Difference between revisions of "Selberg sieve variational problem"

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(New page: 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...)
 
Line 3: Line 3:
 
:<math>\displaystyle M_k := \sup_F \frac{\sum_{m=1}^k J_k^{(m)}(F)}{I_k(F)}</math>
 
:<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
 
where <math>F</math> ranges over square-integrable functions on the simplex
:<math>\displaystyle {\cal R}_k := \{ (t_1,\ldots,t_k) \in [0,+\infty)^k: t_1+\ldots+t_k \leq 1 \}</math>
+
:<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
 
with <math>I_k, J_k^{(m)}</math> being the quadratic forms
:<math>\displaystyle I_k(F) := \int_{{\cal R}_k} F(t_1,\ldots,t_k)^2\ dt_1 \ldots dt_k</math>
+
:<math>\displaystyle I_k(F) := \int_{{\mathcal R}_k} F(t_1,\ldots,t_k)^2\ dt_1 \ldots dt_k</math>
 
and
 
and
:<math>\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>
+
:<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 > \frac{2m}{\theta}</math>.  Thus for instance, <math>M_k > 2</math> implies <math>DHL[k,2]</math> on the Elliott-Halberstam conjecture, and <math>M_k>4</math> implies <math>DHL[k,2]</math> unconditionally.
 
It is known that <math>DHL[k,m+1]</math> holds whenever <math>EH[\theta]</math> holds and <math>M_k > \frac{2m}{\theta}</math>.  Thus for instance, <math>M_k > 2</math> implies <math>DHL[k,2]</math> on the Elliott-Halberstam conjecture, and <math>M_k>4</math> implies <math>DHL[k,2]</math> unconditionally.
Line 20: Line 20:
 
:<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)
 
:<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 $t_2,\ldots,t_k$ to conclude that
+
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>
 
:<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
 
which symmetrises to
Line 28: Line 28:
 
It remains to prove (2).  By Cauchy-Schwarz, it suffices to show that
 
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>
 
:<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 $s = t_2+\ldots+t_k$, the left-hand side evaluates to
+
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>
 
:<math>\frac{1}{k-1} (\log k(1-s) - \log (1-s) ) = \frac{\log k}{k-1}</math>
 
as required.
 
as required.
Line 34: Line 34:
 
== Lower bounds ==
 
== Lower bounds ==
  
 
+
...
  
 
== World records ==
 
== World records ==
Line 51: Line 51:
 
| 2.001162  
 
| 2.001162  
 
| 2.0011797
 
| 2.0011797
 +
|-
 +
| 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
 
| 59
| 3.898
+
| 3.8984021
| 4.148
+
| 4.1479398
 
|}
 
|}

Revision as of 17:04, 8 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 2.0011797
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 3.8984021 4.1479398