Proof. See [Reference Deng, Nahmod and Yue17].

Proof. First, equation (2.18) follows from the definitions in equations (2.9) and (2.17) and an easy induction. We now prove formulas (2.19) and (2.20) inductively, noting also that $(a_k)_{\mathrm {in}}=\sqrt {n_{\mathrm {in}}(k)}\cdot \eta _k(\omega )$ . For $\mathcal{T}\,=\bullet $ , equation (2.19) follows from equation (2.17) with $\mathcal {K}_{\mathcal{T}\,}(\tau ,k_{\mathfrak {r}})=\widetilde {\chi }(\tau )$ that satisfies formula (2.20). Now suppose formulas (2.19) and (2.20) are true for smaller trees; then by formulas (2.7) and (2.17) and Lemma 2.1, up to unimportant coefficients, we can write

$$ \begin{align*}\left(\widetilde{\mathcal J_{\mathcal{T}}}\right)_{k}(\tau)=\frac{i\alpha T}{L^{d}}\sum_{d\in\{0,1\}}\sum_{\left(k_1,k_2,k_3\right)}^*\int_{\mathbb{R}^3}I_d(\tau,\sigma)\prod_{j=1}^3\left[\left(\widetilde{\mathcal J_{\mathcal{T}_j}}\right)_{k_j}\left(\tau_j\right)\right]^{\iota_j}\mathrm{d}\tau_j,\end{align*} $$

where $\sum ^*$ represents summation under the conditions $k_j\in \mathbb {Z}_L^d$ , $k_1-k_2+k_3=k$ and either $k\not \in \{k_1,k_3\}$ or $k=k_1=k_2=k_3$ , the signs $(\iota _1,\iota _2,\iota _3)=(+,-,+)$ , and $\sigma =\tau _1-\tau _2+\tau _3+T\Omega (k_1,k_2,k_3,k)$ . Now applying the induction hypothesis, we can write $\left (\widetilde {\mathcal J_{\mathcal{T}\,}}\right )_{k}(\tau )$ in the form of equation (2.19) with the function

(2.21) $$ \begin{align}\mathcal{K}_{\mathcal{T}}(\tau,k_{\mathfrak{n}}:\mathfrak{n}\in\mathcal{T}\,\,)=\sum_{d\in\{0,1\}}\int_{\mathbb{R}^3}I_d(\tau,\tau_1-\tau_2+\tau_3+T\Omega_{\mathfrak{r}})\prod_{j=1}^3\left[\mathcal{K}_{\mathcal{T}_j}\left(\tau_j,k_{\mathfrak{n}}:\mathfrak{n}\in\mathcal{T}_j\right)\right]^{\iota_j}\mathrm{d}\tau_j, \end{align} $$

where $\mathfrak {r}$ is the root of $\mathcal{T}\,$ with children $\mathfrak {r}_1,\mathfrak {r}_2,\mathfrak {r}_3$ and $\mathcal{T}\,_j$ is the subtree rooted at $\mathfrak {r}_j$ .

It then suffices to prove that $\mathcal {K}_{\mathcal{T}\,}$ defined by equation (2.21) satisfies formula (2.20). By the induction hypothesis, we may fix a choice $d_{\mathfrak {n}}$ for each nonleaf node $\mathfrak {n}$ of each $\mathcal{T}\,_j$ , and let $d_{\mathfrak {r}}=d$ . Then plugging formula (2.20) into equation (2.21), we get

$$ \begin{align*}\left\lvert\partial_{\tau}^a\mathcal{K}_{\mathcal{T}}\right\rvert \lesssim_{a,A}\prod_{\mathfrak{r}\neq\mathfrak{n}\in\mathcal{N}}\frac{1}{\langle T q_{\mathfrak{n}}\rangle}\int_{\mathbb{R}^3}\frac{1}{\left\langle\tau\!-\!d(\tau_1\!-\!\tau_2+\tau_3+T\Omega_{\mathfrak{r}})\right\rangle^A}\frac{1}{\langle \tau_1\!-\!\tau_2+\tau_3+T\Omega_{\mathfrak{r}}\rangle}\prod_{j=1}^3\frac{\mathrm{d}\tau_j}{\left\langle\tau_j\!-\!Td_{\mathfrak{r}_j}q_{\mathfrak{r}_j}\right\rangle^A},\end{align*} $$

which upon integration in $\tau _j$ gives equation (2.20). This completes the proof.

Proof of Proposition 2.4 (assuming Propositions 2.5 and 2.6)

Assume we have already excluded an exceptional set of probability $\lesssim e^{-L^{\theta }}$ . The bound (2.25) follows directly from formulas (2.18) and (2.27); it remains to bound $\mathcal {R}_{N+1}$ . Recall that $\mathcal {R}_{N+1}$ satisfies equation (2.11), so it suffices to prove that the mapping

$$ \begin{align*}v\mapsto \mathcal J_{\sim N}+\mathcal L(v)+\mathcal Q(v)+\mathcal C(v)\end{align*} $$

is a contraction mapping from the set $\mathcal {Z}=\left \{v:\left \lVert v\right \rVert _{h^{s,b}}\leq \rho ^{N}\right \}$ to itself. We will prove only that it maps $\mathcal {Z}$ into $\mathcal {Z}$ , as the contraction part follows in a similar way. Now suppose $\left \lVert v\right \rVert _{h^{s,b}}\leq \rho ^N$ ; then by formulas (2.18) and (2.27), we have

$$ \begin{align*} \left\lVert\mathcal J_{\sim N}\right\rVert_{h^{s,b}}^2\sim L^{-d}\sum_{k\in\mathbb{Z}_L^d}\left\langle k\right\rangle^{2s}\left\lVert(\mathcal J_{\sim N})_k\right\rVert_{h^b}^2\lesssim \left(L^{\theta+C\left(b-\frac{1}{2}\right)}\rho^{N+1}\right)^2\cdot L^{-d}\sum_{k\in\mathbb{Z}_L^d}\left\langle k\right\rangle^{-2s}\ll\rho^{2N},\end{align*} $$

so $\left \lVert \mathcal J_{\sim N}\right \rVert _{h^{s,b}}\ll \rho ^N$ . Next we may use formula (2.29) to bound

$$ \begin{align*}\left\lVert\mathcal L(v)\right\rVert_{h^{s,b}}\leq L^{\theta}\rho^{\frac{1}{2}}\cdot\left\lVert v\right\rVert_{h^{s,b}}\leq L^{\theta}\rho^{\frac{1}{2}}\cdot \rho^N\ll\rho^N.\end{align*} $$

As for the terms $\mathcal Q(v)$ and $\mathcal C(v)$ , we apply the simple bound

(2.30) $$ \begin{align} &\left\lVert\mathcal{IW}(u,v,w)\right\rVert_{h^{s,b}}\lesssim\left\lVert\mathcal{IW} (u,v,w)\right\rVert_{h^{s,1}}\lesssim\left\lVert\mathcal{IW}(u,v,w)\right\rVert_{h^{s,0}}+ \left\lVert\partial_t\mathcal{IW}(u,v,w)\right\rVert_{h^{s,0}}\nonumber\\ &\quad \lesssim\frac{\alpha T}{L^{d}}\sum_{\mathrm{cyc}}\left\lVert u\right\rVert_{h^{s,0}}\left\lVert v_k(t)\right\rVert_{\ell_k^1L_t^{\infty}}\left\lVert w_k(t)\right\rVert_{\ell_k^1L_t^{\infty}}\lesssim \alpha TL^d\left\lVert u\right\rVert_{h^{s,b}}\left\lVert v\right\rVert_{h^{s,b}}\left\lVert w\right\rVert_{h^{s,b}} \end{align} $$

(which easily follows from formula (2.7)), where $\sum _{\mathrm {cyc}}$ means summing in permutations of $(u,v,w)$ . As $\alpha T\leq L^{d}$ , we conclude (also using Proposition 2.5) that

$$ \begin{align*}\left\lVert\mathcal Q(v)\right\rVert_{h^{s,b}}+\left\lVert\mathcal C(v)\right\rVert_{h^{s,b}}\lesssim \alpha TL^{\theta+d+C\left(b-\frac{1}{2}\right)}\rho^{2N}\ll\rho^N,\end{align*} $$

since $\rho \leq L^{-\delta }$ and $N\gg \delta ^{-1}$ . This completes the proof.

