Proof. In [Reference Nathanson and SárközyNS89, Theorem 1], take $h = 2r$ , $z = \lfloor X/2r^2\rfloor$ ; the result is then easily verified.

Proof. Write $I = [x_0, x_0 + L-1]$ , where $x_0 \in \mathbf {Z}_{\geqslant 0}$ . Then $jI = [jx_0, jx_0 + j(L-1)]$ . Note that if $j \geqslant x_0/(L-1)$ , we have $jx_0 + j(L-1) \geqslant (j+1) x_0$ , and so the interval $(j+1) I$ overlaps the interval $jI$ . Therefore, if we set $j_0 := \lceil x_0/(L-1) \rceil$ , for any $j_1 \geqslant j_0$ , the union $I^* := \bigcup _{j_0 \leqslant j \leqslant j_1} jI$ is a discrete interval. Set $j_1 := \lceil 2K/\eta ^2 \rceil$ . We have

\begin{equation*} \min I^* = j_0x_0 \leqslant \big \lceil \tfrac {X}{L-1}\big \rceil X \leqslant \big \lceil \tfrac {2X}{L} \big \rceil X \leqslant \tfrac {4X^2}{L} = \tfrac {4}{\eta } X,\end{equation*}

and

\begin{equation*} \max I^* \geqslant j_1 (L-1) \geqslant \tfrac {2K}{\eta ^2} \tfrac {L}{2} = \tfrac {K}{\eta } X.\end{equation*}

This concludes the proof.

Proof of Theorem 1.1, assuming Proposition 2.1 Let $n$ be some large multiple of $k$ and consider the measure $\mu _n$ as described in § 2. Thus, $\mu _n$ is supported on $\mathcal {S}^k \cap [0, N^k)$ , where $N = b^n$ . Set

(3.1) \begin{equation} t := 8b^{9k^2}, \end{equation}

and write $\mu _n^{(t)}$ for the $t$ -fold convolution power of $\mu _n$ , that is to say,

\begin{equation*} \mu _n^{(t)}(x) = \sum _{x_1 + \cdots + x_{t} = x} \mu _n(x_1) \cdots \mu _n(x_t).\end{equation*}

Then $\widehat {\mu _n^{(t)}} = ( \widehat {\mu _n})^t$ and so by Parseval’s identity and the layer-cake representation,

(3.2) \begin{equation} \sum _x \mu _n^{(t)}(x)^2 = \int ^1_0 |\widehat {\mu _n}(\theta )|^{2t} d\theta = 2t \int ^1_0 \delta ^{2t - 1} \operatorname {meas} \{ \theta : |\widehat {\mu _n}(\theta )| \geqslant \delta \} d \delta = 2t (I_1 + I_2 + I_3), \end{equation}

where $I_1, I_2, I_3$ are the integrals over ranges $[0, 2N^{-1/B}]$ , $[2N^{-1/B}, 1 - c]$ and $[1 - c, 1]$ , respectively, with $c := \frac {1}{4} b^{-3k^2}$ , $B = b^{6k^2}$ (as in Proposition 2.1) and $\operatorname {meas}$ is the Lebesgue measure on the circle ${\mathbf {R}}/\mathbf {Z}$ . We have, for $N$ large,

\begin{equation*} I_1 \leqslant (2N^{-1/B})^{2t - 1} \lt N^{-k}.\end{equation*}

To bound $I_2$ , we use Proposition 2.1, which tells us that the set $\{ \theta \in {\mathbf {R}}/\mathbf {Z} : |\widehat {\mu }_n(\theta )| \geqslant \delta \}$ is contained in the set $\{ \theta \in {\mathbf {R}}/\mathbf {Z} : \Vert \theta q \Vert \leqslant (2/\delta )^B N^{-k} \; \mbox {for some positive} \; q \leqslant (2/\delta )^B\}$ , and so $\operatorname {meas}\{ \theta : |\widehat {\mu _n}(\theta )| \geqslant \delta \} \leqslant 2(2/\delta )^{2B} N^{-k}$ . Since $2t - 1 - 2B \geqslant t$ , we therefore have

\begin{equation*} I_2 \leqslant 2N^{-k} \int ^{1 - c}_0 \delta ^{2t-1}(2/\delta )^{2B} d\delta \leqslant 2N^{-k} (1 - c)^t 2^{2B} \lt N^{-k}.\end{equation*}

For the last inequality, we used the fact that $t = 2B/c$ and so $(1 - c)^t \leqslant e^{-2B}$ .

Finally, to bound $I_3$ , we use Proposition 2.2, which immediately implies that

\begin{equation*} I_3 \leqslant 2b^{k^2} N^{-k} . \end{equation*}

Substituting these bounds for $I_1, I_2$ and $I_3$ into (3.2), we obtain that, for sufficiently large $N$ , $\sum _x \mu _n^{(t)}(x)^2 \leqslant 4t b^{k^2} N^{-k} = 32b^{10k^2} N^{-k}$ . On the other hand, it follows by the Cauchy–Schwarz inequality and the fact that $\sum _x \mu _n^{(t)}(x) = 1$ that $1 \leqslant |\operatorname {Supp}(\mu _n^{(t)})| \sum _x \mu _n^{(t)}(x)^2$ , and so $|\operatorname {Supp}(\mu _n^{(t)})| \geqslant 2^{-5} b^{-10k^2}N^k$ . Thus, since $\mu _n^{(t)}$ is supported on the $t$ -fold sumset of $\mathcal {S}^k \cap [0, N^k)$ , we see that $|t \mathcal {S}^k \cap [0, tN^k)| \geqslant 2^{-5} b^{-10k^2}N^k$ . Applying Theorem3.1 with $X = tN^k$ and $r = 2^8 b^{19k^2}$ , we see that $4rt \mathcal {S}^k \cap [0, 4rt N^k)$ contains an arithmetic progression $P$ of common difference $\lt r$ and length $|P| \geqslant L := 2^{-15} b^{-29k^2} N^k$ .

Since $d_1^k$ and $d_2^k$ are coprime, every number greater than or equal to $(d_1^k - 1)(d_2^k - 1) \lt b^{2k} \lt r$ is a non-negative integer combination of these numbers. Therefore, it is certainly the case that $2r \mathcal {S}^k$ contains $[r, 2r)$ . Since the common difference of $P$ is less than $r$ , $P + [r, 2r)$ contains a discrete interval $I$ of length $\geqslant L$ . This interval is therefore contained in $(4rt + 2r) \mathcal {S}^k \subset 8rt \mathcal {S}^k$ . Note that by construction $I \subset [0, 8rt N^k)$ .

Apply Lemma 3.2, taking $X = X(n) = 8rt N^k$ , $\eta = \frac {L}{X} = 2^{-29} b^{-57k^2}$ and $K = 4b^{k^2}$ . Since $\mathcal {S}$ contains $0$ , we see that $\lfloor 2K/\eta ^2\rfloor 8rt \mathcal {S}^k = 2^{75}b^{142k^2} \mathcal {S}^k$ contains the interval $I_n := [\frac {4}{\eta } X(n), \frac {K}{\eta } X(n)]$ . Remember that here $n$ is any sufficiently large multiple of $k$ . By the choice of $K$ , $\frac {K}{\eta } X(n) = \frac {4}{\eta } X(n+k)$ , and so these intervals overlap. Thus, $\bigcup _{n} I_n$ consists of all sufficiently large integers, and hence, so does $2^{75}b^{142k^2} \mathcal {S}^k$ . Finally, one may note that $2^{75} \lt b^{12k^2}$ for $b \geqslant 3$ and $k \geqslant 2$ .

