Fourier Transforms of Distributions and Hausdorff Measures

Abstract

Consider a distribution whose support has Hausdorff $h$-measure zero. How fast can its Fourier coefficients decay to zero? If $h$ is close to linear, we can improve on known results by a factor of about $(\log n)^{-1/2}$.

1 Introduction

Throughout this paper we work on the circle ${\mathbb T}={\mathbb R}/{\mathbb Z}$. We will use $|I|$ to denote the length of an interval $I$.

Since this is a rather technical paper, the referee has suggested that that I should devote more space than usual to placing it in context. I thank the referee for this and other useful comments.

The subject begins with a famous theorem proved by Cantor using ideas of Riemann and Schwarz.

Theorem 1.1

If

$$\begin{aligned} \sum _{n=-N}^{N}a_{n}\exp (int)\rightarrow 0, \end{aligned}$$

as $N\rightarrow \infty$ for every $t\in {\mathbb T}$ then $a_{n}=0$.

Following on from this, Cantor introduced the notion of a set of uniqueness.

Definition 1.2

A set $E$ is a set of uniqueness if whenever

$$\begin{aligned} \sum _{n=-N}^{N}a_{n}\exp (int)\rightarrow 0, \end{aligned}$$

as $N\rightarrow \infty$ for every $t\in {\mathbb T}{\setminus }E$ then $a_{n}=0$.

Cantor was able to prove that a wide collection of closed countable sets were sets of uniqueness and Lebesgue completed his work by showing that every closed countable set was a set of uniqueness.

Lebesgue also showed that no closed set of positive Lebesgue measure can be of uniqueness. The key to this result and the results that follow lies in an alternative characterisation of sets of uniqueness.

Theorem 1.3

A closed set is of uniqueness if and only if it does not support a distribution $S$ with $\hat{S}(n)\rightarrow 0$ as $|n|\rightarrow \infty$.

(The introduction of this theorem at this point is anachronistic, but not improper since the roots of distribution theory may be found in the work of Riemann referred to above.)

It may have been hoped that every closed set of Lebesgue measure zero was a set of uniqueness, but, if so, the hope was dashed by Mensǒv who proved the following theorem.

Theorem 1.4

There exists a probability measure $\mu$ (that is to say a positive measure of mass $1$) whose support $E$ has Lebesgue measure $0$ but $\hat{\mu }(n)\rightarrow 0$ as $|n|\rightarrow 0$.

At this point the subject bifurcates. One path asks ‘what does a set of uniqueness look like’. The history of this path (together with the pre-history which I summarised above) is well described in [2]. The other path investigates how fast the Fourier transform $\hat{S}(n)$ of a distribution can tend to zero if its support has Lebesgue measure zero. We make the following simple observation.

Lemma 1.5

If $S$ is a distribution with $\sum _{n\ne 0}|\hat{S}(n)|^{2}<\infty$, then $S=f$ for some non-zero $f\in L^{2}$ and so $S$ has a support of strictly positive Lebesgue measure.

Lemma 1.5 can be viewed as yet another manifestation of the Heisenberg uncertainty principle.

Mensǒv’s original construction produces a probability measure $\mu$ whose support has Lebesgue measure zero and such that

$$\begin{aligned} |\hat{\mu }(n)|=O(\log |n|)^{-1/2}. \end{aligned}$$

Later Littlewood constructed a probability measure $\mu$ whose support has Lebesgue measure zero and there exists some constant $c>0$ such that

$$\begin{aligned} |\hat{\mu }(n)|=O(|n|^{-c}), \end{aligned}$$

Wiener and Winter improved Littlewood’s result to obtain

$$\begin{aligned} |\hat{\mu }(n)|=O(|n|^{\epsilon -1/2}) \end{aligned}$$

for any fixed $\epsilon >0$. (References and more detailed information can be found in [1].)

Finally, in [8] and [9], Ivašev-Musatov proved the remarkable result that (subject to certain regularity conditions) the simple bounds of Lemma 1.5 are best possible. Specifically, he proved the following theorem.

Theorem 1.6

If $\phi :{\mathbb Z}^{+}\rightarrow {\mathbb R}^{+}$ is a well behaved decreasing sequence satisfying the condition

$$\begin{aligned} \sum _{n=1}^{\infty }\phi (n)^{2}=\infty , \end{aligned}$$

then we can find a probability measure $\mu$ with support of Lebesgue measure zero such that $|\hat{\mu }(n)|\le \phi (|n|)$ for all $n\ne 0$.

Ivašev-Musatov’s original conditions on $\phi$ can be relaxed to the very natural demand that there exists a $K>0$ such that

$$\begin{aligned} \phi (n)\ge \phi (r)\ge K\phi (n), \end{aligned}$$

for all $2n\ge r\ge n\ge 1$. However, the following result from [5] shows that some condition is required.

Theorem 1.7

Let $q:(0,\infty )\rightarrow {\mathbb R}$ be an increasing strictly positive function. Then we can find a decreasing positive convex sequence $\phi :{\mathbb Z}^{+}\rightarrow {\mathbb R}^{+}$ such that

$$\begin{aligned} \sum _{n=1}^{\infty }q\big (\phi (n)\big )=\infty , \end{aligned}$$

but, if $S$ is a non-zero distribution with $|\hat{S}(n)|\le \phi (n)$ for all $n\ne 0$, then the support of $S$ is the whole circle ${\mathbb T}$.

To say that a closed set has Lebesgue measure zero is to say that it is small. In this paper we consider a more refined criterion for smallness.

Definition 1.8

Let $h:[0,1]\rightarrow {\mathbb R}$ be a continuous strictly increasing function with $h(0)=0$. We say that a closed set $E$ has Hausdorff $h$-measure zero if, given $\epsilon >0$, we can find a countable collection ${\mathcal I}$ of closed intervals such that

$$\begin{aligned} \bigcup _{I\in {\mathcal I}}I\supseteq E \text {and} \ \sum _{I\in {\mathcal I}}h(|I|)\le \epsilon . \end{aligned}$$

If $h(t)=t$ then $h$-measure zero is equivalent to Lebesgue measure zero.

Building on the work of Frostman and others, Beurling proved the following theorem for $h(t)=t^{\alpha }$ with $1\ge \alpha >0$.

Theorem 1.9

If $S$ is a non-zero distribution with $\hat{S}(n)\rightarrow 0$ as $|n|\rightarrow \infty$ and

$$\begin{aligned} \sum _{n\ne 0}\frac{|\hat{S}(n)|^{2}}{nh(n^{-1})}<\infty , \end{aligned}$$

then the support $\mathrm{supp }~S$ of $S$ cannot have Hausdorff $h$-measure zero.

