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 34 of Supplement 3:

Proposition 1Let with and . Thenwhere .

*Proof:* We fix a smooth function with for and for , and allow implied constants to depend on . Let with . From Exercise 34 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.

We are thus interested in getting good bounds on the sum . More generally, we consider normalised exponential sums of the form

where is an interval of length at most for some , and is a smooth function. We will assume smoothness estimates of the form

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 LindelĂ¶f 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 2Let , 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 HĂ¶lder’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.

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

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 (the decay rate in the is allowed to depend on ).

Exercise 4Let 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 .

** â€” 1. Van der Corput estimates â€” **

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 :

Proposition 6Let and , let be an interval of length at most , and let be a smooth function obeying the boundsfor 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:

Proposition 7 (van der Corput inequality)Let , let be an interval of length , and let be a function. Then

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.

If we combine one application of Proposition 7 with Proposition 6, we conclude

Proposition 9Let and , let be an interval of length at most , and let be a smooth function obeying the boundsfor and . Then

*Proof:* If then the claim follows from Proposition 6 (or from (4), if ), so we may assume that . We can also assume that , since otherwise the claim follows from the trivial bound (4).

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:

Proposition 10Let be a sufficiently large constant. Then for any , any , any natural number , any interval of length at most , and any smooth obeying the bounds

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 .

If then the claim follows from (4) (for large enough), so suppose that . Then the quantity is such that . By Proposition 7 we have

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.

Now we can prove part (i) of Theorem 2. We can assume that , since the claim follows from (4) otherwise. For any natural number , we can apply Proposition 10 with and conclude that

Taking infima over , it then suffices to show that

whenever . (The case when can be easily derived from the case, after conceding a multiplicative factor of .)

We prove (8) by induction on . The case is clear, so suppose and that (8) has already been proven for . If

then

and the claim follows from the term of the infimum. If instead

then

and the claim follows from the induction hypothesis.

** â€” 2. Vinogradov estimate â€” **

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

Theorem 11 (Bilinear estimate)Let be a constant, let be a sufficiently large natural number, let , and let be real numbers, with the property that

Let us see how Theorem 2(ii) follows from Theorem 11. By reducing as necessary, we may assume that

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 HĂ¶lder’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 HĂ¶lder’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 HĂ¶lder’s inequality again to spread out the vectors, and also to separate the weight from the phase. Specifically, we have

The quantity has a combinatorial interpretation as the number of solutions to the equation

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

Thus we arrive at the inequality

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 12If , then we havefor 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

and hence

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:

Theorem 13 (Vinogradov mean value theorem)Let be natural numbers such that . Then

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

and , we have , and so it will suffice to show that

whenever and is the normalised quantity

which can be viewed as a measure of irregularity of distribution of the quantity for .

We have the extremely crude bound

coming from the fact that there are choices of , so that

This already establishes the claim when (say). For , what we will do is establish the recursive inequality

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.

*Proof:* By definition, is the number of solutions to

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

and recalling that , we conclude that

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

We recall a classical relation, known as Newton’s identities (or *Newton-Girard identities*), between these power sums and the elementary symmetric polynomials , defined for by the formula

For instance , , and for all . One can also view the as essentially being the coefficients of the polynomial :

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 16If , 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:

Corollary 17Let be a prime with . Then any has at most representations of the formwith 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 systemwith 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 HĂ¶lder’s inequality:

Lemma 19 (HĂ¶lder inequality)We have

*Proof:* It is easiest to prove this by Fourier-analytic means. Observe that we have the Fourier expansion

where

where the asterisk denotes the restriction to those with distinct reductions modulo and is the usual dot product , and

Similarly, we have

where

Since , the claim now follows from HĂ¶lder’s inequality.

Exercise 20Find 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 HĂ¶lder inequality to one involving the number of solutions to systems where each is drawn from a finite multiset (such quantities are known asadditive energies of orderin 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 21If , 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.

Combining (26), (29), and (30) we obtain (24) as required.