Proof of Proposition 4.1. Set $Q := 2k! b^{k (k - 1)/2 +1}$ . Note that, since $2k! \leqslant 2^{k^2/2} \leqslant b^{k^2/2}$ for all $b,k \geqslant 2$ , we have $Q \leqslant b^{k^2}$ . By Dirichlet’s theorem, there is some positive integer $q \leqslant Q$ and an $a$ , coprime to $q$ , such that $|\theta - a/q| \leqslant 1/qQ$ . Set $\eta := \theta - a/q$ , thus $|\eta | \leqslant 1/qQ$ . There is a unique integer $j$ such that

(4.1) \begin{equation} \tfrac {1}{2bq} \lt |(d_2 - d_1) k! b^j \eta | \leqslant \tfrac {1}{2q}. \end{equation}

Now if we had $j \lt k(k-1)/2$ then

\begin{equation*} |(d_2 - d_1) k! b^j \eta | \leqslant (b-1)k! b^{1/2k(k-1) - 1} |\eta | \lt k! b^{1/2k(k-1)}/qQ = 1/2bq,\end{equation*}

contrary to (4.1). If $j \gt kn - k(k+1)/2$ then, using (4.1),

\begin{equation*} \Vert \theta q\Vert = |\eta q| \leqslant \big |2 (d_2 - d_1) k! b^j \big |^{-1} \leqslant (2k!)^{-1} b^{1/2k(k+1) - 1} N^{-k} ,\end{equation*}

in which case the conclusion of the proposition is satisfied.

Suppose, then, that $k(k-1)/2 \leqslant j \leqslant kn -k(k+1)/2$ . Then there is a set $I \subset [0,n)$ , $|I| = k$ , such that $j = \sum _{i \in I} i$ . As usual, write $\mathbf {x} = (x_i)_{i \in [0,n)}$ . It is convenient to write $\mathbf {x}_I$ for the variables $x_i$ , $i \in I$ and $\mathbf {x}_{[0,n) \setminus I}$ for the other variables. For any fixed choice of $\mathbf {x}_{[0,n) \setminus I}$ , we can write, setting $u := {d_1(b^n - 1)}/{b - 1}$ ,

\begin{align*} \big(u + (d_2 - d_1) L_b(\mathbf {x})\big)^k & = \bigg(u + (d_2 - d_1)\sum _{i \in [0, n)} x_i b^i \bigg)^k \\ & = (d_2 - d_1)k! b^j \prod _{i \in I} x_i + \sum _{i \in I} \psi _i(\mathbf {x}_{[0,n) \setminus I}; \mathbf {x}_I) \end{align*}

for some functions $\psi _i$ , where $\psi _i$ does not depend on $x_i$ . It follows that

\begin{align*} |\widehat {\mu }_n(\theta )| & = \big | \mathbf {E}_{\mathbf {x} \in \{0,1\}^{[0,n)}} e \big(\big(u + (d_2 - d_1) L_b(\mathbf {x})\big)^k \big)\big | \\ & \leqslant \mathbf {E}_{\mathbf {x}_{[0,n) \setminus I} \in \{0,1\}^{[0,n) \setminus I}} \bigg | \mathbf {E}_{\mathbf {x}_I \in \{0,1\}^I} \prod _{i \in I} \Psi _i(\mathbf {x}_{[0,n) \setminus I}; \mathbf {x}_I) e \bigg( (d_2 - d_1) k! b^j \theta \prod _{i \in I} x_i\bigg) \bigg |, \end{align*}

where $\Psi _i := e(\psi _i)$ is a 1-bounded function, not depending on $x_i$ . By Proposition A.2 (and the accompanying definition of the Box norm, Definition A.1) it follows that

\begin{equation*} |\widehat {\mu _n}(\theta )|^{2^k} \leqslant \mathbf {E}_{\mathbf {x}_I, \mathbf {x}^{\prime}_I \in \{0,1\}^I} e\bigg((d_2 - d_1) k! b^j \theta \prod _{i \in I} (x_i - x^{\prime}_i)\bigg).\end{equation*}

(The right-hand side here is automatically a non-negative real number). On the right, we now bound all the terms trivially (by $1$ ) except for two: the term with $x_i = x^{\prime}_i = 0$ for all $i \in I$ , and the term with $x_i = 1$ and $x^{\prime}_i = 0$ for all $i \in I$ . This gives, using the inequality $2 - |1 + e(t)| = 4 \sin ^2 {\pi \Vert t \Vert }/{2} \geqslant 4 \Vert t\Vert ^2$ ,

(4.2) \begin{align} \nonumber |\widehat {\mu _n}(\theta )|^{2^k} & \leqslant 1 - \tfrac {2}{4^k} + \tfrac {1}{4^k} |1 + e((d_2 - d_1)k! b^j \theta )| \\ & \leqslant 1 - 2^{2 - 2k} \Vert (d_2 - d_1) k! b^j \theta \Vert ^2. \end{align}

There are now two slightly different cases, according to whether or not $q \mid (d_2 - d_1) k! b^j a$ . If this is the case, then by (4.1),

\begin{equation*} \Vert (d_2 - d_1) k! b^j \theta \Vert = | (d_2 - d_1) k! b^j \eta | \geqslant 1/2bq.\end{equation*}

If, on the other hand, $q \nmid (d_2 - d_1) k! b^j a$ then by (4.1) we have

\begin{equation*} \Vert (d_2 - d_1) k! b^j \theta \Vert \geqslant \tfrac {1}{q} - | (d_2 - d_1) k! b^j \eta | \geqslant \tfrac {1}{2q}.\end{equation*}

In both cases, $\Vert (d_2 - d_1) k! b^j \theta \Vert \geqslant 1/2bQ = (4k!)^{-1} b^{-2-\frac {1}{2}k(k-1)}$ . It follows from (4.2) that

\begin{equation*} |\widehat {\mu _n}(\theta )| \leqslant \big(1 - 2^{2 - 2k} (4k!)^{-2} b^{-4 - k(k-1)}\big)^{1/2^k} \lt 1 - \tfrac {1}{4}b^{-3k^2}, \end{equation*}

that is to say, the hypothesis of the proposition is not satisfied. Here, the second inequality follows from the Bernoulli inequality $(1 - x)^{1/2^k} \leqslant 1 - x/2^k$ and the crude bounds $k! \leqslant b^{k^2/4}$ , $2^{3k} \leqslant b^{2k}$ , both valid for $b \geqslant 3$ and $k \geqslant 2$ .

Proof. By (1.2) and the definition of the measure $\mu _n$ , we have

(5.2) \begin{equation} \widehat {\mu _n}(\theta ) = \mathbf {E}_{\mathbf {x} \in \{0,1\}^n} e \big(\theta \big(u + (d_2 - d_1) L_b(\mathbf {x})\big)^k\big), \end{equation}

where $u := d_1(b^n - 1)/(b - 1)$ . The first stage of the decoupling procedure is to split the variables $\mathbf {x}$ into $k$ disjoint subsets of size $n/k$ . If $\mathbf {x} = (x_i)_{i \in [0, n)} \in \{0,1\}^{[0, n)}$ , for each $j \in [0, k)$ , we write $\mathbf {x}^{(j)} = (x_{ik + j})_{i \in [0, n/k)} \in \{0,1\}^{[0, n/k)}$ . Then

(5.3) \begin{equation} L_b(\mathbf {x}) = \sum _{j \in [0, k)} b^j L_{b^k}(\mathbf {x}^{(j)}). \end{equation}

(Note here that $L_b$ is defined on $\{0,1\}^{[0, n)}$ , whereas $L_{b^k}$ is defined on $\{0,1\}^{[0, n/k)}$ .) By (5.2) we have

\begin{equation*} \widehat {\mu _n}(\theta ) = \mathbf {E}_{\mathbf {x}^{(0)},\dots , \mathbf {x}^{(k-1)} \in \{0,1\}^{[0, n/k)}} e\bigg (\theta \bigg( u + (d_2 - d_1)\sum _{i \in [0, k)} b^i L_{b^k}(\mathbf {x}^{(i)})\bigg)^k\bigg ).\end{equation*}

Expanding out the $k$ th power and collecting terms, this can be written as

\begin{equation*} \mathbf {E}_{\mathbf {x}^{(0)},\dots , \mathbf {x}^{(k-1)} \in \{0,1\}^{[0, n/k)}} \bigg(\prod _{j \in [0, k)} \Psi _j(\mathbf {x}) \bigg)e \bigg(\theta q_0 \prod _{i \in [0, k)} L_{b^k}(\mathbf {x}^{(i)})\bigg),\end{equation*}

where

\begin{equation*} q_0 := k!(d_2 - d_1)^k b^{k(k-1)/2}\end{equation*}

and $\Psi _j$ is some $1$ -bounded function of the variables $\mathbf {x}^{(i)}$ , $i \in [0, k) \setminus \{j\}$ , the precise nature of which does not concern us. The inequality $q_0 \leqslant b^{k^2}$ follows using $|d_1 - d_1| \leqslant b$ and the estimate $k! \leqslant 3^{k(k-1)/2}$ , since $b \geqslant 3$ .

