# The Piatetski-Shapiro theorem

I have just learned that Ilya Piatetski-Shapiro died on February 21, 2009, just a month short of his 80th birthday. Most of his research has been in algebraic number theory and representation theory. I’m not a number theorist, and I know even less about representation theory, so I can’t tell you much about his work in those areas. However, I would like to tell you about an early result of his on the summability of Fourier series, known as the Piatetski-Shapiro theorem in harmonic analysis.

Suppose that $c_k,\ k\in{\mathbb Z}$, is a sequence with the property that $\sum_{k=-\infty}^\infty c_ke^{2\pi i kx}=0$ almost everywhere on $[0,1]$. Does it follow that $c_k=0$ for all $k$? It turns out (due to Menshov) that the answer is negative. Hence the following definition.

A set $E\subset [0,1]$ is called a set of uniqueness if the only sequence $c_k$ such that $\sum_{k=-\infty}^\infty c_ke^{2\pi i kx}=0$ for all $x\in [0,1]\setminus E$ is $c_k=0$ for all $k$. Otherwise, $E$ is called a set of multiplicity.

If $E$ is closed, it is known that $E$ is a set of multiplicity if and only if it supports a distribution whose Fourier coefficients tend to 0 at infinity.

It was thought for a while that the word “distribution” in the last sentence can be replaced by “measure”. This is what Piatetski-Shapiro disproved.

Theorem (Piatetski-Shapiro). There is a closed set $E\subset[0,1]$ such that $E$ is a set of multiplicity, but does not support any measure $\mu$ with $\widehat{\mu}(k)\to 0$ as $|k|\to\infty$.

In other words, $E$ supports a distribution whose Fourier coefficients vanish at infinity, but does not support a measure with the same property!

Piatetski-Shapiro proved that one can take $E$ to be the set of all numbers in $[0,1]$ whose dyadic expansion $\sum_{j=1}^\infty r_j2^{-j}$ obeys $n^{-1}\sum_{j=1}^n r_j\leq r$, where $r$ is a fixed number in $(0,1/2)$.

Alternative proofs of the Piatetski-Shapiro theorem were given by Kaufman, Körner and others. The following brief sketch of the Kaufman-Körner argument is based on an exposition by Nir Lev. See the introduction to his thesis for the full length version.

Let $\nu$ be an integer greater than 2. Consider the Riesz product

$\lambda_s(x)=\prod_{j=1}^N (1+2s\cos (2\pi\nu^jx)),$

where $N$ is a large integer and $s\in (1/2-\epsilon_0,1/2)$ for some small $\epsilon_0$. Note that $\lambda_s$ is non-negative and $\int_0^1\lambda_s=1$. Let also $X(x)= \frac{2}{N} \sum_{j=1}^N \cos (2\pi i\nu^j x)$. A key observation is that $\lambda_s$ concentrates on the set where $X$ is close to 1. Specifically, if

$K=\{x\in[0,1]: \ |X(x)-1|\leq\delta\}$,

then $\int_K \lambda_s(x)dx$ is close to 1. This can be proved by probabilistic arguments: consider $\lambda_s$ as a probability measure and $X(x)$ as the average of the uncorrelated random variables $2\cos(2\pi i\nu^jx)$, each with expectation $2s$ which, we recall, is close to 1.

Ideally, we would like $\lambda_s$ to be “random” in the sense that all its Fourier coefficients with $k\neq 0$ are small. That is generally not true; however, we can find a linear combination $\lambda$ of finitely many $\lambda_s$ with $s$ close to 1/2 which has this property.

Note that $\lambda$ is still (mostly) supported on $K$. This allows us to approximate $\lambda$ by a smooth function $f(x)$ with the following properties:

• $\widehat{f}(0)=1,\ |\widehat{f}(k)|<\epsilon$ for all $k\neq 0$,
• $f$ is supported on $K$ (adjust the $\delta$ slightly if necessary).

Now take a sequence $\nu_j\to\infty$, with the corresponding $X_j, K_j,\ f_j$. It can be proved that, for an appropriate choice of various parameters, the product $f_1\dots f_n$ converges to a distribution $S$, supported on the set $E=\bigcap K_j$, whose Fourier coefficients vanish at infinity. On the other hand, we have $\sup_{x\in E}|X_j(x)-1|\to 0$ as $j\to\infty$, so that for any probability measure $\mu$ supported on $E$ we have $\int X_jd\mu\to 1$ as $j\to\infty$. Hence

$\sum_k \widehat{X_j}(k)\overline{\widehat{\mu}(k)}\to 1. \ (*)$

But $\widehat{X_j}(k)=0$ for all $|k|<\nu_j$, so that $(*)$ cannot hold if $\widehat{\mu}(k)\to 0$ at infinity. We’re done!