If we follow the proof given in Chapter III of [3], it is clear that result holds for all reasonably behaved $h$ such that $h(t)t^{\gamma }\rightarrow \infty$ as $t\rightarrow 0+$ for some $\gamma >0$

Salem [10] showed (in effect) that if $h(t)$ is well behaved and decreases no slower than $t^{\alpha }$ as $t\rightarrow 0+$ for some $\alpha >0$, then we can find a probability measure $\mu$ with support of Hausdorff $h$-measure zero such that

$$\begin{aligned} |\hat{\mu }(n)|\le \big (h(|n|^{-1})\big )^{1/2}\big (\log |n|\big )^{1/2}, \end{aligned}$$

for all $|n|\ge 2$.

By combining Salem’s idea with an idea of Van der Corput used by Ivašev-Musatov, we can improve this result to obtain the following result (see [6]).

Theorem 1.10

Suppose that $h$ is well behaved and that $\phi$ is a well behaved sequence satisfying the condition

$$\begin{aligned} \sum _{n=1}^{\infty }\phi (n)^{2}=\infty . \end{aligned}$$

Then we can find a probability measure $\mu$ with h-measure zero such that $|\hat{\mu }(n)|\le \phi (|n|) (|n|h(|n|^{-1}))^{1/2}(\log |n|)^{1/2}$ for all $n\ne 0$.

The object of this paper is to prove the following result.

Theorem 1.11

Suppose that $h$ is well behaved and that $\phi$ is a well behaved sequence satisfying the condition

$$\begin{aligned} \sum _{n=1}^{\infty }\phi (n)^{2}=\infty . \end{aligned}$$

Then we can find a non-zero distribution $S$ with Hausdorff h-measure zero such that

$$\begin{aligned} |\hat{S}(n)|\le \phi (|n|)(|n|h(|n|^{-1}))^{1/2} \big (\log (|n|\max \{h(|n|^{-1}),\phi (n)^{2}\})\big )^{1/2}, \end{aligned}$$

for all $n\ne 0$.

We specify appropriate good behaviour in the following restatement of Theorem 1.11.

Theorem 1.12

Let $h:[0,1)\rightarrow {\mathbb R}$ be an increasing continuous function with $h(0)=0$ and $h(1)\le 10^{-2}$. Suppose that $\phi :{\mathbb Z}^{+}\rightarrow [0,1/2)$ is a decreasing sequence, with $r\phi (r)^{2}\le 10^{-2}$ for all $r$, satisfying the following conditions.

(A) There exist $1>\kappa _{1}>\kappa _{2}>0$ such that $\kappa _{1}\phi (n)\ge \phi (2n)\ge \kappa _{2}\phi (n)$ for all $n\ge 1$ and $h(t/2)\ge \kappa _{2}h(t)$ for all $1\ge t>0$.

(B) ${\displaystyle \sum _{n=1}^{\infty }\phi (n)^{2}=\infty .}$

Then we can find a non-zero distribution $S$ with support of Hausdorff $h$-measure zero such that $\hat{S}(0)=1$ and

$$\begin{aligned} |\hat{S}(n)|\le \tau (|n|), \end{aligned}$$

for all $n\ne 0$ where

$$\begin{aligned} \bigstar \qquad \qquad \qquad \tau (n)=\phi (n) \big (nh(n^{-1})\big )^{1/2} \big (\log (nh(n^{-1})\phi (n)^{2})\big )^{1/2}, \end{aligned}$$

for all $n\ge 1$.

Since

$$\begin{aligned} 2\log (nh(n^{-1})\phi (n)^{2}) \ge \log (|n|\max \{h(|n|^{-1}),\phi (n)^{2}\}), \end{aligned}$$

and we can replace $\phi$ by $\epsilon \phi$, the conclusions of the first version are unchanged. The conditions $h(1)\le 10^{-2}$ and $n\phi (n)^{2}\le 10^{-2}$ are merely inserted for convenience since the theorem concerns $h(t)$ when $t$ is small and we can replace $\tau (n)$ by $\min \{\phi (n),10^{-2}n^{-1/2}\}$.

The reader will observe that Theorem 1.12 deals with distributions rather than measures and that it only represents an improvement on Theorem 1.10 if $t^{-1}h(t)$ increases rather slowly as $t\rightarrow 0+$. However, it is then a genuine improvement. For example, if $h(t)=t/\log t^{-1}$ for $t$ small, then, taking $\phi (n)=(n\log n\log \log n)^{-1/2}$ for large $n$, we can obtain a non-zero distribution $S$ whose support has $h$-measure $0$ and which satisfies $|\hat{S}(n)|\le |n|^{-1/2}$ for $n\ne 0$.

Even if we were to take $\kappa _{1}=\tfrac{1}{2}$ and $\kappa _{2}=\tfrac{1}{2}-\gamma$ with $\gamma$ small and positive, condition (A) would not rule out any interesting cases. However, our proofs would not be noticeably simpler. Readers will lose nothing if they follow through the proofs in the special case when $h$ and $\phi$ are chosen as in the previous paragraph.

Sections 2 and 3 contain the novel part of the construction. The remainder of the construction can be traced back to the ideas of Ivašev-Musatov.

2 Finite Abelian Groups

Throughout this paper, $G$ will be finite Abelian group with $N$ elements equipped with the counting measure. The dual group $\hat{G}$ will be equipped with Haar measure normalised so that $\hat{G}$ has mass $1$. If $E$ is a subset of $G$ then $|E|$ denotes the number of elements in $E$. We write $C(G)$ for the collection of functions $f:G\rightarrow {\mathbb C}$ and $\chi _{0}$ for the character defined by $\chi _{0}(g)=1$ for all $g\in G$. If $f\in C(G)$ and $F\in C(\hat{G})$, we have

$$\begin{aligned} \hat{f}(\chi )=\sum _{g\in G}f(g)\chi (g), \ \ \hat{F}(g)=\frac{1}{|G|}\sum _{\chi \in {\hat{G}}}F(\chi )\chi (g), \end{aligned}$$

where $|G|$ denotes the number of elements in $G$.

Our object is to prove the following result.

Theorem 2.1

Let $1>\alpha >0$. There exist a constant $K$ with the following property. Given $M$ with $N\ge M\ge N^{\alpha }$ there exists an $g\in C(G)$ with

$$\begin{aligned} \hat{g}(\chi _{0})&=N\\ |\hat{g}(\chi )|&\le K(N/M)^{1/2}\big (\log (N/M)\big )^{1/2} \ \text {otherwise}. \end{aligned}$$

and $|\mathrm{supp }~g|\le M$.