One may now apply the Cauchy–Schwarz inequality $k$ times to eliminate the functions $\Psi _j$ in turn. This procedure is well known from the theory of hypergraph regularity [Reference GowersGow07] or from the proofs of so-called generalised von Neumann theorems in additive combinatorics [Reference Green and TaoGT10]. For a detailed statement, see Proposition A.2. From this it follows that

\begin{equation*} \delta ^{2^k} \leqslant \mathbf {E} e \bigg(\theta q_0 \sum _{\omega \in \{0,1\}^{[0, k)}}(-1)^{|\omega |} \prod _{i \in [0, k)} L_{b^k}(\mathbf {x}_{\omega _i}^{(i)} )\bigg),\end{equation*}

where the average is over $\mathbf {x}^{(0)}_0, \dots, \mathbf {x}_0^{(k-1)}, \mathbf {x}^{(0)}_1, \dots, \mathbf {x}_1^{(k-1)} \in \{0,1\}^{[0, n/k)}$ , and we write $\omega = (\omega _i)_{i \in [0,k)}$ and $|\omega | = \sum _{i = 1}^k |\omega _i|$ . By pigeonhole there is some choice of $\mathbf {x}^{(0)}_1,\dots , \mathbf {x}^{(k-1)}_1$ such that the remaining average over $\mathbf {x}^{(0)}_0, \dots, \mathbf {x}_0^{(k-1)}$ is at least $\delta ^{2^k}$ . This may be written as

\begin{equation*} \delta ^{2^k} \leqslant \big | \mathbf {E}_{\mathbf {x}^{(0)},\dots , \mathbf {x}^{(k-1)} \in \{0,1\}^{[0, n/k)}} e \bigg(\theta q_0 \prod _{i \in [0, k)} (L_{b^k}(\mathbf {x}^{(i)}) + t_i) \bigg)\big |, \end{equation*}

where $t_i := -L_{b^k}(\mathbf {x}_1^{(i)})$ . It follows that, for at least ${1}/{2}\delta ^{2^k} 2^{(k-1)n/k}$ choices of $\mathbf {x}^{(1)},\dots , \mathbf {x}^{(k-1)} \in \{0,1\}^{[0, n/k)}$ , we have

(5.4) \begin{equation} \big | \mathbf {E}_{\mathbf {x}^{(0)}\in \{0,1\}^{[0, n/k)}} e \Bigg(\theta q_0 L_{b^k}(\mathbf {x}^{(0)}) \prod _{i = 1}^{k-1} (L_{b^k}(\mathbf {x}^{(i)}) + t_i) \Bigg)\big | \geqslant \delta ^{2^k}/2. \end{equation}

Let $\alpha \in {\mathbf {R}}/\mathbf {Z}$ be arbitrary. Note that

\begin{align*} \tilde {\text {w}}_n(\alpha ) & = \sum _{i \in [0, n-1)} \Vert \alpha b^{i} \Vert ^2 = \sum _{j \in [0, k)} \sum _{i \in [0, n/k)} \Vert \alpha b^{j + ik}\Vert ^2 \\ & \leqslant \bigg(\sum _{j \in [0, k)} b^{2j} \bigg) \sum _{i \in [0, n/k)} \Vert \alpha b^{ik}\Vert ^2 \leqslant b^{2k} \sum _{i \in [0, n/k)} \Vert \alpha b^{ik}\Vert ^2. \end{align*}

Therefore, using the inequality $|1 + e(t)| = 2|\cos (\pi t)| \leqslant 2 \exp (-\Vert t \Vert ^2)$ , we have

\begin{align*} \big |\mathbf {E}_{\mathbf {y} \in \{0,1\}^{[0, n/k)}} e(\alpha L_{b^k}(\mathbf {y}))\big | &= \prod _{i \in [0, n/k)} \bigg | \frac {1 + e(\alpha b^{ik}) }{2} \bigg | \\ & \leqslant \exp\! \bigg({-}\sum _{i \in [0, n/k)} \Vert \alpha b^{ik} \Vert ^2 \bigg) \leqslant \exp \!\big({-}b^{-2k} \tilde {\text {w}}_n(\alpha )\big). \end{align*}

Combining this with (5.4), Proposition 5.2 follows.

Proof. The conclusion is invariant under applying any of the four involutions $(u,v) \mapsto (\pm u, \pm v)$ to $\Omega$ , so without loss of generality we may suppose that $\Omega \cap ([0, U] \times [0, V])$ has size at least $\varepsilon UV/4$ . It then follows that $\Omega \cap ([\varepsilon U/32, U] \times [\varepsilon V/32, V])$ has size at least $\varepsilon UV/8$ . Covering this box by disjoint dyadic boxes $[2^i, 2^{i+1})\times [2^j, 2^{j+1})$ contained in $[\varepsilon U/64, 2U] \times [\varepsilon V/64, 2V]$ , we see that there is some dyadic box $[U^{\prime}, 2U^{\prime}) \times [V^{\prime}, 2V^{\prime})$ , $\varepsilon U/64 \leqslant U^{\prime} \leqslant U$ , $\varepsilon V/64 \leqslant V^{\prime} \leqslant V$ , on which the density of $\Omega$ is at least $\varepsilon /32$ . Without loss of generality, suppose that $U^{\prime} \leqslant V^{\prime}$ , and set $X := U^{\prime} V^{\prime} \geqslant 1$ . Set $\Omega ^{\prime}:= \Omega \cap \big([U^{\prime}, 2U^{\prime}) \times [V^{\prime}, 2V^{\prime})\big)$ .

For $n \in \mathbf {Z}$ , denote by $r(n)$ the number of representations of $n$ as $u_1 v_1 + u_2 v_2$ with $(u_1, v_1)$ and $(u_2, v_2)$ in $\Omega ^{\prime}$ , and by $\tilde r(n)$ the number of representations as $u_1 v_1 + u_2 v_2$ with $(u_1, v_1), (u_2, v_2) \in [U^{\prime}, 2U^{\prime}) \times [V^{\prime}, 2V^{\prime})$ . Thus, $r(n) \leqslant \tilde r(n)$ . By the Cauchy–Schwarz inequality,

(6.1) \begin{equation} (\varepsilon X/32)^4 \leqslant |\Omega ^{\prime}|^4 = \bigg(\sum _n r(n)\bigg)^2 \leqslant |\operatorname {Supp}(r)| \sum _n \tilde r(n)^2. \end{equation}

Now, denoting by $\nu (n)$ the number of divisors of $n$ in the range $[U^{\prime}, 2U^{\prime})$ ,

\begin{equation*} \tilde r(n) \leqslant \sum _{m \leqslant 4X} \nu (m) \nu (n - m) = \sum _{\substack {d, e \in [U^{\prime}, 2U^{\prime}) \\ (d,e) | n}} \sum _{\substack {m \leqslant 4X \\ d \mid m, e \mid n - m}} 1 \leqslant 8X \!\!\!\sum _{\substack {d,e \in [U^{\prime}, 2U^{\prime}) \\ (d,e) | n}} \tfrac {1}{[d,e]} .\end{equation*}

Here, in the last step we used the fact that the set of $m$ satisfying $d \mid m$ and $e \mid n - m$ is a single residue class modulo $[d,e]$ (the lowest common multiple of $d$ and $e$ ), whose intersection with the interval $[1, 4X)$ has size $\leqslant 1 + 4X /[d, e] \leqslant 8 X/[d,e]$ since $[d, e] \leqslant (2U^{\prime})^2 \leqslant 4X$ .

