Proof. If $\sum _{a\in A}w_{a}^{2}=0$ the claim of the lemma is trivial. Otherwise, define new weights

$$\begin{eqnarray}w_{a}^{\prime }:=\frac{w_{a}}{(\mathop{\sum }_{a\in A}w_{a}^{2})^{1/2}}\end{eqnarray}$$

which satisfy (8). It thus suffices to show that

$$\begin{eqnarray}\biggl\|\mathop{\sum }_{a\in A}w_{a}^{\prime }a^{it}\biggr\|_{2k}^{2}\leqslant \unicode[STIX]{x1D6EC}_{k}(A)+o_{T\rightarrow \infty }(1),\end{eqnarray}$$

which is a straightforward consequence of our definition of $\unicode[STIX]{x1D6EC}_{k}(A)$ and Lemma 2.1.◻

Proof. Apply Lemma 3.2 with $w_{a}=1$ if $a\in A^{\prime }$ and $w_{a}=0$ otherwise.◻

Proof. Expanding the $k$ th power of the left-hand side of (12) gives

(13) $$\begin{eqnarray}\mathop{\sum }_{j_{1},\ldots ,j_{2k}}\lim _{T\rightarrow \infty }\frac{1}{T}\int _{0}^{T}f_{j_{1}}(t)\cdots f_{j_{k}}(t)\overline{f_{j_{k+1}}(t)}\cdots \overline{f_{j_{2k}}(t)}\,dt.\end{eqnarray}$$

For fixed $j_{1},\ldots ,j_{2k}$ , the quantity

$$\begin{eqnarray}\lim _{T\rightarrow \infty }\frac{1}{T}\int _{0}^{T}f_{j_{1}}(t)\cdots f_{j_{k}}(t)\overline{f_{j_{k+1}}(t)}\cdots \overline{f_{j_{2k}}(t)}\,dt\end{eqnarray}$$

gives a weighted count of the number of solutions to equations of the form

(14) $$\begin{eqnarray}(m_{1}p^{j_{1}}+u)\cdots (m_{k}p^{j_{k}}+u)=(m_{k+1}p^{j_{k+1}}+u)\cdots (m_{2k}p^{j_{2k}}+u),\end{eqnarray}$$

with each of the $m_{i}\in \mathbb{Q}$ coprime to $p$ . We claim that there are no solutions to (14) if all of the $j_{i}$ are distinct. Indeed, suppose that all of these powers are distinct, and in particular there is a unique smallest power, say $j_{1}<j_{2},j_{3},\ldots ,j_{2k}$ . Then we get (expanding the brackets, subtracting $u^{k}$ from both sides, and multiplying out all denominators of the $m_{i}$ ) an equation

(15) $$\begin{eqnarray}Mp^{j_{1}}+N_{1}p^{j_{1}+1}=N_{2}p^{j_{1}+1}\end{eqnarray}$$

for some integers $M,N_{1}$ and $N_{2}$ , such that $(M,p)=1$ . This is a contradiction, since the right-hand side is divisible by $p^{j_{1}+1}$ and the left-hand side is not. Therefore, there are no contributions to (13) coming from those terms where all of the $j_{i}$ are distinct.

It remains to consider the cases where two or more of the powers $j_{1},\ldots ,j_{2k}$ are the same. There are three kinds of ways in which this can happen:

1. (1) $j_{i}=j_{i}^{\prime }$ with $1\leqslant i\leqslant k$ and $k+1\leqslant i^{\prime }\leqslant 2k$ (there are $k^{2}$ possible positions for such a pair $(i,i^{\prime })$ );

2. (2) $j_{i}=j_{i}^{\prime }$ with $1\leqslant i,i^{\prime }\leqslant k$ (there are $\binom{k}{2}$ possible positions for such a pair $(i,i^{\prime })$ );

3. (3) $j_{i}=j_{i}^{\prime }$ with $k+1\leqslant i,i^{\prime }\leqslant 2k$ (there are $\binom{k}{2}$ possible positions for such a pair $(i,i^{\prime })$ ).

Suppose we are in situation (1) above. Specifically, suppose that $j_{1}=j_{2k}$ . The other $k^{2}-1$ cases can be dealt with by the same argument. Then these terms in (13) can be rewritten as

$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{j_{1}}\lim _{T\rightarrow \infty }\frac{1}{T}\int _{0}^{T}f_{j_{1}}(t)\overline{f_{j_{1}}(t)}\mathop{\sum }_{j_{2},\ldots ,j_{2k-1}}f_{j_{2}}(t)\cdots f_{j_{k}}(t)\overline{f_{j_{k+1}}(t)}\cdots \overline{f_{j_{2k-1}}(t)}\,dt\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{j}\lim _{T\rightarrow \infty }\frac{1}{T}\int _{0}^{T}|f_{j}(t)|^{2}\biggl|\mathop{\sum }_{j}f_{j}(t)\biggr|^{2(k-1)}\,dt.\nonumber\end{eqnarray}$$

Suppose we are in situation (2). Specifically, suppose that $j_{1}=j_{2}$ . The other $\binom{k}{2}$ cases can be dealt with by the same argument. Then these terms in (13) can be rewritten as