Notice that the condition $M\ge N^{\alpha }$ is effectively irrelevant, since the theorem only produces useful information when $M$ is close to $N$. The reader should think in terms of $M=N/\log N$.

Theorem 2.1 follows at once from the following more general (but, I suspect, no more useful) result.

Theorem 2.2

Let $1>\alpha >0$. There exists a constant $K_{1}$ with the following property. Given $M$ with $N\ge M\ge N^{\alpha }$ and $f\in C(G)$, there exists an $F\in C(G)$ such that

$$\begin{aligned} |\hat{F}(\chi )-\hat{f}(\chi )|\le K_{1}\Vert f\Vert _{2}(N/M)^{1/2}, \big (\log (N/M)\big )^{1/2} \end{aligned}$$

for all $\chi \in \hat{G}$ and $|\mathrm{supp }~F|\le M$.

Proof

Proof of Theorem 2.1 from Theorem 2.2 Let $f(t)=1$ for all $t\in G$ and choose $F$ as in Theorem 2.2. Provided that $N$ is sufficiently large, we will have $|\hat{F}(0)|>N/2$. Set $g=N\big (\hat{F}(0)\big )^{-1}F$. $\square$

We obtain Theorem 2.2 by repeated use of the following result.

Theorem 2.3

Let $1>\beta >0$. There exists a constant $L\ge 1$ with the following property. Given $M$ with $N\ge M\ge N^{\beta }$ and $f\in C(G)$, we can find $u,\,v\in C(G)$ satisfying the following conditions.

• (i) $|\hat{f}(\chi )-\hat{u}(\chi )-\hat{v}(\chi )|\le L(N/M)^{1/2}\Vert f\Vert _{2}\big (\log (N/M)\big )^{1/2}$ for all $\chi \in \hat{G}$.

• (ii) $|\mathrm{supp }~u|\le 3M$.

• (iii) $\Vert v\Vert _{2}\le 2^{-8}\Vert f\Vert _{2}$

Proof

Proof of Theorem 2.2 from Theorem 2.3 Take $\beta =\alpha /2$. Let $r$ be the integer such that $2^{r-1}\le N^{\beta }<2^{r}$ and let $P$ be an integer with $2^{4}P\le M<(2^{4}+1)P$. If $f\in C(G)$, set $v_{0}=f$. By Theorem 2.3, we can find $u_{j},\,v_{j}\in C(G)$ $[1\le j\le r]$ such that the following conditions hold.

• (i) $_{j}$ We have

  $$\begin{aligned} |\hat{v}_{j-1}(\chi )-\hat{u}_{j}(\chi )-\hat{v}_{j}(\chi )|&\le L\left( \frac{N}{2^{-j}P}\right) ^{1/2}\Vert v_{j}\Vert _{2} \left( \log \left( \frac{N}{2^{-j}P}\right) \right) ^{1/2}\\&\le L_{1}2^{2j}\left( \frac{N}{P}\right) ^{1/2}\Vert v_{j}\Vert _{2} \left( \log \left( \frac{N}{P}\right) \right) ^{1/2} \end{aligned}$$

  for all $\chi \in \hat{G}$ all $1\le j\le r$ and some $L_{1}\ge 1$ depending on $\beta$ and so on $\alpha$.

• (ii) $_{j}$ $|\mathrm{supp }~u_{j}|\le 2^{4-j}P$.

• (iii) $_{j}$ $\Vert v_{j}\Vert _{2}\le 2^{-8}\Vert v_{j-1}\Vert _{2}$.

We observe that $\Vert v_{j}\Vert _{2}\le 2^{-8j}\Vert f\Vert _{2}$ and so we have

$$\begin{aligned} (i)_{j}' \ |\hat{v}_{j-1}(\chi )-\hat{u}_{j}(\chi )-\hat{v}_{j}(\chi )| \le 2^{-j}L_{1}\left( \frac{N}{P}\right) ^{1/2}\Vert f\Vert _{2} \left( \log \left( \frac{N}{P}\right) \right) ^{1/2}, \end{aligned}$$

for all $\chi \in \hat{G}$.

Taking $j=r$, we have $\Vert v_{r}\Vert _{2}\le 2^{-8r}\Vert f\Vert _{2}$ and so

$$\begin{aligned} |\hat{v}_{r}(\chi )|\le \Vert v\Vert _{1}\le 2^{-8r}N^{1/2}\Vert f\Vert _{2}, \end{aligned}$$

so, automatically,

$$\begin{aligned} |\hat{v}_{r}(\chi )|\le 2^{-r-1} 2^{-j}\left( \frac{N}{P}\right) ^{1/2}\Vert f\Vert _{2}. \end{aligned}$$

Now set $F=\sum _{j=1}^{r-1}u_{j}$. Conditions (ii)$_{j}$ yield

$$\begin{aligned} |\mathrm{supp }~f|\le \sum _{j=1}^{r-1}|\mathrm{supp }~u_{j}| \le \sum _{j=1}^{r-1}2^{4-j}P\le M. \end{aligned}$$

Conditions (i)$_{j}'$ and the observation of the previous paragraph yield

$$\begin{aligned} |\hat{F}(\chi )-\hat{f}(\chi )|&=\left| \sum _{j=1}^{r-1}\hat{u}_{j}(\chi )-\hat{v_{0}}(\chi )\right| \\&=\left| \hat{v}_{r}(\chi )+ \sum _{j=1}^{r}\big (\hat{u}_{j}(\chi )+\hat{v}_{j}(\chi ) -\hat{v}_{j-1}(\chi )\big )\right| \\&\le |\hat{v}_{r}(\chi )|+ \sum _{j=1}^{r}\big |\hat{u}_{j}(\chi )+\hat{v}_{j}(\chi ) -\hat{v}_{j-1}(\chi )\big |\\&\le L_{1}\sum _{j=1}^{r+1}2^{-j} \left( \frac{N}{P}\right) ^{1/2}\Vert f\Vert _{2} \left( \log \left( \frac{N}{P}\right) \right) ^{1/2}\\&\le L_{1}\left( \frac{N}{P}\right) ^{1/2}\Vert f\Vert _{2} \left( \log \left( \frac{N}{P}\right) \right) ^{1/2}\\&\le K_{1}(N/M)^{1/2}\Vert f\Vert _{2}\big (\log (N/M)\big )^{1/2} \end{aligned}$$

for an appropriate $K_{1}$. $\square$

3 A Probabilistic Argument

Our proof of Theorem 2.3 is probabilistic. The idea of using probabilistic methods in our context goes back to Kaufman [4]. We shall use a version on the following well known inequality of Bernstein.

