theorem 2 is deduced in turn from facts about the distribution of zeroes of ${L}$-functions. We first need a version of the truncated explicit formula that does not lose unnecessary logarithms:

Exercise 5 (Log-free truncated explicit formula) With the hypotheses as above, show that

$\displaystyle \sum_{n \leq x} \Lambda(n) \chi(n) = - \sum_{\hbox{Re}(\rho) > 3/4; |\hbox{Im}(\rho)| \leq q^2; L(\rho,\chi)=0} \frac{x^\rho}{\rho} + O( \frac{x}{q^2} \log^2 q)$

for any non-principal character ${\chi}$ of modulus ${q}$, where we assume ${x \geq q^C}$ for some large ${C}$; for the principal character establish the same formula with an additional term of ${x}$ on the right-hand side. (Hint: this is almost immediate from Exercise 45(iv) and Theorem 21 ofNotes 2) with (say) ${T := q^2}$, except that there is a factor of ${\log^2 x}$ in the error term instead of ${\log^2 q}$ when ${x}$ is extremely large compared to ${q}$. However, a closer inspection of the proof (particularly with regards to the truncated Perron formula in Proposition 12 of Notes 2) shows that the ${\log^2 x}$ factor can be replaced fairly easily by ${\log x \log q}$. To get rid of the final factor of ${\log x}$, note that the proof of Proposition 12 used the rather crude bound ${\Lambda(n) = O(\log n)}$. If one replaces this crude bound by more sophisticated tools such as the Brun-Titchmarsh inequality, one will be able to remove the factor of ${\log x}$.

Using the Fourier inversion formula

$\displaystyle 1_{n = a\ (q)} = \frac{1}{\phi(q)} \sum_{\chi\ (q)} \overline{\chi(a)} \chi(n)$

(see Theorem 69 of Notes 1), we thus have

$\displaystyle \psi(x;a,q) = \frac{x}{\phi(q)} ( 1 - \sum_{\chi\ (q)} \overline{\chi(a)} \sum_{\hbox{Re}(\rho) > 3/4; |\hbox{Im}(\rho)| \leq q^2; L(\rho,\chi)=0} \frac{x^{\rho-1}}{\rho}$

$\displaystyle + O( \frac{\log^2 q}{q} ) )$

and so it suffices by the triangle inequality (bounding ${1/\rho}$ very crudely by ${O(1)}$, as the contribution of the low-lying zeroes already turns out to be quite dominant) to show that

$\displaystyle \sum_{\chi\ (q)} \sum_{\sigma > 3/4; |t| \leq q^2; L(\sigma+it,\chi)=0} x^{\sigma-1} \ll \exp( - c \frac{\log x}{\log q} ) \ \ \ \ \ (2)$

when no exceptional zero is present, and

$\displaystyle \sum_{\chi\ (q)} \sum_{\sigma > 3/4; |t| \leq q^2; L(\sigma+it,\chi)=0; \sigma+it \neq \beta} x^{\sigma-1} \ll \exp( - c \frac{\log x}{\log q} \log \frac{1}{\varepsilon} ) \ \ \ \ \ (3)$

when an exceptional zero is present.

To handle the former case (2), one uses two facts about zeroes. The first is the classical zero-free region (Proposition 51 from Notes 2), which we reproduce in our context here:

Proposition 6 (Classical zero-free region) Let ${q, T \geq 2}$. Apart from a potential exceptional zero ${\beta}$, all zeroes ${\sigma+it}$ of ${L}$-functions ${L(\cdot,\chi)}$ with ${\chi}$ of modulus ${q}$ and ${|t| \leq T}$ are such that

$\displaystyle \sigma \leq 1 - \frac{c}{\log qT}$

for some absolute constant ${c>0}$.