$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{j_{1}}\lim _{T\rightarrow \infty }\frac{1}{T}\int _{0}^{T}f_{j_{1}}^{2}(t)\mathop{\sum }_{j_{3},\ldots ,j_{2k}}f_{j_{3}}(t)\cdots f_{j_{k}}(t)\overline{f_{j_{k+1}}(t)}\cdots \overline{f_{j_{2k}}(t)}\,dt\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant \mathop{\sum }_{j}\lim _{T\rightarrow \infty }\frac{1}{T}\int _{0}^{T}|f_{j}(t)|^{2}\biggl|\mathop{\sum }_{j}f_{j}(t)\biggr|^{k-2}\biggl|\mathop{\sum }_{j}\overline{f_{j}(t)}\biggr|^{k}\,dt\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{j}\lim _{T\rightarrow \infty }\frac{1}{T}\int _{0}^{T}|f_{j}(t)|^{2}\biggl|\mathop{\sum }_{j}f_{j}(t)\biggr|^{2(k-1)}\,dt.\nonumber\end{eqnarray}$$

The same argument also works in case (3). Returning to (13), we then have

$$\begin{eqnarray}\displaystyle & & \displaystyle \lim _{T\rightarrow \infty }\biggl(\frac{1}{T}\int _{0}^{T}\biggl|\mathop{\sum }_{j}f_{j}(t)\biggr|^{2k}\,dt\biggr)\nonumber\\ \displaystyle & & \displaystyle \qquad \times \mathop{\sum }_{j_{1},\ldots ,j_{2k}}\lim _{T\rightarrow \infty }\frac{1}{T}\int _{0}^{T}f_{j_{1}}(t)\cdots f_{j_{k}}(t)\overline{f_{j_{k+1}}(t)}\cdots \overline{f_{j_{2k}}(t)}\,dt\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant \biggl(k^{2}+2\binom{k}{2}\biggr)\mathop{\sum }_{j}\lim _{T\rightarrow \infty }\frac{1}{T}\int _{0}^{T}|f_{j}(t)|^{2}\biggl|\mathop{\sum }_{j}f_{j}(t)\biggr|^{2(k-1)}dt\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant 2k^{2}\mathop{\sum }_{j}\lim _{T\rightarrow \infty }\frac{1}{T}\int _{0}^{T}|f_{j}(t)|^{2}\biggl|\mathop{\sum }_{j}f_{j}(t)\biggr|^{2(k-1)}\,dt.\nonumber\end{eqnarray}$$

Here we have used the fact that the weights are positive and real, which allows us to overcount solutions which may occur in more than one of the three cases above. Finally, an application of Hölder’s inequality gives

$$\begin{eqnarray}\displaystyle & & \displaystyle \lim _{T\rightarrow \infty }\biggl(\frac{1}{T}\int _{0}^{T}\biggl|\mathop{\sum }_{j}f_{j}(t)\biggr|^{2k}\,dt\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant 2k^{2}\mathop{\sum }_{j}\lim _{T\rightarrow \infty }\biggl(\frac{1}{T}\int _{0}^{T}|f_{j}(t)|^{2k}\,dt\biggr)^{1/k}\biggl(\frac{1}{T}\int _{0}^{T}\biggl|\mathop{\sum }_{j}f_{j}(t)\biggr|^{2k}\,dt\biggr)^{1-1/k}.\nonumber\end{eqnarray}$$

A rearrangement of this inequality completes the proof. ◻

Proof. We establish the convention that $f_{\boldsymbol{j}}$ is identically zero for all $\boldsymbol{j}\notin {\mathcal{J}}$ . Therefore, we can complete the sum in (16), and the aim is to prove that

$$\begin{eqnarray}\displaystyle & & \displaystyle \lim _{T\rightarrow \infty }\biggl(\frac{1}{T}\int _{0}^{T}\biggl|\mathop{\sum }_{\boldsymbol{ j}\in \mathbb{Z}_{{\geqslant}0}^{K}}f_{\boldsymbol{j}}(t)\biggr|^{2k}\,dt\biggr)^{1/k}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant (2k^{2})^{K}\mathop{\sum }_{\boldsymbol{j}\in \mathbb{Z}_{{\geqslant}0}^{K}}\lim _{T\rightarrow \infty }\biggl(\frac{1}{T}\int _{0}^{T}|f_{\boldsymbol{ j}}(t)|^{2k}\,dt\biggr)^{1/k}.\nonumber\end{eqnarray}$$