Lemma 3.1

If $X_{1}$, $X_{2}$, ..., $X_{N}$ are independent real valued random variables with $|X_{n}|\le B$, ${\mathbb E}X_{n}=0$ for each $1\le n\le N$ and $\sigma ^{2}=\sum _{n=1}^{N}{\mathbb E}|X_{n}|^{2}$, then

$$\begin{aligned} \Pr \big ( X_{1}+X_{2}+\ldots +X_{N} \ge \lambda \big ) \le \exp \left( -\frac{\lambda ^{2}}{2(\sigma ^{2}+B\lambda /3)}\right) , \end{aligned}$$

for all $\lambda >0$.

Proof

This is given in chapter 7, section 4 of [11]. $\square$

We shall use the following immediate modification.

Lemma 3.2

If $Z_{1}$, $Z_{2}$, ..., $Z_{N}$ are independent complex valued random variables with $|Z_{n}|\le C$, ${\mathbb E}X_{n}=0$ for each $1\le n\le N$ and $\sigma ^{2}=\sum _{n=1}^{N}{\mathbb E}|X_{n}|^{2}$ then

$$\begin{aligned} \Pr \big ( |Z_{1}+Z_{2}+\ldots +Z_{N}| \ge \lambda \big ) \le 4\exp \left( -\frac{\lambda ^{2}}{16\sigma ^{2}}\right) , \end{aligned}$$

for all $C^{-1}\sigma ^{2}>\lambda >0$.

Proof

Consider real and imaginary parts separately. $\square$

We now start our proof of Theorem 2.3. By inspection, there is no loss of generality in supposing $M\le N/2$. Suppose that $G$ is a finite Abelian group with $N$ elements, that $N/2\ge M\ge N^{\beta }$ and $f\in C(G)$. We define

$$\begin{aligned} E_{1}=\{t\in G\,:\,|f(t)|\ge M^{-1/2}\Vert f\Vert _{2}\}, \end{aligned}$$

and observe that

$$\begin{aligned} M^{-1}\Vert f\Vert _{2}^{2}|E_{1}|\le \sum _{t\in E_{1}}|f(t)|^{2} \le \Vert f\Vert _{2}^{2}, \end{aligned}$$

so that $|E_{1}|\le M$.

We now define independent random variables $U(t)$ by the rule

$$\begin{aligned} U(t)&=f(t)&\text {if } t\in E_{1},\\ \Pr \{U(t)=0\}&=1-M/N&\text {if } t\notin E_{1},\\ \Pr \{U(t)=Nf(t)/M\}&=M/N&\text {if } t\notin E_{1}. \end{aligned}$$

If we write

$$\begin{aligned} E_{2}=\{t\notin E\,:\,W(t)\ne 0\}, \end{aligned}$$

then Tchebychev’s inequality tells us that, provided $N$ is large enough (independent of the choice of $M\ge N^{\beta }$), we have

$$\begin{aligned} \Pr \big (|E_{2}|\ge 2M\big )<1/4. \end{aligned}$$

Thus

$$\begin{aligned} \Pr \big (|\mathrm{supp }~U|\ge 3M)<1/4. \end{aligned}$$

Let $L$ be a strictly positive constant, to be determined later. The Fourier inversion theorem for finite Abelian groups tells us that we can find $V\in C(G)$ such that

$$\begin{aligned} \hat{V}(\chi )=0 \ \text {if }|\hat{U}(\chi )-\hat{f}(\chi )|\le L(N/M)^{1/2}\big (\log N/M\big )^{1/2}\Vert f\Vert _{2}, \end{aligned}$$

and

$$\begin{aligned} \hat{V}(\chi )=\hat{f}(\chi )-\hat{U}(\chi ) \ \text {otherwise}. \end{aligned}$$

We are interested in estimating $\Vert V\Vert _{2}$. We start by estimating $|\hat{V}(\chi )|^{2}$. Let us set

$$\begin{aligned} Z(t)=(f(t)-U(t)\big )\chi (t), \end{aligned}$$

so that

$$\begin{aligned} \sum _{t\in G}Z(t) =\hat{f}(\chi )-\hat{U}(\chi ). \end{aligned}$$

We observe that ${\mathbb E}Z(t)=0$ and $|Z(t)|\le M^{-1/2}\Vert f\Vert _{2}$, so we may apply Lemma 3.2. We note also that

$$\begin{aligned} {\mathbb E}|Z(t)|^{2}\le \frac{M}{N}\left( 1-\frac{M}{N}\right) \left( \frac{N|f(t)|}{M}\right) ^{2} \le \frac{N}{M}|f(t)|^{2}, \end{aligned}$$

so

$$\begin{aligned} \sum _{t\in G}{\mathbb E}|Z(t)|^{2}\le \frac{N}{M}\Vert f\Vert _{2}^{2}. \end{aligned}$$

Lemma 3.2 thus gives

$$\begin{aligned} \Pr \big (\big |\hat{f}(\chi )-\hat{U}(\chi )\big | \ge \lambda \big )&=\Pr \left( \left| \sum _{t\in G}Z(t)\right| \ge \lambda \right) \\&\le 4\exp \left( -\frac{M\lambda ^{2}}{16N\Vert f\Vert _{2}^{2}}\right) , \end{aligned}$$

for all $\lambda _{0}>\lambda >0$ where $\lambda _{0}=NM^{-1/2}\Vert f\Vert _{2}$.

Since $\Vert f\Vert _{1}\le N^{1/2}\Vert f\Vert _{2}$ we have

$$\begin{aligned} |\hat{V}(\chi )|\le 2NM^{-1}\Vert f\Vert _{1}\le 2N^{3/2}\Vert f\Vert _{2}, \end{aligned}$$

and

$$\begin{aligned} {\mathbb E}\hat{V}^{2}(\chi ){\mathbb I}_{|\hat{V}(\chi )|\ge \lambda _{0}}&\le 4N^{3}\Vert f\Vert _{2}^{2}\Pr (|V(\chi )|\ge \lambda _{0})\\&\le 16N^{3}\Vert f\Vert _{2}^{2}\exp \left( -\frac{N}{16N}\right) \\&\le 2^{-10}\Vert f\Vert _{2}^{2}, \end{aligned}$$

provided only that $N$ is large enough.