Using this zero-free region, we have

$\displaystyle x^{\sigma-1} \ll \log x \int_{1/2}^{1-c/\log q} 1_{\alpha < \sigma} x^{\alpha-1}\ d\alpha$

whenever ${\sigma}$ contributes to the sum in (2), and so the left-hand side of (2) is bounded by

$\displaystyle \ll \log x \int_{1/2}^{1 - c/\log q} N( \alpha, q, q^2 ) x^{\alpha-1}\ d\alpha$

where we recall that ${N(\alpha,q,T)}$ is the number of zeroes ${\sigma+it}$ of any ${L}$-function of a character ${\chi}$ of modulus ${q}$ with ${\sigma \geq \alpha}$ and ${0 \leq t \leq T}$ (here we use conjugation symmetry to make ${t}$ non-negative, accepting a multiplicative factor of two).

In Exercise 25 of Notes 6, the grand density estimate

$\displaystyle N(\alpha,q,T) \ll (qT)^{4(1-\alpha)} \log^{O(1)}(qT) \ \ \ \ \ (4)$

is proven. If one inserts this bound into the above expression, one obtains a bound for (2) which is of the form

$\displaystyle \ll (\log^{O(1)} q) \exp( - c \frac{\log x}{\log q} ).$

Unfortunately this is off from what we need by a factor of ${\log^{O(1)} q}$ (and would lead to a weak form of Linnik’s theorem in which ${p}$ was bounded by ${O( \exp( \log^{O(1)} q ) )}$ rather than by ${q^{O(1)}}$). In the analogous problem for prime number theorems in short intervals, we could use the Vinogradov-Korobov zero-free region to compensate for this loss, but that region does not help here for the contribution of the low-lying zeroes with ${t = O(1)}$, which as mentioned before give the dominant contribution. Fortunately, it is possible to remove this logarithmic loss from the zero-density side of things:

Theorem 7 (Log-free grand density estimate) For any ${q, T > 1}$ and ${1/2 \leq \alpha \leq 1}$, one has

$\displaystyle N(\alpha,q,T) \ll (qT)^{O(1-\alpha)}.$

The implied constants are effective.

We prove this estimate below the fold. The proof follows the methods of the previous section, but one inserts various sieve weights to restrict sums over natural numbers to essentially become sums over “almost primes”, as this turns out to remove the logarithmic losses. (More generally, the trick of restricting to almost primes by inserting suitable sieve weights is quite useful for avoiding any unnecessary losses of logarithmic factors in analytic number theory estimates.)

Exercise 8 Use Theorem 7 to complete the proof of (2).

Now we turn to the case when there is an exceptional zero (3). The argument used to prove (2) applies here also, but does not gain the factor of ${\log \frac{1}{\varepsilon}}$ in the exponent. To achieve this, we need an additional tool, a version of the Deuring-Heilbronn repulsion phenomenon due to Linnik:

Theorem 9 (Deuring-Heilbronn repulsion phenomenon) Suppose ${q \geq 2}$ is such that there is an exceptional zero ${\beta = 1 - \frac{\varepsilon}{\log q}}$ with ${\varepsilon}$ small. Then all other zeroes ${\sigma+it}$ of ${L}$-functions of modulus ${q}$ are such that

$\displaystyle \sigma \leq 1 - c \frac{\log \frac{1}{\varepsilon}}{\log(q(2+|t|))}.$

In other words, the exceptional zero enlarges the classical zero-free region by a factor of ${\log \frac{1}{\varepsilon}}$. The implied constants are effective.

Exercise 10 Use Theorem 7 and Theorem 9 to complete the proof of (3), and thus Linnik’s theorem.

Exercise 11 Use Theorem 9 to give an alternate proof of (Tatuzawa’s version of) Siegel’s theorem (Theorem 62 of Notes 2). (Hint: if two characters have different moduli, then they can be made to have the same modulus by multiplying by suitable principal characters.)