Proof. First assume $\eta _k$ is Gaussian. Then by the standard hypercontractivity estimate for an Ornstein–Uhlenbeck semigroup (see, e.g., [Reference Oh and Thomann40]), we know that formula (3.2) holds with M replaced by $\mathbb {E}\left \lvert F(\omega )\right \rvert ^2$ . Now to estimate $\mathbb {E}\left \lvert F(\omega )\right \rvert ^2$ , by dividing the sum (3.1) into finitely many terms and rearranging the subscripts, we may assume in a monomial of equation (3.1) that

(3.4) $$ \begin{align}k_{1}=\cdots=k_{j_1}, k_{j_1+1}=\cdots=k_{j_2},\cdots,k_{j_{r-1}+1}=\cdots=k_{j_r},\quad 1\leq j_1<\cdots <j_r=n, \end{align} $$

and the $k_{j_s}$ are different for $1\leq s\leq r$ . Such a monomial has the form

$$ \begin{align*}\prod_{s=1}^r\eta_{k_{j_s}}^{b_s}\left(\overline{\eta_{k_{j_s}}}\right)^{c_s},\qquad b_s+c_s=j_s-j_{s-1}\ (j_0=0),\end{align*} $$

where the factors for different s are independent. We may also assume $b_s=c_s$ for $1\leq s\leq q$ and $b_s\neq c_s$ for $q+1\leq s\leq r$ , and for $1\leq j\leq j_q$ we may assume $\iota _j$ has the same sign as $(-1)^j$ . Then we can further rewrite this monomial as a linear combination of

$$ \begin{align*}\prod_{s=1}^pb_s!\prod_{s=p+1}^q\left(\left\lvert\eta_{k_{j_s}}\right\rvert^{2b_s}-b_s!\right)\prod_{s=q+1}^r\eta_{k_{j_s}}^{b_s}\left(\overline{\eta_{k_{j_s}}}\right)^{c_s}\end{align*} $$

for $1\leq p\leq q$ . Therefore, $F(\omega )$ is a finite linear combination of expressions of the form

$$ \begin{align*}\sum_{k_{j_1},\ldots,k_{j_r}}a_{k_{j_1},\ldots,k_{j_1},\ldots,k_{j_r},\ldots k_{j_r}}\prod_{s=1}^pb_s!\prod_{s=p+1}^q\left(\left\lvert\eta_{k_{j_s}}\right\rvert^{2b_s}-b_s!\right)\prod_{s=q+1}^r\eta_{k_{j_s}}^{b_s}\left(\overline{\eta_{k_{j_s}}}\right)^{c_s}.\end{align*} $$

Due to independence and the fact that $\mathbb {E}\left (\left \lvert \eta \right \rvert ^{2b}-b!\right )=\mathbb {E}\left (\eta ^b\left (\overline {\eta }\right )^c\right )=0$ for a normalised Gaussian $\eta $ and $b\neq c$ , we conclude that

(3.5) $$ \begin{align}\mathbb{E}\left\lvert F(\omega)\right\rvert^2\lesssim\sum_{k_{j_{p+1}},\ldots, k_{j_r}}\left(\sum_{k_{j_1},\ldots,k_{j_p}}\left\lvert a_{k_{j_1},\ldots,k_{j_1},\ldots,k_{j_r},\ldots k_{j_r}}\right\rvert\right)^2, \end{align} $$

which is bounded by the right-hand side of equation (3.3), by choosing $X=\left \{1,3,\ldots ,j_p-1\right \}$ and $Y=\left \{2,4,\ldots ,j_p\right \}$ , as under our assumptions $(k_{2i-1},k_{2i})$ is a pairing for $2i\leq j_p$ .

Now assume $\eta _k$ is uniformly distributed on the unit circle. Let $\{g_k(\omega )\}$ be independent, identically distributed normalised Gaussians as in the first part, and consider the random variable

$$ \begin{align*}H(\omega)=\sum_{k_1,\ldots,k_n}\left\lvert a_{k_1\cdots k_n}\right\rvert\prod_{j=1}^ng_{k_j}^{\iota_j}.\end{align*} $$

We can calculate

(3.6) $$ \begin{align}\mathbb{E}\left(\left\lvert F(\omega)\right\rvert^{2q}\right)=\sum_{\left(k_j^i,\ell_j^i\right)}\prod_{i=1}^qa_{k_1^i\cdots k_n^i}\overline{a_{\ell_1^i\cdots\ell_n^i}}\mathbb{E}\left(\prod_{i=1}^q\prod_{j=1}^n\eta_{k_j^i}^{\iota_j}\overline{\eta_{\ell_j^i}^{\iota_j}}\right), \end{align} $$

where $1\leq i\leq q$ and $1\leq j\leq n$ , and similarly for H,

(3.7) $$ \begin{align}\mathbb{E}\left(\left\lvert H(\omega)\right\rvert^{2q}\right)=\sum_{\left(k_j^i,\ell_j^i\right)}\prod_{i=1}^q\left\lvert a_{k_1^i\cdots k_n^i}\right\rvert\left\lvert a_{\ell_1^i\cdots\ell_n^i}\right\rvert\mathbb{E}\left(\prod_{i=1}^q\prod_{j=1}^ng_{k_j^i}^{\iota_j}\overline{g_{\ell_j^i}^{\iota_j}}\right). \end{align} $$

The point is that we always have

$$ \begin{align*}\left\lvert\mathbb{E}\left(\prod_{i=1}^q\prod_{j=1}^r\eta_{k_j^i}^{\iota_j}\overline{\eta_{\ell_j^i}^{\iota_j}}\right)\right\rvert\leq \mathrm{Re}\mathbb{E}\left(\prod_{i=1}^q\prod_{j=1}^rg_{k_j^i}^{\iota_j}\overline{g_{\ell_j^i}^{\iota_j}}\right).\end{align*} $$

In fact, in order for either side to be nonzero, for any particular k we must have

$$ \begin{gather*}\#\left\{(i,j):k_j^i=k,\iota_j=+\right\}+\#\left\{(i,j):\ell_j^i=k,\iota_j=-\right\}\\=\#\left\{(i,j):k_j^i=k,\iota_j=-\right\}+\#\left\{(i,j):\ell_j^i=k,\iota_j=+\right\}. \end{gather*} $$

Let both be equal to m; then by independence, the factor that the $\eta _k^{\pm }$ s contribute to the expectation on the left-hand side will be $\mathbb {E}\left \lvert \eta _k\right \rvert ^{2m}=1$ , while for the right-hand side it will be $\mathbb {E}\left \lvert g_k\right \rvert ^{2m}=m!\geq 1$ .

This implies that $\mathbb {E}\left (\left \lvert F\right \rvert ^{2q}\right )\leq \mathbb {E}\left (\left \lvert H\right \rvert ^{2q}\right )$ for any positive integer q; since formula (3.2) holds for H, we have

$$ \begin{align*}\mathbb{E}\left(\left\lvert H\right\rvert^{2q}\right)\leq (Cq)^{nq}M^q\end{align*} $$

with an absolute constant C. This gives an upper bound for $\mathbb {E}\left (\left \lvert F\right \rvert ^{2q}\right )$ , and by Chebyshev inequality, we deduce formula (3.2) for F.

Proof. We first consider $S_3$ . Let $k-x=p$ and $k-z=q$ ; then we may write $p=\left (L^{-1}u_1,\ldots , L^{-1}u_d\right )$ and similarly for q, where each $u_i$ and $v_i$ is an integer and belongs to a fixed interval of length $O\left (L^{1+\theta }\right )$ . Moreover, from $(x,y,z)\in S_3$ we deduce that

$$ \begin{align*}\left|\sum_{i=1}^d\beta_iu_iv_i+\frac{L^2m}{2}\right|\leq\frac{L^2T^{-1}}{2}.\end{align*} $$

We may assume $u_iv_i=0$ for $1\leq i\leq r$ , and $\sigma _i:=u_iv_i\neq 0$ for $r+1\leq i\leq d$ ; then the number of choices for $(u_i,v_i:1\leq i\leq r)$ is $O\left (L^{r+\theta }\right )$ . It is known (see [Reference Deng, Nahmod and Yue17, Reference Deng, Nahmod and Yue18]) that given $\sigma \neq 0$ , the number of integer pairs $(u,v)$ such that u and v each belongs to an interval of length $O\left (L^{1+\theta }\right )$ and $uv=\sigma $ is $O\left (L^{\theta }\right )$ . Therefore, if $\left \lvert k\right \rvert ,\left \lvert a\right \rvert ,\left \lvert b\right \rvert \leq L^{\theta }$ , then $\#S_3$ is bounded by $O\left (L^{r+\theta }\right )$ times the number of choices for $(\sigma _{r+1},\ldots ,\sigma _d)$ that satisfy

(3.12) $$ \begin{align}\left\lvert\sigma_j\right\rvert\leq L^{2+\theta}\ (r+1\leq j\leq d),\qquad \sum_{j=r+1}^d\beta_i\sigma_i=-\frac{L^2m}{2}+O\left(L^2T^{-1}\right).\end{align} $$

Using the assumption $T\leq L^{d}$ , it suffices to show that the number of choices for $(\sigma _{r+1},\ldots ,\sigma _d)$ satisfying formula (3.12) is at most $O\left (1+L^{2(d-r)+\theta }T^{-1}\right )$ . This latter bound is trivial if $d-r=1$ or $L^2T^{-1}\geq 1$ , so we may assume $d-r\geq 2$ , $T\geq L^{2}$ and $\beta _i$ is generic. It is well known in Diophantine approximation (see, e.g., [Reference Cassels9]) that for generic $\beta _i$ we have

$$ \begin{align*}\left\lvert\sum_{i=r+1}^d\beta_i\eta_i\right\rvert\gtrsim\left(\max_{r+1\leq i\leq d}\left\langle\eta_i\right\rangle\right)^{-(d-r-1)-\theta}\quad\text{if }\eta_i\text{ are not all }0,\end{align*} $$

so the distance between any two points $(\sigma _i:r+1\leq i\leq d)$ and $(\sigma _i':r+1\leq i\leq d)$ satisfying formula (3.12) is at least $\left (L^2T^{-1}\right )^{-\frac {1}{d-r-1}-\theta }$ . Since all these points belong to a box which has size $O(1)$ in one direction and size $O\left (L^{2+\theta }\right )$ in other orthogonal directions, we deduce that the number of solutions to formula (3.12) is at most $1+L^{\theta } L^{2(d-r-1)}L^2T^{-1}$ , as desired.

