Fourier reduction: Difference between revisions
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:<math>|F(x+\pi(1)) + \ldots + F(x+\pi(n))| \leq C</math> | :<math>|F(x+\pi(1)) + \ldots + F(x+\pi(n))| \leq C</math> | ||
for all (1-O_N(1/M)) of the x in <math>({\Bbb Z}/M{\Bbb Z})^d</math>, and all <math>1 \leq n \leq N</math>. Applying Plancherel to this, we obtain | for all <math>(1-O_N(1/M))</math> of the x in <math>({\Bbb Z}/M{\Bbb Z})^d</math>, and all <math>1 \leq n \leq N</math>. Applying Plancherel to this, we obtain | ||
:<math> \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> | :<math> \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> | ||
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for all n up to N. Taking a weak limit of these probability measures (using Prokhorov's theorem) we can in fact get this for all n. So to solve EDP, it in fact suffices to show that | for all n up to N. Taking a weak limit of these probability measures (using Prokhorov's theorem) we can in fact get this for all n. So to solve EDP, it in fact suffices to show that | ||
:<math> \sup_n {\Bbb E} |g(1)+\ldots+g(n)|^2 = \infty</math> | :<math> \sup_n {\Bbb E} |g(1)+\ldots+g(n)|^2 = \infty</math> (*) | ||
for all probabilistic completely multiplicative functions taking values in S^1. This should be compared to the completely multiplicative special case of EDP, in which g takes values in {-1,+1} and is deterministic. | for all probabilistic completely multiplicative functions taking values in <math>S^1</math>. This should be compared to the completely multiplicative special case of EDP, in which g takes values in {-1,+1} and is deterministic. | ||
If one is interested in square-invariant functions only (so <math>f(q^2 x) = f(x)</math> for all rational q) then we can restrict g to be {-1,+1} valued (basically because <math>({\Bbb Z}/M{\Bbb Z})^d</math> can now be replaced with <math>({\Bbb Z}/2{\Bbb Z})^d</math> in the above analysis. | If one is interested in square-invariant functions only (so <math>f(q^2 x) = f(x)</math> for all rational q) then we can restrict g to be {-1,+1} valued (basically because <math>({\Bbb Z}/M{\Bbb Z})^d</math> can now be replaced with <math>({\Bbb Z}/2{\Bbb Z})^d</math> in the above analysis. | ||
Actually, (*) is equivalent to the following Hilbert-space version of EDP: if <math>f: {\Bbb N} \to S</math> takes values in the unit sphere of a (real or complex) Hilbert space, then the discrepancy of f is unbounded (note that discrepancy can be defined in arbitrary normed vector spaces). The derivation of Hilbert-space EDP from (*) follows the argument above (f is now vector-valued instead of scalar, but Plancherel's theorem and all the other tools used above go through without difficulty). | |||
Conversely, if (*) failed, then we have a probability distribution <math>\mu</math> on the space of completely multiplicative functions for which the left-hand side of (*) is bounded. If we then set H to be the complex Hilbert space <math>L^2(d\mu)</math> and let <math>f(n) \in L^2(d\mu)</math> be the evaluation map <math>g \mapsto g(n)</math> for each n, we see that f has bounded discrepancy. |
Latest revision as of 16:46, 12 May 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 [math]\displaystyle{ (1-O_N(1/M)) }[/math] 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_i) = e(\xi_i/M) }[/math] for all i=1,...,d, we have
- [math]\displaystyle{ e( \xi \cdot \pi(j) / M ) = g(j) }[/math]
for all j=1,...,N, and thus
- [math]\displaystyle{ \frac{1}{N} \sum_{n=1}^N |\sum_{j=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 |\sum_{j=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!
Actually, one can do a little better than this. From (1) we see that [math]\displaystyle{ |\hat F(\xi)|^2 }[/math] induces a probability measure (depending on N,M) on completely multiplicative functions [math]\displaystyle{ g: {\Bbb Q} \to S^1 }[/math] (strictly speaking, this function is only defined on rationals that involve the primes [math]\displaystyle{ p_1,\ldots,p_d }[/math], but one can extend to the rationals by setting g to equal 1 on all other primes), and for g drawn from this probability measure the above arguments in fact show that
- [math]\displaystyle{ {\Bbb E} |g(1)+\ldots+g(n)|^2 \ll C }[/math]
for all n up to N. Taking a weak limit of these probability measures (using Prokhorov's theorem) we can in fact get this for all n. So to solve EDP, it in fact suffices to show that
- [math]\displaystyle{ \sup_n {\Bbb E} |g(1)+\ldots+g(n)|^2 = \infty }[/math] (*)
for all probabilistic completely multiplicative functions taking values in [math]\displaystyle{ S^1 }[/math]. This should be compared to the completely multiplicative special case of EDP, in which g takes values in {-1,+1} and is deterministic.
If one is interested in square-invariant functions only (so [math]\displaystyle{ f(q^2 x) = f(x) }[/math] for all rational q) then we can restrict g to be {-1,+1} valued (basically because [math]\displaystyle{ ({\Bbb Z}/M{\Bbb Z})^d }[/math] can now be replaced with [math]\displaystyle{ ({\Bbb Z}/2{\Bbb Z})^d }[/math] in the above analysis.
Actually, (*) is equivalent to the following Hilbert-space version of EDP: if [math]\displaystyle{ f: {\Bbb N} \to S }[/math] takes values in the unit sphere of a (real or complex) Hilbert space, then the discrepancy of f is unbounded (note that discrepancy can be defined in arbitrary normed vector spaces). The derivation of Hilbert-space EDP from (*) follows the argument above (f is now vector-valued instead of scalar, but Plancherel's theorem and all the other tools used above go through without difficulty).
Conversely, if (*) failed, then we have a probability distribution [math]\displaystyle{ \mu }[/math] on the space of completely multiplicative functions for which the left-hand side of (*) is bounded. If we then set H to be the complex Hilbert space [math]\displaystyle{ L^2(d\mu) }[/math] and let [math]\displaystyle{ f(n) \in L^2(d\mu) }[/math] be the evaluation map [math]\displaystyle{ g \mapsto g(n) }[/math] for each n, we see that f has bounded discrepancy.