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Revision of Getting rid of nasty cutoffs from Fri, 01/05/2009 - 23:56

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When faced with an unpleasant looking cutoff condition in a quantity one wishes to estimate, try expanding that condition using Fourier analysis.

Prerequisites

Basic harmonic analysis

Example 1

The classic example of this technique is the Pólya-Vinogradov theorem. Let p be a prime, and write (x | p) for the Legendre symbol: thus (x | p) = 1 if x is a quadratic residue modulo p and is -1 otherwise. The Pólya-Vinogradov theorem states that if m < p then

 \sum_{x = 1}^m (x | p) = O(\sqrt{p}\log p).

What this means is that, provided m > C\sqrt{p}\log p, about 50 percent of the numbers less than m are quadratic residues modulo p.

There is a particular value of m for which the result is very easy: when m = p-1, one is just counting the number of quadratic residues and nonresidues modulo p (excluding zero), and it is well-known that there are (p-1)/2. Thus

 \sum_{x = 1}^{p-1} (x |p) = 0.

The only obstacle lying between this easy result and the Pólya-Vinogradov theorem is a cutoff hiding in the sum \sum_{x=1}^m. To see this more clearly let us write the quantity we are to bound as

 \sum_{x = 1}^{p-1} 1_{\{1,\dots,m\}}(x)(x | p) .

The notation 1_A(x) means a function which equals 1 on the set A and is zero elsewhere. Now we apply the trick under discussion: we claim that the cutoff f = 1_{\{1,\dots,m\}} may be expanded as a Fourier series

 f(x) = \sum_{r = 0}^{p-1} c_{r,m} e^{2\pi i xr/p}

where \sum_r |c_{r,m}| = O(\log p). To see this one uses (discrete) Fourier analysis on \Z/p\Z. The Fourier coefficients

 \hat{f}(r) = \frac{1}{p}\sum_{x = 0}^{p-1} f(x) e^{-2\pi i xr/p}

of f(x) are just geometric series, the common ratio of series one must sum for \hat{f}(r) being e^{2\pi i r/p}. An easy exercise shows that |\hat{f}(r)| \leq C/|r|, where |r| is the size of r when reduced to lie between -p/2 and p/2. The claim now follows from the Fourier inversion formula.

Applying this decomposition of 1_{\{1,\dots,m\}}(x) and the triangle inequality, our task now follows if we can establish that

 \sum_{x = 1}^{p-1} e^{2\pi i xr/p} (x | p) = O(\sqrt{p})

for all r = 0,\dots,p-1. We have now completed the sum to the whole range x = 1,\dots,p-1, though we have paid the price of introducing the exponential e^{2\pi i xr/p}.

It turns out that this is not a huge price to pay: these new sums are Gauss sums and it is possible to show that they are O(\sqrt{p}). We leave the details as an exercise (the solution to which may be found in any number of places).

Example 2

This trick is ubiquitous in analytic number theory. It is very important, for example, in the treatment of sums \sum_{p \leq N} f(p) using the so-called method of bilinear forms; see B.~Green's article Three topics in additive prime number theory, Chapter 2, for more information.

General discussion

Sometimes this principle is described as all cutoffs are the same, or as completing exponential sums. It can lead to extra logarithmic factors as in the Pólya-Vinogradov inequality mentioned above; in this example reducing the size of these factors is a major unsolved problem. Often it is very helpful to smooth the cutoffs before decomposing them into Fourier modes.

See also "When controlling an oscillatory integral bump functions and bounded phase corrections are not very important".