Next, without any assumption on $(k,a,b)$ , we need to prove formula (3.11). By definition (2.24) we can check that $Q^2\geq L^{2d}\left (\min \left (T,L^2\right )\right )^{-1}$ , so it suffices to prove the first inequality of formula (3.10), assuming $T\leq L^2$ . But this again follows from formula (3.12), noting that now $\left \lvert \sigma _j\right \rvert \leq L^{2+\theta }$ is no longer true, but each $\sigma _j$ still has at most $L^{2+\theta }$ possible values.

Finally we consider $S_2$ , which is much easier. In fact, formula (3.11) follows from formula (3.10), so we only need to prove the latter. Now if $T\leq L$ , we trivially have $\#S_2\leq L^{d+\theta }$ , as y will be fixed once x is; if $T\geq L$ , then we may assume $x_d-y_d\neq 0$ if the sign $\pm $ is $-$ , and then fix the first coordinates $x_j (1\leq j\leq d-1)$ and hence $y_j (1\leq j\leq d-1)$ . Then we have that $x_d\pm y_d$ is fixed, and $x_d^2\pm y_d^2$ belongs to a fixed interval of length $O\left (T^{-1}\right )$ . Since $x_d,y_d\in L^{-1}\mathbb {Z}$ , we know that $x_d$ has at most $1+L^2T^{-1}$ choices, which implies what we want to prove.

Proof. We proceed by induction. The base cases directly follow from formula (3.11). Now suppose the desired bound holds for all smaller trees, and consider $\mathcal{T}\,$ . Let $\mathfrak {r}_1,\mathfrak {r}_2,\mathfrak {r}_3$ be the children of the root node $\mathfrak {r}$ and $\mathcal{T}\,_j$ be the subtree rooted at $\mathfrak {r}_j$ . Let $l_j$ be the number of leaves in $\mathcal{T}\,_j$ , $p_j$ the number of pairs within $\mathcal{T}\,_j$ and $p_{ij}$ the number of pairings between $\mathcal{T}\,_i$ and $\mathcal{T}\,_j$ , and let $r_j=\left \lvert \mathcal {R}\cap \mathcal{T}\,_j\right \rvert $ ; then we have

$$ \begin{align*}l=l_1+l_2+l_3,\qquad p=p_1+p_2+p_3+p_{12}+p_{13}+p_{23},\qquad r=r_1+r_2+r_3+\mathbf{1}_{\mathfrak{r}\in\mathcal R}.\end{align*} $$

Also note that $\lvert k_{\mathfrak {n}}\rvert \lesssim L^{\theta }$ for all $\mathfrak {n}\in \mathcal{T}\,$ .

The proof will be completely algorithmic, with the discussion of a lot of cases. The general strategy is to perform the following four operations, which we refer to as $\mathcal {O}_j (0\leq j\leq 3)$ , in a suitable order. Here in operation $\mathcal {O}_0$ we apply formula (3.11) to count the number of choices for the values among $\left \{k_{\mathfrak {r}},k_{\mathfrak {r}_1},k_{\mathfrak {r}_2},k_{\mathfrak {r}_2}\right \}$ that are not already fixed (this step may be trivial if three of these four vectors are already fixed –i.e., coloured – or if one of them is already fixed and $k_{\mathfrak {r}}=k_{\mathfrak {r}_1}=k_{\mathfrak {r}_2}=k_{\mathfrak {r}_3}$ ). In operations $\mathcal {O}_j (1\leq j\leq 3)$ , we apply the induction hypothesis to one of the subtrees $\mathcal{T}\,_j$ and count the number of choices for $\left (k_{\mathfrak {n}}:\mathfrak {n}\in \mathcal{T}\,_j\right )$ . Let the number of choices associated with $\mathcal {O}_j (0\leq j\leq 3)$ be $M_j$ , with superscripts indicating different cases. In the whole process we may colour new nodes $\mathfrak {n}$ red if $k_{\mathfrak {n}}$ is already fixed during the previous operations, namely when $\mathfrak {n}=\mathfrak {r}$ and we have performed $\mathcal {O}_0$ before, when $\mathfrak {n}=\mathfrak {r}_j$ and we have performed $\mathcal {O}_0$ or $\mathcal {O}_j$ before or when $\mathfrak {n}$ is a leaf that has a partner in $\mathcal{T}\,_j$ and we have performed $\mathcal {O}_j$ before.

(1) Suppose $\mathfrak {r}\not \in \mathcal R$ ; then we may assume that there is a red leaf from $\mathcal{T}\,_1$ .Footnote ¹¹ We first perform $\mathcal {O}_1$ and get a factor

$$ \begin{align*}M_1^{(1)}:= L^{\theta} Q^{l_1-p_1-r_1}.\end{align*} $$

Now $\mathfrak {r}_1$ is coloured red, as is any leaf in $\mathcal{T}\,_2\cup \mathcal{T}\,_3$ which has a partner in $\mathcal{T}\,_1$ . There are then two cases.

(1.1) Suppose now there is a leaf in $\mathcal{T}\,_2\cup \mathcal{T}\,_3$ , say from $\mathcal{T}\,_2$ , that is red. Then we perform $\mathcal {O}_2$ and get a factor

$$ \begin{align*}M_2^{(1.1)}:=L^{\theta} Q^{l_2-p_2-r_2-p_{12}}.\end{align*} $$

Now $\mathfrak {r}_2$ is coloured red, as is any leaf of $\mathcal{T}\,_3$ which has a partner in $\mathcal{T}\,_2$ . There are again two cases.

(1.1.1) Suppose now there is a red leaf in $\mathcal{T}\,_3$ ; then we perform $\mathcal {O}_3$ and get a factor

$$ \begin{align*}M_3^{(1.1.1)}:=L^{\theta} Q^{l_3-p_3-r_3-p_{13}-p_{23}},\end{align*} $$

then colour $\mathfrak {r}_3$ red and apply $\mathcal {O}_0$ to get a factor $M_0^{(1.1.1)}:=1$ . Thus

$$ \begin{align*}M\leq M_1^{(1)}M_2^{(1.1)}M_3^{(1.1.1)}M_0^{(1.1.1)}=L^{l-p-r+\theta},\end{align*} $$

which is what we need.

(1.1.2) Suppose after step (1.1) there is no red leaf in $\mathcal{T}\,_3$ ; then $r_3=p_{13}=p_{23}=0$ . We perform $\mathcal {O}_0$ and get a factor $M_0^{(1.1.2)}:=L^{\theta } Q^{1}$ (perhaps with slightly enlarged $\theta $ ; the same applies later). Now we may colour $\mathfrak {r}_3$ red and perform $\mathcal {O}_3$ to get a factor

$$ \begin{align*}M_3^{(1.1.2)}:=L^{\theta} Q^{l_3-p_3-1}.\end{align*} $$

Thus

$$ \begin{align*}M\leq M_1^{(1)}M_2^{(1.1)}M_0^{(1.1.2)}M_3^{(1.1.2)}=L^{\theta} Q^{l-p-r},\end{align*} $$

which is what we need.

(1.2) Now suppose that after step (1) there is no red leaf in $\mathcal{T}\,_2\cup \mathcal{T}\,_3$ ; then $r_2=r_3=p_{12}=p_{13}=0$ . There are two cases.