## 49 comments

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7 February, 2015 at 1:23 pm

AnthonyUnder equation (1), should it be ‘setting x = 1’ instead of ‘s = 1’?

[Corrected, thanks – T.]7 February, 2015 at 1:24 pm

Eytan PaldiIn the line below (1), it seems that “” should be ““.

7 February, 2015 at 1:30 pm

Anthony…although this then gives s/(s-1), not 1/(s-1).

[The factor of can be absorbed into the error; alternatively, one can take to be slightly less than rather than exactly equal to . -T.]7 February, 2015 at 2:41 pm

AnonymousMy browser can not parse three formulas: the last line of proof of prop 9, two lines above (21), and two lines before lemma 18.

[Corrected, thanks – T.]7 February, 2015 at 10:16 pm

AnonymousUse “\colon” instead of “:” for maps. Example: .

8 February, 2015 at 1:47 am

AnonymousIn proposition 1, it may be added that is positive.

8 February, 2015 at 4:25 am

AnonymousIn the line above proposition 1, it should be exercise 33 (instead of exercise 34).

[Corrected, thanks – T.]8 February, 2015 at 7:39 am

AnonymousIn proposition 1, is the implied constant of “” (under the supremum) independent of ?

[Yes – T.]8 February, 2015 at 1:48 pm

AnonymousIn the line below (3), it should be “derivative of “.

[Corrected, thanks – T.]9 February, 2015 at 12:55 am

AnonymousIt seems that in proposition 5, the constant can be replaced by any constant .

[Fair enough; I had thought there was possibly a factor of to take care of, but there isn’t, so I switched the constant to 2 instead. -T.]9 February, 2015 at 2:01 pm

MrCactu5 (@MonsieurCactus)I would like to know why the assumptions 3 are reasonable. Proposition 1 seems to say that very far away from the real axis, you can ignore the first terms, and the zeta function is not more oscillatory than the thing you have written down.

9 February, 2015 at 2:18 pm

MrCactu5 (@MonsieurCactus)Yeah I am definitely not sure. and yet “ is a polynomial of degree roughly “

9 February, 2015 at 3:48 pm

Terence TaoThe intuition here is that analytic functions behave like polynomials (a similar philosophy was adopted in https://terrytao.wordpress.com/2014/12/05/245a-supplement-2-a-little-bit-of-complex-and-fourier-analysis/ ), particularly if the radius of convergence is larger than the interval on which one is trying to exploit polynomial-type behaviour. In the case of the function for , the radius of convergence is about , which is reflected in the increasing powers of N in the denominator of (3). In particular, the error term in a k-term Taylor expansion is about , which can be significantly less than 1 if is not too small compared with .

10 February, 2015 at 2:45 am

Trevor WooleyGreat to see Vinogradov’s methods poularised!

Just a comment that one can make a case now for saying that the main conjecture in Vinogradov’s mean value theorem (your equation (20)) has nearly been proved. By using the multigrade enhancement of my efficient congruencing method, one obtains the main conjecture in the cubic case in full, as well as for arbitrary when (and also when ). Of course, if one proves (20) for , then it follows for all , so in a sense we are only variables (out of ) away from finishing off the conjecture. [There’s an account of this kind of thing in http://arxiv.org/pdf/1404.3508v1.pdf ]

[Reference added – T.]11 February, 2015 at 2:36 am

AnonymousIt seems that it took more than 20 years to discover the application of Vinogradov mean value theorem (from 1935) to his estimate (theorem 2(ii)).

13 February, 2015 at 10:16 pm

254A, Notes 6: Large values of Dirichlet polynomials, zero density estimates, and primes in short intervals | What's new[…] the previous set of notes, we studied upper bounds on sums such as for that were valid for all in a given range, such as ; […]

14 February, 2015 at 12:54 pm

Eytan PaldiIn exercise 4, it seems that (in order to use proposition 1), the condition is also needed.

