Proof. Indeed, let $Y^{\prime}=\{y\in Y:|N_X(y)|\geq |X|/2k\}$ . Then

\begin{equation*}|Y^{\prime}|\leq \tfrac {2k}{|X|}\sum _{y\in Y}|N_X(y)|=\tfrac {2k}{|X|}\sum _{x\in X}|N_Y(x)|\leq 2k^2.\end{equation*}

Let $Y^{\prime\prime}=Y\setminus Y^{\prime}$ and observe that

\begin{equation*}\sum _{x,x^{\prime}\in X}|N_{Y^{\prime\prime}}(x)\cap N_{Y^{\prime\prime}}(x^{\prime})|=\sum _{y\in Y^{\prime\prime}}|N_X(y)|^2\leq \tfrac {|X|}{2k}\sum _{x\in X}|N_Y(x)|\leq \tfrac {|X|^2}{2},\end{equation*}

and so at most half of the pairs $(x,x^{\prime})\in X\times X$ have a common neighbour outside of $Y^{\prime}$ .

Proof. Let $X=A$ and $Y=\{p:p|a\text { for some }a\in A\}$ , and let $G=\{(a,p)\in X\times Y:p|a\}$ . Applying Lemma 3.1, we find $P$ with $|P|=r\leq 2k^2$ such that for at least half of all pairs $(a,a^{\prime})\in A\times A$ , all primes dividing both $a$ and $a^{\prime}$ belong to $P$ . Each $a\in A$ factors as

\begin{equation*}a=b_a\prod _{p\in P}p^{v_p(a)}\end{equation*}

with $b_a$ coprime to each $p\in P$ , and we thus partition $A$ according to the valuations $v_{p}(a)$ with $p\in P$ and we get

\begin{equation*}A=\bigcup _{{{\boldsymbol {v}}}\in {\mathbb {N}}_0^r}{{\boldsymbol {p}}}^{{\boldsymbol {v}}}\cdot B_{{\boldsymbol {v}}}\end{equation*}

for some finite sets of integers $B_{{\boldsymbol {v}}}$ which are coprime to each $p\in P$ . Furthermore, if $(a,a^{\prime})$ is a pair for which all common prime factors do belong to $P$ , then $b_a$ and $b_{a^{\prime}}$ are coprime.

Proof. For convenience, replace $A$ with $A\cup -A$ so as to assume $A=-A$ . Let $p=p_1$ . Consider the equation

\begin{equation*}a_1-a_2=a_3-a_4\,\end{equation*}

with variables in $A$ . Suppose, $a_1$ has the minimum $p$ -adic valuation $v_p(a_1)$ among all $v_p(a_i)$ . Then, by reducing $a_1=a_2+a_3-a_4$ modulo $p^{v_p(a_1)+1}$ , we see a second term in the equation must have the same $p$ -adic valuation, and at the cost of a constant factor (from rearrangement), this term is $a_2$ .

It follows that

\begin{equation*}E_+(A,A) \ll \sum _v E(p^v B_v, A)\ll \sum _v E_+ (B_v, B_v)^{1/2} E_+(A,A)^{1/2}\end{equation*}

by Cauchy–Schwarz (applied to the additive energy) and cancelling $p^v$ . Rearranging, and applying the resulting estimate for each prime from $\{p_1,\ldots ,p_r\}$ , we find

\begin{equation*}E_+(A,A)^{1/2}\ll _r\sum _{{{\boldsymbol {v}}}}E(B_{{\boldsymbol {v}}},B_{{\boldsymbol {v}}})^{1/2}.\end{equation*}

The final claim comes from applying the trivial estimate

\begin{equation*}E_+(B_{{\boldsymbol {v}}},B_{{\boldsymbol {v}}})\leq |B_{{\boldsymbol {v}}}|^3\end{equation*}

to each summand, whence

\begin{equation*}E_+(A,A)^{1/2}\ll _r \max _{{{\boldsymbol {v}}}}|B_{{\boldsymbol {v}}}|^{1/2}\sum _{{{\boldsymbol {v}}}}|B_{{\boldsymbol {v}}}|=\max _{{{\boldsymbol {v}}}}|B_{{\boldsymbol {v}}}|^{1/2}|A|.\end{equation*}