(1.2.1) Suppose there is a single leaf in $\mathcal{T}\,_2\cup \mathcal{T}\,_3$ , say from $\mathcal{T}\,_2$ . Then we will perform $\mathcal {O}_0$ and get a factor $M_0^{(1.2.1)}:=L^{\theta } Q^{2}$ . Now we may colour $\mathfrak {r}_2$ and $\mathfrak {r}_3$ red and perform $\mathcal {O}_3$ to get a factor

$$ \begin{align*}M_3^{(1.2.1)}:=L^{\theta} Q^{l_3-p_3-1}.\end{align*} $$

Now any leaf of $\mathcal{T}\,_2$ which has a partner in $\mathcal{T}\,_3$ is coloured red, so we may perform $\mathcal {O}_2$ and get a factor

$$ \begin{align*}M_2^{(1.2.1)}:=L^{\theta} Q^{l_2-p_2-p_{23}-1}.\end{align*} $$

Thus

$$ \begin{align*}M\leq M_1^{(1)}M_0^{(1.2.1)}M_3^{(1.2.1)}M_2^{(1.2.1)}=L^{\theta} Q^{l-p-r},\end{align*} $$

which is what we need.

(1.2.2) Suppose there is no single leaf in $\mathcal{T}\,_2\cup \mathcal{T}\,_3$ ; then all leaves in $\mathcal{T}\,_2\cup \mathcal{T}\,_3$ are paired to one another, which implies that $k_{\mathfrak {r}_2}=k_{\mathfrak {r}_3}$ and that $\mathfrak {r}_2$ and $\mathfrak {r}_3$ have opposite signs, and hence by the admissibility condition we must have $k_{\mathfrak {r}}=k_{\mathfrak {r}_1}=k_{\mathfrak {r}_2}=k_{\mathfrak {r}_3}$ . This allows us to perform $\mathcal {O}_0$ and colour $\mathfrak {r}_2$ and $\mathfrak {r}_3$ red with $M_0^{(1.2.2)}:=1$ , then perform $\mathcal {O}_3$ and colour red any leaf of $\mathcal{T}\,_2$ which has a partner in $\mathcal{T}\,_3$ , then perform $\mathcal {O}_2$ (for which we use the second bound in formula (3.15)). This leads to the factors

$$ \begin{align*}M_3^{(1.2.2)}:=L^{\theta} Q^{l_3-p_3-1},\qquad M_2^{(1.2.2)}\leq L^{\theta} Q^{l_2-p_2-p_{23}-1+1},\end{align*} $$

and thus

$$ \begin{align*}M\leq M_1^{(1)}M_0^{(1,2,2)}M_3^{(1.2.2)}M_2^{(1.2.2)}=L^{\theta} Q^{l-p-r-1},\end{align*} $$

which is better than what we need.

(2) Now suppose $\mathfrak {r}\in \mathcal R$ ; then $r=r_1+r_2+r_3+1$ . There are two cases.

(2.1) Suppose there is one single leaf that is not red, say from $\mathcal{T}\,_1$ . There are again two cases.

(2.1.1) Suppose there is a red leaf in $\mathcal{T}\,_2\cup \mathcal{T}\,_3$ , say $\mathcal{T}\,_2$ . Then we perform $\mathcal {O}_2$ and get a factor

$$ \begin{align*}M_2^{(2.1.1)}:=L^{\theta} Q^{l_2-p_2-r_2}.\end{align*} $$

We now colour red $\mathfrak {r}_2$ and any leaf in $\mathcal{T}\,_1\cup \mathcal{T}\,_3$ which has a partner in $\mathcal{T}\,_2$ . There are a further two cases.

(2.1.1.1) Suppose now there is a red leaf in $\mathcal{T}\,_3$ ; then we perform $\mathcal {O}_3$ and get a factor

$$ \begin{align*}M_3^{(2.1.1.1)}:= L^{\theta} Q^{l_3-p_3-r_3-p_{23}}.\end{align*} $$

Now we perform $\mathcal {O}_0$ and get a factor $M_0^{(2.1.1.1)}:=1$ , then colour red $\mathfrak {r}_1$ as well as any leaf of $\mathcal{T}\,_1$ which has a partner in $\mathcal{T}\,_3$ , and perform $\mathcal {O}_1$ to get a factor

$$ \begin{align*}M_1^{(2.1.1.1)}:=L^{\theta} Q^{l_1-p_1-r_1-p_{12}-p_{13}-1}.\end{align*} $$

Thus

$$ \begin{align*}M\leq M_2^{(2.1.1)}M_3^{(2.1.1.1)}M_0^{(2.1.1.1)}M_{1}^{(2.1.1.1)}=L^{\theta} Q^{l-p-r},\end{align*} $$

which is what we need.

(2.1.1.2) Suppose after step (2.1.1) there is no red leaf in $\mathcal{T}\,_3$ ; then $r_3=p_{23}=0$ . We perform $\mathcal {O}_0$ and get a factor $M_0^{(2.1.1.2)}:=L^{\theta } Q^{1}$ . Then we colour $\mathfrak {r}_1$ and $\mathfrak {r}_3$ red and perform $\mathcal {O}_3$ to get a factor

$$ \begin{align*}M_3^{(2.1.1.2)}:=L^{\theta} Q^{l_3-p_3-1}.\end{align*} $$

Finally we colour red any leaf of $\mathcal{T}\,_1$ which has a partner in $\mathcal{T}\,_3$ , and perform $\mathcal {O}_1$ to get a factor

$$ \begin{align*}M_1^{(2.1.1.2)}:=L^{\theta} Q^{l_1-p_1-r_1-p_{12}-p_{13}-1}.\end{align*} $$

Thus

$$ \begin{align*}M\leq M_2^{(2.1.1)}M_0^{(2.1.1.2)}M_3^{(2.1.1.2)}M_{1}^{(2.1.1.2)}=L^{\theta} Q^{l-p-r},\end{align*} $$

which is what we need.

(2.1.2) Suppose in the beginning there is no red leaf in $\mathcal{T}\,_2\cup \mathcal{T}\,_3$ ; then $r_2=r_3=0$ . There are again two cases.

(2.1.2.1) Suppose there is a leaf in $\mathcal{T}\,_2\cup \mathcal{T}\,_3$ , say from $\mathcal{T}\,_2$ , that is either single or paired with a leaf in $\mathcal{T}\,_1$ . Then we perform $\mathcal {O}_0$ and get a factor $M_0^{(2.1.2.1)}:=L^{\theta } Q^{2}$ . After this we colour $\mathfrak {r}_1,\mathfrak {r}_2,\mathfrak {r}_3$ red and perform $\mathcal {O}_3$ to get a factor

$$ \begin{align*}M_3^{(2.1.2.1)}:= L^{\theta} Q^{l_3-p_3-1}.\end{align*} $$

We then colour red any leaf of $\mathcal{T}\,_1$ and $\mathcal{T}\,_2$ which has a partner in $\mathcal{T}\,_3$ , and perform $\mathcal {O}_2$ to get a factor

$$ \begin{align*}M_2^{(2.1.2.1)}:= L^{\theta} Q^{l_2-p_2-p_{23}-1}.\end{align*} $$

Finally we colour red any leaf of $\mathcal{T}\,_1$ which has a partner in $\mathcal{T}\,_2$ , and perform $\mathcal {O}_1$ to get a factor

$$ \begin{align*}M_{1}^{(2.1.2.1)}:= L^{\theta} Q^{l_1-p_1-r_1-p_{12}-p_{13}-1}.\end{align*} $$

Thus

$$ \begin{align*}M\leq M_0^{(2.1.2.1)}M_3^{(2.1.2.1)}M_2^{(2.1.2.1)}M_{1}^{(2.1.2.1)}=L^{\theta} Q^{l-p-r},\end{align*} $$

which is what we need.

(2.1.2.2) Suppose there is no leaf in $\mathcal{T}\,_2\cup \mathcal{T}\,_3$ that is either single or paired with a leaf in $\mathcal{T}\,_1$ ; then in the same way as in case (1.2.2), we must have $k_{\mathfrak {r}}=k_{\mathfrak {r}_1}=k_{\mathfrak {r}_2}=k_{\mathfrak {r}_3}$ . Moreover, we have $p_{12}=p_{13}=0$ . Then we perform $\mathcal {O}_0$ and get a factor $M_0^{(2.1.2.2)}:=1$ . After this we colour $\mathfrak {r}_1,\mathfrak {r}_2,\mathfrak {r}_3$ red and perform $\mathcal {O}_3$ to get a factor

$$ \begin{align*}M_3^{(2.1.2.2)}:= L^{\theta} Q^{l_3-p_3-1}.\end{align*} $$

We then colour red any leaf of $\mathcal{T}\,_2$ which has a partner in $\mathcal{T}\,_3$ and perform $\mathcal {O}_2$ to get a factor

$$ \begin{align*}M_2^{(2.1.2.2)}\leq L^{\theta} Q^{l_2-p_2-p_{23}-1+1}.\end{align*} $$

Finally, we perform $\mathcal {O}_1$ , again using the second part of estimate (3.15), to get a factor

$$ \begin{align*}M_{1}^{(2.1.2.2)}:= L^{\theta} Q^{l_1-p_1-r_1-1}.\end{align*} $$

Thus

$$ \begin{align*}M\leq M_0^{(2.1.2.2)}M_3^{(2.1.2.2)}M_2^{(2.1.2.2)}M_{1}^{(2.1.2.2)}=L^{\theta} Q^{l-p-r-1},\end{align*} $$

which is better than what we need.