We proceed by induction on $K$ , the base case $K=1$ being given by Lemma 4.2. Then

$$\begin{eqnarray}\displaystyle & & \displaystyle \lim _{T\rightarrow \infty }\biggl(\frac{1}{T}\int _{0}^{T}\biggl|\mathop{\sum }_{\boldsymbol{ j}\in \mathbb{Z}_{{\geqslant}0}^{K}}f_{\boldsymbol{j}}(t)\biggr|^{2k}\,dt\biggr)^{1/k}\nonumber\\ \displaystyle & & \displaystyle \quad =\lim _{T\rightarrow \infty }\biggl(\frac{1}{T}\int _{0}^{T}\biggl|\mathop{\sum }_{j_{K}\in \mathbb{Z}_{{\geqslant}0}}\biggl(\mathop{\sum }_{\boldsymbol{j}^{\prime }\in \mathbb{Z}_{{\geqslant}0}^{K-1}}f_{\boldsymbol{j}^{\prime },j_{K}}(t)\biggr)\biggr|^{2k}\,dt\biggr)^{1/k}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant 2k^{2}\mathop{\sum }_{j_{K}\in \mathbb{Z}_{{\geqslant}0}}\lim _{T\rightarrow \infty }\biggl(\frac{1}{T}\int _{0}^{T}\biggl|\mathop{\sum }_{\boldsymbol{ j}^{\prime }\in \mathbb{Z}_{{\geqslant}0}^{K-1}}f_{\boldsymbol{j}^{\prime },j_{K}}(t)\biggr|^{2k}\,dt\biggr)^{1/k}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant 2k^{2}\mathop{\sum }_{j_{K}\in \mathbb{Z}_{{\geqslant}0}}(2k^{2})^{K-1}\mathop{\sum }_{\boldsymbol{j}^{\prime }\in \mathbb{Z}_{{\geqslant}0}^{K-1}}\lim _{T\rightarrow \infty }\biggl(\frac{1}{T}\int _{0}^{T}|f_{\boldsymbol{ j}^{\prime },j_{K}}(t)|^{2k}\,dt\biggr)^{1/k}\nonumber\\ \displaystyle & & \displaystyle \quad =(2k^{2})^{K}\mathop{\sum }_{\boldsymbol{j}\in \mathbb{Z}_{{\geqslant}0}^{K}}\lim _{T\rightarrow \infty }\biggl(\frac{1}{T}\int _{0}^{T}|f_{\boldsymbol{ j}}(t)|^{2k}\,dt\biggr)^{1/k}.\nonumber\end{eqnarray}$$

The first inequality above follows from an application of Lemma 4.2, using the fact that

$$\begin{eqnarray}\mathop{\sum }_{\boldsymbol{j}^{\prime }\in \mathbb{Z}_{{\geqslant}0}^{K-1}}f_{\boldsymbol{j}^{\prime },j_{K}}(t)\in {\mathcal{F}}_{p_{K},j_{K},u}.\end{eqnarray}$$

The second inequality follows from the induction hypothesis. ◻

Proof. For each $\boldsymbol{j}=(j_{1},\ldots ,j_{K})\in {\mathcal{J}}$ , define $a_{\boldsymbol{j}}=p_{1}^{j_{1}}\ldots p_{K}^{j_{K}}x(\boldsymbol{j})$ , so that $A=\{a_{\boldsymbol{j}}:\boldsymbol{j}\in {\mathcal{J}}\}$ . Define $f_{\boldsymbol{j}}(t)=w_{\boldsymbol{j}}(a_{\boldsymbol{j}}+u)^{it}$ for some weights $w=\{w_{\boldsymbol{j}}:\boldsymbol{j}\in {\mathcal{J}}\}$ satisfying

$$\begin{eqnarray}\mathop{\sum }_{\boldsymbol{j}\in {\mathcal{J}}}w_{\boldsymbol{j}}^{2}=1.\end{eqnarray}$$

Note that $f_{\boldsymbol{j}}(t)\in {\mathcal{F}}_{\boldsymbol{p},\,\boldsymbol{j},u}$ , because of the divisibility conditions in the statement of the lemma. Note also that

$$\begin{eqnarray}\lim _{T\rightarrow \infty }\frac{1}{T}\int _{0}^{T}|f_{\boldsymbol{ j}}(t)|^{2k}\,dt=w_{\boldsymbol{ j}}^{2k}.\end{eqnarray}$$

Using this, as well as Lemma 4.3 and Corollary 3.1, we conclude that

$$\begin{eqnarray}\displaystyle E_{k,w}(A+u)^{1/k} & = & \displaystyle \lim _{T\rightarrow \infty }\biggl(\frac{1}{T}\int _{0}^{T}\biggl|\mathop{\sum }_{\boldsymbol{ j}\in {\mathcal{J}}}w_{\boldsymbol{j}}(a_{\boldsymbol{j}}+u)^{it}\biggr|^{2k}\,dt\biggr)^{1/k}\nonumber\\ \displaystyle & = & \displaystyle \lim _{T\rightarrow \infty }\biggl(\frac{1}{T}\int _{0}^{T}\biggl|\mathop{\sum }_{\boldsymbol{ j}\in {\mathcal{J}}}f_{\boldsymbol{j}}(t)\biggr|^{2k}\,dt\biggr)^{1/k}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle (2k^{2})^{K}\mathop{\sum }_{\boldsymbol{j}\in {\mathcal{J}}}\lim _{T\rightarrow \infty }\biggl(\frac{1}{T}\int _{0}^{T}|f_{\boldsymbol{ j}}(t)|^{2k}\,dt\biggr)^{1/k}\nonumber\\ \displaystyle & = & \displaystyle (2k^{2})^{K}.\nonumber\end{eqnarray}$$