Next we observe that, if $2^{r}L(N/M)^{1/2}\big (\log N/M\big )^{1/2}\le \lambda _{0}$, then

$$\begin{aligned} {\mathbb E}&|\hat{V}(\chi ){\mathbb I}_{2^{r}L(N/M)^{1/2}(\log N/M)^{1/2}\Vert f\Vert _{2} \le |V(\chi )|<2^{r+1}L(N/M)^{1/2}(\log N/M)^{1/2}\Vert f\Vert _{2}}|^{2}\\&\le 2^{2r+2}L(N/M)(\log N/M)\Pr \big (|\hat{V}(\chi )| \ge 2^{r}L(N/M)^{1/2}(\log N/M)^{1/2}\big )\\&\le 2^{2r+6}L(N/M)(\log N/M)\Vert f\Vert _{2}^{2}\exp \left( -\frac{2^{2r}L(N/M)(\log N/M)\Vert f\Vert _{2}^{2}M}{16N\Vert f\Vert _{2}^{2}}\right) \\&=2^{2r+6}L(N/M)(\log N/M)\Vert f\Vert _{2}^{2}\exp (-2^{2r-4}L\log N/M)\\&=2^{2r+6}L(M/N)^{1-2^{2r-4}L}\Vert f\Vert _{2}^{2}\\&\le L2^{2r+7-2^{2r-4}L}\Vert f\Vert _{2}^{2}\\&\le 2^{-20-2r}\Vert f\Vert _{2}^{2}, \end{aligned}$$

for an appropriate $L$ (for example $L=100000$).

By addition,

$$\begin{aligned} {\mathbb E}|\hat{V}(\chi )|^{2}={\mathbb E} |\hat{V}(\chi ){\mathbb I}_{L(N/M)^{1/2}(\log N/M)^{1/2}\Vert f\Vert _{2}\le |V|}|^{2} \le 2^{-20}\Vert f\Vert _{2}^{2}. \end{aligned}$$

Thus

$$\begin{aligned} {\mathbb E}\Vert V\Vert _{2}^{2}= N^{-1}{\mathbb E}\Vert \hat{V}\Vert _{2}^{2} =N^{-1}{\mathbb E}\sum _{\chi \in \hat{G}}|V(\chi )|^{2} =N^{-1}\sum _{\chi \in \hat{G}}{\mathbb E}|V(\chi )|^{2}\le 2^{-20}\Vert f\Vert _{2}^{2}, \end{aligned}$$

and

$$\begin{aligned} \Pr (\Vert V\Vert _{2}^{2}\le 2^{-18}\Vert f\Vert _{2}^{2})\le 1/4. \end{aligned}$$

Thus there is a probability at least $1/2$ that both

$$\begin{aligned} \Vert V\Vert _{2}\le 2^{-9}\Vert f\Vert _{2}, \end{aligned}$$

and

$$\begin{aligned} |\mathrm{supp }~U |\le 3M. \end{aligned}$$

Since there must be an example of any event with strictly positive probability, Theorem 2.3 follows.

4 Measures on the Circle

The move from finite Abelian groups to the circle is very easy. We first apply Theorem 2.1 to the case when $G$ is cyclic.

Lemma 4.1

If $\alpha >0$ there exists a constant $K$ with the following property. Given $M$ with $N\ge M\ge N^{\alpha }$, we can find a measure $\mu$ on the circle ${\mathbb T}$ whose support consists of at most $M$ points each of which has the form $v/N$ for some integer $v$ and which has the following properties.

• (i) $|\mathrm{supp }~\mu |\le M$.

• (ii) $\hat{\mu }(0)=1$.

• (iii) $|\hat{\mu }(r)|\le KM^{-1/2}\big (\log (N/M)\big )^{1/2}$ for all $r\not \equiv 0\pmod {N}$.

Proof

Let $G$ be the subgroup of ${\mathbb T}$ consisting of points of the form $v/N$ with $v$ an integer. Take $g$ as in Theorem 2.1 and set

$$\begin{aligned} \mu =N^{-1}\sum _{t\in G}g(t)\delta _{t}. \end{aligned}$$

$\square$

We now smooth the measure. of Lemma 4.1 in a standard manner. The next lemma just establishes the appropriate notation for this.

Lemma 4.2

Let $k:{\mathbb R}\rightarrow {\mathbb R}$ be an infinitely differentiable function such that

• (i) $k(t)\ge 0$ for all $t\in {\mathbb R}$.

• (ii) $k(t)=0$ for $|t|\ge 1/2$.

• (iii) ${\displaystyle \int _{\mathbb R}k(t)\,dt=1.}$

Then, if we define $k_{N}:{\mathbb T}\rightarrow {\mathbb R}$ by

$$\begin{aligned}k_{N}(t)= {\left\{ \begin{array}{ll} Nk(Nt)&{}\text {if } |t|\le N^{-1}/2,\\ 0&{}\text {otherwise}, \end{array}\right. }\end{aligned}$$

it follows that $k_{N}$ is an infinitely differentiable positive function with $\mathrm{supp }~k_{N}\subseteq [-N^{-1}/2,N^{-1}/2]$ and $\Vert k_{N}\Vert _{1}=1$.

If $m$ is an integer with $m\ge 1$ there exists a constant $A(m)$ such that

$$\begin{aligned}|\hat{k}_{N}(r)|\le {\left\{ \begin{array}{ll} 1&{}\text {if } |r|<N \\ A(m)(N/|r|)^{m}&{}\text {if } |r|\ge N \end{array}\right. }.\end{aligned}$$

Proof

Immediate. $\square$

Here is the smoothed version of Lemma 4.1.

Lemma 4.3

There exist constants $K$ and $K_{m}$ with the following property. Given $M$ with $N\ge M\ge N^{\alpha }$ we can find a smooth function $f:{\mathbb T}\rightarrow {\mathbb C}$ whose support is contained in the union at most $M$ intervals each of length $2N^{-1}$ with the following properties.

• (i) $\hat{f}(0)=1$.

• (ii) $|\hat{f}(r)|\le KM^{-1/2}\big (\log (N/M)\big )^{1/2}$ for all $r\ne 0$.

• (iii) $|\hat{f}(r)|\le K_{m}M^{-1/2}\big (\log (N/M)\big )^{1/2} (N/|r|)^{m}$ for all $|r|\ge N$.

