A function with support of finite measure and "small" spectrum

TL;DR

This paper constructs a real-valued function supported on a finite measure set with a spectrum that has zero density, advancing understanding of spectral properties of functions.

Contribution

It introduces a novel example of a function with finite measure support and a spectrum of zero density, which was previously unknown.

Findings

Function supported on finite measure set

Spectrum with zero density

Advances spectral analysis understanding

Abstract

We construct a function on the real line supported on a set of finite measure whose spectrum has density zero.

Equations

$$F=0\text{ almost everywhere on }\mathbb{R}\setminus S.$$

$$|S|<+\infty ,|\mathbb{R}\setminus Q|=+\infty$$

$$|Q\cap (-R,R)|=o(R)\text{ as }R\to \infty .$$

$$0⩽F_0(t)⩽1(t\in \mathbb{R})$$

$$F_n:=F_{n-1}+G_n(n=1,2,...),$$

$$G_n(t):=F_{n-1}(t)[1-F_{n-1}(t)]\cos k_{n}t$$

$$|G_n(t)|⩽\max \{F_{n-1}(t),1-F_{n-1}(t)\}$$

$$0⩽F_n(t)⩽1.$$

$${\int }_{\mathbb{R}}{G}_n={\int }_{\mathbb{R}}{F}_{n-1}{G}_n=0$$

$${\int }_{\mathbb{R}}{G}_n^2=\frac{1}{2}{\int }_{\mathbb{R}}{F}_{n-1}^2[1-F_{n-1}{]}^2,$$

$${\int }_{\mathbb{R}}{F}_n={\int }_{\mathbb{R}}{F}_0=:C$$

$$I_n:={\int }_{\mathbb{R}}{F}_n(1-F_n)⩽C$$

$$I_n={\int }_{\mathbb{R}}[F_{n-1}+G_n][1-F_{n-1}-G_n]=I_{n-1}-{\int }_{\mathbb{R}}{G}_n^2,$$

$$\sum _{n\in [1,N]}{\int }_{\mathbb{R}}{G}_n^2\le I_0-I_N⩽C,$$

$$\sum _n{\int }_{\mathbb{R}}{G}_n^2⩽C$$

$$Q_0$$

$$Q_n$$

$$\mathrm{spec}{F}_{n-1}$$

$$\mathrm{spec}{G}_n$$

$$|Q\cap (-R,R)|=o(R)\text{ as }R\to \infty .$$

$$F_{n_{\ell }}\to F\text{ almost everywhere on }\mathbb{R}\text{ as }\ell \to \infty .$$

$$F⩾0\text{ almost everywhere and }{\int }_{\mathbb{R}}F<\infty .$$

$$\sum _n{\int }_{\mathbb{R}}[F_n(1-F_n){]}^2=2\sum _n{\int }_{\mathbb{R}}{G}_n^2<+\infty ,$$

$$F(1-F)=\underset{\ell \to \infty }{\lim }{F}_{n_{\ell }}(1-F_{n_{\ell }})=0\text{ almost everywhere,}$$

$$h(R):=|Q\cap (-R,R)|.$$

$$\underset{R\to \infty }{\lim }\underset{x\in \mathbb{R}}{\sup }\frac{1}{2R}|Q\cap (x-R,x+R)|=0.$$

Full text

1. The result

Let $F$ be a function in $L^{2}(\mathbb{R})$. We say that it is supported on $S$ if

[TABLE]

Suppose the set $S\subset\mathbb{R}$ is of finite Lebesgue measure. Then the Fourier transform $\widehat{F}$ of $F$ is a continuous function, so the spectrum of $F$ is naturally defined as the closure of the set where $\widehat{F}$ takes non-zero values.

According to the uncertianty principle, the support and the spectrum of a (non-trivial) function $F$ cannot be both “small sets” . This principle has various versions (see e.g. [HAJ94]).

In particular, the classic uniqueness theorem for analytic functions implies that if $F$ is supported on an interval and it has a “spectral gap” (that is, $\widehat{F}=0$ on an interval ) then $F=0$.

Another important result says that if the support $S$ and the spectrum $Q$ of $F$ are both of finite measure then $F=0$ [Ben74/85], [AB77].

On the other hand, $F$ may have a support of finite measure and a spectral gap; see [Kr82], where such an example was constructed with $F=1_{S}$.

Answering a question posed by Benedicks, Kargaev and Volberg [KV92] constructed an example of a function $F$ such that

[TABLE]

(here and below by $|A|$ we denote the Lebesgue measure of the set $A$).

The goal of this note is to prove the following

Theorem**.**

There is a function $F\in L^{2}(\mathbb{R})$ supported by a set $S$ of finite measure, such that

[TABLE]

In addition, $F$ can be chosen as the indicator function of $S$.

The proof below is based on a simple construction, completely different from the ones in the cited papers.

2. Proof

2.1.

Take a Schwartz function $F_{0}$ such that

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and its Fourier transform $\widehat{F}_{0}$ is positive on $(-1,1)$ and vanishes outside that interval. Define a sequence of functions $F_{n}$ recursively by

[TABLE]

where

[TABLE]

We are going to prove that if the numbers $k_{n}$ grow sufficiently fast, then the sequence $F_{n}$ converges to a function $F$ satisfying the requirements of the theorem.

2.2.

Clearly, $F_{n}$ and $G_{n}$ are Schwartz functions.

A simple induction shows that for every $t\in\mathbb{R}$, we have

[TABLE]

and

[TABLE]

The Fourier transforms of $F_{n-1}[1-F_{n-1}]$, $F_{n-1}^{2}[1-F_{n-1}]$, and $F_{n-1}^{2}[1-F_{n-1}]^{2}$ vanish outside a compact interval, so for each $n\geqslant 1$, we have:

[TABLE]

and

[TABLE]

provided that $k_{n}$ is chosen sufficiently large. It follows that

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and, thereby,

[TABLE]

(here, as usual, by $C$ we denote a positive constant that may vary from line to line).

Observe also that

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which implies that

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and so

[TABLE]

2.3.

Define the sequence $Q_{n}$ of intervals on (another copy of) $\mathbb{R}$ recursively as follows:

[TABLE]

(here $\operatorname{conv}E$ denotes the convex hull of a set $E\subset\mathbb{R}$). Clearly, for every n,

[TABLE]

Set $Q:=Q_{0}\cup\bigcup_{n}([k_{n}+2Q_{n-1}]\cup[-k_{n}+2Q_{n-1}])$.

Choosing $k_{n}$ growing sufficiently fast we can ensure that the spectra of $G_{n}$ are parewise disjoint and

[TABLE]

2.4.

Consider the series $F_{0}+G_{1}+G_{2}+\dots$ Since the spectra of the terms are pairwise disjoint, this series is orthogonal in $L^{2}(\mathbb{R})$. Then (3) implies that it converges in $L^{2}(\mathbb{R})$ to some non-trivial function $F$. The partial sums of this series are $F_{n}$. Take a subsequence $F_{n_{\ell}}$ such that

[TABLE]

Recall that all $F_{n}$ are non-negative functions, so (2) implies that

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It follows from (1) and (3) that

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so we must have

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which implies that $F$ is the indicator-function of a set $S$. According to (5), this set has finite measure. Clearly the spectrum of $F$ is a subset of $Q$. Due to (4) it has density zero. This finishes the proof.

Remark**.**

Consider the function

[TABLE]

In the conditions of the Theorem, it can not be bounded. However the proof above shows that it may increase arbitrarily slowly. It remains an open question, however, if $Q$ can have uniform density [math], i.e., if it is possible that

[TABLE]