Selberg sieve variational problem: Difference between revisions

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:<math>{\mathcal R}'_k = \{ (t_1,\ldots,t_k) \in [0,1]^k: t_1+\ldots+t_k \leq 1 + \min(t_1,\ldots,t_k) \}</math>
:<math>{\mathcal R}'_k = \{ (t_1,\ldots,t_k) \in [0,1]^k: t_1+\ldots+t_k \leq 1 + \min(t_1,\ldots,t_k) \}</math>


provided that one works with a generalisation of <math>EH[\theta]</math> which controls more general Dirichlet convolutions than the von Mangoldt function (a precise assertion in this regard may be found in BFI).  In fact one should be able to work in the even larger region
provided that one works with a generalisation of <math>EH[\theta]</math> which controls more general Dirichlet convolutions than the von Mangoldt function (a precise assertion in this regard may be found in BFI).  In fact one should be able to work in any larger region <math>R</math> for which


:<math>{\mathcal R}''_{k,\theta} = \{ (t_1,\ldots,t_k) \in [0,1/\theta]^k: t_1+\ldots+t_k \leq 1 + \max(t_1,\ldots,t_k) \} \cup \frac{1}{\theta} \cdot {\mathcal R}_k</math>
:<math>R + R \subset \{ (t_1,\ldots,t_k) \in [0,2/\theta]^k: t_1+\ldots+t_k \leq 2 + \max(t_1,\ldots,t_k) \} \cup \frac{2}{\theta} \cdot {\mathcal R}_k</math>


provided that all the marginal distributions of F are supported on <math>{\mathcal R}_{k-1}</math>, thus (assuming F is symmetric)
provided that all the marginal distributions of F are supported on <math>{\mathcal R}_{k-1}</math>, thus (assuming F is symmetric)


:<math>\int_0^\infty F(t_1,\ldots,t_{k-1},t_k)\ dt_k = 0 </math> when <math>t_1+\ldots+t_{k-1} > 1.</math>
:<math>\int_0^\infty F(t_1,\ldots,t_{k-1},t_k)\ dt_k = 0 </math> when <math>t_1+\ldots+t_{k-1} > 1.</math>
For instance, one can take <math>R = \frac{1}{\theta} \cdot {\mathcal R}_k</math>, or one can take <math>R = \{ (t_1,\ldots,t_k) \in [0,1/\theta]^k: t_1 +\ldots +t_{k-1} \leq 1 </math> (although the latter option breaks the symmetry for F).  Perhaps other choices are also possible.

Revision as of 22:07, 12 December 2013

Let [math]\displaystyle{ M_k }[/math] be the quantity

[math]\displaystyle{ \displaystyle M_k := \sup_F \frac{\sum_{m=1}^k J_k^{(m)}(F)}{I_k(F)} }[/math]

where [math]\displaystyle{ F }[/math] ranges over square-integrable functions on the simplex

[math]\displaystyle{ \displaystyle {\mathcal R}_k := \{ (t_1,\ldots,t_k) \in [0,+\infty)^k: t_1+\ldots+t_k \leq 1 \} }[/math]

with [math]\displaystyle{ I_k, J_k^{(m)} }[/math] being the quadratic forms

[math]\displaystyle{ \displaystyle I_k(F) := \int_{{\mathcal R}_k} F(t_1,\ldots,t_k)^2\ dt_1 \ldots dt_k }[/math]

and

[math]\displaystyle{ \displaystyle J_k^{(m)}(F) := \int_{{\mathcal R}_{k-1}} (\int_0^{1-\sum_{i \neq m} t_i} F(t_1,\ldots,t_k)\ dt_m)^2 dt_1 \ldots dt_{m-1} dt_{m+1} \ldots dt_k. }[/math]

It is known that [math]\displaystyle{ DHL[k,m+1] }[/math] holds whenever [math]\displaystyle{ EH[\theta] }[/math] holds and [math]\displaystyle{ M_k \gt \frac{2m}{\theta} }[/math]. Thus for instance, [math]\displaystyle{ M_k \gt 2 }[/math] implies [math]\displaystyle{ DHL[k,2] }[/math] on the Elliott-Halberstam conjecture, and [math]\displaystyle{ M_k\gt 4 }[/math] implies [math]\displaystyle{ DHL[k,2] }[/math] unconditionally.

