Difference between revisions of "Selberg sieve variational problem"

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| [http://terrytao.wordpress.com/2013/11/22/polymath8b-ii-optimising-the-variational-problem-and-the-sieve/#comment-255681 3.95608]
 
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All upper bounds come from (1).
 
All upper bounds come from (1).

Revision as of 12:32, 9 December 2013

Let [math]M_k[/math] be the quantity

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

where [math]F[/math] ranges over square-integrable functions on the simplex

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

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

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

and

[math]\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]DHL[k,m+1][/math] holds whenever [math]EH[\theta][/math] holds and [math]M_k \gt \frac{2m}{\theta}[/math]. Thus for instance, [math]M_k \gt 2[/math] implies [math]DHL[k,2][/math] on the Elliott-Halberstam conjecture, and [math]M_k\gt4[/math] implies [math]DHL[k,2][/math] unconditionally.

Upper bounds

We have the upper bound

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

that is proven as follows.

The key estimate is

[math] \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]t_2,\ldots,t_k[/math] to conclude that

[math]\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]\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 \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]s = t_2+\ldots+t_k[/math], the left-hand side evaluates to

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

as required.

Lower bounds

...

World records

[math]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).