Theorem 9 is proven by similar methods to that of Theorem 7, the basic idea being to insert a further weight of ${1 * \chi_1}$ (in addition to the sieve weights), the point being that the exceptional zero causes this weight to be quite small on the average. There is a strengthening of Theorem 9 due to Bombieri that is along the lines of Theorem 7, obtaining the improvement

$\displaystyle N'(\alpha,q,T) \ll \varepsilon (1 + \frac{\log T}{\log q}) (qT)^{O(1-\alpha)} \ \ \ \ \ (5)$

with effective implied constants for any ${1/2 \leq \alpha \leq 1}$ and ${T \geq 1}$ in the presence of an exceptional zero, where the prime in ${N'(\alpha,q,T)}$ means that the exceptional zero ${\beta}$ is omitted (thus ${N'(\alpha,q,T) = N(\alpha,q,T)-1}$ if ${\alpha \leq \beta}$). Note that the upper bound on ${N'(\alpha,q,T)}$ falls below one when ${\alpha > 1 - c \frac{\log \frac{1}{\varepsilon}}{\log(qT)}}$ for a sufficiently small ${c>0}$, thus recovering Theorem 9. Bombieri’s theorem can be established by the methods in this set of notes, and will be given as an exercise to the reader.

Remark 12 There are a number of alternate ways to derive the results in this set of notes, for instance using the Turan power sums method which is based on studying derivatives such as

$\displaystyle \frac{L'}{L}(s,\chi)^{(k)} = (-1)^k \sum_n \frac{\Lambda(n) \chi(n) \log^k n}{n^s}$

$\displaystyle \approx (-1)^{k+1} k! \sum_\rho \frac{1}{(s-\rho)^{k+1}}$

for ${\hbox{Re}(s)>1}$ and large ${k}$, and performing various sorts of averaging in ${k}$ to attenuate the contribution of many of the zeroes ${\rho}$. We will not develop this method here, but see for instance Chapter 9 of Montgomery’s book. See the text of Friedlander and Iwaniec for yet another approach based primarily on sieve-theoretic ideas.

Remark 13 When one optimises all the exponents, it turns out that the exponent in Linnik’s theorem is extremely good in the presence of an exceptional zero – indeed Friedlander and Iwaniec showed can even get a bound of the form ${p \ll q^{2-c}}$ for some ${c>0}$, which is even stronger than one can obtain from GRH! There are other places in which exceptional zeroes can be used to obtain results stronger than what one can obtain even on the Riemann hypothesis; for instance, Heath-Brown used the hypothesis of an infinite sequence of Siegel zeroes to obtain the twin prime conejcture.

In the previous set of notes, we studied upper bounds on sums such as ${|\sum_{N \leq n \leq N+M} n^{-it}|}$ for ${1 \leq M \leq N}$ that were valid for all ${t}$ in a given range, such as ${[T,2T]}$; this led in turn to upper bounds on the Riemann zeta ${\zeta(\sigma+it)}$ for ${t}$ in the same range, and for various choices of ${\sigma}$. While some improvement over the trivial bound of ${O(N)}$ was obtained by these methods, we did not get close to the conjectural bound of ${O( N^{1/2+o(1)})}$ that one expects from pseudorandomness heuristics (assuming that ${T}$ is not too large compared with ${N}$, e.g. ${T = O(N^{O(1)})}$.

However, it turns out that one can get much better bounds if one settles for estimating sums such as ${|\sum_{N \leq n \leq N+M} n^{-it}|}$, or more generally finite Dirichlet series (also known as Dirichlet polynomials) such as ${|\sum_n a_n n^{-it}|}$, for most values of ${t}$ in a given range such as ${[T,2T]}$. Equivalently, we will be able to get some control on the large values of such Dirichlet polynomials, in the sense that we can control the set of ${t}$ for which ${|\sum_n a_n n^{-it}|}$ exceeds a certain threshold, even if we cannot show that this set is empty. These large value theorems are often closely tied with estimates for mean values such as ${\frac{1}{T}\int_T^{2T} |\sum_n a_n n^{-it}|^{2k}\ dt}$ of a Dirichlet series; these latter estimates are thus known as mean value theorems for Dirichlet series. Our approach to these theorems will follow the same sort of methods used in Notes 3, in particular relying on the generalised Bessel inequality from those notes.