Since the bound above does not depend on the weights, we are done. ◻

Proof. By shifting if necessary, we may assume $L$ is a linear subspace without loss of generality. Let $v_{1},\ldots ,v_{d}$ be $d$ linearly independent vectors in $L$ which we place as rows in a matrix $M$ of dimension $d\times l$ and rank $d$ . Performing row reduction on $M$ and permuting columns if necessary, we get a matrix $M^{\prime }$ in a row echelon form. Since the rank of $M^{\prime }$ is $d$ , the left-most minor of $M^{\prime }$ is a $d\times d$ identity matrix. The rows of $M^{\prime }$ span $L$ , and this is exactly the claim of the lemma.◻

Proof (of Theorem 4.1).

It follows from Freiman’s Lemma that if $|AA|\leqslant K|A|$ with $|A|$ sufficiently large, then $A$ has multiplicative dimension at most $K$ . Thus, there is an affine subspace of $\mathbb{R}^{t}$ containing ${\mathcal{P}}(A)$ and of dimension at most $K$ . Permuting the coordinates if necessary, it follows from an application of Lemma 4.5 that each $a\in A$ can be written as $a=p_{1}^{v_{p_{1}}(a)}\cdots p_{K}^{v_{p_{K}}(a)}n_{a}$ where $n_{a}$ is not divisible by any $p_{i}$ with $1\leqslant i\leqslant K$ and the vector $(v_{p_{1}}(a),\ldots ,v_{p_{K}}(a))$ is unique to $a$ . This is precisely the situation of Lemma 4.4. This proves (10).

From the Cauchy–Schwarz inequality and Corollary 3.3

$$\begin{eqnarray}|A|^{2k}\leqslant |(A+u)^{(k)}|E_{k}(A+u)\leqslant (2k^{2})^{kK}|A|^{k}|(A+u)^{(k)}|,\end{eqnarray}$$

whence

$$\begin{eqnarray}|A|^{k}\leqslant (2k^{2})^{kK}|(A+u)^{(k)}|.\Box\end{eqnarray}$$

Proof. It suffices to prove the inequality for all sufficiently large $T$ , which we assume fixed for now. Then

(18) $$\begin{eqnarray}\biggl(\frac{1}{T}\int _{0}^{T}\biggl|\mathop{\sum }_{\boldsymbol{j}\in {\mathcal{J}}}f_{\boldsymbol{j}}(t)\biggr|^{2k}\,dt\biggr)^{1/k}=\biggl(\mathop{\biggl\|\mathop{\sum }_{n=1}^{N}\mathop{\sum }_{\boldsymbol{j}\in {\mathcal{J}}_{n}}f_{\boldsymbol{j}}\biggr\|}\nolimits_{2k}\biggr)^{2}\leqslant \biggl(\mathop{\sum }_{n=1}^{N}\biggl\|\mathop{\sum }_{\boldsymbol{j}\in {\mathcal{J}}_{n}}f_{\boldsymbol{j}}\biggr\|_{2k}\biggr)^{2},\end{eqnarray}$$

by the triangle inequality. By the Cauchy–Schwarz inequality, (18) is bounded by

(19) $$\begin{eqnarray}N\mathop{\sum }_{n=1}^{N}\biggl\|\mathop{\sum }_{j\in {\mathcal{J}}_{n}}f_{j}\biggr\|_{2k}^{2}.\end{eqnarray}$$

Letting $T\rightarrow \infty$ we get the claim of the lemma.◻

Proof. First we partition

$$\begin{eqnarray}{\mathcal{J}}=\mathop{\bigcup }_{S\subseteq \{1,\ldots ,K\}}{\mathcal{J}}_{S}.\end{eqnarray}$$

By Lemma 5.1, we have

$$\begin{eqnarray}\displaystyle & & \displaystyle \lim _{T\rightarrow \infty }\biggl(\frac{1}{T}\int _{0}^{T}\biggl|\mathop{\sum }_{\boldsymbol{ j}\in {\mathcal{J}}}f_{\boldsymbol{j}}(t)\biggr|^{2k}\,dt\biggr)^{1/k}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant 2^{K}\mathop{\sum }_{S\subseteq \{1,\ldots ,K\}}\lim _{T\rightarrow \infty }\biggl(\frac{1}{T}\int _{0}^{T}\biggl|\mathop{\sum }_{\boldsymbol{ j}\in {\mathcal{J}}_{S}}f_{\boldsymbol{j}}(t)\biggr|^{2k}\,dt\biggr)^{1/k}.\nonumber\end{eqnarray}$$