Proof. Let $P_A$ denote the set of primes which divide some element of $A$ , and let

\begin{equation*}P=P_A\setminus \{p_1,\ldots ,p_j\}.\end{equation*}

Consider the bipartite graph

\begin{equation*}G=\{(a,p)\in A\times P:p|a\}\end{equation*}

and observe that $\mathrm {g.c.d.}(a,a^{\prime})\not \in \left \langle p_1,\ldots ,p_j\right \rangle _+$ if and only if $a$ and $a^{\prime}$ have a common neighbour in $P$ . Now, denoting $N_P(a)$ the neighbours of $a$ in $P$ , we have $|N_P(a)|\leq k-j$ , since each $a$ has at most $k$ prime factors, and $j$ of those are $p_1,\ldots ,p_j$ . Thus, writing $N_A(p)$ for the neighbours of $p$ in $A$ , we find by double counting that

\begin{equation*}(k-j)|A|\geq \sum _{a\in A}|N_P(a)|= \sum _{p\in P}|N_A(p)|\geq \frac {1}{\max _p |N_A(p)|}\sum _{p\in P}|N_A(p)|^2.\end{equation*}

If (1) fails, then the rightmost sum above is at least $|A|^2/2$ and so there is some $p_{j+1}\in P$ with $|N_A(p_{j+1})|\geq |A|/2(k-j)$ . Let $A^{\prime}=N_A(p_{j+1})$ so that $A^{\prime}$ further decomposes (by factoring out the appropriate powers of $p_{j+1}$ ) as

\begin{equation*}A^{\prime}=\bigcup _{b^{\prime}\in B^{\prime}}b^{\prime}\cdot \Gamma _{b^{\prime}}\end{equation*}

where, denoting $\boldsymbol p = (p_1,\ldots ,p_{j+1})$ , we have

\begin{equation*}B^{\prime}=\{{a}/\boldsymbol v_{\boldsymbol p}(a):\;a\in A^{\prime}\}.\end{equation*}

Proof. Set $A_0=B_0=A$ . By the upper bound on $K$ and Lemma 1.3, at least $|A_0|^2/2$ of pairs in $A_0\times A_0$ are not coprime. Then there is some $a\in A_0$ , not coprime with at least $|A_0|/2$ elements of $A_0$ , hence a prime $p_1$ that divides at least $\frac {|A_0|}{2k}$ elements of $A_0$ ; the subset of these elements of $A_0$ is denoted as $A_1$ . The corresponding ‘base set’ $B_1 = \{ {a}/{p_1^{v_{p_1}(a)}}:\;a\in A_1\}$ that is $B_1$ arises by dividing each $a\in A_1$ by the maximum power of $p_1$ that divides $a$ .

Now, beginning with $j=1$ and each $\Gamma _b\subset \langle p_1\rangle _+$ , repeatedly apply the Iteration Lemma to $A_j$ until its outcome becomes (1); otherwise denote the resulting from outcome (2) set $ A^{\prime}$ as $A_{j+1}$ . Since each $a\in A$ has at most $k$ prime factors, the iteration will terminate. We obtain a sequence of distinct primes $p_1,\ldots ,p_r$ , a sequence of sets $A_1\supseteq \cdots \supseteq A_r$ such that $p_1\cdots p_j$ divides each element of $A_{j}$ , and sets $B_1,\ldots ,B_r$ of integers such that $B_j$ is coprime to $p_1\cdots p_j$ . Namely $B_j = \{{a}/{ \boldsymbol p^{\boldsymbol v_{\boldsymbol p}(a)}}:\,a\in A_j\}$ where $\boldsymbol p=(p_1,\ldots ,p_j)$ .

Since $p_1\cdots p_j$ divides every element of $A_j$ , and $B_j$ is coprime to $p_1\cdots p_j$ , we have $\omega (b)\leq k-j$ for $b\in B_j$ , and $r\leq k$ .

We set $B^{\prime}=B_r$ and $A^{\prime}=A_r$ , so that we have

\begin{equation*}A^{\prime}=\bigcup _{b\in B^{\prime}}b \cdot \Gamma _b\end{equation*}

for some sets $\Gamma _b\subseteq \left \langle p_1,\ldots , p_r\right \rangle _+$ .

