### 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.

26 weeks 5 daysago47 weeks 3 daysago2 years 18 weeksago2 years 42 weeksago3 years 7 weeksago