[Fair enough -T.]17 February, 2015 at 11:46 am

Alastair IrvingI’m slightly confused. IN the Vinogradov estimate theorem 2(ii) you gain by over the trivial bound (the same as in Iwaniec-Kowalski). However, many papers which consider estimating Weyl sums using the Vinogradov method only gain by . I understood that one of the consequences of Wooley’s efficient congruencing method is that the can be removed. Please can someone explain what I’m missing, in particular why there is no in the above theorem?

17 February, 2015 at 11:54 am

Terence TaoI believe the results you are referring to are for a slightly different exponential sum, namely sums of the form where is a polynomial of degree k. The sums here are a little different: where is a smooth function (not a polynomial) which obeys the derivative bound (3) for some . In particular the parameter “k” is slightly different in the two contexts. (Both results use Vinogradov’s mean value theorem, but in slightly different ways; in the situation considered in this post, the error exponent in the mean value theorem only needs to be less than a small multiple of , so one can take to be a large multiple of , but when dealing with the polynomial sums it seems that the usual methods need this error exponent to be much smaller, like , leading to being chosen as large as .)

1 March, 2015 at 1:13 pm

254A, Supplement 7: Normalised limit profiles of the log-magnitude of the Riemann zeta function (optional) | What's new[…] limit profiles are known unconditionally, beyond (i)-(iv). For instance, from Exercise 3 of Notes 5 we have as , which implies that any normalised limit profile for is bounded by on the critical […]

6 March, 2015 at 6:31 am

AnonymousIs it possible to generalize the bilinear estimate (10) to trilinear (or even multilinear) estimates ?

If so, can Vinogradov estimate (6) be improved ?

6 March, 2015 at 11:06 am

Terence TaoThis is theoretically possible, however in practice we have far fewer efficient techniques for dealing with trilinear or higher estimates than for bilinear ones. For instance, bilinear estimates can often be recast as operator norm estimates for a linear operator, for which one can hope to use spectral theory methods to help analyse (e.g. the identity for any natural number ). Trilinear estimates correspond to bilinear operators, for which we do not have spectral theory methods at our disposal.

11 December, 2015 at 5:04 am

AnonymousIs the new decoupling method (used in the proof of Vinogradov main conjecture) seems to have the potential to prove such trilinear (or even higher) estimates?

11 December, 2015 at 8:23 am

Terence TaoThis is potentially possible, and certainly worth exploring. Certainly the Bourgain-Demeter-Guth method relies heavily on multilinear estimates (though not exactly of the type under discussion here).

30 March, 2015 at 12:49 pm

254A, Notes 8: The Hardy-Littlewood circle method and Vinogradov’s theorem | What's new[…] not be discussed further here, save to note that the Vinogradov mean value theorem (Theorem 13 from Notes 5) and its variants are particularly useful for getting good bounds on ; see for instance the ICM […]

11 September, 2015 at 6:27 am

The Erdos discrepancy problem via the Elliott conjecture | What's new[…] from the Vinogradov-Korobov zero-free region for (see this previous blog post) it is not difficult to show […]

2 October, 2015 at 10:55 am

valuevarDear Terry,

The exponent in Exercise 3(ii) is off (obviously a typo).

[Corrected, thanks – T.]3 October, 2015 at 5:14 am

valuevarThe error is still there.

3 October, 2015 at 8:26 am

Terence TaoThe min-exponent was corrected to a max. Was there another error you were referring to?

[Edit: see it now, the second should be . -T.]8 December, 2015 at 9:25 am

Terence TaoNews flash: the Vinogradov main conjecture (estimate (20) in the blog post) has now been proven by Bourgain, Demeter, and Guth, using (a modification of) the decoupling restriction theorem of Bourgain and Demeter! http://arxiv.org/abs/1512.01565 Presumably this will impact the best known bounds on the zeta function and the Waring problem, though I don’t know the precise implications in this regard.

8 December, 2015 at 1:39 pm

AnonymousIs it possible that this has the potential to improve the current (Vinogradov-Korobov) exponent in the PNT error term ?