By the Iteration Lemma,

\begin{equation*} |A^{\prime}|\geq \tfrac {|A|}{2^k k!}. \end{equation*}

We now run a popularity argument, thinning the set $A^{\prime}$ by a factor of at most $8$ to get the lower bound one the corresponding base set $|B|$ . Set

\begin{equation*} L=\tfrac {|A^{\prime}|^2}{4^{k+1}K|A|}, \end{equation*}

and consider the subset $A^{\prime\prime}$ of $A^{\prime}$ , which is the union of ‘poor’ fibres, namely

\begin{equation*}A^{\prime\prime}=\bigcup _{b\in B^{\prime}: \,|\Gamma _b|\lt L }b \cdot \Gamma _b.\end{equation*}

Let $B^{\prime\prime} = \{b\in B^{\prime}:\, |\Gamma _b|\lt L\}$ be the corresponding subset of $B^{\prime}$ . Suppose, for contradiction, that $|A^{\prime\prime}|\geq \frac {7}{8}|A^{\prime}|.$ Then, since at least $|A^{\prime}|^2/2$ pairs $(a,a^{\prime}) \in A^{\prime}\times A^{\prime}$ have g.c.d. in $\langle p_1,\ldots ,p_r\rangle _+$ , by the assumption on the cardinality of $A^{\prime\prime}$ , at least $|A^{\prime}|^2/4$ pairs $(a,a^{\prime}) \in A^{\prime\prime}\times A^{\prime\prime}$ have g.c.d. in $\langle p_1,\ldots ,p_r\rangle _+$ . Let $G\subseteq A^{\prime\prime}\times A^{\prime\prime}$ be the set of these pairs.

By Lemma 1.3 and the definition of $B^{\prime\prime}$ , a product $q=aa^{\prime}$ must have fewer than $4^k L$ representations with $(a,a^{\prime})\in G$ . It follows that

\begin{equation*} (4^kL)(K|A|) \gt |A^{\prime}|^2/4, \end{equation*}

which contradicts the definition of $L$ .

Hence, there are at least $|A^{\prime}|/8$ elements in the set $\tilde A:= A^{\prime}\setminus A^{\prime\prime}$ , which is the union of the fibres $\Gamma _b$ , with $b\in B^{\prime}\setminus B^{\prime\prime}$ that have size at least $L$ . Note that $B := B^{\prime}\setminus B^{\prime\prime} = \{{a}/{ \boldsymbol p^{\boldsymbol v_{\boldsymbol p}(a)}}:\,a\in \tilde A\}$ where $\boldsymbol p=(p_1,\ldots ,p_r)$ .

Observing that $|B|\leq |A^{\prime}|/L$ , together with the lower bound for $|A^{\prime}|$ and the definition of $L$ completes the proof.

Proof. Let

\begin{equation*}f_\varepsilon (t)=\sum _n\varepsilon (n)\widehat {f}(n)e(nt).\end{equation*}

Then assuming clause (1) of the lemma,

\begin{equation*} \|f\|_{L^{q}}\leq \tfrac {1}{c_q}\left \|S_{\mathcal {P}} f\right \|_{L^{q}}=\tfrac {1}{c_q}\left \|S_{\mathcal {P}} f_\varepsilon \right \|_{L^{q}}\leq \tfrac {C_q}{c_q}\|f_\varepsilon \|_{L^{q}}. \end{equation*}

Conversely, if we assume clause (2) of the lemma, then we can write $f=(f_{\varepsilon })_\varepsilon$ so the reverse inequality

\begin{equation*}\|f_\varepsilon \|_{L^q}^q\leq C_q^q \|f\|_{L^q}^q\end{equation*}

holds, and taking expectation over all choices of $\varepsilon$ ,

\begin{equation*}C_q^q\|f\|_{L^q}^q\geq \mathbb {E}_\varepsilon \|f_\varepsilon \|_{L^{q}}^{q}= \int _0^1 \mathbb {E}_\varepsilon \biggl|\sum _{P\in \mathcal {P}}\varepsilon (P)\sum _{n\in P}\widehat {f}(n)e(nt)\biggr|^q dt.\end{equation*}

We get from Khintchine’s inequality (see, for instance, Lemma 5.5 of [Reference Muscalu and SchlagMS13]) that

\begin{equation*}\mathbb {E}_\varepsilon \biggl|\sum _{P\in \mathcal {P}}\varepsilon (P)\sum _{n\in P}\widehat {f}(n)e(nt)\biggr|^q \geq c_q^{\prime}\biggl(\sum _{P\in \mathcal {P}}\biggl|\sum _{n\in P}\widehat {f}(n)e(nt)\biggr|^2\biggr )^{\!q/2}=c_q^{\prime}(S_{\mathcal {P}} f)^q,\end{equation*}

which proves the second inequality from clause (1) upon integration over $t\in [0,1]$ .