Upper bounds

We have the upper bound

[math]\displaystyle{ \displaystyle M_k \leq \frac{k}{k-1} \log k }[/math] (1)

that is proven as follows.

The key estimate is

[math]\displaystyle{ \displaystyle \int_0^{1-t_2-\ldots-t_k} F(t_1,\ldots,t_k)\ dt_1)^2 \leq \frac{\log k}{k-1} \int_0^{1-t_2-\ldots-t_k} F(t_1,\ldots,t_k)^2 (1 - t_1-\ldots-t_k+ kt_1)\ dt_1. }[/math]. (2)

Assuming this estimate, we may integrate in [math]\displaystyle{ t_2,\ldots,t_k }[/math] to conclude that

[math]\displaystyle{ \displaystyle J_k^{(1)}(F) \leq \frac{\log k}{k-1} \int F^2 (1-t_1-\ldots-t_k+kt_1)\ dt_1 \ldots dt_k }[/math]

which symmetrises to

[math]\displaystyle{ \sum_{m=1}^k J_k^{(m)}(F) \leq k \frac{\log k}{k-1} \int F^2\ dt_1 \ldots dt_k }[/math]

giving the desired upper bound (1).

It remains to prove (2). By Cauchy-Schwarz, it suffices to show that

[math]\displaystyle{ \displaystyle \int_0^{1-t_2-\ldots-t_k} \frac{dt_1}{1 - t_1-\ldots-t_k+ kt_1} \leq \frac{\log k}{k-1}. }[/math]

But writing [math]\displaystyle{ s = t_2+\ldots+t_k }[/math], the left-hand side evaluates to

[math]\displaystyle{ \frac{1}{k-1} (\log k(1-s) - \log (1-s) ) = \frac{\log k}{k-1} }[/math]

as required.

Lower bounds

...

World records

[math]\displaystyle{ k }[/math] Lower bound Upper bound
4 1.845 1.848
5 2.001162 2.011797
10 2.53 2.55842
20 3.05 3.1534
30 3.34 3.51848
40 3.52 3.793466
50 3.66 3.99186
59 3.95608 4.1479398

All upper bounds come from (1).

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 [math]\displaystyle{ J_k^{(m)}(F) }[/math] accordingly). Firstly, one can enlarge the simplex [math]\displaystyle{ {\mathcal R}_k }[/math] to the larger region

[math]\displaystyle{ {\mathcal R}'_k = \{ (t_1,\ldots,t_k) \in [0,1]^k: t_1+\ldots+t_k \leq 1 + \min(t_1,\ldots,t_k) \} }[/math]

provided that one works with a generalisation of [math]\displaystyle{ EH[\theta] }[/math] which controls more general Dirichlet convolutions than the von Mangoldt function (a precise assertion in this regard may be found in BFI). In fact one should be able to work in any larger region [math]\displaystyle{ R }[/math] for which

[math]\displaystyle{ R + R \subset \{ (t_1,\ldots,t_k) \in [0,2/\theta]^k: t_1+\ldots+t_k \leq 2 + \max(t_1,\ldots,t_k) \} \cup \frac{2}{\theta} \cdot {\mathcal R}_k }[/math]

provided that all the marginal distributions of F are supported on [math]\displaystyle{ {\mathcal R}_{k-1} }[/math], thus (assuming F is symmetric)

[math]\displaystyle{ \int_0^\infty F(t_1,\ldots,t_{k-1},t_k)\ dt_k = 0 }[/math] when [math]\displaystyle{ t_1+\ldots+t_{k-1} \gt 1. }[/math]

For instance, one can take [math]\displaystyle{ R = \frac{1}{\theta} \cdot {\mathcal R}_k }[/math], or one can take [math]\displaystyle{ R = \{ (t_1,\ldots,t_k) \in [0,1/\theta]^k: t_1 +\ldots +t_{k-1} \leq 1 }[/math] (although the latter option breaks the symmetry for F). Perhaps other choices are also possible.