Fourier reduction: Difference between revisions
New page: Let f be an arbitrary function from <math>{\Bbb Z}</math> to {-1,+1} of discrepancy at most C. Let N be a moderately large integer, let <math>p_1,\ldots,p_d</math> be the primes in [N], a... |
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and thus | and thus | ||
:<math>\frac{1}{N} \sum_{n=1}^N | :<math>\frac{1}{N} |\sum_{n=1}^N g(j)|^2 \ll C</math>. | ||
So, if one can show a uniform bound | So, if one can show a uniform bound | ||
:<math>\frac{1}{N} \sum_{n=1}^N | :<math>\frac{1}{N} |\sum_{n=1}^N g(j)|^2 \geq \omega(N)</math> | ||
where <math>\omega(N)</math> goes to infinity as <math>N \to \infty</math>, for arbitrary <math>S^1</math>-valued multiplicative functions, one is done! | where <math>\omega(N)</math> goes to infinity as <math>N \to \infty</math>, for arbitrary <math>S^1</math>-valued multiplicative functions, one is done! | ||
Revision as of 23:15, 26 January 2010
Let f be an arbitrary function from [math]\displaystyle{ {\Bbb Z} }[/math] to {-1,+1} of discrepancy at most C. Let N be a moderately large integer, let [math]\displaystyle{ p_1,\ldots,p_d }[/math] be the primes in [N], and let M be a huge integer (much larger than N). Then we can define a function [math]\displaystyle{ F: ({\Bbb Z}/M{\Bbb Z})^d \to \{-1,+1\} }[/math] by the formula
- [math]\displaystyle{ F(a_1,\ldots,a_d) := f( p_1^{a_1} \ldots p_d^{a_d} ). }[/math]
whenever [math]\displaystyle{ a_1,\ldots,a_d \in [M] }[/math]. Note that F has a normalised L^2 norm of 1, so by the Plancherel identity
- [math]\displaystyle{ \sum_{\xi \in ({\Bbb Z}/M{\Bbb Z})^d} |\hat F(\xi)|^2 = 1. }[/math] (1)
Let [math]\displaystyle{ \pi: [N] \to {\Bbb Z}^d }[/math] be the map
- [math]\displaystyle{ \pi(p_1^{a_1} \ldots p_d^{a_d}) := (a_1,\ldots,a_d) }[/math]
then by hypothesis one has
- [math]\displaystyle{ |F(x+\pi(1)) + \ldots + F(x+\pi(n))| \leq C }[/math]
for all (1-O_N(1/M)) of the x in [math]\displaystyle{ ({\Bbb Z}/M{\Bbb Z})^d }[/math], and all [math]\displaystyle{ 1 \leq n \leq N }[/math]. Applying Plancherel to this, we obtain
- [math]\displaystyle{ \sum_{\xi \in ({\Bbb Z}/M{\Bbb Z})^d} |\hat F(\xi)|^2 |\sum_{j=1}^n e( \xi \cdot \pi(j) / M ) |^2 \ll C }[/math]
for each such n, and so on averaging in n we have
- [math]\displaystyle{ \sum_{\xi \in ({\Bbb Z}/M{\Bbb Z})^d} |\hat F(\xi)|^2 \frac{1}{N} \sum_{n=1}^N |\sum_{j=1}^n e( \xi \cdot \pi(j) / M )|^2 \ll C. }[/math]
Comparing this with (1) and using the pigeonhole principle, we conclude that there exists [math]\displaystyle{ \xi }[/math] such that
- [math]\displaystyle{ \frac{1}{N} \sum_{n=1}^N |\sum_{j=1}^n e( \xi \cdot \pi(j) / M )|^2 \ll C }[/math].
If we let [math]\displaystyle{ g: {\Bbb N} \to S^1 }[/math] be a completely multiplicative function such that [math]\displaystyle{ g(p_j) = e(\xi_j/M) }[/math] for all j=1,...,d, we have
- [math]\displaystyle{ e( \xi \cdot \pi(j) / M ) = g(j) }[/math]
and thus
- [math]\displaystyle{ \frac{1}{N} |\sum_{n=1}^N g(j)|^2 \ll C }[/math].
So, if one can show a uniform bound
- [math]\displaystyle{ \frac{1}{N} |\sum_{n=1}^N g(j)|^2 \geq \omega(N) }[/math]
where [math]\displaystyle{ \omega(N) }[/math] goes to infinity as [math]\displaystyle{ N \to \infty }[/math], for arbitrary [math]\displaystyle{ S^1 }[/math]-valued multiplicative functions, one is done!