Selberg sieve variational problem

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 {\cal 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_{{\cal R}_k} F(t_1,\ldots,t_k)^2\ dt_1 \ldots dt_k }[/math]

and

[math]\displaystyle{ \displaystyle J_k^{(m)}(F) := \int_{{\cal 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 $t_2,\ldots,t_k$ 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 $s = t_2+\ldots+t_k$, 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.

World records

[math]\displaystyle{ k }[/math]	Lower bound	Upper bound
4	1.845	1.848
5	2.001162	2.0011797
59	3.898	4.148