Now consider the set $\unicode[STIX]{x1D70B}_{S}({\mathcal{J}}_{S})$ which consists of vectors in $\mathbb{Z}^{|S|}$ , each with non-negative entries. Thus, applying Lemma 4.3 to these, we get

$$\begin{eqnarray}\displaystyle & & \displaystyle \lim _{T\rightarrow \infty }\biggl(\frac{1}{T}\int _{0}^{T}\biggl|\mathop{\sum }_{\boldsymbol{ j}\in {\mathcal{J}}_{S}}f_{\boldsymbol{j}}(t)\biggr|^{2k}\,dt\biggr)^{1/k}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant (2k^{2})^{|S|}\mathop{\sum }_{\boldsymbol{j}_{S}\in \unicode[STIX]{x1D70B}_{S}({\mathcal{J}}_{S})}\lim _{T\rightarrow \infty }\biggl(\frac{1}{T}\int _{0}^{T}\biggl|\mathop{\sum }_{\boldsymbol{ j}\in {\mathcal{J}}_{S}(\boldsymbol{j}_{S})}f_{\boldsymbol{j}}(t)\biggr|^{2k}\,dt\biggr)^{1/k}\nonumber\end{eqnarray}$$

and the theorem follows. ◻

Proof. Suppose that $A\subset \mathbb{Q}$ and $|AA|\leqslant K|A|$ and let $\unicode[STIX]{x1D6F1}=\{p_{1},\ldots ,p_{l}\}$ be the set of all primes that appear in the prime decomposition of the elements of $A$ . So, each element $a\in A$ can be expressed uniquely as $a=p_{1}^{v_{p_{1}}(a)}\cdots p_{l}^{v_{p_{l}}(a)}$ . We have no control over the size of $\unicode[STIX]{x1D6F1}$ , but we do know that $\unicode[STIX]{x1D6F1}$ is finite.

As in the proof of Theorem 4.1, we also let ${\mathcal{P}}:A\rightarrow \mathbb{Z}^{n}$ denote the prime evaluation map defined by ${\mathcal{P}}(p_{1}^{v_{p_{1}}}(a)\cdots p_{n}^{v_{p_{t}}}(a))=(v_{p_{1}}(a),\ldots ,v_{p_{t}}(a))$ and note that $|{\mathcal{P}}(A)+{\mathcal{P}}(A)|=|AA|$ .

By Freiman’s Lemma ${\mathcal{P}}(A)$ has affine dimension at most $K$ . Then, by Lemma 4.5, we have that

$$\begin{eqnarray}\displaystyle {\mathcal{P}}(A) & = & \displaystyle \{(j_{1},\ldots ,j_{K},f_{K+1}(j_{1},\ldots ,j_{K}),\ldots ,f_{l}(j_{1},\ldots ,j_{K})):\nonumber\\ \displaystyle & & \displaystyle ~~(j_{1},\ldots ,j_{K})\in \unicode[STIX]{x1D70B}({\mathcal{P}}(A))\},\nonumber\end{eqnarray}$$

where $\unicode[STIX]{x1D70B}$ in the projection from $\mathbb{R}^{l}$ to the first $K$ coordinates, and each $f_{i}$ is either constant, or linear in variables $j_{1},\ldots ,j_{K}$ .

We then apply the inverse map ${\mathcal{P}}^{-1}$ and see that $A$ has the form

$$\begin{eqnarray}A=\{p_{1}^{j_{1}}\ldots p_{K}^{j_{K}}x(j_{1},\ldots ,j_{K}):(j_{1},\ldots ,j_{K})\in \unicode[STIX]{x1D70B}({\mathcal{P}}(A))\},\end{eqnarray}$$

where $x(j_{1},\ldots ,j_{K})=p_{K+1}^{f_{K+1}(j_{1},\ldots ,j_{K})}\cdots p_{l}^{f_{l}(j_{1},\ldots ,j_{K})}$ . Note that $x$ is coprime to $p_{1}\cdots p_{K}$ . Let ${\mathcal{J}}=\unicode[STIX]{x1D70B}({\mathcal{P}}(A))$ and for $\boldsymbol{j}=(j_{1},\ldots ,j_{K})\in {\mathcal{J}}$ , let

$$\begin{eqnarray}a_{\boldsymbol{j}}=p_{1}^{j_{1}}\ldots p_{K}^{j_{K}}x(\boldsymbol{j}).\end{eqnarray}$$

Now suppose $w$ is any weight function defined on $A+1$ and we consider

$$\begin{eqnarray}\displaystyle \mathop{\sum }_{a\in A}w_{a+1}(a+1)^{it} & = & \displaystyle \mathop{\sum }_{\boldsymbol{j}\in {\mathcal{J}}}w_{a_{\boldsymbol{j}}+1}(a_{\boldsymbol{j}}+1)^{it}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{S\subset \{1,\ldots ,K\}}\mathop{\sum }_{\boldsymbol{j}_{S}\in \unicode[STIX]{x1D70B}_{S}({\mathcal{J}}_{S})}~\mathop{\sum }_{\boldsymbol{j}\in {\mathcal{J}}_{S}(\boldsymbol{j}_{S})}w_{a_{\boldsymbol{j}}+1}(a_{\boldsymbol{j}}+1)^{it}.\nonumber\end{eqnarray}$$