Our main application of the large value theorems for Dirichlet polynomials will be to control the number of zeroes of the Riemann zeta function ${\zeta(s)}$ (or the Dirichlet ${L}$-functions ${L(s,\chi)}$) in various rectangles of the form ${\{ \sigma+it: \sigma \geq \alpha, |t| \leq T \}}$ for various ${T > 1}$ and ${1/2 < \alpha < 1}$. These rectangles will be larger than the zero-free regions for which we can exclude zeroes completely, but we will often be able to limit the number of zeroes in such rectangles to be quite small. For instance, we will be able to show the following weak form of the Riemann hypothesis: as ${T \rightarrow \infty}$, a proportion ${1-o(1)}$ of zeroes of the Riemann zeta function in the critical strip with ${|\hbox{Im}(s)| \leq T}$ will have real part ${1/2+o(1)}$. Related to this, the number of zeroes with ${|\hbox{Im}(s)| \leq T}$ and ${|\hbox{Re}(s)| \geq \alpha}$ can be shown to be bounded by ${O( T^{O(1-\alpha)+o(1)} )}$ as ${T \rightarrow \infty}$ for any ${1/2 < \alpha < 1}$.

In the next set of notes we will use refined versions of these theorems to establish Linnik’s theorem on the least prime in an arithmetic progression.

Our presentation here is broadly based on Chapters 9 and 10 in Iwaniec and Kowalski, who give a number of more sophisticated large value theorems than the ones discussed here.

We return to the study of the Riemann zeta function ${\zeta(s)}$, focusing now on the task of upper bounding the size of this function within the critical strip; as seen in Exercise 43 of Notes 2, such upper bounds can lead to zero-free regions for ${\zeta}$, which in turn lead to improved estimates for the error term in the prime number theorem.

In equation (21) of Notes 2 we obtained the somewhat crude estimates

$\displaystyle \zeta(s) = \sum_{n \leq x} \frac{1}{n^s} - \frac{x^{1-s}}{1-s} + O( \frac{|s|}{\sigma} \frac{1}{x^\sigma} ) \ \ \ \ \ (1)$

for any ${x > 0}$ and ${s = \sigma+it}$ with ${\sigma>0}$ and ${s \neq 1}$. Setting ${x=1}$, we obtained the crude estimate

$\displaystyle \zeta(s) = \frac{1}{s-1} + O( \frac{|s|}{\sigma} )$

in this region. In particular, if ${0 < \varepsilon \leq \sigma \ll 1}$ and ${|t| \gg 1}$ then we had ${\zeta(s) = O_\varepsilon( |t| )}$. Using the functional equation and the Hadamard three lines lemma, we can improve this to ${\zeta(s) \ll_\varepsilon |t|^{\frac{1-\sigma}{2}+\varepsilon}}$; see Supplement 3.
Now we seek better upper bounds on ${\zeta}$. We will reduce the problem to that of bounding certain exponential sums, in the spirit of Exercise 34 of Supplement 3:

Proposition 1 Let ${s = \sigma+it}$ with ${0 < \varepsilon \leq \sigma \ll 1}$ and ${|t| \gg 1}$. Then

$\displaystyle \zeta(s) \ll_\varepsilon \log(2+|t|) \sup_{1 \leq M \leq N \ll |t|}$

$\displaystyle N^{1-\sigma} |\frac{1}{N} \sum_{N \leq n < N+M} e( -\frac{t}{2\pi} \log n)|$

where ${e(x) := e^{2\pi i x}}$.

