We return to the study of the Riemann zeta function , 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 , 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
for any and with and . Setting , we obtained the crude estimate
in this region. In particular, if and then we had . Using the functional equation and the Hadamard three lines lemma, we can improve this to ; see Supplement 3.
Now we seek better upper bounds on . We will reduce the problem to that of bounding certain exponential sums, in the spirit of Exercise 33 of Supplement 3:
Proof: We fix a smooth function with for and for , and allow implied constants to depend on . Let with . From Exercise 33 of Supplement 3, we have
for some sufficiently large absolute constant . By dyadic decomposition, we thus have
We can absorb the first term in the second using the case of the supremum. Writing , where
it thus suffices to show that
for each . But from the fundamental theorem of calculus, the left-hand side can be written as
and the claim then follows from the triangle inequality and a routine calculation.
for some , all , and all , where is the -fold derivative of ; in the case , of interest for the Riemann zeta function, we easily verify that these estimates hold with . (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 of 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 .
The trivial bound for (2) is
and we will seek to obtain significant improvements to this bound. Pseudorandomness heuristics predict a bound of for (2) for any if ; 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 when and depends on . We present two such results here, which perform well for small and large values of respectively:
Theorem 2 Let , let be an interval of length at most , and let be a smooth function obeying (3) for all and .
The factor of 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 . The estimate (6) is superior to (5) when for large, since (after optimising in ) (5) gives a gain of the form over the trivial bound, while (6) gives . We have not attempted to obtain completely optimal estimates here, settling for a relatively simple presentation that still gives good bounds on , 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 like a polynomial of degree roughly ; 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 times, which is why the gain is only exponentially small in . The Vinogradov estimate achieves linearisation by raising the exponential sum to a significantly smaller power – on the order of – by using Hölder’s inequality in combination with the fact that the discrete curve becomes roughly equidistributed in the box after taking the sumset of about 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 “-adic” version of this equidistribution to reduce the claim at a given scale to a smaller scale with , and then proceeding by induction.
Exercise 3 (Subconvexity bound)
- (i) Show that for all . (Hint: use the case of the Van der Corput estimate.)
- (ii) For any , show that as .
Exercise 4 Let be such that , and let .
- (i) (Littlewood bound) Use the van der Corput estimate to show that whenever .
- (ii) (Vinogradov-Korobov bound) Use the Vinogradov estimate to show that whenever .
As noted in Exercise 43 of Notes 2, the Vinogradov-Korobov bound leads to the zero-free region , which in turn leads to the prime number theorem with error term
for . If one uses the weaker Littlewood bound instead, one obtains the narrower zero-free region
(which is only slightly wider than the classical zero-free region) and an error term
in the prime number theorem.
Exercise 5 (Vinogradov-Korobov in arithmetic progressions) Let be a non-principal character of modulus .
- (i) (Vinogradov-Korobov bound) Use the Vinogradov estimate to show that whenever and
(Hint: use the Vinogradov estimate and a change of variables to control for various intervals of length at most and residue classes , in the regime (say). For , do not try to capture any cancellation and just use the triangle inequality instead.)
- (ii) Obtain a zero-free region
for , for some (effective) absolute constant .
- (iii) Obtain the prime number theorem in arithmetic progressions with error term
whenever , , is primitive, and depends (ineffectively) on .
— 1. Van der Corput estimates —
In this section we prove Theorem 2(i). To motivate the arguments, we will use an analogy between the sums and the integrals (cf. Exercise 11 from Notes 1). This analogy can be made rigorous by the Poisson summation formula (after applying some smoothing to the integral over to truncate the frequency summation), but we will not need to do so here.
Write . We can control the integral by integration by parts:
If obeys (3) for , we thus have
An analogous argument, using summation by parts instead of integration by parts, controls the sum , but only in the regime when :
for and . If , then
Proof: We may assume that for some integers , and after deleting the right endpoint it suffices to show that
From the case of (3) and the mean value theorem one has for all . Thus we have
and we can write
From the cases of (3) and the mean value theorem we see that has size and derivative on . The claim then follows from a summation by parts.