Proof

Take $\mu$ as in Lemma 4.1 and $k_{N}$ as in Lemma 4.2 Set $J_{N}=N{\mathbb I}_{[-N^{-1}/2,N^{-1}/2]}$ and $f=J_{N}*k_{N}*\mu$. Observe that

$$\begin{aligned} \hat{f}(r)=\hat{J}_{N}(r)\hat{k}_{N}(r)\hat{\mu }(r), \end{aligned}$$

and that $\hat{J}_{N}(r)=0$ whenever $r\equiv 0 \pmod {N}$, $r\ne 0$. $\square$

5 The Ivašev-Musatov Argument

We now combine the functions of the previous section.

It may be helpful to explain how we use Conditions (A) and (B) of Theorem 1.12. Condition (B) is used in a form suggested by the Cauchy condensation test.

Lemma 5.1

Suppose that $\phi$ satisfy the conditions of Theorem 1.12. Then

$$\begin{aligned} \sum _{j=1}^{\infty }2^{j}\phi (2^{j})^{2}=\infty . \end{aligned}$$

Proof

Immediate. $\square$

Condition (A) is used in three ways. Firstly it tells us that $\tau (r)$ does not change too much as $r$ increases from $2^{j}$ to $2^{j+1}$ so, instead of looking at $\tau (r)$ for general $r$, we can consider the case $r=2^{j}$. Secondly it tells us that $\tau (2^{j})$ decreases no faster than some geometric series. Finally, in the opposite direction, it tells us that $\tau (2^{j})$ decreases at least as fast as some geometric series.

The arguments that follow use lots of estimates involving constants whose value is unimportant. What is important is that these constants, although they may depend on $\kappa _{1}$ and $\kappa _{2}$ do not depend on the particular $\phi$ and $h$ considered in the hypotheses of Theorem 1.12.

We start by choosing appropriate $M$ and $N$ in Lemma 4.3.

Lemma 5.2

Let $h$ and $\phi$ satisfy condition (A). If $m\ge 1$ We can find a constants $K_{1}$ and $K_{2}$ and $n_{1}$ with the following property. If $n\ge n_{1}$ we can find a smooth function $g_{n}:{\mathbb T}\rightarrow {\mathbb C}$ whose support is contained in the union of a finite collection ${\mathcal I}_{n}$ of intervals with the following properties.

• (i) $\hat{g}_{n}(0)=1$.

• (ii) $|\hat{g_{n}}(r)|\le K_{1}2^{-n}\phi (2^{n})^{-2}\tau (n)$ for all $r\ne 0$.

• (iii) $|\hat{g}_{n}(r)|\le K_{1}2^{-n}\phi (2^{n})^{-2}\tau (n) (2^{n}/|r|)^{m}$ for all $|r|\ge N$.

• (iv) ${\displaystyle \sum _{I\in {\mathcal I}_{n}}h(|I|) \le K_{1}2^{n}\phi (2^{n})^{2}}$

Proof

Write $N(n)=2^{n}$ and take $M(n)$ to be the integer part of

$$\begin{aligned} \frac{2^{n}\phi (2^{n})^{2}}{h(2^{-n})}. \end{aligned}$$

Condition (A) tells us that there exists an $\alpha >0$ such that (provided $n$ is large enough) $M(n)>N(n)^{\alpha }$. We now apply Lemma 4.3 with $N=N(n)$ and $M=M(n)$. If we choose $K_{1}$ appropriately, Conditions (i), (ii) and (iii) can now be read off from Lemma 4.3 whilst condition (iv) follows on recalling that, by condition (A), $\kappa _{1}h(2n^{-1})\ge h(n^{-1})$. $\square$

Lemma 5.3

There exists a constant $C$ with the following property. Suppose $h$ and $\phi$ satisfy the conditions of Theorem 1.12. Then we can find an infinitely differentiable function $g:{\mathbb T}\rightarrow {\mathbb R}$ with the following properties.

• (i) $\hat{g}(0)=1$.

• (ii) $|\hat{g}(r)|\le C\tau (r)$ for all $r\ne 0$ where $\tau$ is defined in $\bigstar$ (see Theorem 1.12).

• (iii) There exists a finite collection ${\mathcal I}$ of intervals such that

  $$\begin{aligned} \bigcup _{I\in {\mathcal I}}I\supseteq \mathrm{supp }~g\text {but} \ \sum _{I\in \mathcal I}h(|I|)\le C. \end{aligned}$$

Proof

By condition (B) we can find $K_{2}>0$ and $\alpha >0$ and $m\ge 2$ (depending only on $\kappa _{1}$ and $\kappa _{2}$ such that

$$\begin{aligned} K_{2}\tau (2^{n})\ge \tau (r)(2^{n}/r)^{\alpha } \end{aligned}$$

for all $1\le r\le 2^{n}$ and

$$\begin{aligned} K_{2}\tau (r)\ge \tau (2^{n})(2^{n}/r)^{m-2} \end{aligned}$$

for all $r\ge 2^{n}$.

Let $n_{1}$ be chosen so that the conclusions of Lemma 5.2 hold. By Lemma 5.1 we can find an $n_{2}$ such that

$$\begin{aligned} 1\le \sum _{n=n_{1}}^{n_{2}}2^{n}\phi (2^{n})^{2}\le 2. \end{aligned}$$

We set

$$\begin{aligned} G=\sum _{n=n_{1}}^{n_{2}}2^{n}\phi (2^{n})^{2}g_{n} \end{aligned}$$

so $\hat{G}(0)=\sum _{n=n_{1}}^{n_{2}}2^{n}\phi (2^{n})^{2}\ge 1$ and, setting $g=(\hat{G}(0))^{-1}G$, we have $\hat{g}(0)=1$ automatically.

Observe first that $\mathrm{supp }~g=\mathrm{supp }G$ is covered by a finite collection ${\mathcal I}= \bigcup _{n=n_{1}}^{n_{2}}{\mathcal I}_{n}$ of intervals with

$$\begin{aligned} \sum _{I\in {\mathcal I}}h(|I|)\le \sum _{n=n_{1}}^{n_{2}} \sum _{I\in {\mathcal I}_{n}}h(|I|) \le \sum _{n=n_{1}}^{n_{2}}K_{1}2^{n}\phi (2^{n})^{2} \le 2K_{1}\le C, \end{aligned}$$

provided only that $C$ is large enough.

Lemma 5.2 also tells us that

• (ii) $|2^{n}\phi (2^{n})^{2}\hat{g}_{n}(r)|\le K_{1}\tau (2^{n})$ for all $r\ne 0$.

• (iii) $|2^{n}\phi (2^{n})^{2}\hat{g}_{n}(r)|\le K_{1}\tau (2^{n}) (2^{n}/|r|)^{m}$ for all $|r|\ge 2^{n}$.