Setting $\delta := (d,e)$ and $d = \delta d^{\prime}$ , $e = \delta e^{\prime}$ , so that $[d,e] = \delta d^{\prime} e^{\prime}$ , it then follows that

\begin{equation*} \tilde r(n) \leqslant 8 X \sum _{ \delta \mid n} \tfrac {1}{\delta } \sum _{d^{\prime}, e^{\prime} \in [U^{\prime}/\delta , 2U^{\prime}/\delta )} \tfrac {1}{d^{\prime} e^{\prime}} \leqslant 8 X \sum _{\delta | n} \tfrac {1}{\delta }.\end{equation*}

Since $\tilde r(n)$ is supported where $n \leqslant 8X$ , we have

\begin{align*} \sum _{n} \tilde r(n)^2 & \leqslant (8X)^2 \sum _{n \leqslant 8X} \bigg(\sum _{\delta | n} \frac {1}{\delta }\bigg)^2 = (8X)^2 \sum _{\delta _1, \delta _2 \leqslant 8X} \frac {1}{\delta _1 \delta _2} \sum _{n \leqslant 8X} 1_{[\delta _1, \delta _2] | n} \\ & \leqslant (8X)^2 \sum _{\delta _1, \delta _2 \leqslant 8X} \tfrac {1}{\delta _1 \delta _2} \bigg(\frac {8X}{[\delta _1, \delta _2]} + 1\bigg) . \end{align*}

The contribution from the $+1$ term is $\leqslant (8X)^2 (1 + \log 8X)^2 \lt 2^{10} X^3$ , since $X \geqslant 1$ . Since $[\delta _1, \delta _2] \geqslant \sqrt {\delta _1 \delta _2}$ , the contribution from the main term is $\leqslant 2^8 \zeta (\frac {3}{2})^2 X^3 \lt 2^{11}X^3$ . It follows that $\sum _n \tilde r(n)^2 \leqslant 2^{12} X^3$ . Comparing with (6.1), we obtain $|\operatorname {Supp}(r)| \geqslant 2^{-32} \varepsilon ^4 X \geqslant 2^{-44} \varepsilon ^6 UV$ . Since we are assuming that $\varepsilon \leqslant 2^{-44}$ , this is at least $\varepsilon ^7 UV$ , and the proof is complete.

Proof. We have

\begin{equation*} \sum _{h_2,\dots , h_t} \bigg( \sum _x 1_{S_1}(x) 1_{S_2} (x+h_2) \cdots 1_{S_t} (x + h_t) \bigg) = \prod _{i = 1}^t |S_i| \geqslant \eta ^t X^t.\end{equation*}

Since the $h_i$ may be restricted to range over $[-2X, 2X]$ , which contains at most $5X$ integers, there is some choice of $h_2,\dots , h_t$ so that $\sum _x 1_{S_1}(x) 1_{S_2} (x+h_2) \cdots 1_{S_t} (x + h_t) \geqslant (\eta /5)^t X$ . That is, there is a set $S$ , $|S| \geqslant (\eta /5)^t X$ , such that $S \subseteq S_1 \cap (S_2 - h_2) \cap \cdots \cap (S_t - h_t)$ . But then $S - S \subseteq \bigcap _{i=1}^t (S_i - S_i)$ , and the result is proved.

Proof. It is convenient to write $\phi _j(\mathbf {y}) := L_d(\mathbf {y}) + t_j$ , $j = 1,\dots , r$ . Note, for further use, the containment

(6.2) \begin{equation} \phi _j(\{0,1\}^{[0, m)}) \subset [-2N, 2N], \end{equation}

which follows from the fact that $|t_j| \leqslant N$ .

Turning to the proof, we proceed by induction on $r$ . In the case $r = 1$ , we can apply Theorem2.3. Noting that $L_d(\{0,1,\dots , d-1\}^m) = \{0,1,\dots , N - 1\}$ , we see that at least $\alpha ^{\log _2 d} N$ elements of $\{0,1,\dots , N-1\}$ are the sum of $d-1$ elements $L_d(\mathbf {y}_1)$ , $\mathbf {y}_1 \in A$ . Since, for any $\mathbf {y}_1^{(1)},\dots , \mathbf {y}_1^{(d-1)} \in A$ , we have

\begin{equation*} \sum _{i = 1}^{d-1} \phi _1(\mathbf {y}_1^{(i)}) = \sum _{i = 1}^{d-1} L_d(\mathbf {y}_1^{(i)}) + (d - 1) t_1,\end{equation*}

we see that at least $\alpha ^{\log _2 d} N$ elements of $[-d N, d N]$ are the sum of $d - 1$ elements $\phi _1(\mathbf {y}_1)$ , $\mathbf {y}_1 \in A$ , which gives the required result in this case.

Now suppose that $r \geqslant 2$ , and that we have proven the result for smaller values of $r$ . For each $\mathbf {y}_r \in \{0,1\}^{[0,m)}$ , denote by $A(\mathbf {y}_r) \subseteq (\{0,1\}^{[0,m)})^{r-1}$ the maximal set such that $A(\mathbf {y}_r) \times \{\mathbf {y}_r\} \subseteq A$ . By a simple averaging argument there is a set $Y$ of at least $(\alpha /2) 2^{m}$ values of $\mathbf {y}_r$ such that $|A(\mathbf {y}_r)| \geqslant (\alpha /2) 2^{m(r-1)}$ . By the inductive hypothesis, for each $\mathbf {y}_r \in Y$ , there is a set

(6.3) \begin{equation} B(\mathbf {y}_r) \subseteq [- (8d N)^{r-1}, (8d N)^{r-1}], \end{equation}

with

(6.4) \begin{equation} |B(\mathbf {y}_r)| \geqslant (\alpha /4)^{(32d)^{r-1}} N^{r-1}, \end{equation}

such that everything in $B(\mathbf {y}_r)$ is a $\pm$ sum of at most $(4d)^{r-1}$ elements $\phi _1(\mathbf {y}_1) \cdots \phi _{r-1}(\mathbf {y}_{r-1})$ with $(\mathbf {y}_1,\dots , \mathbf {y}_{r-1}) \in A(\mathbf {y}_r)$ . Observe that everything in $(B(\mathbf {y}_r) - B(\mathbf {y}_r)) \phi _r(\mathbf {y}_r)$ is then a $\pm$ combination of at most $2 (4d)^{r-1}$ elements $\phi _1(\mathbf {y}_1) \cdots \phi _{r}(\mathbf {y}_r)$ with $(\mathbf {y}_1,\dots , \mathbf {y}_r) \in A$ .