To prove the first inequality in clause (1), we appeal to duality. Let $q^{\prime}$ be the dual exponent to $q$ and take a trigonometric polynomial $g$ such that $\|g\|_{L^{q^{\prime}}}=1$ . By orthogonality and the triangle inequality,

\begin{equation*}|\langle f,g\rangle | \leq \int _0^1 \sum _{P\in \mathcal {P}}\biggl|\sum _{n\in P}\widehat {f}(n)e(nt)\biggr|\biggl|\sum _{n\in P}\widehat {g}(n)e(nt)\biggr|dt.\end{equation*}

By the Cauchy–Schwarz inequality, the right-hand side is at most

\begin{equation*}\left \langle S_{\mathcal {P}} f,S_{\mathcal {P}} g\right \rangle \leq \|S_{\mathcal {P}} f\|_{L^q}\|S_{\mathcal {P}} g\|_{L^{q^{\prime}}}\leq C_{q^{\prime}}\left \|S_{\mathcal {P}} f\right \|_{L^q},\end{equation*}

where in the last estimate we have applied Hölder’s inequality and the second inequality from clause (1) to $g$ .

Proof. First, assume $q\geq 2$ . Let $\varepsilon :{\mathbb {Z}}\to \{-1,1\}$ be a function which is constant on the parts of $\mathcal {P}_2$ , and suppose $f$ is a trigonometric polynomial. By hypothesis and Lemma 5.4, there is a positive constant $C$ such that

\begin{equation*}\|f_\varepsilon \|_{L^q}\leq C\|f\|_{L^q},\end{equation*}

whence

\begin{equation*}\|S_{\mathcal {P}_1}f_\varepsilon \|_{L^q}\leq C_{q}(\mathcal {P}_1)\|f_\varepsilon \|_{L^q}\leq C_{q}(\mathcal {P}_1)C\|f\|_{L^q}.\end{equation*}

Now

\begin{equation*}(S_{\mathcal {P}_1}f_\varepsilon (t))^2=\sum _{P_1\in \mathcal {P}_1}\sum _{n,m\in P_1}\varepsilon (n)\varepsilon (m)\widehat {f}(n)\overline {\widehat {f}(m)}e((n-m)t),\end{equation*}

and taking expectation over $\varepsilon$ yields

\begin{align*} \mathbb {E}_\varepsilon ((S_{\mathcal {P}_1}f_\varepsilon (t))^2) =\sum _{P_1\in \mathcal {P}_1}\sum _{P_2,P_2^{\prime}\in \mathcal {P}_2}\sum _{\substack {n,m\in P_1\\[3pt] n\in P_2,m\in P_2^{\prime}}}\mathbb {E}_\varepsilon (\varepsilon (n)\varepsilon (m))\widehat {f}(n)\overline {\widehat {f}(m)}e((n-m)t). \end{align*}

The expectation vanishes unless $P_2=P_2^{\prime}$ , in which case it is 1, and hence

\begin{equation*}\mathbb {E}_\varepsilon ((S_{\mathcal {P}_1}f_\varepsilon (t))^2)=\sum _{P_1\in \mathcal {P}_1}\sum _{P_2\in \mathcal {P}_2}\sum _{n,m\in P_1\cap P_2}\widehat {f}(n)\overline {\widehat {f}(m)}e((n-m)t)=(S_{\mathcal {P}}f(t))^2.\end{equation*}

Raising to the power $q/2$ , we find

\begin{equation*}(S_{\mathcal {P}} f(t))^q=(\mathbb {E}_\varepsilon ((S_{\mathcal {P}_1}f_\varepsilon (t))^2)^{q/2}\leq \mathbb {E}_{\varepsilon }((S_{\mathcal {P}_1}f_\varepsilon (t))^q)\end{equation*}

by Jensen’s inequality, and integrating over $t$ shows

\begin{equation*}\|S_{\mathcal {P}}f\|_{L^q}^q\leq (C_{\mathcal {P}_1}C)^q\|f\|_{L^q}^q\end{equation*}

as required.