To use this proposition in the regime , we will use the following inequality of van der Corput, which is a basic application of the Cauchy-Schwarz inequality:
The point of this proposition is that it effectively replaces the phase by the differenced phase for some medium-sized ; it is an oscillatory version of the trivial observation that if is close to constant, then is also close to constant. From the fundamental theorem of calculus, we see that if obeys the estimates (3), then obeys a variant of (3) in which has been replaced by . Since , this reduces , and so one can hope to then iterate this proposition to the point where one can apply Proposition 6.
Proof: By rounding down, we may assume that is an integer. For any , we have
and thus on averaging in
There are only values of for which the inner sum is non-vanishing. By the Cauchy-Schwarz inequality, we thus have
and so it will suffice to show that
The left-hand side may be expanded as
The contribution of the diagonal terms is which is acceptable. For the off-diagonal terms, we may use symmetry to restrict to the case , picking up a factor of . After shifting by , we may thus bound this contribution by
Since each occurs times as a difference , the claim follows.
Exercise 8 (Qualitative Weyl exponential sum estimates) Let be a polynomial with real coefficients .
- (i) If and is irrational, show that as . (Hint: induct on , using geometric series for the base case and Proposition 7 for the induction step.)
- (ii) If and at least one of is irrational, show that as .
- (iii) If all of the are rational, show that converges as to a limit that is not necessarily zero.
One can obtain more quantitative estimates on the decay rate of in terms of how badly approximable by rationals one or more of the coefficients are; see for instance Chapter 8 of Iwaniec-Kowalski for some estimates of this type.
for and . Then
Set , then . By Proposition 7 we have
where . From the fundamental theorem of calculus we have
for . By Proposition 6 we then have
so on summing in
and the claim follows from the choice of .
We can iterate this:
for and , one has
We have avoided the use of asymptotic notation for this proposition because we need to use induction on .
Proof: We induct on . The case follows from Proposition 9. Now suppose that and the claim has already been proven for .
From the fundamental theorem of calculus, obeys the bounds (7) for with replaced by . Thus by the induction hypothesis,
Performing the sum over , we obtain
which by the choice of simplifies to
and the claim follows for large enough.
whenever . (The case when can be easily derived from the case, after conceding a multiplicative factor of .
and the claim follows from the term of the infimum. If instead
and the claim follows from the induction hypothesis.
— 2. Vinogradov estimate —
We now prove Theorem 2(ii), loosely following the treatment in Iwaniec and Kowalski. We first observe that for bounded values of , part (ii) of this theorem follows from the already proven part (i) (replacing by, say, ), so we may assume now that is larger than any specified absolute constant.
The first step of Vinogradov’s argument is to use a Taylor expansion and a bilinear averaging to reduce the problem to a bilinear exponential sum estimate involving polynomials with medium-sized coefficients. More precisely, we will derive Theorem 2(ii) from
for some depending only on .
for some large constant , as the claim is trivial otherwise.
Set . For any , we have
and hence on averaging in
The condition can only be satisfied if lies within of , and if lies in and is further than from the endpoints of then the constraint may be dropped. It thus suffices to show that
for all in that are further than from the endpoints of .
Fix . From (3) and Taylor expansion we see that
where . Since and , we see from (11) that the error term is (say) if is large enough, so
and it thus suffices to show that
From (3) we have
which (since and is large) implies from (11) that
if (say). The claim now follows from Theorem 11 (with replaced by ).
It remains to prove Theorem 11. We fix and allow all implied constants to depend on . We may assume that
for some large constant (depending on ), as the claim is trivial otherwise.
We write the left-hand side of (10) as
where is the bilinear form
and is the polynomial curve
lies in the box , but occupies only a very sparse subset of this box. However, if one takes iterated sumsets
of this curve, one expects (if is large compared with ) to fill out a much denser subset of this box (or more precisely, of a slightly larger box in which all sides have been multiplied by ), due to the “curvature” of (13). By using the device of Hölder’s inequality, we will be able to estimate the sparse sum with a sum over such a sumset in a box, which will be significantly easier to estimate, provided one can establish the expected density property of the sumset, or at least something reasonably close to that density property. This latter claim will be accomplished by a deep result known as the Vinogradov mean value theorem.