8 December, 2015 at 1:58 pm

Terence TaoMy initial understanding is that one may need to know how the implied constant in (20) depends on k and epsilon (and maybe also l) in order to get such an improvement. For instance to my knowledge the deep recent results of Wooley have not led to any improvement on Vinogradov-Korobov, in part due to the poor dependence of constants in this regard. But it is plausible that some breakthrough in this direction is now close at hand.

12 December, 2015 at 2:51 pm

Eytan PaldiIn the proof that theorem 11 implies theorem 2(ii), in the sixth line above (12), it is assumed (inside parentheses) that “ is large” but this assumption is not(!) part of theorem 2(ii) (which allows any natural ).

12 December, 2015 at 6:48 pm

Terence TaoOne can use the van der Corput estimates to assume that k is large (see the first paragraph of Section 2).

13 December, 2015 at 2:16 am

Eytan PaldiThanks! (In fact, it seems by first proving (6) for where is sufficiently large, that for the bounded case , (6) follows with some smaller – as the condition is implied by ).

13 December, 2015 at 6:46 pm

AnonymousIt seems that (defined below (12)) is (in general) a complex number, hence in all the inequalities (below (13)) involving powers of (which are also complex), should be replaced by .

(or alternatively, should be defined as the absolute value of the LHS of (10)).

[Corrected, thanks – T.]16 December, 2015 at 1:50 pm

AnonymousIt seems that the first and second lines in the RHS of the estimate above (14) are due to bounding with (due to Holder inequality) by the Riesz-Thorin interpolation bound between and (in terms of ).

(i.e. it seems that not only Holder inequality is used in this estimate.)

16 December, 2015 at 2:35 pm

AnonymousI made a mistake (there are no operator norms here), it seems that a generalization of Holder inequality (interpolation between norms) is used.

19 December, 2015 at 1:29 pm

AnonymousThe needed norm interpolation is Littlewood’s inequality (see e.g. in the interpolation section in the Wikipedia article on Holder’s inequality) applied to with and .

20 December, 2015 at 4:20 pm

AnonymousIn the third line above (18), should be .

[Corrected, thanks – T.]24 December, 2015 at 4:03 am

AnonymousIt seems that the coefficients in (16) are not used in the transition to (18). Hence (for simplicity) they can be eliminated before(16).

24 December, 2015 at 4:44 am

AnonymousIn fact, it is not clear if these coefficients can be eliminated before (16).

26 December, 2015 at 7:52 am

AnonymousIt is interesting to observe that the summations in (14)-(15) are not necessarily restricted to the box (because is supported in this box) – this fact is actually used in the application of C-S inequality to get (18).

27 December, 2015 at 2:58 am

AnonymousIn the proof of lemma 12, it seems clearer to remark that the inequality in the line above “and hence” is valid for the above chosen subset of indices that satisfy (9).

[Clarification added, thanks – T.]27 December, 2015 at 4:43 am

AnonymousIn the proof of lemma 12, it is assumed (in the last two lines) that for some large . But this assumption is not(!) present in the statement of lemma 12.

[This bound comes from the ambient hypothesis (12). -T.]28 March, 2017 at 5:25 pm

Claudeh5Hello professor Tao,

In “Vinogradov estimate” the first equation (which is on 3 lines) has a small error in the line 3: a square is lost in the integral : |f”(t)/|f'(t)|^2

[Corrected, thanks – T.]8 July, 2018 at 4:09 am

AnonymousDear Prof. Tao,

I’m confused with the last two sentences of the proof of Prop. 6. The size may be $O(AN/T)$? And the bound in Prop. 6 seems too strong comparing with Cor. 8.13 in Iwaniec–Kowalski’s book. Thanks.

1 June, 2019 at 1:11 pm

Sungjin KimDear Professor Tao,

In the Exercise 5 (iii), shouldn’t there be on the right side, not ?

[Corrected, thanks – T.]4 September, 2019 at 1:47 pm

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