Proof: We fix a smooth function ${\eta: {\bf R} \rightarrow {\bf C}}$ with ${\eta(t)=1}$ for ${t \leq -1}$ and ${\eta(t)=0}$ for ${t \geq 1}$, and allow implied constants to depend on ${\eta}$. Let ${s=\sigma+it}$ with ${\varepsilon \leq \sigma \ll 1}$. From Exercise 34 of Supplement 3, we have

$\displaystyle \zeta(s) = \sum_n \frac{1}{n^s} \eta( \log n - \log C|t| ) + O_\varepsilon( 1 )$

for some sufficiently large absolute constant ${C}$. By dyadic decomposition, we thus have

$\displaystyle \zeta(s) \ll_{\varepsilon} 1 + \log(2+|t|) \sup_{1 \leq N \ll |t|} |\sum_{N \leq n < 2N} \frac{1}{n^s} \eta( \log n - \log C|t| )|.$

We can absorb the first term in the second using the ${N=1}$ case of the supremum. Writing ${\frac{1}{n^s} \eta( \log n - \log|C| t ) = N^{-\sigma} e( - \frac{t}{2\pi} \log n ) F_N(n)}$, where

$\displaystyle F_N(n) := (N/n)^\sigma \eta(\log n - \log C|t| ),$

it thus suffices to show that

$\displaystyle \sum_{N \leq n < 2N} e(-\frac{t}{2\pi} \log N) F_N(n) \ll \sup_{1 \leq M \leq N} |\sum_{N \leq n < N+M} e(-\frac{t}{2\pi} \log n)|$

for each ${N}$. But from the fundamental theorem of calculus, the left-hand side can be written as

$\displaystyle F_N(2N) \sum_{N \leq n < 2N} e(-\frac{t}{2\pi} \log n)$

$\displaystyle - \int_0^{N} (\sum_{N \leq n < N+M} e(-\frac{t}{2\pi} \log n)) F'_N(M)\ dM$

and the claim then follows from the triangle inequality and a routine calculation. $\Box$
We are thus interested in getting good bounds on the sum ${\sum_{N \leq n < N+M} e( -\frac{t}{2\pi} \log n )}$. More generally, we consider normalised exponential sums of the form

$\displaystyle \frac{1}{N} \sum_{n \in I} e( f(n) ) \ \ \ \ \ (2)$

where ${I \subset {\bf R}}$ is an interval of length at most ${N}$ for some ${N \geq 1}$, and ${f: {\bf R} \rightarrow {\bf R}}$ is a smooth function. We will assume smoothness estimates of the form

$\displaystyle |f^{(j)}(x)| = \exp( O(j^2) ) \frac{T}{N^j} \ \ \ \ \ (3)$

for some ${T>0}$, all ${x \in I}$, and all ${j \geq 1}$, where ${f^{(j)}}$ is the ${j}$-fold derivative of ${f}$; in the case ${f(x) := -\frac{t}{2\pi} \log x}$, ${I \subset [N,2N]}$ of interest for the Riemann zeta function, we easily verify that these estimates hold with ${T := |t|}$. (One can consider exponential sums under more general hypotheses than (3), but the hypotheses here are adequate for our needs.) We do not bound the zeroth derivative ${f^{(0)}=f}$ of ${f}$ directly, but it would not be natural to do so in any event, since the magnitude of the sum (2) is unaffected if one adds an arbitrary constant to ${f(n)}$.
The trivial bound for (2) is

$\displaystyle \frac{1}{N} \sum_{n \in I} e(f(n)) \ll 1 \ \ \ \ \ (4)$

and we will seek to obtain significant improvements to this bound. Pseudorandomness heuristics predict a bound of ${O_\varepsilon(N^{-1/2+\varepsilon})}$ for (2) for any ${\varepsilon>0}$ if ${T = O(N^{O(1)})}$; this assertion (a special case of the exponent pair hypothesis) would have many consequences (for instance, inserting it into Proposition 1 soon yields the Lindelöf hypothesis), but is unfortunately quite far from resolution with known methods. However, we can obtain weaker gains of the form ${O(N^{1-c_K})}$ when ${T \ll N^K}$ and ${c_K > 0}$ depends on ${K}$. We present two such results here, which perform well for small and large values of ${K}$ respectively:

Theorem 2 Let ${2 \leq N \ll T}$, let ${I}$ be an interval of length at most ${N}$, and let ${f: I \rightarrow {\bf R}}$ be a smooth function obeying (3) for all ${j \geq 1}$ and ${x \in I}$.

• (i) (van der Corput estimate) For any natural number ${k \geq 2}$, one has

$\displaystyle \frac{1}{N} \sum_{n \in I} e( f(n) ) \ll (\frac{T}{N^k})^{\frac{1}{2^k-2}} \log^{1/2} (2+T). \ \ \ \ \ (5)$

• (ii) (Vinogradov estimate) If ${k}$ is a natural number and ${T \leq N^{k}}$, then

$\displaystyle \frac{1}{N} \sum_{n \in I} e( f(n) ) \ll N^{-c/k^2} \ \ \ \ \ (6)$

for some absolute constant ${c>0}$.

The factor of ${\log^{1/2} (2+T)}$ can be removed by a more careful argument, but we will not need to do so here as we are willing to lose powers of ${\log T}$. The estimate (6) is superior to (5) when ${T \sim N^K}$ for ${K}$ large, since (after optimising in ${k}$) (5) gives a gain of the form ${N^{-c/2^{cK}}}$ over the trivial bound, while (6) gives ${N^{-c/K^2}}$. We have not attempted to obtain completely optimal estimates here, settling for a relatively simple presentation that still gives good bounds on ${\zeta}$, and there are a wide variety of additional exponential sum estimates beyond the ones given here; see Chapter 8 of Iwaniec-Kowalski, or Chapters 3-4 of Montgomery, for further discussion.

We now briefly discuss the strategies of proof of Theorem 2. Both parts of the theorem proceed by treating ${f}$ like a polynomial of degree roughly ${k}$; in the case of (ii), this is done explicitly via Taylor expansion, whereas for (i) it is only at the level of analogy. Both parts of the theorem then try to “linearise” the phase to make it a linear function of the summands (actually in part (ii), it is necessary to introduce an additional variable and make the phase a bilinear function of the summands). The van der Corput estimate achieves this linearisation by squaring the exponential sum about ${k}$ times, which is why the gain is only exponentially small in ${k}$. The Vinogradov estimate achieves linearisation by raising the exponential sum to a significantly smaller power – on the order of ${k^2}$ – by using Hölder’s inequality in combination with the fact that the discrete curve ${\{ (n,n^2,\dots,n^k): n \in \{1,\dots,M\}\}}$ becomes roughly equidistributed in the box ${\{ (a_1,\dots,a_k): a_j = O( M^j ) \}}$ after taking the sumset of about ${k^2}$ copies of this curve. This latter fact has a precise formulation, known as the Vinogradov mean value theorem, and its proof is the most difficult part of the argument, relying on using a “${p}$-adic” version of this equidistribution to reduce the claim at a given scale ${M}$ to a smaller scale ${M/p}$ with ${p \sim M^{1/k}}$, and then proceeding by induction.

One can combine Theorem 2 with Proposition 1 to obtain various bounds on the Riemann zeta function:

Exercise 3 (Subconvexity bound)

• (i) Show that ${\zeta(\frac{1}{2}+it) \ll (1+|t|)^{1/6} \log^{O(1)}(2+|t|)}$ for all ${t \in {\bf R}}$. (Hint: use the ${k=3}$ case of the Van der Corput estimate.)
• (ii) For any ${0 < \sigma < 1}$, show that ${\zeta(\sigma+it) \ll (1+|t|)^{\max( \frac{1-\sigma}{3}, \frac{1}{2} - \frac{2\sigma}{3}) + o(1)}}$ as ${|t| \rightarrow \infty}$ (the decay rate in the ${o(1)}$ is allowed to depend on ${\sigma}$).