To get the claim for $1\leq q\lt 2$ , we use duality and the random multiplier formulation. Indeed, let $q^{\prime}\geq 2$ be the exponent conjugate to $q$ , let $\varepsilon :{\mathbb {Z}}\to \{-1,1\}$ be a function which is constant on the parts of $\mathcal {P}$ , and let $g$ be a trigonometric polynomial with $\|g\|_{L^{q^{\prime}}}=1$ . Then by Parseval and Hölder’s inequality,

\begin{equation*}|\langle f_\varepsilon ,g\rangle |=| \langle f,g_\varepsilon\rangle |\leq \|f\|_{L^q}\|g_\varepsilon \|_{L^{q^{\prime}}}\leq C_{q^{\prime}}(\mathcal {P})\|f\|_{L^q},\end{equation*}

which shows $\|f_\varepsilon \|_{L^q}\leq C_{q}\|f\|_{L^q}$ for any $\varepsilon :{\mathbb {Z}}\to \{-1,1\}$ which is constant on the parts of $\mathcal {P}$ , and hence the boundedness of $S_{\mathcal {P}}$ follows from Lemma 5.4.

Proof of Theorem 5.1. To each prime $p_i$ with $1\leq i\leq r$ we associate the partition $\mathcal {P}_i$ of $\mathbb {Z}$ into $p_i$ -adic scales. From Theorem 5.3 and Lemma 5.4, we see that for each prime $p_i$ with $1\leq i\leq r$ , we find a constant $C_q$ such that $\|S_{\mathcal {P}_i}f\|_{L^q}\leq C_q\|f\|_{L^q}$ . From duality, it suffices to show that the common refinement of the partitions $\mathcal {P}_i$ yields a bounded square function. This in turn follows from Lemma 5.5 applied $r-1$ times after we note, from the proof of the lemma, that its constants $C_q(\mathcal {P}_j)$ can be taken to ensure that in the notation of the lemma $C_q(\mathcal {P}) = C_q(\mathcal {P}_1)\cdot C_q(\mathcal {P}_2).$

Proof. Let $p$ be a fixed prime. It will be convenient to normalize $B$ as follows. First, if $p$ divides $n$ then we write $n=p^{v}n^{\prime}$ with $\mathrm {g.c.d.}(n^{\prime},p)=1$ . Then $X_p(B,n)\subseteq X_p(B,n^{\prime})$ and so there is no loss of generality in assuming $\mathrm {g.c.d.}(n,p)=1$ . Next, if $B$ lies in a single congruence class modulo $p^{r_0}$ for some $r_0\gt 0$ , then we may replace $B$ with $B-\min B$ (or any other element of $B$ ), without affecting $|X_p(B)|$ . The result would be that $B-\min B$ consists of multiples of $p^{r_0}$ , and since $p$ does not divide $n$ , we can then bound $|X_p(B,n)|$ by $|X_p(p^{-r_0}B,n)|$ . So we may further assume the elements of $B$ are not all congruent modulo $p$ .

We proceed by induction on $|B|$ . When $|B|=2$ , suppose $B=\{b,b^{\prime}\}$ is a set and $p$ is a prime. Then $b-b^{\prime}=p^rn$ for at most one value of $r$ , so $|X_p(B)|\leq 1$ and this establishes the base case. For larger $B$ , we condition on the value of $b\ (\text {mod }p)$ . To do that, we write

\begin{equation*}B_u=\{b\in B:b\equiv u\, (\text {mod }p)\}\end{equation*}

and let $\mu (u)=\frac {|B_u|}{|B|}$ be the accompanying probability measure. Given $b\equiv u\, (\text {mod }p)$ , we either have $b-b^{\prime}=n$ in which case $b^{\prime}\equiv u-n\ (\text {mod }p)$ , and such solutions contribute at most

\begin{align*} \sum _{u\, (\text {mod }p)}\min \{|B_u|,|B_{u-n}|\}&= |B|\sum _{u\, (\text {mod }p)}\min \{\mu (u),\mu (n-u)\}\\[3pt] &\leq |B|\sum _{u\, (\text {mod }p)}\min \{\mu (u),1-\mu (u)\}\\[3pt] &=|B|\min \{1,2-2\mu (u_{\max })\} \end{align*}

solutions, where $u_{\max }$ is the residue class for which $\mu (u)$ is largest. Otherwise $b-b^{\prime}=p^rn$ for some $r\gt 0$ in which case $b$ and $b^{\prime}$ agree modulo $p$ . Thus,

\begin{equation*}|X_p(B,n)|\leq \sum _{u\, (\text {mod }p)}|X_p(B_u,n)|+|B|\min \{1,2-2\mu (u_{\max })\}.\end{equation*}