Suppose now that $z \in (d - 1) \phi _r(Y) = \phi _r(Y) + \cdots + \phi _r(Y)$ . Note that, by (6.2),

(6.5) \begin{equation} |z| \lt 2dN. \end{equation}

For each such $z$ , pick a representation $z = \phi _r(\mathbf {y}_r^{(1)}) + \cdots + \phi _r(\mathbf {y}_r^{(d-1)})$ with $\mathbf {y}_r^{(i)} \in Y$ for $i = 1,\dots , d-1$ , and define $S(z) := \bigcap _{i=1}^{d-1} (B(\mathbf {y}_r^{(i)}) - B(\mathbf {y}_r^{(i)}))$ . By (6.3), (6.4) and Lemma 6.3 (taking $X := (8dN)^{r-1}$ , $\eta := (8d)^{-(r-1)} (\alpha /4)^{(32d)^{r-1}}$ and $t := d - 1$ in that lemma), we have

(6.6) \begin{align} \nonumber |S(z)| & \geqslant 5^{-(d-1)} (8d)^{-(r-1)(d-2)} (\alpha /4)^{(32 d)^{r-1} (d - 1)} N^{r-1} \\ & \geqslant (\alpha /2)^{4d (32d)^{r-1}} N^{r-1}. \end{align}

Here, the second bound is crude and uses the inequality

\begin{equation*} (2d+2)(32d)^{r-1} \geqslant (d-1)\log _2 5 + (r-1)(d-2) \log _2(8d),\end{equation*}

valid for $d \geqslant 2$ and $r \geqslant 1$ (by a large margin if $r \gt 1$ ).

Note that everything in $S(z) z$ is a $\pm$ combination of at most $2(d-1) (4d)^{r-1}$ elements $\phi _1(\mathbf {y}_1) \cdots \phi _r( \mathbf {y}_r)$ with $(\mathbf {y}_1,\dots , \mathbf {y}_r) \in A$ . Set $\Omega := \bigcup _{z \in (d-1) \phi _r(Y)} (S(z) \times \{z\})$ . Then $\Omega \subset [-U, U] \times [-V, V]$ where by (6.3) and (6.5) we can take $U := 2(8dN)^{r-1}$ and $V := 2 dN$ . Now by Theorem2.3, and recalling that $|Y| \geqslant (\alpha /2) 2^m$ , we have $| (d - 1) \phi _r(Y) | = |(d - 1) L_d(Y) | \geqslant (\alpha /2)^{\log _2 d} N$ . From this and (6.6), we have $|\Omega | \geqslant (\alpha /2)^{4d(32d)^{r-1} + \log _2 d} N^r$ . Thus, noting that $UV = 2^{3r -1 + r \log _2 d} N^r$ , it follows that $|\Omega | \geqslant \varepsilon UV$ with

(6.7) \begin{equation} \varepsilon := (\alpha /2)^{4d(32d)^{r-1} + 3r + (r+1) \log _2 d}. \end{equation}

Now we aim to apply Lemma 6.2. For such an application to be valid, we require $\varepsilon \lt 2^{-44}$ , which is comfortably a consequence of (6.7). We also need that $U,V \geqslant 64/\varepsilon$ , which follows from (6.7) and the lower bound on $N$ in the hypotheses of the proposition. Note that if $(u_1, v_1) = (S(z), z)$ , $(u_2, v_2) = (S(z^{\prime}), z^{\prime}) \in \Omega$ then $u_1 v_1 + u_2 v_2 = S(z) z^{\prime} + S(z^{\prime}) z^{\prime}$ is a $\pm$ combination of at most $(4d)^r$ elements $\phi _1(\mathbf {y}_1) \cdots \phi _r( \mathbf {y}_r)$ with $(\mathbf {y}_1,\dots , \mathbf {y}_r) \in A$ , and by Lemma 6.2 there are $\geqslant \varepsilon ^7 UV \gt \varepsilon ^7 N^r$ such elements. To conclude the argument, we need only check that $\varepsilon ^7 \geqslant (\alpha /2)^{(32 d)^r}$ , which, using (6.7), comes down to checking that $4d (32 d)^{r-1} \geqslant 7(3r + (r + 1) \log _2 d)$ , which is comfortably true for all $d, r \geqslant 2$ .

Proof of Proposition 6.1. In the following proof we suppress a number of short calculations, showing that various constants are bounded by $C = b^{7k^2/2}$ . These calculations are all simple finger exercises using the assumption that $b \geqslant 3$ and $k \geqslant 2$ .

First apply Proposition 5.2. As in the statement of that Proposition, we obtain $t_1,\dots , t_{k-1} \in \mathbf {Z}$ , $|t_j| \leqslant N$ such that, for at least ${1}/{2} \delta ^{2^k} 2^{(k-1) n/k}$ choices of $\mathbf {x}^{(1)},\dots , \mathbf {x}^{(k-1)} \in \{0,1\}^{[0, n/k)}$ , we have

(6.8) \begin{equation} \tilde {\text {w}}_n \Bigg( \theta q_0 \prod _{i=1}^{k-1} (L_{b^k}(\mathbf {x}^{(i)}) + t_i) \Bigg) \leqslant 2^k b^{2k} \log (2/\delta ) \end{equation}

for some positive integer $q_0 \leqslant b^{k^2}$ . (For the definition of $\tilde {\text {w}}_n$ , see Definition 5.1.) To this conclusion, we apply Proposition 6.4, taking $m := n/k$ , $r := k-1$ and $d := b^k$ in that proposition, and taking $A$ to be the set of all $(\mathbf {x}^{(1)},\dots , \mathbf {x}^{(k-1)})$ as just described; thus, we may take $\alpha := \delta ^{2^k}/2$ . Note that $N = d^m = b^n$ is the same quantity. The reader may check that the lower bound on $N$ required for this application of Proposition 6.4 is a consequence of the assumption on $N$ in Proposition 6.1.

We conclude that at least $(\delta ^{2^k}/4)^{(32 b^k)^{k-1}} N^{k-1} \gt (\delta /2)^{C} N^{k-1}$ integers $x$ with $|x| \leqslant (8b^k N)^{k-1}$ may be written as a $\pm$ sum of at most $(4b^k)^{k-1}$ numbers of the form $\prod _{i=1}^{k-1} (L_{b^k}(\mathbf {x}^{(i)}) + t_i)$ , with $(\mathbf {x}^{(1)},\dots , \mathbf {x}^{(k-1)}) \in A$ . By (6.8), the fact that $\tilde {\text {w}}_n(-\alpha ) = \tilde {\text {w}}_n(\alpha )$ , as well as the (easily verified) subadditivity property

\begin{equation*} \tilde {\text {w}}_n(\alpha _1 + \cdots + \alpha _s) \leqslant s (\tilde {\text {w}}_n(\alpha _1) + \dots + \tilde {\text {w}}_n(\alpha _s)),\end{equation*}

we see that, for all such $x$ , we have

\begin{equation*} \tilde {\text {w}}_n(\theta q_0 x) \leqslant (4b^k)^{2(k-1)} 2^k b^{2k} \log (2/\delta ) \lt C \log (2/\delta ).\end{equation*}

Finally, note that for all these $x$ , we have $|q_0 x| \leqslant b^{k^2}(8 b^k)^{k-1} N^{k-1}$ , which is less than $C N^{k-1}$ . This concludes the proof.

Proof. Let the centred expansion of $\alpha {(\operatorname {mod}\, 1)}$ be (7.1), and suppose that $\alpha _i$ is a nonzero digit. We have $\alpha b^{i-1} \equiv \sum _{j \geqslant 0} \alpha _{i +j} b^{-j - 1} {(\operatorname {mod}\, 1)}$ . However,

\begin{equation*} \bigg|\sum _{j \geqslant 0} \alpha _{i +j} b^{-j - 1} \bigg| \leqslant \tfrac {b}{2} \sum _{j \geqslant 0} b^{-j-1} = \tfrac {b}{2(b-1)} \leqslant \tfrac {3}{4},\end{equation*}

and, since $\alpha _i \neq 0$ ,

\begin{equation*} \bigg|\sum _{j \geqslant 0} \alpha _{i +j} b^{-j - 1} \bigg| \geqslant \tfrac {1}{b} - \tfrac {b}{2}\sum _{j \geqslant 1} b^{-j-1} = \tfrac {b-2}{2b(b-1)} \geqslant \tfrac {1}{4b}.\end{equation*}

Thus, $\Vert \alpha b^{i-1} \Vert \geqslant 1/4b$ and the upper bound follows.