We estimate $|\hat{g}(r)|$ using the observation that

$$\begin{aligned} |\hat{g}(r)|\le |\hat{G}(r)|\le \sum _{n=n_{1}}^{n_{2}}2^{n}\phi (2^{n})^{2}|\hat{g}_{n}(r)|. \end{aligned}$$

There are three cases according as $|r|<2^{n_{1}}$, $2^{n_{2}}<|r|$ or $2^{n_{1}}\le |r|\le 2^{n_{2}}$. The proofs of the first two cases are similar to, but simpler than, the third, so we only consider the third case.

If $2^{n_{1}}\le |r|\le 2^{n_{2}}$ then we can find an integer $u$ with $n_{1}\le u<n_{2}$ and $2^{u}\le |r|\le 2^{u+1}$. It follows (choosing appropriate $C_{j}$ and $C$ depending ultimately only on $\kappa _{1}$, $\kappa _{2}$) that

$$\begin{aligned} |\hat{g}(r)|&\le \sum _{n=n_{1}}^{u-1}2^{n}\phi (2^{n})^{2}|\hat{g}_{n}(r)| +\sum _{n=u}^{n_{2}}2^{n}\phi (2^{n})^{2}|\hat{g}_{n}(r)|\\&\le \sum _{n=n_{1}}^{u-1} K_{1}\tau (2^{n})(2^{n}/|r|)^{m} +\sum _{n=u}^{n_{2}} K_{1}\tau (2^{n})\\&\le \sum _{n=n_{1}}^{u-1} K_{1}K_{2}\tau (r) (2^{n}/|r|)^{m-2}(2^{n}/|r|)^{m} +\sum _{n=u}^{n_{2}} K_{1}K_{2}\tau (2^{u})(|r|/2^{n})^{\alpha }\\&\le C_{1}\tau (|r|)+C_{2}\tau (2^{u})\le C\tau (|r|) \end{aligned}$$

and we are done. $\square$

Theorem 5.4

Suppose $h$ and $\phi$ satisfy the conditions of Theorem 1.12. Then, given $\epsilon >0$, we can find an infinitely differentiable function $f_{\epsilon }:{\mathbb T}\rightarrow {\mathbb R}$ with the following properties.

• (i) $\hat{f}_{\epsilon }(0)=1$.

• (ii) $|\hat{f}_{\epsilon }(r)|\le \epsilon$ for all $r\ne 0$.

• (iii) There exists a finite collection ${\mathcal I}$ of intervals such that

  $$\begin{aligned} \bigcup _{I\in {\mathcal I}}I\supseteq \mathrm{supp }~f_{\epsilon } \text {but} \sum _{I\in \mathcal I}h(|I|)\le \epsilon . \end{aligned}$$

• (iv) $|\hat{f}_{\epsilon }(n)|\le \epsilon \tau (n)$ for all $n\ne 0$.

Proof

Let $\eta >0$ be a small number to be chosen later. If we set $\tilde{\phi }=\eta \phi$ and $\tilde{h}=\eta h$, then $\tilde{\phi }$ and $\tilde{h}$ also satisfy the conditions of Theorem 1.12 with the same $h$, $\kappa _{1}$ and $\kappa _{2}$. Thus we can we can find an infinitely differentiable function $G:{\mathbb T}\rightarrow {\mathbb R}$ with the following properties.

• (i) $\hat{G}(0)=1$.

• (ii) If we write

  $$\begin{aligned} \tilde{\tau }(n)=\tilde{\phi }(n) \big (n\tilde{h}(n^{-1})\big )^{1/2} \big (\log (n\tilde{h}(n^{-1})\tilde{\phi }(n)^{2})\big )^{1/2} \end{aligned}$$

  for $n\ge 1$, then

  $$\begin{aligned} |\hat{G}(r)|\le C\tilde{\tau }(|r|) \end{aligned}$$

  for all $r\ne 0$.

• (iii) There exists a finite collection ${\mathcal I}$ of intervals such that

  $$\begin{aligned} \bigcup _{I\in {\mathcal I}}I\supseteq \mathrm{supp }~G \ \text {but} \sum _{I\in \mathcal I}h(|I|)\le \epsilon . \end{aligned}$$

  Provided that $\eta$ is small enough, $f_{\epsilon }=G$ satisfies the required conditions.$\square$

6 A Baire Category Argument

Theorem 5.4 contains the essence of Theorem 1.12. One way of completing the proof is to consider the distributional limit $\prod _{j=1}^{\infty }f_{\epsilon (j)}$ with $f_{\epsilon (j)}$ as in Theorem 5.4 and $\epsilon (j)\rightarrow 0$ extremely fast. We shall use a Baire category argument instead, but the main work for both methods involves arguments along the lines of the proof of our Lemma 6.4.

All Baire category arguments require an appropriate complete metric space.

Lemma 6.1

• (i) Consider the space ${\mathcal F}$ of non-empty closed subsets of ${\mathbb T}$. If we set

  $$\begin{aligned} d_{\mathcal F}(E,F)=\sup _{e\in E}\inf _{f\in F}|e-f| + \sup _{f\in F}\inf _{e\in E}|e-f|, \end{aligned}$$

  then $({\mathcal F},d_{\mathcal F})$ is a complete metric space.

• (ii) Suppose $h$ and $\phi$ satisfy the conditions of Theorem 1.12. Consider the space ${\mathcal E}$ consisting of ordered pairs $(E,S)$ where $E\in {\mathcal F}$ and $S$ is a distribution with $\mathrm{supp }~S\subseteq E$, and $\tau (|n|)^{-1}\hat{S}(n)\rightarrow 0$ as $|n|\rightarrow \infty$. If we take

  $$\begin{aligned} d_{\mathcal E}\big ((E,S),(F,T)\big ) =d_{\mathcal F}(E,F)+|\hat{S}(0)-\hat{T}(0)| +\sup _{n\ne 0}\tau (n)^{-1}|\hat{S}(n)-\hat{T}(n)|, \end{aligned}$$

  then $({\mathcal E},d_{\mathcal E})$ is a non-empty complete metric space.

• (iii) The collection ${\mathcal D}$ of $(E,S)$, where $S=g$ in the distributional sense, $g$ is an infinitely differentiable function and $E$ is the union of a finite set of intervals, is dense in ${\mathcal E}$.

Proof

• (i) This metric is called the Hausdorff metric and is discussed, with proofs, in [7] (see Chapter II §21 VII and Chapter III §33 IV).

• (ii) Use weak compactness. The space ${\mathcal E}$ is non-empty because $({\mathbb T},\mu )$ (with $\mu$ Haar measure) lies in ${\mathcal E}$.

