Selberg sieve variational problem
(→More general variational problems)
|Line 331:||Line 331:|
Revision as of 22:47, 9 January 2014
Let Mk be the quantity
where F ranges over square-integrable functions on the simplex
with being the quadratic forms
It is known that DHL[k,m + 1] holds whenever EH[θ] holds and . Thus for instance, Mk > 2 implies DHL[k,2] on the Elliott-Halberstam conjecture, and Mk > 4 implies DHL[k,2] unconditionally.
We have the upper bound
that is proven as follows.
The key estimate is
- . (2)
Assuming this estimate, we may integrate in to conclude that
which symmetrises to
giving the desired upper bound (1).
It remains to prove (2). By Cauchy-Schwarz, it suffices to show that
But writing , the left-hand side evaluates to
We will need some parameters c,T,τ > 0 and a > 1 to be chosen later (in practice we take c close to 1 / logk, T a small multiple of c, and τ a small multiple of c/k.
For any symmetric function F on the simplex , one has
and so by scaling, if F is a symmetric function on the dilated simplex , one has
after adjusting the definition of the functionals suitably for this rescaled simplex.
Now let us apply this inequality r in the interval [1,1 + τ] and to truncated tensor product functions
for some bounded measurable , not identically zero, with . We have the probabilistic interpretations
where , and are iid random variables in [0,T] with law , and we adopt the convention that vanishes when b < a. We thus have
for any r.
We now introduce the random function h = hr by
Observe that if Sk − 1 < r, then
and hence by the Legendre identity
We also note that (using the iid nature of the Xi to symmetrise)
Inserting these bounds into (*) and rearranging, we conclude that
where is the defect from the upper bound. Splitting the integrand into regions where s or t is larger than or less than T, we obtain
We now focus on Y1. It is only non-zero when . Bounding , we see that
where log + (x) is equal to logx when and zero otherwise. We can rewrite this as
We write and . Using the bound we have
and thus (bounding ).
Symmetrising, we conclude that
For Z2, which is a tiny term, we use the crude bound
For Z1, we use the bound
valid for any a > 1, which can be verified because the LHS is concave for , while the RHS is convex and is tangent to the LHS as x=a. We then have
A good choice for a = a[r] here is (assuming ), in which case the formula simplifies to
Thus far, our arguments have been valid for arbitrary functions g. We now specialise to functions of the form
Note the identity
on [0,min(r − Sk − 1,T)]. Thus
Bounding and using symmetry, we conclude
Since , we conclude that
where Z4 = Z4[r] is the quantity
Putting all this together, we have
At this point we encounter a technical problem that Z4 diverges logarithmically (up to a cap of logk) as Sk − 1 approaches r. To deal with this issue we average in r, and specifically over the interval [1,1 + τ]. One can calculate that
for all x if we have (because the two expressions touch at x = 1 + τ, with the RHS being convex with slope at least (1 + τ / 2) / τ there. and the LHS lying underneath this tangent line). Assuming this, we conclude that
provided that kμ < 1 − τ, and hence
Now we deal with the Z4 integral. We split into two contributions, depending on whether or not. If , then we may bound
when , so this portion of may be bounded by
and so this portion of is bounded by
As for the portion when 1 + uτ − Sk − 1 > 2c, we bound , and so this portion of may be bounded by
- = VU
We thus arrive at the final bound
provided that kμ < 1 − τ and the denominator is positive.
We work in the asymptotic regime . Setting
for some absolute constants and β,γ > 0, one calculates
and so the constraint 1 − kμ > τ becomes
With for , one has
and one also calculates
- Z2,Z3 = o(1)
assuming that α + logβ − 1 > γ and
and where .
In particular, by setting α,β,γ as absolute constants obeying these constraints, we have , and so
- Mk = logk + O(1).
If we set c = τ: = 1 / logk, with T a small multiple of c, then for a large absolute constant A, and σ2 is a small multiple of . This makes the denominator comparable to 1; one can check that all the terms in the numerator are O(1), finally giving the bound
- Δk = O(1)
and thus we have the lower bound
More general variational problems
It appears that for the purposes of establish DHL type theorems, one can increase the range of F in which one is taking suprema over (and extending the range of integration in the definition of accordingly). Firstly, one can enlarge the simplex to the larger region
provided that one works with a generalisation of EH[θ] which controls more general Dirichlet convolutions than the von Mangoldt function (a precise assertion in this regard may be found in BFI). In fact (as shown here) one can work in any larger region R for which
provided that all the marginal distributions of F are supported on , thus (assuming F is symmetric)
For instance, one can take , or one can take (although the latter option breaks the symmetry for F). See this blog post for more discussion.
If the marginal distributions of F are supported in instead of , one still has a usable lower bound in which is replaced by the slightly smaller quantity ; see this blog post for more discussion.
For k>2, all upper bounds on Mk come from (1). Upper bounds on M'k come from the inequality that follows from an averaging argument, and upper bounds on M''k (on EH, using the prism as the domain) come from the inequality by comparing M''k with a variational problem on the prism (details here). The 1D bound is the optimal value for Mk when the underlying function F is restricted to be of the "one-dimensional" form .
For k=2, M'2 = M''2 = 2 can be computed exactly by taking F to be the indicator function of the unit square (for the lower bound), and by using Cauchy-Schwarz (for the upper bound). can be computed exactly as the solution to the equation .
The quantity is defined as the supremum of where F is now supported on , and is defined as but now restricted to , and is a parameter to be optimised in. The crude upper bound of k for any of the Mk type quantities comes from the parity problem obstruction that each separate event "n + hi prime" can occur with probability at most 1/2.