By Theorem 5.2,

(22) $$\begin{eqnarray}\displaystyle & & \displaystyle \lim _{T\rightarrow \infty }\biggl(\frac{1}{T}\int _{0}^{T}\biggl|\mathop{\sum }_{a\in A}w_{a+1}(a+1)^{it}\biggr|^{2k}\,dt\biggr)^{1/k}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant (4k^{2})^{K}\mathop{\sum }_{S\subset \{1,\ldots ,K\}}\mathop{\sum }_{\boldsymbol{j}_{S}\in \unicode[STIX]{x1D70B}_{S}({\mathcal{J}}_{S})}\lim _{T\rightarrow \infty }\biggl(\frac{1}{T}\int _{0}^{T}\biggl|\mathop{\sum }_{\boldsymbol{ j}\in {\mathcal{J}}_{S}(\boldsymbol{j}_{S})}w_{a_{\boldsymbol{j}}+1}(a_{\boldsymbol{j}}+1)^{it}\biggr|^{2k}\,dt\biggr)^{1/k}.\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

If we define

$$\begin{eqnarray}A_{S,\,\boldsymbol{j}_{S}}=\{a_{\boldsymbol{j}}:\boldsymbol{j}\in {\mathcal{J}}_{S}(\boldsymbol{j}_{S})\}\end{eqnarray}$$

and $w_{S,\,\boldsymbol{j}_{S}}$ to be the weights $w$ restricted to $A_{S,\,\boldsymbol{j}_{S}}+1$ then the innermost quantity above is

$$\begin{eqnarray}\lim _{T\rightarrow \infty }\biggl(\frac{1}{T}\int _{0}^{T}\biggl|\mathop{\sum }_{\boldsymbol{j}\in {\mathcal{J}}_{S}(\boldsymbol{j}_{S})}w_{a_{\boldsymbol{j}}+1}(a_{\boldsymbol{j}}+1)^{it}\biggr|^{2k}\,dt\biggr)^{1/k}=E_{k,w_{S,\,\boldsymbol{j}_{S}}}(A_{S,\,\boldsymbol{j}_{S}}+1)^{1/k}.\end{eqnarray}$$

This energy is a weighted count of solutions to the equation

(23) $$\begin{eqnarray}(a_{\boldsymbol{j}_{1}}+1)\cdots (a_{\boldsymbol{j}_{k}}+1)=(a_{\boldsymbol{j}_{k+1}}+1)\cdots (a_{\boldsymbol{j}_{2k}}+1).\end{eqnarray}$$

Define $\boldsymbol{l}_{S}$ to be the $K$ -dimensional vector obtained from $\boldsymbol{j}_{S}$ by adding zero entries to those positions not corresponding to $S$ . We can then write $\boldsymbol{j}\in {\mathcal{J}}_{S}(\boldsymbol{j}_{S})$ as $\boldsymbol{l}_{S}+\boldsymbol{j}^{\prime }$ where $\boldsymbol{j}^{\prime }$ has negative entries on coordinates from $S^{\prime }=\{1,\ldots ,K\}\setminus S$ and zeros on coordinates in $S$ . We let $\boldsymbol{p}_{S}$ denote the vector of primes with indices from $S$ and $\boldsymbol{q}$ denote the vector of primes with indices from $S^{\prime }$ . Finally, if we use the notation $\boldsymbol{z}^{\boldsymbol{m}}=z_{1}^{m_{1}}\cdots z_{K}^{m_{K}}$ , then for $\boldsymbol{j}_{i}=\boldsymbol{l}_{S}+\boldsymbol{j}_{i}^{\prime }$ we have

$$\begin{eqnarray}a_{\boldsymbol{j}_{i}}=\boldsymbol{p}_{S}^{\boldsymbol{l}_{S}}\boldsymbol{q}^{\,\boldsymbol{j}_{i}^{\prime }}x(\boldsymbol{j}_{i}).\end{eqnarray}$$

Now we claim the following.

Claim 6.2. If (23) holds then

(24) $$\begin{eqnarray}\boldsymbol{j}_{1}^{\prime }+\cdots +\boldsymbol{j}_{k}^{\prime }=\boldsymbol{j}_{k+1}^{\prime }+\cdots +\boldsymbol{j}_{2k}^{\prime }\end{eqnarray}$$

and

(25) $$\begin{eqnarray}x(\boldsymbol{j}_{1})\cdots x(\boldsymbol{j}_{k})=x(\boldsymbol{j}_{k+1})\cdots x(\boldsymbol{j}_{2k}).\end{eqnarray}$$

In particular

$$\begin{eqnarray}a_{\boldsymbol{j}_{1}}\cdots a_{\boldsymbol{j}_{k}}=a_{\boldsymbol{j}_{k+1}}\cdots a_{\boldsymbol{j}_{2k}}.\end{eqnarray}$$

Proof of claim.

