After that whirlwind tour through Khintchine’s inequality, let’s take a look at the less well-known Rudin’s inequality. This says (at least in one form) that, given a finite abelian group and a function if the support of the Fourier transform of is entirely contained inside a dissociated set (we’ll come to what that means later) then for any we have

Oh ho ho, this looks suspiciously similar to our form of Khintchine’s inequality from the last post:

What are the differences? Well, in Rudin’s inequality we deal with functions with Fourier support contained inside a dissociated set, and we’re taking norms over the group . In Khintchine’s inequality, however, the norms are now being taken over the probability space, and our functions take in something from our probability space and spit out a distribution of signs.

Using the Fourier inversion formula, we can write out Rudin’s inequality in a way which highlights the similarity even more. Let denote the support of the Fourier transform of . The left hand side of Rudin’s inequality becomes

while the left hand side of Khintchine’s inequality is, if we denote our coefficients as instead, indexed over (all we’re doing here is changing our notation),

The difference is now staring us in the face. All we’ve done in moving from Khintchine’s inequality to Rudin’s inequality is swapped our random variable to a function .

In other words, the characters from a dissociated set behave like independent random variables taking values in . We will now try to make this heuristic precise, and deduce Rudin’s inequality as a corollary of Khintchine’s inequality. The following follows the proof of Rudin’s inequality given in [Gr1].

We first randomise the sum by introducing a random assignment of signs to get , where are independent random variables taking the values . Khintchine’s inequality then gives, for any fixed ,

Taking the expectation over all we get (taking norms over now)

By the pigeonhole principle there is some such that

The left hand side is nearly what we want; if we could replace the by we would have Rudin’s inequality. To do this, I’ll borrow a trick I found in [Gr1], but first I should say what dissociated means: the set is dissociated if and only if there are no non-trivial equations of the form . This is useful because it allows us to write

for *any *distribution of signs , as can be checked by expanding out the product and convolution. Furthermore, by the same argument we check that the product on the right has norm equal to 1. It follows that we can bound for any and we have proven Rudin’s inequality. Notice that here was where we really needed , to pass from a randomised version of to itself.

The ‘randomisation’ technique we used here was to use probabilistic arguments and the pigeonhole principle to show that the desired result was true for some random twist of the original function , and then to argue that the original function is suitably controlled by *every *random twist of . This latter condition is where we needed to invoke dissociativity.

To recap: Rudin’s inequality can be proved using randomisation plus Khintchine’s inequality, and dissociativity is invoked the control the randomisation part. We will soon see how an analogous sort of argument in physical space gives the powerful new method of Croot and Sisask.