(2.2) Now suppose that in the beginning all single leaves are red – that is, $\mathcal R=\mathcal {S}\cup \{\mathfrak {r}\}$ . Then we can argue in exactly the same way as in case (2.1), except that in the last step where we perform $\mathcal {O}_1$ , it may happen that the root $\mathfrak {r}_1$ as well as all leaves of $\mathcal{T}\,_1$ are red at that time, so we lose one power of Q in view of the weaker bound from the induction hypothesis. However, since $\mathcal R=\mathcal {S}\cup \{\mathfrak {r}\}$ , we are in fact allowed to lose this power, so we can still close the inductive step, in the same way as in case (2.1). This completes the proof.

Proof. In the proof of Proposition 3.5, we are now in case (2.1.2). In each subcase, either (2.1.2.1) or (2.1.2.2), we perform the operation $\mathcal {O}_0$ first. In case (2.1.2.1), by formula (3.10) – noting that the extra conditions are satisfied – we can replace the bound $M_0^{(2.1.2.1)}$ by $M_0':= L^{\theta } L^{2d}T^{-1}$ , so we get

$$ \begin{align*}M\leq M_0'M_3^{(2.1.2.1)}M_2^{(2.1.2.1)}M_{1}^{(2.1.2.1)}=L^{\theta} Q^{l-p-3}L^{2d}T^{-1}.\end{align*} $$

In case (2.1.2.2) we get an improvement: we have $M\leq L^{\theta } Q^{l-p-2}$ , which also implies formula (3.16), since we can check $Q\leq L^{2d}T^{-1}\leq Q^2$ by definition.

Proof of Proposition 2.5. We start with equation (2.19). Let $\lvert \mathcal{T}\,\rvert =3n+1$ . Due to the rapid decay of $\sqrt {n_{\mathrm {in}}}$ , we may assume in the summation that $\lvert k_{\mathfrak {l}}\rvert \leq L^{\theta }$ for any $\mathfrak {l}\in \mathcal L$ , and so $\lvert k\rvert \leq L^{\theta }$ also. For any fixed value of $\tau $ , we may apply Lemma 3.1 to the L-certain estimate $\left (\widetilde {\mathcal J_{\mathcal{T}\,}}\right )_k(\tau )$ . Namely, L-certainly, we have, for some choice of pairing and with colouring $\mathcal R=\{\mathfrak {r}\}$ and $n_{\mathfrak {r}}=k$ ,

(3.17) $$ \begin{align}\left\langle k\right\rangle^{4s}\left\lvert\left(\widetilde{\mathcal J_{\mathcal{T}}}\right)_k(\tau)\right\rvert^2\leq L^{\theta}\left(\frac{\alpha T}{L^{d}}\right)^{2n}\sum_{\left(k_{\mathfrak{l}}:\mathfrak{l}\in\mathcal{S}\right)}\left[\sum_{\left(k_{\mathfrak{l}}:\mathfrak{l}\in\mathcal L\backslash\mathcal{S}\right)}^{**}\left\lvert\mathcal{K}_{\mathcal{T}}(\tau,k_{\mathfrak{n}}:\mathfrak{n}\in\mathcal{T}\,\,)\right\rvert\right]^2, \end{align} $$

where $\sum ^{**}$ represents summation under the condition that the unique admissible assignment determined by $(k_{\mathfrak {l}}:\mathfrak {l}\in \mathcal {S})$ and $(k_{\mathfrak {l}}:\mathfrak {l}\in \mathcal L\backslash \mathcal {S})$ is strongly admissible. Next we would like to assume formula (3.17) for all $\tau $ , which can be done by the following trick. First, due to the decay factor in formula (2.20) and the assumption $\lvert k_{\mathfrak {l}}\rvert \leq L^{\theta }$ , we may assume $\lvert \tau \rvert \leq L^{d+\theta }$ ; moreover, choosing a large power D, we may divide the interval $\left [-L^{\theta },L^{\theta }\right ]$ into subintervals of length $L^{-D}$ and pick one point $\tau _j$ from each interval. Due to the differentiability of $\mathcal {K}_{\mathcal{T}\,}$ (see formula (2.20)), we can bound the difference

$$ \begin{align*}\left\lvert\mathcal{K}_{\mathcal{T}}(\tau,k_{\mathfrak{n}}:\mathfrak{n}\in\mathcal{T}\,\,)-\mathcal{K}_{\mathcal{T}}\left(\tau_j,k_{\mathfrak{n}}:\mathfrak{n}\in\mathcal{T}\,\right)\right\rvert\end{align*} $$

by a large negative power of L, provided $\tau $ is in the same interval as $\tau _j$ . Therefore, as long as formula (3.17) is true for each $\tau _j$ , we can assume it is true for each $\tau $ up to negligible errors. Since the number of $\tau _j$ s is at most $O\left (L^{2D}\right )$ and formula (3.17) holds L-certainly for each fixed $\tau _j$ , we conclude that, L-certainly, formula (3.17) holds for all $\tau $ .

Now, by expanding the square in formula (3.17), it suffices to bound the quantity

$$ \begin{align*}\int_{\mathbb{R}}\left\langle\tau\right\rangle^{2b}\sum_{\left(k_{\mathfrak{l}}:\mathfrak{l}\in\mathcal{S}\right)}\sum_{\left(k_{\mathfrak{l}}:\mathfrak{l}\in\mathcal L\backslash\mathcal{S}\right)}^{**}\sum_{\left(k_{\mathfrak{l}}':\mathfrak{l}\in\mathcal L\backslash\mathcal{S}\right)}^{**'}\left\lvert\mathcal{K}_{\mathcal{T}}(\tau,k_{\mathfrak{n}}:\mathfrak{n}\in\mathcal{T}\,\,)\right\rvert\cdot\left\lvert\mathcal{K}_{\mathcal{T}}\left(\tau,k_{\mathfrak{n}}':\mathfrak{n}\in\mathcal{T}\right)\right\rvert\mathrm{d}\tau,\end{align*} $$

where $(k_{\mathfrak {n}}:\mathfrak {n}\in \mathcal{T}\,\,)$ is the unique admissible assignment determined by $(k_{\mathfrak {l}}:\mathfrak {l}\in \mathcal L)$ and $(k_{\mathfrak {l}}:\mathfrak {l}\in \mathcal L\backslash \mathcal {S})$ , and $\left (k_{\mathfrak {n}}':\mathfrak {n}\in \mathcal{T}\,\right )$ is the one determined by $(k_{\mathfrak {l}}:\mathfrak {l}\in \mathcal L)$ and $\left (k_{\mathfrak {l}}':\mathfrak {l}\in \mathcal L\backslash \mathcal {S}\right )$ . The conditions in the summations $\sum ^{**}$ and $\sum ^{**'}$ correspond to these two assignments being strongly admissible. By formula (2.20), we have (for some choice of $d_{\mathfrak {n}}$ )

$$ \begin{align*}&\left\langle\tau\right\rangle^{2b}\left\lvert\mathcal{K}_{\mathcal{T}}(\tau,k_{\mathfrak{n}}:\mathfrak{n}\in\mathcal{T}\,\,)\right\rvert\cdot\left\lvert\mathcal{K}_{\mathcal{T}}\left(\tau,k_{\mathfrak{n}}':\mathfrak{n}\in\mathcal{T}\right)\right\rvert\\ &\quad\lesssim\left\langle\tau\right\rangle^{2b}\left\langle \tau-Td_{\mathfrak{r}}q_{\mathfrak{r}}\right\rangle^{-10}\left\langle \tau-Td_{\mathfrak{r}}q_{\mathfrak{r}}'\right\rangle^{-10}\prod_{\mathfrak{n}\in\mathcal{N}}\left\langle Tq_{\mathfrak{n}}\right\rangle^{-1}\left\langle Tq_{\mathfrak{n}}'\right\rangle^{-1},\end{align*} $$

where $q_{\mathfrak {n}}$ and $q_{\mathfrak {n}}'$ are defined from the assignments $(k_{\mathfrak {n}})$ and $\left (k_{\mathfrak {n}}'\right )$ , respectively, via equation (2.16). Thus the integral in $\tau $ gives

$$ \begin{align*}\max\left(\langle Tq_{\mathfrak{r}}\rangle,\left\langle Tq_{\mathfrak{r}}'\right\rangle\right)^{-2+2b}\left\langle T\left(q_{\mathfrak{r}}-q_{\mathfrak{r}}'\right)\right\rangle^{-5}\prod_{\mathfrak{r}\neq\mathfrak{n}\in\mathcal{N}}\left\langle Tq_{\mathfrak{n}}\right\rangle^{-1}\left\langle Tq_{\mathfrak{n}}'\right\rangle^{-1},\end{align*} $$

and it suffices to bound

$$ \begin{align*}\sum_{\left(k_{\mathfrak{l}}:\mathfrak{l}\in\mathcal{S}\right)}\sum_{\left(k_{\mathfrak{l}}:\mathfrak{l}\in\mathcal L\backslash\mathcal{S}\right)}^{**}\sum_{\left(k_{\mathfrak{l}}':\mathfrak{l}\in\mathcal L\backslash\mathcal{S}\right)}^{**'}\max\left(\langle Tq_{\mathfrak{r}}\rangle,\left\langle Tq_{\mathfrak{r}}'\right\rangle\right)^{-2+2b}\left\langle T\left(q_{\mathfrak{r}}-q_{\mathfrak{r}}'\right)\right\rangle^{-5}\prod_{\mathfrak{r}\neq\mathfrak{n}\in\mathcal{N}}\left\langle Tq_{\mathfrak{n}}\right\rangle^{-1}\left\langle Tq_{\mathfrak{n}}'\right\rangle^{-1}.\end{align*} $$