We begin with (24). We have

$$\begin{eqnarray}a_{\boldsymbol{j}_{i}}+1=\frac{\boldsymbol{p}_{S}^{\boldsymbol{l}_{S}}x(\boldsymbol{j}_{i})+\boldsymbol{q}^{-\boldsymbol{j}_{i}^{\prime }}}{\boldsymbol{q}^{-\boldsymbol{j}_{i}^{\prime }}}.\end{eqnarray}$$

The non-zero entries of $-\boldsymbol{j}_{i}^{\prime }$ are positive and correspond to indices in $S^{\prime }$ . Furthermore $\boldsymbol{p}_{S}^{\boldsymbol{l}_{S}}x(\boldsymbol{j}_{i})$ is coprime to $p_{t}$ for any $t\in S^{\prime }$ . Thus

$$\begin{eqnarray}v_{p_{t}}(a_{\boldsymbol{j}_{i}}+1)=v_{p_{t}}(\boldsymbol{q}^{\,\boldsymbol{j}_{i}^{\prime }}),\end{eqnarray}$$

and from this (24) follows.

Next, we recall that $x(\boldsymbol{j}_{i})$ is of the form $p_{K+1}^{f_{K+1}(\boldsymbol{j}_{i})}\cdots p_{l}^{f_{l}(\boldsymbol{j}_{i})}$ for primes $p_{K+1},\ldots ,p_{l}$ and functions $f_{i}$ which are linear or constant. Thus

$$\begin{eqnarray}v_{p_{t}}(x(\boldsymbol{j}_{1})\cdots x(\boldsymbol{j}_{k}))=f_{t}(\boldsymbol{j}_{1})+\cdots +f_{t}(\boldsymbol{j}_{k})\end{eqnarray}$$

and

$$\begin{eqnarray}v_{p_{t}}(x(\boldsymbol{j}_{k+1})\cdots x(\boldsymbol{j}_{2k}))=f_{t}(\boldsymbol{j}_{k+1})+\cdots +f_{t}(\boldsymbol{j}_{2k})\end{eqnarray}$$

for any $t\in \{K+1,\ldots ,l\}$ . Writing $f_{t}=H_{t}+c_{t}$ for a linear form $H_{t}$ and a constant $c_{t}$ we see

$$\begin{eqnarray}v_{p_{t}}(x(\boldsymbol{j}_{1})\cdots x(\boldsymbol{j}_{k}))=H_{t}(\boldsymbol{j}_{1}+\cdots +\boldsymbol{j}_{k})+kc_{t}\end{eqnarray}$$

and

$$\begin{eqnarray}v_{p_{t}}(x(\boldsymbol{j}_{k+1})\cdots x(\boldsymbol{j}_{2k}))=H_{t}(\boldsymbol{j}_{k+1}+\cdots +\boldsymbol{j}_{2k})+kc_{t}.\end{eqnarray}$$

In view of the first part of the claim and the fact that $\boldsymbol{j}_{i}=\boldsymbol{l}_{S}\oplus \boldsymbol{j}_{i}^{\prime }$ , we have

$$\begin{eqnarray}\boldsymbol{j}_{1}+\cdots +\boldsymbol{j}_{k}=\boldsymbol{j}_{k+1}+\cdots +\boldsymbol{j}_{2k}\end{eqnarray}$$

and (25) follows. ◻

Having established this claim, we can multiply both sides of (23) by

$$\begin{eqnarray}\mathop{\prod }_{i=1}^{k}a_{\boldsymbol{ j}_{i}}^{-1}=\mathop{\prod }_{i=k+1}^{2k}a_{\boldsymbol{ j}_{i}}^{-1}\end{eqnarray}$$

to get the equation

$$\begin{eqnarray}(1+a_{\boldsymbol{j}_{1}}^{-1})\cdots (1+a_{\boldsymbol{ j}_{k}}^{-1})=(1+a_{\boldsymbol{ j}_{k+1}}^{-1})\cdots (1+a_{\boldsymbol{ j}_{2k}}^{-1}).\end{eqnarray}$$

Counting such equations with the same weights, we are now evaluating

$$\begin{eqnarray}E_{k,w_{S,\,\boldsymbol{j}_{S}}}(A_{S,\,\boldsymbol{j}_{S}}^{-1}+1)^{1/k}=\lim _{T\rightarrow \infty }\biggl(\frac{1}{T}\int _{0}^{T}\biggl|\mathop{\sum }_{\boldsymbol{j}\in {\mathcal{J}}_{S}(\boldsymbol{j}_{S})}w_{a_{\boldsymbol{j}}+1}(a_{\boldsymbol{j}}^{-1}+1)^{it}\biggr|^{2k}\,dt\biggr)^{1/k}.\end{eqnarray}$$

Crucially, the powers of primes indexed by $S^{\prime }$ are now all positive in the above quantity, and we can apply Lemma 4.3. The above is thus at most

$$\begin{eqnarray}\displaystyle & & \displaystyle (2k^{2})^{K-|S|}\mathop{\sum }_{\boldsymbol{j}\in {\mathcal{J}}_{S}(\boldsymbol{j}_{S})}\lim _{T\rightarrow \infty }\biggl(\frac{1}{T}\int _{0}^{T}|w_{a_{\boldsymbol{j}}+1}(a_{\boldsymbol{j}}^{-1}+1)^{it}|^{2k}\,dt\biggr)^{1/k}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant (2k^{2})^{K}\mathop{\sum }_{\boldsymbol{j}\in {\mathcal{J}}_{S}(\boldsymbol{j}_{S})}|w_{a_{\boldsymbol{j}}+1}|^{2}.\nonumber\end{eqnarray}$$

