# Selberg sieve variational problem

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 Revision as of 15:31, 9 December 2013 (view source) (→World records)← Older edit Revision as of 20:32, 9 December 2013 (view source) (→World records)Newer edit → Line 73: Line 73: |- |- | 59 | 59 - | [http://terrytao.wordpress.com/2013/11/22/polymath8b-ii-optimising-the-variational-problem-and-the-sieve/#comment-255681 3.96508] + | [http://terrytao.wordpress.com/2013/11/22/polymath8b-ii-optimising-the-variational-problem-and-the-sieve/#comment-255681 3.95608] | 4.1479398 | 4.1479398 |} |} All upper bounds come from (1). All upper bounds come from (1).

## Revision as of 20:32, 9 December 2013

Let Mk be the quantity

$\displaystyle M_k := \sup_F \frac{\sum_{m=1}^k J_k^{(m)}(F)}{I_k(F)}$

where F ranges over square-integrable functions on the simplex

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

with $I_k, J_k^{(m)}$ being the quadratic forms

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

and

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

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

## Upper bounds

We have the upper bound

$\displaystyle M_k \leq \frac{k}{k-1} \log k$ (1)

that is proven as follows.

The key estimate is

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

Assuming this estimate, we may integrate in $t_2,\ldots,t_k$ to conclude that

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

which symmetrises to

$\sum_{m=1}^k J_k^{(m)}(F) \leq k \frac{\log k}{k-1} \int F^2\ dt_1 \ldots dt_k$

giving the desired upper bound (1).

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

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

But writing $s = t_2+\ldots+t_k$, the left-hand side evaluates to

$\frac{1}{k-1} (\log k(1-s) - \log (1-s) ) = \frac{\log k}{k-1}$

as required.

...

## World records

k 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).