Quick description
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 be a prime, and write for the Legendre symbol: thus if is a quadratic residue modulo and is otherwise. The Pólya-Vinogradov theorem states that if then
What this means is that, provided , about 50 percent of the numbers less than are quadratic residues modulo .
There is a particular value of for which the result is very easy: when , one is just counting the number of quadratic residues and nonresidues modulo (excluding zero), and it is well-known that there are . Thus
The only obstacle lying between this easy result and the Pólya-Vinogradov theorem is a cutoff hiding in the sum . To see this more clearly let us write the quantity we are to bound as
The notation means a function which equals on the set and is zero elsewhere. Now we apply the trick under discussion: we claim that the cutoff may be expanded as a Fourier series
where . To see this one uses (discrete) Fourier analysis on . The Fourier coefficients
of are just geometric series, the common ratio of series one must sum for being . An easy exercise shows that , where is the size of when reduced to lie between and . The claim now follows from the Fourier inversion formula.
Applying this decomposition of and the triangle inequality, our task now follows if we can establish that
for all . We have now completed the sum to the whole range , though we have paid the price of introducing the exponential .
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 . 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 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.
Example 3
Suppose one has an integral operator
which one knows to be bounded from to for some . Then, given any bump function , the smoothly truncated bump function
is also bounded from to (and in fact the operator norm of is bounded by that of , times a constant depending only on ). To see this, observe that the support of can be placed in a box for some sufficiently large . Performing a Fourier series expansion, one can write
for some rapidly decreasing Fourier coefficients . Interchanging sums and integrals (neglecting for now the issue of how to justify this; in practice, one can create an epsilon of room and regularize and as needed), we obtain
Applying the triangle inequality, we conclude that
Thus if is bounded from to with operator norm , one has
and the claim then follows from the rapid decrease of the coefficients .
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.
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