We turn to the details. Let be a large natural number (depending on ) to be chosen later; in fact we will eventually take for a large absolute constant . From Hölder’s inequality we have
We remove the absolute values to write this as
for some coefficients of magnitude . The right-hand side can be rearranged using the bilinearity of as
We collect some terms to obtain the inequality
where for each , is the number of representations of of the form with . Note that is supported in the box
We now use Hölder’s inequality again to spread out the vectors, and also to separate the weight from the phase. Specifically, we have
with ; its estimation is the deepest and most difficult part of Vinogradov’s argument, and will be discussed presently. Leaving this aside for now, we return to (16) and expand out the right-hand side, using the triangle inequality and bilinearity to bound it by
The quantity lies in the box . Furthermore, from (15) and the Cauchy-Schwarz inequality, every has at most representations of the form
The point here is that the phase is now a linear function of the variables , in contrast to the polynomial function of the variables in the original definition of . In particular, the exponential sum is now fairly easy to estimate:
Lemma 12 If , then we have
for some (depending only on ).
Proof: We can factor the left-hand side as
Since and we have the trivial bound
it will suffice to show that
for at least choices of , for some .
By (9), we can find at least choices of with and
Henceforth we restrict to one of these choices. By summing the geometric series, or by using the trivial bound, we see that
where denotes the distance of to the nearest integer. On any interval of length , we see from the quantitative integral test (Lemma 2 of Notes 1) that
for all of the under consideration. The claim follows since , , and for some large (the latter hypothesis coming from (12)).
From this lemma and (18) we have
for some depending only on . To conclude the desired bound , it thus suffices to establish the following result:
for an absolute constant .
If we apply this theorem with equal to a sufficiently large multiple of , we obtain the required bound .
Actually, Vinogradov proved a slightly weaker estimate than this; the claim above (in a sharper form) was obtained by later refinements of the argument due to Stechkin and Karatsuba. This result has a number of applications beyond that of controlling the Riemann zeta function; for instance it has application to the Waring problem of expressing large natural numbers as the sum of powers, which we will not discuss further here. The reason for the name “mean value theorem” is that the quantity has a Fourier-analytic interpretation as a mean
of the exponential sum
The estimate (19) should be compared with the lower bound
where the term comes from the diagonal contribution to (17), and the term coming from (14) and the Cauchy-Schwarz inequality. Informally, (19) is an assertion that the -fold sum of the discrete curve (13) is somewhat close to being uniformly distributed on . The main conjecture of Vinogradov asserts the near-optimal bound
for any choice of and . In the recent work of Wooley and Ford-Wooley, an improved version of the congruencing method given below known as efficient congruencing has been developed and used to establish the main conjecture (20) in many cases. See this recent ICM proceedings article of Wooley for a survey of the latest developments in this direction.
We now turn to the proof of the Vinogradov mean value theorem. Since
which can be viewed as a measure of irregularity of distribution of the quantity for .
We have the extremely crude bound
whenever and some ; iterating this times and then using (23), we will obtain the claim. Note how it is important here that no powers of are lost in the estimate (24). However, we can be quite generous in losing factors such as , , or as these are easily absorbed in the term. To prove (24), we begin with a technical reduction. For a prime , let us first define a restricted version of to be the number of solutions to
with , with having distinct reductions mod , and also having distinct reductions mod . As we are taking to be rather large, one should think of the constraint of having distinct reductions mod to be a mild condition, so that is morally of the same size as . This is confirmed by the following proposition:
The reason why we need the prime to be somewhat comparable to will be clearer later.
with . If the set has cardinality less than , then there are ways to choose the , and ways to choose , leading to a contribution of at most to . Similarly if has cardinality less than . Thus we may restrict attention to the case when and have cardinality at least . By paying a factor of , we may then restrict to the case where are all distinct, and are all distinct. In particular, the quantity
is non-zero and has cardinality at most . In particular, there are at most primes larger than that define this quantity. On the other hand, from the prime number theorem we can find distinct primes in the range . We thus see that for each solution to (25) with , distinct, and distinct, there is a such that has distinct reductions mod , and also has distinct reductions mod , so that this tuple contributes to . Thus we have
and the claim follows.