Since all the qs are bounded by $L^{\theta }$ , and $T\leq L^d$ , we may fix the integer parts of each $Tq_{\mathfrak {n}}$ and $Tq_{\mathfrak {n}}'$ for each $\mathfrak {n}\in \mathcal {N}$ , and reduce the foregoing sum to a counting bound, at the price of losing a power $L^{C\left (b-\frac {1}{2}\right )}$ . Now by definition (2.16), each $q_{\mathfrak {n}}$ is a linear combination of $\Omega _{\mathfrak {n}}$ s, and conversely, each $\Omega _{\mathfrak {n}}$ is a linear combination of $q_{\mathfrak {n}}$ s. So once the integer parts of each $Tq_{\mathfrak {n}}$ and $Tq_{\mathfrak {n}}'$ are fixed, we have also fixed $\sigma _{\mathfrak {n}}\in \mathbb {R}$ and $\sigma _{\mathfrak {n}}'\in \mathbb {R}$ , such that

(3.18) $$ \begin{align}\lvert\Omega_{\mathfrak{n}}-\sigma_{\mathfrak{n}}\rvert\leq T^{-1},\qquad \left\lvert\Omega_{\mathfrak{n}}'-\sigma_{\mathfrak{n}}'\right\rvert\leq T^{-1}.\end{align} $$

Therefore we are reduced to counting the number of $(k_{\mathfrak {l}}:\mathfrak {l}\in \mathcal {S})$ , $(k_{\mathfrak {l}}:\mathfrak {l}\in \mathcal L\backslash \mathcal {S})$ and $\left (k_{\mathfrak {l}}':\mathfrak {l}\in \mathcal L\backslash \mathcal {S}\right )$ such that the assignments $(k_{\mathfrak {n}})$ and $\left (k_{\mathfrak {n}}'\right )$ are both strongly admissible and satisfy formula (3.18). Now let $\lvert \mathcal L\rvert =l=2n+1$ and p be the number of pairs; then $\lvert \mathcal {S}\rvert =2n+1-2p$ . First we count the number of choices for $(k_{\mathfrak {l}}:\mathfrak {l}\in \mathcal {S})$ and $(k_{\mathfrak {l}}:\mathfrak {l}\in \mathcal L\backslash \mathcal {S})$ , where we apply Corollary 3.6 with $\mathcal R=\{\mathfrak {r}\}$ and get the factor $M:=L^{\theta } Q^{2n-p-2}L^{2d}T^{-1}$ ; then, with $k_{\mathfrak {l}}$ fixed for all $\mathfrak {l}\in \mathcal {S}$ , we count the number of choices for $\left (k_{\mathfrak {l}}':\mathfrak {l}\in \mathcal L\backslash \mathcal {S}\right )$ by applying Proposition 3.5 with $\mathcal R=\mathcal {S}\cup \{\mathfrak {r}\}$ and get the factor $M':= L^{\theta }Q^p$ . In the end we have, L-certainly,

$$ \begin{align*}\sup_k\left\langle k\right\rangle^{4s}\left\lvert\left(\widetilde{\mathcal J_{\mathcal{T}}}\right)_k(\tau)\right\rvert^2 & \leq L^{\theta+C\left(b-\frac{1}{2}\right)}\left(\frac{\alpha T}{L^{d}}\right)^{2n}MM'\\ & \leq L^{\theta+C\left(b-\frac{1}{2}\right)}\left(\frac{\alpha T}{L^{d}}\right)^{2n}Q^{2n-2}L^{2d}T^{-1}=L^{\theta+C\left(b-\frac{1}{2}\right)}\rho^{2n-2}\left(\alpha^2T\right)\end{align*} $$

by the definition of Q in formula (3.11), as desired.

Proof of Proposition 2.6. There are three steps.

Step 1: First reductions. We start with some simple observations. The operator $\mathcal {P}_+(v)=\mathcal {IW}\left (\mathcal J_{\mathcal{T}\,_1},\mathcal J_{\mathcal{T}\,_2},v\right )$ , where $\mathcal {I}$ and $\mathcal {W}$ are defined in formulas (2.6) and (2.7). Now in formula (2.7) we may assume $\lvert k_1\rvert ,\lvert k_2\rvert \leq L^{\theta }$ , for the same reason as in the proof of Proposition 2.5. We thus have

$$ \begin{align*}L^{-\theta}\leq\frac{\left\langle k\right\rangle^s}{\left\langle k_3\right\rangle^s}\leq L^{\theta},\end{align*} $$

so instead of $h^{s,b}$ bounds we only need to consider $h^{0,b}$ bounds. Next notice that if $\mathcal {I}$ is defined by equation (2.6) and $\mathcal {I}_1$ is defined by $\mathcal {I}_1F=\chi \cdot (\mathrm {sgn}*(\chi \cdot F))$ , then we have the identity $2\mathcal {I}F(t)=\mathcal {I}_1F(t)-\chi (t)\mathcal {I}_1F(0)$ , so for $b>\frac {1}{2}$ we have $\left \lVert \mathcal {I}F\right \rVert _{h^{s,b}}\lesssim \left \lVert \mathcal {I}_1F\right \rVert _{h^{s,b}}$ . Therefore, in estimating $\mathcal {P}_+$ we may replace the operator G that appears in the formula for $\mathcal {I}$ by $\mathcal {I}_1$ . The advantage is that $\mathcal {I}_1$ has a formula

$$ \begin{align*}\widetilde{\mathcal{I}_1F}(\tau)=\int_{\mathbb{R}}I_1(\tau,\sigma)\widetilde{F}(\sigma)\mathrm{d}\sigma,\end{align*} $$

where $I_1$ is as in Lemma 2.1, so we may get rid of the $I_0$ term. From now on we will stick to the renewed definition of $\mathcal {I}$ . Next, by Proposition 2.5 we have the trivial bound

$$ \begin{align*} &\left\lVert\mathcal{P}_+v\right\rVert_{h^{0,1}} \sim\left\lVert\mathcal{P}_+v\right\rVert_{\ell_k^2L_t^2}+\left\lVert\partial_t\mathcal{P}_+v\right\rVert_{\ell_k^2L_t^2}\\ \lesssim&\left\lVert v\right\rVert_{\ell_k^2L_t^2}\cdot\frac{\alpha T}{L^{d}}\sum_{k_1,k_2}\left\lVert\left(\mathcal J_{\mathcal{T}_1}\right)_{k_1}\right\rVert_{L_t^{\infty}}\left\lVert\left(\mathcal J_{\mathcal{T}_2}\right)_{k_2}\right\rVert_{L_t^{\infty}}\lesssim\alpha TL^{d+\theta+C\left(b-\frac{1}{2}\right)}\rho^{n_1+n_2} \cdot\left\lVert v\right\rVert_{h^{0,0}}. \end{align*} $$

Note also that $\alpha T\leq L^{d}$ and $\rho \leq L^{-\varepsilon }$ , so by interpolation it suffices to L-certainly bound the $h^{0,b}\to h^{0,1-b}$ norm of (the renewed version of) $\mathcal {P}_+$ by $L^{\theta }\rho ^{n_1+n_2+1}$ .