Since $B_u\neq B$ by our normalization, we apply induction and find

\begin{align*} |X_p(B_u,n)|&\leq |B_u|(1+4\log _2|B_u|)\\[3pt] &=\mu (u)|B|(1+4\log _2|B|)-4\mu (u)|B|\log (1/\mu (u)). \end{align*}

Putting this all together,

\begin{equation*}|X_p(B,n)|\leq |B|(1+4\log _2|B|)-4|B|(H(\mu )-\tfrac {1}{4}\min \{1,2-2\mu (u_{\max })\}).\end{equation*}

Here,

\begin{equation*}H(\mu )=\sum _{u\, (\text {mod }p)}\mu (u)\log _2\tfrac {1}{\mu (u)}\end{equation*}

is the entropy of the measure $\mu$ . We claim

\begin{equation*}H(\mu )-\tfrac {1}{4}\min \{1,2-2\mu (u_{\max })\})\geq 0.\end{equation*}

If the minimum is achieved on the second term, the inequality is true since

\begin{align*} \tfrac {1-\mu (u_{\max })}{2}\leq \log _2\tfrac {1}{\mu (u_{\max })}&= \sum _{u\, (\text {mod }p)}\mu (u)\log _2\left (\frac {1}{\mu (u_{\max })}\right )\\[3pt] & \leq \sum _{u\, (\text {mod }p)}\mu (u)\log _2\left (\frac {1}{\mu (u)}\right ). \end{align*}

Otherwise $\mu (u_{\max })\leq 1/2$ , in which case the minimum value of entropy is $1$ by its convexity properties (corresponding to the least uncertain case when there are only two values of $u$ with equal probabilities $1/2$ ).

Proof. For ease of notation, let $|B|=n$ , and if necessary, augment $\Gamma$ by adjoining $-1$ to it. Consider the undirected graph $G$ on the vertex set $B$ , whose edges are those $\{b_1,b_2\}$ satisfying $b_1-b_2\in \Gamma$ . Let the number of edges be denoted by $nf(n)$ , for some function $f(n)$ . One can assume that $f(n)$ is increasing and larger than $(\log n)^r$ , or else there is nothing to prove.

Let $d=f(n)/2$ . We first prune $G$ by iteratively removing vertices with degree less than $d$ , updating the degrees of the vertices (but not the threshold $d$ ) after each stage to reflect any removal. This process must terminate after at most $n$ steps as there are at most $n$ vertices that can be removed, and when it does terminate, we can have removed no more than $n\cdot f(n)/2$ edges. If necessary, we redefine $G$ to be the pruned graph, in which each vertex has degree at least $f(n)/2$ .

Fix a vertex $b_0$ , and consider a non-degenerate path of length $l$ in $G$ , starting from $b_0$ . The path $b_0,b_1,\ldots ,b_l$ corresponds to the telescopic sum

\begin{equation*} (b_1-b_0)+(b_2-b_1)+\cdots +(b_l-b_{l-1}) = \gamma _1 + \gamma _2 +\cdots +\gamma _l,\end{equation*}

and we call the path non-degenerate if no subsum of the right-hand side vanishes. Given a non-degenerate path of length $l$ , one can append to it at least $f(n)/2 - (2^l-1)$ edges and get a non-degenerate path of length $l+1$ . Indeed, there are only $2^l-1$ edges that could lead to a degeneracy. Hence, we find by way of induction that the number of non-degenerate paths of length $l$ is at least $f(n)^l/4^l,$ provided that $f(n)\geq 2^{l+2}$ .

So, there are at least $f(n)^{l-1}/(4^{l-1} n)$ non-degenerate paths between $b_0$ and some other element $b\in B$ , and so $f(n)^{l}/(4^{l} n)$ paths from $b_0$ to some $b_1\in B$ , by appending an edge $b-b_1$ to the said path. On the other hand, Theorem 6.2 provides the upper bound for this number of paths, once one chooses $a_i = 1/(b_1-b_0)$ for $1\leq i\leq l$ . Taking logarithms and assuming that $r\geq 2$ and, say $l\geq 100$ , it simplifies the upper bound to

\begin{equation*}l\log f(n) \leq l^{\frac{11}{2}} r + \log n .\end{equation*}

Upon choosing $l \approx \left (\frac {\log n}{r}\right )^{\frac {2}{11}}$ to balance the terms of the right-hand side, we conclude that

\begin{equation*}\log f(n) \ll {r}^{\frac{2}{11}} (\log n)^{\frac{9}{11}},\end{equation*}

which completes the proof in view of the bound on $r$ assumed in the statement of the lemma.