Introducing the normalised quantity
for some prime .
The next step is to analyse the multiplicity properties of the -fold sum . Clearly we have
where the power sums are defined by
Lemma 15 (Newton’s identities) For any , one has the polynomial identity
Thus for instance
Proof: We use the method of generating functions. We rewrite the polynomial identity (27) as
On the other hand, taking logarithmic derivatives we have (as formal power series)
From the geometric series formula we have (as formal power series)
and the claim then follows by equating coefficients.
Corollary 16 If , then the quantity determines the multiset up to permutations. In particular, any element of has at most representations of the form .
Proof: The quantity determines the quantities for . By Newton’s identities and induction, this determines the quantities for , and thus determines the polynomial . The claim now follows from the unique factorisation of polynomials.
This particular result turns out to not give particularly good bounds on ; the sums are so sparsely distributed that the number of representations of a given (in, say, the box ) is typically zero. However, if we localise “-adically”, by replacing the integers with the ring , we get a more usable result:
with having distinct reductions mod , where by abuse of notation we localise to the ring in the obvious fashion.
To put it another way, this corollary asserts that the map from to is at most -to-one, if one excludes those which do not have distinct reductions mod .
Proof: Suppose we have two representations
with and each having distinct reductions mod . By Newton’s identities as before, we have for (note that as there is no difficulty dividing by for ). Therefore we have the polynomial identity
as polynomials over . In particular, each is a root of and thus must equal one of the since the reductions mod are all distinct (and so all but one of the factors will be invertible in ). This implies that is a permutation of , and the claim follows.
Returning to the integers, and specialising to the range of primes produced by Lemma 14, we conclude
Lemma 18 (Linnik’s lemma) Let be a prime with , and let . Then the number of solutions to the system
with and , with having distinct reductions modulo , is at most .
A version of this lemma also applies for primes outside of the range , but this is basically the range where the lemma is the most efficient. Indeed, probabilistic heuristics suggest that the number of solutions here should be approximately , which equals in the range .
Proof: Each residue class mod consists of residue classes mod . Since
it thus suffices (replacing with different elements of ) to show that for any , the number of solutions to the lifted system
with and is at most . But each residue class mod meets in at most representatives, so the claim follows from Corollary 17, crudely bounding by .
Now we need to consider sums of terms, rather than just terms. Recall that is the number of solutions to
with , with having distinct reductions mod , and also having distinct reductions mod . We also need an even more restricted version of , which is defined similarly to but with the additional constraint that
Probabilistic heuristics suggest that should be about as large as . One side of this prediction is confirmed by the following application of Hölder’s inequality:
Proof: It is easiest to prove this by Fourier-analytic means. Observe that we have the Fourier expansion
where the asterisk denotes the restriction to those with distinct reductions modulo and is the usual dot product , and
Similarly, we have
Since , the claim now follows from Hölder’s inequality.
Exercise 20 Find a non-Fourier-analytic proof of the above lemma, based on the Cauchy-Schwarz inequality, in the case when is a power of two. (Hint: you may wish to generalise the Hölder inequality to one involving the number of solutions to systems where each is drawn from a finite multiset (such quantities are known as additive energies of order in the additive combinatorics literature).) For an additional challenge: find a Cauchy-Schwarz proof that works for arbitrary values of .
Next, we use the Linnik lemma and some elementary arguments to bound :
Lemma 21 If , then
Proof: By definition, is the number of solutions to the system
where , with and each having distinct reductions mod , and also
for some . From the binomial theorem, is a linear transform of , so
Writing and for and some , we conclude that
In particular, taking the coefficient
There are choices for , and then by Linnik’s lemma once are chosen, there are at most choices for . Once these are chosen, we still have to select subject to a constraint of the form
for some depending on . The number of solutions to this system is
which by the Cauchy-Schwarz inequality is bounded by . The claim follows.