Inserting this into (22), we conclude that

$$\begin{eqnarray}\displaystyle & & \displaystyle \lim _{T\rightarrow \infty }\biggl(\frac{1}{T}\int _{0}^{T}\biggl|\mathop{\sum }_{a\in A}w_{a+1}(a+1)^{it}\biggr|^{2k}\,dt\biggr)^{1/k}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant (4k^{2})^{K}\mathop{\sum }_{S\subset \{1,\ldots ,K\}}\mathop{\sum }_{\boldsymbol{j}_{S}\in \unicode[STIX]{x1D70B}_{S}({\mathcal{J}}_{S})}(2k^{2})^{K}\mathop{\sum }_{\boldsymbol{j}\in {\mathcal{J}}_{S}(\boldsymbol{j}_{S})}|w_{a_{\boldsymbol{j}}+1}|^{2}\nonumber\\ \displaystyle & & \displaystyle \quad =(8k^{4})^{K}\mathop{\sum }_{\boldsymbol{j}\in {\mathcal{J}}}|w_{a_{\boldsymbol{j}}+1}|^{2}.\nonumber\end{eqnarray}$$

Since the inequality above is true for any set of weights on $A+1$ , (20) then follows from the definition of $\unicode[STIX]{x1D6EC}_{k}(A+1)$ .

To prove (21), we apply the Cauchy–Schwarz inequality and Corollary 3.3, just as we did in the conclusion of the proof of Theorem 4.1. This yields

$$\begin{eqnarray}\displaystyle |A|^{2k} & {\leqslant} & \displaystyle E_{k}(A+1)|(A+1)^{(k)}|\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \unicode[STIX]{x1D6EC}_{k}^{k}(A+1)|A|^{k}|(A+1)^{(k)}|\nonumber\\ \displaystyle & {\leqslant} & \displaystyle (8k^{4})^{k}K|A|^{k}|(A+1)^{(k)}|,\nonumber\end{eqnarray}$$

and a rearrangement of this inequality completes the proof of (21). ◻

Corollary.

Let $q_{1},\ldots ,q_{r}$ be rationals, $I$ an integer interval of length $H$ and $S$ be a set of rational numbers of the form

$$\begin{eqnarray}s=q_{1}^{\unicode[STIX]{x1D6FC}_{1}}\ldots q_{r}^{\unicode[STIX]{x1D6FC}_{r}}\end{eqnarray}$$

such that $\unicode[STIX]{x1D6FC}_{i}\in I$ . Then for any rational $c_{1},c_{2}\neq 0$ the number of solutions $(s_{1},s_{2})\in S\times S$ to

(26) $$\begin{eqnarray}c_{1}s_{1}+c_{2}s_{2}=1\end{eqnarray}$$

is bounded by $(\log H)^{Cr}$ with some absolute effective constant $C>0$ .

Proof. We may assume without loss of generality that none of $q_{i},i=1,\ldots ,r$ divide $c_{1}$ or $c_{2}$ . Let $S_{1}$ be the set of $s_{1}\in S$ such that $(s_{1},s_{2})$ is a solution to (26) for some $s_{2}\in S$ . By the hypothesis $S_{1}$ is contained in the generalized geometric progression

$$\begin{eqnarray}G:=\{q_{1}^{i_{1}}\ldots q_{r}^{i_{r}}:\,i_{1},\ldots i_{r}\in I\}.\end{eqnarray}$$

By Lemma 4.5, there exists a set of primes $p_{1},\ldots ,p_{r}$ such that each element in $x\in c_{1}S_{1}$ is uniquely defined by the powers of $p_{1},\ldots ,p_{r}$ in the prime decomposition of $x$ . Thus, we can apply the argument of Theorem 6.1 with $K=r$ (see Remark 6.3) and then Corollary 3.3 for $c_{1}S_{1}$ and write

(27) $$\begin{eqnarray}E_{k}(c_{1}S_{1}-1)\leqslant \exp (Crk\log k)|S_{1}|^{k}.\end{eqnarray}$$

On the other hand, the shifted set $c_{1}S_{1}-1$ is contained in $-c_{2}G$ , so

(28) $$\begin{eqnarray}|(c_{1}S_{1}-1)^{(k)}|\leqslant |G^{(k)}|<\exp (r\log (2kH)).\end{eqnarray}$$

Applying Cauchy–Schwarz to (27) and combining with (28) we have

$$\begin{eqnarray}|S_{1}|^{k}\exp (-Crk\log k)<\exp (r\log (2kH))\end{eqnarray}$$

so

$$\begin{eqnarray}|S_{1}|\leqslant \exp (Cr\log k+r\log (2kH)/k).\end{eqnarray}$$

Taking

$$\begin{eqnarray}k=\frac{\log H}{\log \log H}\end{eqnarray}$$

we get

$$\begin{eqnarray}|S_{1}|\leqslant \exp (C^{\prime }r\log \log H)=(\log H)^{C^{\prime }r}\end{eqnarray}$$

for some absolute $C^{\prime }>0$ .◻