The lower bound is not needed elsewhere in the paper, but we sketch the proof for completeness. Let $I := \{ i : \alpha _i \neq 0\}$ . Given $j$ , denote by $i(j)$ the distance from $j$ to the smallest element of $I$ that is greater than $j$ . Then

\begin{equation*} \Vert \alpha b^j \Vert = \bigg \Vert \sum _{i \in I, i \gt j} \alpha _i b^{-i + j} \bigg \Vert \leqslant \tfrac {b}{2} \sum _{m \geqslant i(j)} b^{-m} = \tfrac {b^2}{2(b-1)} b^{-i(j)} .\end{equation*}

Now square this and sum over $j$ , and use the fact that $\# \{ j : i(j) = i\} \leqslant |I| = \operatorname {w}_n(\alpha )$ for all $i$ .

Proof. We proceed by induction on $\sum _{j = 1}^4 |A_j| + \sum _{j = 1}^4 r_j$ , the result being obvious when this quantity is zero. Suppose now that $\sum _{j = 1}^4 |A_j| + \sum _{j = 1}^4 r_j = n \gt 0$ and that the result has been proven for all smaller values of $n$ . If any of the $A_j$ are empty, or if $A_1 = A_2 = A_3 = A_4 = \{0\}$ , the result is obvious.

Suppose this is not the case, but that $b$ divides every element of $\bigcup _{j = 1}^4 A_j$ . Let $b^m$ be the largest power of $b$ that divides every element of $\bigcup _{j = 1}^4 A_j$ , this being well defined since this set contains at least one nonzero element. Then, if the number of quadruples in $A_1 \times A_2 \times A_3 \times A_4$ with $a_1 + a_2 = a_3 + a_4 + e$ is nonzero, we must have $e = 0$ , and the number of such quadruples is the same as the number in ${1}/{b^m}A_1 \times {1}/{b^m} A_2 \times {1}/{b^m} A_3 \times {1}/{b^m} A_4$ . Thus, replacing $A_j$ by $\frac {1}{b^m} A_j$ , we may assume that not all the elements of $\bigcup _{j=1}^4 A_j$ are divisible by $b$ .

For each $j \in \{1,2,3,4\}$ and for each $i \in (-{b}/{2}, {b}/{2}]$ , write $A_j^{(i)}$ for the set of $x \in A_j$ whose first digit $x_0$ (in the centred base $b$ expansion (7.2)) is $i$ . Write $\alpha _j(i)$ for the relative density of $A_j^{(i)}$ in $A_j$ , that is to say, $|A_j^{(i)}| = \alpha _j(i) |A_j|$ . Any quadruple $(a_1, a_2, a_3, a_4)$ with $a_1 + a_2 = a_3 + a_4 + e$ must have $a_j \in A_j^{(i_j)}$ , where $i_1 + i_2 \equiv i_3 + i_4 + e {(\operatorname {mod}\, b)}$ . Let us estimate the number of such quadruples $(a_1, a_2, a_3, a_4)$ for each quadruple $(i_1, i_2, i_3, i_4) \in (-{b}/{2}, {b}/{2}]^4$ satisfying this condition.

First note that $i_1 + i_2 = i_3 + i_4 + e + e^{\prime} b$ for some integer $e^{\prime}$ , where

\begin{equation*} |e^{\prime}| \leqslant \tfrac {1}{b} ( |i_1 + i_2 - i_3 - i_4| + |e|) \leqslant \tfrac {3(b-1)}{b} \lt b,\end{equation*}

where here we noted that $|i_1 - i_3|, |i_2 - i_4|, |e| \leqslant b - 1$ . We then have ${1}/{b}(a_1 - i_1) + {1}/{b}(a_2 - i_2) - {1}/{b}(a_3 - i_3) - {1}/{b}(a_4 - i_4) = - e^{\prime}$ . Now the set $A^{\prime}_j := \frac {1}{b}(A_j^{(i_j)} - i_j)$ is a finite set of integers, all of whose elements $x$ have $\operatorname {d}_b(x) \leqslant r^{\prime}_j := r_j - 1_{i_j \neq 0}$ . Note that $\sum _{j = 1}^4 |A^{\prime}_j| + \sum _{j = 1}^4 r^{\prime}_j \lt \sum _{j = 1}^4 |A_j| + \sum _{j = 1}^4 r_j$ ; if any $i_j$ is not zero, this follows from the fact that $r^{\prime}_j = r_j - 1$ , whereas if $i_1 = i_2 = i_3 = i_4 = 0$ we have $\sum _{j = 1}^4 |A^{\prime}_j| = \sum _{j = 1}^4 |A_j^{(0)}| \lt \sum _{j = 1}^4 |A_j|$ , since not every element of $\bigcup _{j = 1}^4 A_j$ is a multiple of $b$ .

It follows from the inductive hypothesis that the numbers of quadruples $(a_1, a_2, a_3, a_4)$ with $a_1 + a_2 = a_3 + a_4 + e$ , and with $a_j \in A_j^{(i_j)}$ , $j = 1,\dots , 4$ , is bounded above by

\begin{equation*} (2b)^{r_1 + r_2 + r_3 + r_4 - \#\{j : i_j \neq 0\}} \prod _{j = 1}^4 |A_j^{(i_j)}|^{1/2}.\end{equation*}

To complete the inductive step, it is therefore enough to show that

(7.3) \begin{equation} \sum _{i_1 + i_2 \equiv i_3 + i_4 + e {(\operatorname {mod}\, b)}} (2b)^{-\# \{j : i_j \neq 0\}} \prod _{j=1}^4 \alpha _j(i_j)^{1/2} \leqslant 1. \end{equation}

If $e \not \equiv 0 {(\operatorname {mod}\, b)}$ then we have $\# \{j : i_j \neq 0\} \geqslant 1$ for all $(i_1, i_2, i_3, i_4)$ in this sum, and moreover (where all congruences are $(\operatorname {mod}\, b)$ )

\begin{align*} & \sum _{i_1 + i_2 \equiv i_3 + i_4 + e} \prod _{j=1}^4 \alpha _j(i_j)^{1/2} \\ & \quad = \sum _{x \in \mathbf {Z}/b\mathbf {Z}} \bigg(\sum _{i_1 + i_2 \equiv x+e } \alpha _1(i_1)^{1/2} \alpha _2(i_2)^{1/2} \bigg) \bigg( \sum _{i_3 + i_4 \equiv x } \alpha _3(i_3)^{1/2} \alpha _4(i_4)^{1/2}\bigg) \\ & \quad \leqslant \sum _{x \in \mathbf {Z}/b\mathbf {Z}} \bigg(\sum _{i_1 + i_2 \equiv x+e } \frac {\alpha _1(i_1) + \alpha _2(i_2)}{2} \bigg) \bigg(\sum _{i_3 + i_4 \equiv x } \frac {\alpha _3(i_3) + \alpha _4(i_4)}{2}\bigg) = b, \end{align*}

since $\sum _i \alpha _j(i) = 1$ for each $j$ . Therefore, (7.3) holds in this case.