• (iii) Let $k_{m}:{\mathbb T}\rightarrow {\mathbb R}$ be as in Lemma 4.2. Then

  $$\begin{aligned} (E+[-m^{-1},m^{-1}],S*k_{m})\in {\mathcal D} \end{aligned}$$

  and

  $$\begin{aligned} d_{\mathcal E}\big ((E+[-m^{-1},m^{-1}],S*k_{m}),(E,S)\big ) \rightarrow 0 \end{aligned}$$

  as $m\rightarrow \infty$.

$\square$

Theorem 1.12 follows in the usual way from the following Baire category version.

Theorem 6.2

Let $({\mathcal E},d_{\mathcal E})$ be as in Lemma 6.1. Then there is a subset of first category $\Gamma$ with the following property. If $(E,S)\in {\mathcal E}{\setminus }\Gamma$ and $\eta >0$, we can find a finite collection ${\mathcal I}$ of intervals such that

$$\begin{aligned} \bigcup _{I\in {\mathcal I}}I\supseteq E, \ \text {but} \ \sum _{I\in \mathcal I}h(|I|)<\eta . \end{aligned}$$

Proof

Proof of Theorem 1.12 from Theorem 6.2 Let $\mu$ be Haar measure. By Theorem 6.2, we can find $(E,T)\in {\mathcal E}{\setminus }\Gamma$ with

$$\begin{aligned} d\big ((E,T),({\mathbb T},\mu )\big )<1/2. \end{aligned}$$

We thus have

$$\begin{aligned} |\hat{T}(0)|>1/2\ \text {and}\ |\hat{T}(n)|<\tau (|n|)/2 \ \text {for all } n\ne 0. \end{aligned}$$

Set $S=\hat{T}(0)^{-1}T$. $\square$

Lemma 6.3

Let $({\mathcal E},d_{\mathcal E})$ be as in Lemma 6.1. If $q$ is a positive integer, we consider the set $\Gamma _{q}$ consisting of those $(E,S)$ such that we can find a finite collection ${\mathcal I}$ of intervals such that

$$\begin{aligned} \bigcup _{I\in {\mathcal I}}I\supseteq E \ \text {but} \ \sum _{I\in \mathcal I}h(|I|)<1/q \end{aligned}$$

is dense.

Proof

Proof of Theorem 6.2 from Lemma 6.3 The set $\Gamma _{q}$ is open, so $\bigcap _{q=1}^{\infty }\Gamma _{q}$ has a complement of first category. $\square$

Using Lemma 6.1 (iii), we see that Lemma 6.3 follows at once from the next lemma

Lemma 6.4

We use the notation and hypotheses of Lemmas 6.3 and 6.1. Given $(E,S)\in {\mathcal D}$, $q\ge 1$ and $\eta >0$ we can find $(F,T)\in \Gamma _{q}$ with

$$\begin{aligned} d_{\mathcal E}\big ((E,S),(F,T)\big )<\eta . \end{aligned}$$

Proof

We have the distributional equality $S=g$ where $g$ is infinitely differentiable. We choose a finite subset $E'$ of $E$ such that $d_{\mathcal F}(E,E')<\eta /3$. Let $f_{\epsilon }$ have the properties set out in Theorem 5.4. I claim that, provided only that $\epsilon$ is small enough, taking $T=f_{\epsilon }g$ and

$$\begin{aligned} F=E\cap (E'\cup \mathrm{supp }~f_{\epsilon }) \end{aligned}$$

will do. We observe that $\mathrm{supp }~T\subseteq E\cap \mathrm{supp }~f_{\epsilon }$, so $d_{\mathcal F}(E,F)<\eta /3$ automatically.

We now need to bound

$$\begin{aligned} |\hat{S}(n)-\hat{T}(n)| =|\hat{g}(n)-(f_{\epsilon }g)\hat{}(n)|. \end{aligned}$$

Condition (A) tells us that there exists an integer $m\ge 1$ and a $\delta >0$ such that $\tau (r)\ge \delta r^{-m}$ for $r\ge 1$. Since $g$ is infinitely differentiable, we can find an $A$ such that $|\hat{g}(r)|\le A|r|^{-m-1}$ for $r\ne 0$. Condition (A) also tells us that there is a constant $B$ such that $\tau (u)\le B \tau (n)$ whenever $|u-n|\le n/2$ and $n\ge 1$. If $n\ne 0$,

$$\begin{aligned} |\hat{g}(n)&-(f_{\epsilon }g)\hat{}(n)| =\left| \sum _{r\ne n}\hat{g}(r)\hat{f}_{\epsilon }(n-r)\right| \le \sum _{r\ne n}|\hat{g}(r)\hat{f}_{\epsilon }(n-r)|\\&=\sum _{|r|\le |n|/2}|\hat{g}(r)||\hat{f}_{\epsilon }(n-r)| +\sum _{|r|> |n|/2,\,r\ne n}|\hat{g}(r)||\hat{f}_{\epsilon }(n-r)|\\&\le \sum _{|r|\le |n|/2}\epsilon |\hat{g}(r)|\tau (|n-r|) +\sum _{|r|> |n|/2,\,r\ne n}\epsilon |\hat{g}(r)|\\&\le \epsilon B\sum _{|r|\le n/2}\epsilon |\hat{g}(r)|\tau (|n|) +\sum _{|r|> n/2}\epsilon |\hat{g}(r)|\\&\le \epsilon B\left( |\hat{g}(0)| +2A\sum _{r=1}^{\infty }r^{-m-1}\right) \tau (|n|) +2\epsilon A\sum _{r\ge |n|/2}^{\infty }r^{-m-1}\\&\le \epsilon B\left( |\hat{g}(0)|+4A\right) \tau (|n|) +2^{m+4}\epsilon An^{-m}. \end{aligned}$$

Thus, provided only that $\epsilon$ is small enough,

$$\begin{aligned} |\hat{S}(n)-\hat{T}(n)|< 3^{-1}\eta \tau (|n|). \end{aligned}$$

for all $n\ne 0$.

Finally,

$$\begin{aligned} |\hat{g}(0)-(f_{\epsilon }g)\hat{}(0)|\le \sum _{r\ne 0}|\hat{g}(r)||\hat{f}_{\epsilon }(-r)| \le \epsilon \sum _{r\ne 0}|\hat{g}(r)|, \end{aligned}$$

so, provided only that $\epsilon$ is small enough,

$$\begin{aligned} |\hat{S}(0)-\hat{T}(0)|< 3^{-1}\eta \end{aligned}$$

and we are done. $\square$