Now, using Lemma 2.1 and noticing that the bound (2.8) is symmetric in $\sigma $ and $\tau $ , we have the formula

(3.19) $$ \begin{align} \left(\widetilde{\mathcal{P}_+v}\right)_k(\tau)=\frac{i\alpha T}{L^{d}}\left\langle\tau\right\rangle^{-1} &\sum_{\left(m_1,m_2,k'\right)}^*\int_{\mathbb{R}^3}J\left(\tau,\sigma_1-\sigma_2+ \tau'+T\Omega(m_1,m_2,k',k)\right)\nonumber\\ &\times\left(\widetilde{\mathcal J_{\mathcal{T}_1}}\right)_{m_1}(\sigma_1)\overline{\left(\widetilde{\mathcal J_{\mathcal{T}_2}}\right)_{m_2}(\sigma_2)}\cdot\widetilde{v}_{k'}(\tau')\mathrm{d}\sigma_1\mathrm{d}\sigma_2\mathrm{d}\tau', \end{align} $$

where $J=J(\tau ,\eta )$ and all its derivatives are bounded by $\left \langle \tau -\eta \right \rangle ^{-10}$ . By elementary estimates we have

(3.20) $$ \begin{align}\left\lVert\widetilde{w}_{k}(\tau)\right\rVert_{L_{\tau}^1\ell_k^2}\lesssim\left\lVert\left\langle \tau\right\rangle^b\widetilde{w}_{k}(\tau)\right\rVert_{\ell_k^2L_{\tau}^2},\qquad\left\lVert\left\langle\tau\right\rangle^{1-b}\left\langle\tau\right\rangle^{-1}w_k(\tau)\right\rVert_{\ell_k^2L_{\tau}^2}\lesssim \left\lVert\widetilde{w}_{k}(\tau)\right\rVert_{L_{\tau}^{\infty}\ell_k^2},\end{align} $$

and thus it suffices to L-certainly bound the $\ell ^2\to \ell ^2$ norm of the operator

(3.21) $$ \begin{align} \mathcal{X}:(\mathcal{X}v)_k=\frac{\alpha T}{L^{d}} \sum_{\left(m_1,m_2,k'\right)}^*v_{k'}\cdot\int_{\mathbb{R}^2} &J \left(\tau,\sigma_1-\sigma_2+\tau'+T\Omega(m_1,m_2,k',k)\right)\nonumber\\ &\times \left(\widetilde{\mathcal J_{\mathcal{T}_1}}\right)_{m_1}(\sigma_1)\overline{\left(\widetilde{\mathcal J_{\mathcal{T}_2}}\right)_{m_2}(\sigma_2)}\mathrm{d}\sigma_1\mathrm{d}\sigma_2 \end{align} $$

uniformly in $\tau $ and $\tau '$ .

Step 2: Second reductions. At this point we apply similar arguments as in the proof of Proposition 2.5. Namely, we first restrict $\lvert \tau \rvert ,\lvert \tau '\rvert \leq L^{\theta ^{-1}}$ (otherwise we can gain a power of either $\lvert \tau \rvert ^{\frac {1}{2}\left (b-\frac {1}{2}\right )}$ or $\lvert \tau '\rvert ^{\frac {1}{2}\left (b-\frac {1}{2}\right )}$ from the extra room when applying formula (3.20), which turns into a large power of L and closes the whole estimate), and then divide this interval into subintervals of length $L^{-\theta ^{-1}}$ and apply differentiability to reduce to $O\left (L^{C\theta ^{-1}}\right )$ choices of $(\tau ,\tau ')$ . Therefore, it suffices to fix $\tau $ and $\tau '$ and L-certainly bound $\left \lVert \mathcal {X}\right \rVert _{\ell ^2\to \ell ^2}$ . Let $\tau -\tau '=\zeta $ be fixed.

Now use equation (2.19) for the $\mathcal J_{\mathcal{T}\,_j}$ factors, assuming also $\left \lvert k_{\mathfrak {l}}\right \rvert \leq L^{\theta }$ in each tree, and integrate in $(\sigma _1,\sigma _2)$ . This leads to further reduced expression for $\mathcal {X}$ , which can be described as follows. First let the tree $\mathcal{T}\,$ be defined such that its root is $\mathfrak {r}$ and three subtrees from left to right are $\mathcal{T}\,_1$ , $\mathcal{T}\,_2$ and a single node $\mathfrak {r}'$ . Then we have

$$ \begin{align*}(\mathcal{X}v)_k=\sum_{k'}\mathcal{X}_{kk'}v_{k'},\end{align*} $$

where the matrix coefficients are given by

$$ \begin{align*}\mathcal{X}_{kk'}=\left(\frac{\alpha T}{L^{d}}\right)^{n_1+n_2+1}\sum_{\left(k_{\mathfrak{n}}:\mathfrak{n}\in\mathcal{T}\right)}\mathcal{K}\left(\left\lvert k\right\rvert_{\beta}^2-\left\lvert k'\right\rvert_{\beta}^2,\ k_{\mathfrak{l}}:\mathfrak{r}'\neq\mathfrak{l}\in\mathcal L\right)\cdot\frac{1}{\left\langle Tq_{\mathfrak{r}}-\zeta\right\rangle^5}\prod_{\mathfrak{n}\in\mathcal{N}\setminus \{\mathfrak{r}\}}\frac{1}{\langle Tq_{\mathfrak{n}}\rangle}\prod_{\mathfrak{l}\in\mathcal L\setminus \{\mathfrak{r}'\}}\eta_{k_{\mathfrak{l}}}^{\iota_{\mathfrak{l}}},\end{align*} $$

where the sum is taken over all admissible assignments $(k_{\mathfrak {n}}:\mathfrak {n}\in \mathcal{T}\,\,)$ which satisfy $k_{\mathfrak {r}}=k$ , $k_{\mathfrak {r}'}=k'$ and $\left \lvert k_{\mathfrak {n}}\right \rvert \leq L^{\theta }$ for $\mathfrak {n}\not \in \{\mathfrak {r},\mathfrak {r}'\}$ , and the coefficient satisfies $\left \lvert \mathcal {K}\right \rvert \leq L^{\theta }$ and $\left \lvert \partial \mathcal {K}\right \rvert \leq L^{\theta } T$ . Moreover, we observe that $\mathcal {K}$ and $q_{\mathfrak r}$ depend on the variables $k_{\mathfrak {r}}=k$ and $k_{\mathfrak {r}'}=k'$ only through the quantity $\left \lvert k\right \rvert _{\beta }^2-\left \lvert k'\right \rvert _{\beta }^2$ .

Next we argue in the same way as in the proof of Proposition 2.5 and fix the integer parts of $Tq_{\mathfrak {n}}$ for $\mathfrak {n}\in \mathcal {N}\setminus \{\mathfrak {r}\}$ , as well as the integer part of $Tq_{\mathfrak {r}}-\zeta $ , at a cost of $(\log L)^{O(1)}$ . All these can be assumed to be $\leq L^{\theta ^{-1}}$ due to the decay $\left \langle Tq_{\mathfrak {r}}-\zeta \right \rangle ^{-5}$ and the bounds on $\tau $ and $\tau '$ . This is equivalent to fixing some real numbers $\sigma _{\mathfrak {n}}=O\left (L^{\theta ^{-1}}\right )$ and requiring the assignment $(k_{\mathfrak {n}}:\mathfrak {n}\in \mathcal{T}\,\,)$ to satisfy $\lvert \Omega _{\mathfrak {n}}-\sigma _{\mathfrak {n}}\rvert \leq T^{-1}$ for each $\mathfrak {n}\in \mathcal {N}$ . Let this final operator, obtained by all the previous reductions, be $\mathcal {G}$ . Schematically, the operator $\mathcal {G}$ can be viewed as ‘attaching’ two trees $\mathcal{T}\,_1$ and $\mathcal{T}\,_2$ to a single node $\mathfrak {r}'$ .

Step 3: The high-order $\mathcal {G}\mathcal {G}^*$ argument. For this, we consider the adjoint operator $\mathcal {G}^*$ . A similar argument gives a formula for $\mathcal {G}^*$ , which is associated with a tree $\mathcal{T}\,^*$ formed by attaching the two trees $\mathcal{T}\,_2$ and $\mathcal{T}\,_1$ (with $\mathcal{T}\,_2$ on the left of $\mathcal{T}\,_1$ ) to a single node $\mathfrak {r}'$ , in the same way that $\mathcal {G}$ is associated with $\mathcal{T}\,$ . Given a large positive integer D, we will consider $(\mathcal {G}\mathcal {G}^*)^{D}$ , which is associated with a tree $\mathcal{T}\,^D$ . The precise description is as follows.