Suppose, then, that $e \equiv 0{(\operatorname {mod}\, b)}$ , which means that $e = 0$ . Then, if $i_1 + i_2 \equiv i_3 + i_4 {(\operatorname {mod}\, b)}$ we either have $(i_1, i_2, i_3, i_4) = (0,0,0,0)$ , or else $\# \{j : i_j \neq 0\} \geqslant 2$ , and so to establish (7.3) it suffices to show that

(7.4) \begin{equation} \prod _{j = 1}^4 \alpha _j(0)^{1/2} + (2b)^{-2} \sum _{\substack {i_1 + i_2 \equiv i_3 + i_4 {(\operatorname {mod}\, b)} \\ (i_1, i_2, i_3, i_4) \neq (0,0,0,0)}} \prod _{j=1}^4 \alpha _j(i_j)^{1/2} \leqslant 1. \end{equation}

Write $\varepsilon _j := 1 - \alpha _j(0)$ . We first estimate the contribution to the sum where none of $i_1, i_2, i_3, i_4$ are zero. We have, similarly to the above (and again with congruences being $(\operatorname {mod}\, b)$ ),

\begin{align*} &\!\!\!\! \sum _{\substack {i_1 + i_2 \equiv i_3 + i_4 \\ i_1i_2i_3i_4 \neq 0}} \prod _{j=1}^4 \alpha _j(i_j)^{1/2} \\ & \quad = \sum _{x \in \mathbf {Z}/b\mathbf {Z}} \left(\sum _{\substack {i_1 + i_2 \equiv x \\ i_1i_2 \neq 0}} \alpha _1(i_1)^{1/2} \alpha _2(i_2)^{1/2} \right) \left( \sum _{\substack {i_3 + i_4 \equiv x \\ i_3i_4 \neq 0}} \alpha _3(i_3)^{1/2} \alpha _4(i_4)^{1/2}\right) \\ & \quad \leqslant \sum _{x \in \mathbf {Z}/b\mathbf {Z}} \sum _{\substack {i_1 + i_2 \equiv x\\ i_1i_2 \neq 0}} \bigg(\frac {\alpha _1(i_1) + \alpha _2(i_2)}{2} \bigg)\left(\sum _{\substack {i_3 + i_4 \equiv x\\i_3i_4 \neq 0}} \frac {\alpha _3(i_3) + \alpha _4(i_4)}{2}\right) \\ & \quad \leqslant b \bigg(\frac {\varepsilon _1 + \varepsilon _2}{2}\bigg)\bigg(\frac {\varepsilon _3 + \varepsilon _4}{2}\bigg) \lt b\sum _{j = 1}^4 \varepsilon _j. \end{align*}

Next we estimate the contribution to the sum in (7.4) from the terms where at least one, but not all, of $i_1, i_2, i_3, i_4$ are zero. In each such term, at least two $i_j, i_{j^{\prime}}$ are not zero, say with $j \lt j^{\prime}$ . Fix a choice of $j,j^{\prime}$ . Then, for each $i_j, i_{j^{\prime}}$ , there are at most two choices of the other $i_{t}$ , $t \in \{1,2,3,4\} \setminus \{j,j^{\prime}\}$ , one of which must be zero and the other then being determined by the relation $i_1 + i_2 \equiv i_3 + i_4 {(\operatorname {mod}\, b)}$ . It follows that the contribution to the sum in (7.4) from this choice of $j,j^{\prime}$ is

\begin{equation*} \leqslant 2 \sum _{i_j, i_{j^{\prime}} \neq 0} \alpha _j(i_j)^{1/2} \alpha _{j^{\prime}}(i_{j^{\prime}})^{1/2} = 2 \bigg( \sum _{i \neq 0} \alpha _j(i)^{1/2} \bigg) \bigg ( \sum _{i \neq 0} \alpha _{j^{\prime}}(i)^{1/2}\bigg) \leqslant 2b \varepsilon _j^{1/2}\varepsilon _{j^{\prime}}^{1/2} \leqslant b\big(\varepsilon _j + \varepsilon _{j^{\prime}}\big) ,\end{equation*}

where in the middle step we used the Cauchy–Schwarz inequality and the fact that $\sum _{i \neq 0} \alpha _j(i) = \varepsilon _j$ . Summing over the six choices of $j,j^{\prime}$ gives an upper bound of $3b \sum _{j = 1}^4 \varepsilon _j$ . Putting all this together, we see that the left-hand side of (7.4) is bounded above by $\prod _{j = 1}^4 (1 - \varepsilon _j)^{1/2} + {1}/{b} \sum _{j = 1}^4 \varepsilon _j$ . Using $\prod _{j = 1}^4 (1 - \varepsilon _j)^{1/2} \leqslant 1 - {1}/{2}\sum _{j = 1}^4 \varepsilon _j$ , it follows that this is at most $1$ . This completes the proof of (7.4), and, hence, of Lemma 7.5.

Proof of Proposition 7.3. Consider the map $\psi : {\mathbf {R}} \rightarrow \mathbf {Z}$ defined as follows. If $\alpha {(\operatorname {mod}\, 1)}$ has centred base $b$ expansion as in (7.1), set $\psi (\alpha ) := \alpha _0 b^{n-1} + \dots + \alpha _{n-2} b + \alpha _{n-1}$ . Observe that

(7.5) \begin{equation} \operatorname {d}_b(\psi (\alpha )) = \operatorname {w}_n(\alpha ). \end{equation}

Note that

(7.6) \begin{equation} \Vert \alpha - b^{1-n} \psi (\alpha ) \Vert \leqslant \sum _{i \geqslant n} \tfrac {b}{2} b^{-i} \leqslant \tfrac {3}{4} b^{1-n}. \end{equation}

Thus, if $\alpha _1 + \alpha _2 = \alpha _3 + \alpha _4$ then

\begin{equation*} \Vert b^{1-n} (\psi (\alpha _1) + \psi (\alpha _2) - \psi (\alpha _3) - \psi (\alpha _4)) \Vert \leqslant 3 b^{1-n}.\end{equation*}

Note also that, since $\psi$ takes values in $\mathbf {Z} \cap [-\frac {3}{4} b^n, \frac {3}{4} b^n]$ , we have

\begin{equation*} |\psi (\alpha _1) + \psi (\alpha _2) - \psi (\alpha _3) - \psi (\alpha _4)| \leqslant 3 b^n.\end{equation*}

Now if $x \in \mathbf {Z}$ is an integer with $\Vert b^{1-n} x \Vert \leqslant 3 b^{1-n}$ and $|x| \leqslant 3 b^n$ then $x$ takes (at most) one of the $7(6b+1)$ values $\lambda b^{n-1} + \lambda ^{\prime}$ , $\lambda \in \{-3b, \dots , 3b\}$ , $\lambda ^{\prime} \in \{0, \pm 1, \pm 2, \pm 3\}$ . Denoting by $\Sigma$ the set consisting of these $7(6b + 1)$ values, we see that $\psi$ has the following almost-homomorphism property: if $\alpha _1 + \alpha _2 = \alpha _3 + \alpha _4$ then

\begin{equation*} \psi (\alpha _1) + \psi (\alpha _2) - \psi (\alpha _3) - \psi (\alpha _4) \in \Sigma .\end{equation*}

With parameters as in the statement of Proposition 7.3, consider the map $\pi : [-M, M] \rightarrow \mathbf {Z}$ given by

(7.7) \begin{equation} \pi (m) := \psi (\theta m). \end{equation}

Since the map $m \mapsto \theta m$ is a homomorphism from $\mathbf {Z}$ to $\mathbf {R}$ , we see that $\pi$ also has an almost-homomorphism property, namely that if $m_1 + m_2 = m_3 + m_4$ then

(7.8) \begin{equation} \pi (m_1) + \pi (m_2) - \pi (m_3) - \pi (m_4) \in \Sigma . \end{equation}

Denote by $\mathscr {M}$ the set of all $m \in [-M, M]$ such that $\operatorname {w}_n(\theta m) \leqslant r$ . Thus, by the assumptions of Proposition 7.3, $|\mathscr {M}| \geqslant \eta M$ . Denote $A := \pi (\mathscr {M})$ . By the definition (7.7) of $\pi$ , (7.5) and the definition of $\mathscr {M}$ , we see that $\operatorname {d}_b(a) \leqslant r$ for all $a \in A$ . For $a \in A$ , denote by $X_a := \pi ^{-1}(a) \cap \mathscr {M}$ the $\pi$ -fibre above $a$ . Decompose $A$ according to the dyadic size of these fibres, thus, for $j \in \mathbf {Z}_{\geqslant 0}$ set,