Exercise 4 Let ${t}$ be such that ${|t| \geq 100}$, and let ${\sigma \geq 1/2}$.

• (i) (Littlewood bound) Use the van der Corput estimate to show that ${\zeta(\sigma+it) \ll \log^{O(1)} |t|}$ whenever ${\sigma \geq 1 - O( \frac{(\log\log |t|)^2}{\log |t|} ))}$.
• (ii) (Vinogradov-Korobov bound) Use the Vinogradov estimate to show that ${\zeta(\sigma+it) \ll \log^{O(1)} |t|}$ whenever ${\sigma \geq 1 - O( \frac{(\log\log |t|)^{2/3}}{\log^{2/3} |t|} )}$.

As noted in Exercise 43 of Notes 2, the Vinogradov-Korobov bound leads to the zero-free region ${\{ \sigma+it: \sigma > 1 - c \frac{1}{(\log |t|)^{2/3} (\log\log |t|)^{1/3}}; |t| \geq 100 \}}$, which in turn leads to the prime number theorem with error term

$\displaystyle \sum_{n \leq x} \Lambda(n) = x + O\left( x \exp\left( - c \frac{\log^{3/5} x}{(\log\log x)^{1/5}} \right) \right)$

for ${x > 100}$. If one uses the weaker Littlewood bound instead, one obtains the narrower zero-free region

$\displaystyle \{ \sigma+it: \sigma > 1 - c \frac{\log\log|t|}{\log |t|}; |t| \geq 100 \}$

(which is only slightly wider than the classical zero-free region) and an error term

$\displaystyle \sum_{n \leq x} \Lambda(n) = x + O( x \exp( - c \sqrt{\log x \log\log x} ) )$

in the prime number theorem.

Exercise 5 (Vinogradov-Korobov in arithmetic progressions) Let ${\chi}$ be a non-principal character of modulus ${q}$.

• (i) (Vinogradov-Korobov bound) Use the Vinogradov estimate to show that ${L(\sigma+it,\chi) \ll \log^{O(1)}(q|t|)}$ whenever ${|t| \geq 100}$ and

$\displaystyle \sigma \geq 1 - O( \min( \frac{\log\log(q|t|)}{\log q}, \frac{(\log\log(q|t|))^{2/3}}{\log^{2/3} |t|} ) ).$

(Hint: use the Vinogradov estimate and a change of variables to control ${\sum_{n \in I: n = a\ (q)} \exp( -it \log n)}$ for various intervals ${I}$ of length at most ${N}$ and residue classes ${a\ (q)}$, in the regime ${N \geq q^2}$ (say). For ${N < q^2}$, do not try to capture any cancellation and just use the triangle inequality instead.)

• (ii) Obtain a zero-free region

$\displaystyle \{ \sigma+it: \sigma > 1 - c \min( \frac{1}{(\log |t|)^{2/3} (\log\log |t|)^{1/3}}, \frac{1}{\log q} );$

$\displaystyle |t| \geq 100 \}$

for ${L(s,\chi)}$, for some (effective) absolute constant ${c>0}$.

• (iii) Obtain the prime number theorem in arithmetic progressions with error term

$\displaystyle \sum_{n \leq x: n = a\ (q)} \Lambda(n) = \frac{x}{\phi(q)} + O\left( x \exp\left( - c_A \frac{\log^{3/5} x}{(\log\log x)^{1/5}} \right) \right)$

whenever ${x > 100}$, ${q \leq \log^A x}$, ${a\ (q)}$ is primitive, and ${c_A>0}$ depends (ineffectively) on ${A}$.