(7.9) \begin{equation} A_j := \{ a \in A : 2^{-j-1} M \lt |X_a| \leqslant 2^{-j} M\}. \end{equation}

Denote by $\mathscr {M}_j \subset \mathscr {M}$ the points of $\mathscr {M}$ lying above $A_j$ , that is to say, $\mathscr {M}_j := \bigcup _{a \in A_j} X_{a}$ . Define $\eta _j$ by $|\mathscr {M}_j| = \eta _j M$ . Since $\mathscr {M}$ is the disjoint union of the $\mathscr {M}_j$ , we have

(7.10) \begin{equation} \sum _j \eta _j \geqslant \eta . \end{equation}

By (7.9) we have $2^{-j-1} M |A_j| \leqslant |\mathscr {M}_j| \leqslant 2^{-j} M |A_j|$ , and so

(7.11) \begin{equation} 2^j \eta _j \leqslant |A_j| \leqslant 2^{j+1} \eta _j. \end{equation}

Now by a simple application of the Cauchy–Schwarz inequality any subset of $[-M, M]$ of size at least $\varepsilon M$ has at least $\varepsilon ^4 M^3/4$ additive quadruples. In particular, for any $j \in \mathbf {Z}_{\geqslant 0}$ , there are $\geqslant \eta _j^4 M^3/4$ additive quadruples in $\mathscr {M}_j$ . By (7.8), there is some $\sigma _j \in \Sigma$ such that, for $\geqslant 2^{-10} b^{-1}\eta _j^4 M^3$ additive quadruples in $\mathscr {M}_j$ , we have

(7.12) \begin{equation} \pi (m_1) + \pi (m_2) = \pi (m_3) + \pi (m_4) + \sigma _j. \end{equation}

For each $j$ , fix such a choice of $\sigma _j$ . Now the number of such quadruples with $\pi (m_i) = a_i$ for $i = 1,2,3,4$ is, for a fixed choice of $a_1,\dots , a_4$ , satisfying

(7.13) \begin{equation} a_1 + a_2 = a_3 + a_4 + \sigma _j, \end{equation}

the number of additive quadruples in $X_{a_1} \times X_{a_2} \times X_{a_3} \times X_{a_4}$ , which is bounded above by $|X_{a_1}| |X_{a_2}| |X_{a_3}| \leqslant 2^{-3j} M^3$ since three elements of an additive quadruple determine the fourth. It follows that the number of $(a_1, a_2, a_3, a_4) \in A_j^4$ satisfying (7.13) is $\geqslant 2^{-10} b^{-1} 2^{3j} \eta _j^4$ . By (7.11), this is $\geqslant 2^{-13} b^{-1} \eta _j |A_j|^3$ .

Now if $S_1, S_2, S_3, S_4$ are additive sets then $E(S_1, S_2, S_3, S_4)$ , the number of solutions to $s_1 + s_2 = s_3 + s_4$ with $s_i \in S_i$ , is bounded by $\prod _{i = 1}^4 E(S_i)^{1/4}$ , where $E(S_i)$ is the number of additive quadruples in $S_i$ . This is essentially the Gowers–Cauchy–Schwarz inequality for the $U^2$ -norm; it may be proven by two applications of the Cauchy–Schwarz inequality or alternatively from Hölder’s inequality on the Fourier side. Applying this with $S_1 = S_2 = S_3 = A_j$ and $S_4 = A_j + \sigma _j$ , and noting that $E(A_j + \sigma _j) = E(A_j)$ , we see that $E(A_j) \geqslant 2^{-13} b^{-1}\eta _j |A_j|^3$ .

By Proposition 2.4, we have $|A_j| \leqslant 2^{4r + 13} b^{4r+1} \eta _j^{-1}$ . Comparing with (7.11) gives $\eta _j \leqslant 2^{2r + 7-j/2} b^{2r + 1/2}$ . Take $J$ to be the least integer such that $2^{J/2} \geqslant 2^{2r + 9} b^{2r+1/2} \eta ^{-1}$ ; then $\sum _{j \geqslant J} \eta _j \lt \eta$ , and so by (7.10), some $\mathscr {M}_j$ , $j \leqslant J-1$ , is nonempty. In particular, by (7.9) there is some value of $a$ such that $|X_a| \geqslant 2^{-J} M \geqslant 2^{-4r - 20} b^{-4r-1} \eta ^2 M$ . Fix this value of $a$ and set $\mathscr {M}^{\prime} := X_a$ . Thus, to summarise,

(7.14) \begin{equation} |\mathscr {M}^{\prime}| \geqslant 2^{-4r - 20} b^{-4r-1} \eta ^2 M \end{equation}

and if $m \in \mathscr {M}^{\prime}$ then $\pi (m) = a$ . Note that the condition on $M$ in the statement of Proposition 7.3 implies (comfortably) that $|\mathscr {M}^{\prime}| \geqslant 2$ .

Note that, by (7.6) and the definition (7.7) of $\pi$ , we have that if $m \in \mathscr {M}^{\prime}$ then

(7.15) \begin{equation} \Vert \theta m - b^{1-n} a \Vert \leqslant \tfrac {3}{4} b^{1-n}. \end{equation}

Pick some $m_0 \in \mathscr {M}^{\prime}$ , and set $\mathscr {M}^{\prime\prime} := \mathscr {M}^{\prime} - m_0 \subset [-2M, 2M]$ . By the triangle inequality and (7.15), we have

(7.16) \begin{equation} \Vert \theta m \Vert \leqslant \tfrac {3}{2} b^{1-n} \lt 2b N^{-1} \end{equation}

for all $m \in \mathscr {M}^{\prime\prime}$ . (Recall that, by definition, $N = b^n$ .) Replacing $\mathscr {M}^{\prime\prime}$ by $-\mathscr {M}^{\prime\prime}$ if necessary (and since $|\mathscr {M}^{\prime\prime}| \geqslant 2$ ), it follows that there are at least $2^{-4r - 22} b^{-4r-1} \eta ^2 M$ integers $m \in \{1,\dots , 2M\}$ satisfying (7.16).

Now we apply Lemma C.1, taking $L = 2M$ , $\delta _1 = 2b N^{-1}$ and $\delta _2= 2^{-4r - 22} b^{-4r-1} \eta ^2$ in that result. The conditions of the lemma hold under the assumptions that $M, N \geqslant b^{20r} \eta ^{-2}$ (using here the fact that $b \geqslant 3$ ). The conclusion implies that there is some positive integer $q \leqslant b^{20r} \eta ^{-2}$ such that $\Vert \theta q \Vert \leqslant b^{20r} \eta ^{-2} N^{-1} M^{-1}$ , which is what we wanted to prove.

Proof. First apply Proposition 6.1. The conclusion is that, for at least $(\delta /2)^{C}N^{k-1}$ values of $m$ , $ |m| \leqslant C N^{k-1}$ , we have $\tilde {\text {w}}_n(\theta m) \leqslant C\log (2/\delta )$ , where $C := b^{7k^2/2}$ . By Lemma 7.2, for these values of $m$ , we have $\operatorname {w}_n(\theta m) \leqslant 16b^2 C \log (2/\delta )$ . (For the definitions of $\tilde {\text {w}}_n$ and $\operatorname {w}_n$ , see Definitions 5.1 and 7.1 respectively.) Now apply Proposition 7.3 with $\eta := (\delta /2)^{C}C^{-1}$ , $r = \lceil 16 b^2 C\log (2/\delta )\rceil$ , $N = b^n$ (as usual) and $M := C N^{k-1}$ .

To process the resulting conclusion, note that $b^{20r} \eta ^{-2} \leqslant (2/\delta )^{C^{\prime}}$ , with

\begin{equation*} C^{\prime} := 2C + 320 b^2 C \log b + \log _2(C^2 b^{20}) \lt 321 b^2C \log b \lt b^{8}C \lt B.\end{equation*}

Proposition 2.1 then follows.