First, $\mathcal{T}\,^D$ is a tree with root node $\mathfrak {r}_0=\mathfrak {r}$ , and its first two subtrees (from the left) are $\mathcal{T}\,_1$ and $\mathcal{T}\,_2$ . The third subtree has root $\mathfrak {r}_1$ , and its first two subtrees (from the left) are $\mathcal{T}\,_2$ and $\mathcal{T}\,_1$ . The third subtree has root $\mathfrak {r}_2$ , and its first two subtrees (from the left) are $\mathcal{T}\,_1$ and $\mathcal{T}\,_2$ , and so on. This process repeats and eventually stops at $\mathfrak {r}_{2D}=\mathfrak {r}'$ , which is a single node, finishing the construction of $\mathcal{T}\,^D$ . As usual, denote by $\mathcal L^D$ and $\mathcal N^D$ the set of leaves and branching nodes, respectively. Then the kernel of $(\mathcal {G}\mathcal {G}^*)^{D}$ is given by

(3.22) $$ \begin{align} \left((\mathcal{G}\mathcal{G}^*)^{D}\right)_{kk'}=\sum_{\left(k_{\mathfrak{n}}:\mathfrak{n}\in\mathcal{T}^D\right)}\mathcal{K}^{(D)}&\left(\left\lvert k_{\mathfrak{r}_j}\right\rvert_{\beta}^2-\left\lvert k_{\mathfrak{r}_{j+1}}\right\rvert_{\beta}^2:0\leq j\leq 2D-1,\ k_{\mathfrak{l}}:\mathfrak{r}'\neq\mathfrak{l}\in\mathcal L^D\right)\nonumber\\ &\quad\quad\quad\quad\quad\quad\times\left(\frac{\alpha T}{L^{d}}\right)^{2D\left(n_1+n_2+1\right)}\prod_{\mathfrak{r}'\neq\mathfrak{l}\in\mathcal L^D}\eta_{k_{\mathfrak{l}}}^{\iota_{\mathfrak{l}}}, \end{align} $$

where $\left \lvert \mathcal {K}^{(D)}\right \rvert \leq L^{\theta }$ and $\left \lvert \partial \mathcal {K}^{(D)}\right \rvert \leq L^{\theta } T$ , and the sum is taken over all admissible assignments $\left (k_{\mathfrak {n}}:\mathfrak {n}\in \mathcal{T}\,^D\right )$ that satisfy $(k_{\mathfrak {r}},k_{\mathfrak {r}'})=(k,k')$ , $\lvert k_{\mathfrak {l}}\rvert \leq L^{\theta }$ for $\mathfrak {r}'\neq \mathfrak {l}\in \mathcal L$ and $\lvert \Omega _{\mathfrak {n}}-\sigma _{\mathfrak {n}}\rvert \leq T^{-1}$ for $\mathfrak {n}\in \mathcal {N}^D$ , where $\sigma _{\mathfrak {n}}=O\left (L^{\theta ^{-1}}\right )$ are fixed. Moreover, $\mathcal {K}^{(D)}$ depends on the variables $k_{\mathfrak {r}}=k$ and $k_{\mathfrak {r}'}=k'$ only through the quantities $\left \lvert k_{\mathfrak {r}_j}\right \rvert _{\beta }^2-\left \lvert k_{\mathfrak {r}_{j+1}}\right \rvert _{\beta }^2$ for $0\leq j\leq 2D-1$ .

Now note that each $\left ((\mathcal {G}\mathcal {G}^*)^{D}\right )_{kk'}$ is an explicit multilinear Gaussian expression. Since for fixed k (or $k'$ ) the number of choices for $k'$ (or k) is $O\left (L^{d+\theta }\right )$ , by Schur’s estimate we know

$$ \begin{align*} \left\lVert(\mathcal{G}\mathcal{G}^*)^{D}\right\rVert_{\ell^2\to\ell^2}\lesssim L^{d+\theta}\sup_{k,k'}\left\lvert\left((\mathcal{G}\mathcal{G}^*)^{D}\right)_{kk'}\right\rvert. \end{align*} $$

So it suffices to L-certainly bound $\left \lvert \left ((\mathcal {G}\mathcal {G}^*)^{D}\right )_{kk'}\right \rvert $ uniformly in k and $k'$ . We first consider this estimate with fixed $(k,k')$ . Applying Lemma 3.1, we can fix some pairings of $\mathcal{T}\,^D$ and the set $\mathcal {S}^D$ of single leaves, and argue as in the proof of Proposition 2.5 to conclude L-certainly that

$$ \begin{align*} \left\lvert\left((\mathcal{G}\mathcal{G}^*)^{D}\right)_{kk'}\right\rvert^2\lesssim L^{\theta}\left(\frac{\alpha T}{L^{d}}\right)^{4D\left(n_1+n_2+1\right)}\sum_{\left(k_{\mathfrak{l}}:\mathfrak{l}\in\mathcal{S}^D\right)}\sum_{\left(k_{\mathfrak{l}}:\mathfrak{l}\in\mathcal L^D\backslash\mathcal{S}^D\right)}^{**}\sum_{\left(k_{\mathfrak{l}'}:\mathfrak{l}\in\mathcal L^D\backslash\mathcal{S}^D\right)}^{**'}1, \end{align*} $$

where the condition for summation, as in the proof of Proposition 2.5, is that the unique admissible assignment $\left (k_{\mathfrak {n}}:\mathfrak {n}\in \mathcal{T}\,^D\right )$ determined by $\left (k_{\mathfrak {l}}:\mathfrak {l}\in \mathcal {S}^D\right )$ and $\left (k_{\mathfrak {l}}:\mathfrak {l}\in \mathcal L^D\backslash \mathcal {S}^D\right )$ satisfies all the conditions already listed, and the same happens for $\left (k_{\mathfrak {n}}':\mathfrak {n}\in \mathcal{T}\,^D\right )$ corresponding to $\left (k_{\mathfrak {l}}:\mathfrak {l}\in \mathcal {S}^D\right )$ and $\left (k_{\mathfrak {l}}':\mathfrak {l}\in \mathcal L^D\backslash \mathcal {S}^D\right )$ . We know that $\mathcal{T}\,^D$ is a tree of scale $2D(n_1+n_2+1)$ ,, and so $\left \lvert \mathcal L^D\right \rvert =4D(n_1+n_2+1)+1$ ; let the number of pairings be p, and then $\left \lvert \mathcal {S}^D\right \rvert =4D(n_1+n_2+1)-2p$ . By Proposition 3.5 we can bound the number of choicesFootnote ¹² for $\left (k_{\mathfrak {l}}:\mathfrak {l}\in \mathcal {S}^D\right )$ and $\left (k_{\mathfrak {l}}':\mathfrak {l}\in \mathcal L^D\backslash \mathcal {S}^D\right )$ by $M=L^{\theta } Q^{4D\left (n_1+n_2+1\right )-p}$ , and bound the number of choices for $\left (k_{\mathfrak {l}}':\mathfrak {l}\in \mathcal L^D\backslash \mathcal {S}^D\right )$ given $\left (k_{\mathfrak {l}}:\mathfrak {l}\in \mathcal {S}^D\right )$ by $M'=L^{\theta } Q^{p}$ . In the end, for any fixed $(k,k')$ , we have that L-certainly,

$$ \begin{align*} L^{d+\theta}\sup_{k,k'}\left\lvert\left((\mathcal{G}\mathcal{G}^*)^{D}\right)_{kk'}\right\rvert\leq L^{d+\theta}\left(\frac{\alpha T}{L^{d}}\right)^{2D\left(n_1+n_2+1\right)}\left(MM'\right)^{1/2}\leq L^{d+\theta}\rho^{2D\left(n_1+n_2+1\right)}. \end{align*} $$

Finally we need to L-certainly make this bound uniform in all choices of $(k,k')$ . This is not obvious, since we impose no upper bound on $\lvert k\rvert $ and $\lvert k'\rvert $ , so the number of exceptional sets we remove in the L-certain condition could presumably be infinite. However, note that the coefficient $\boldsymbol {\mathcal {K}}$ depends on k and $k'$ only through the quantities $\left \lvert k\right \rvert _{\beta }^2-\left \lvert k_{\mathfrak {r}_{j}}\right \rvert _{\beta }^2$ . Let $\mathcal {D}=\mathcal L\backslash \{\mathfrak {r'}\}$ ; then $\lvert k_{\mathfrak {l}}\rvert \leq L^{\theta }$ for $\mathfrak {l}\in \mathcal {D}$ , and the condition for summation creates the restriction that $\left \lvert \left \lvert k\right \rvert _{\beta }^2-\left \lvert k_{\mathfrak {r}_{j}}\right \rvert _{\beta }^2\right \rvert \leq L^{\theta ^{-1}}$ . The reduction from infinitely many possibilities for k (and hence $k'$ ) to finitely many is done by invoking the following result, whose proof will be left to the end:

Claim 3.7. Let $k\in \mathbb {Z}_L^d$ , and consider the function

$$ \begin{align*}f_{(k)}:m\mapsto \left\lvert k\right\rvert_{\beta}^2-\left\lvert k+m\right\rvert_{\beta}^2,\qquad \mathrm{Dom}\left(f_{(k)}\right)=\left\{m\in\mathbb{Z}_L^d:\lvert m\rvert\leq L^{\theta},\left\lvert\left\lvert k\right\rvert_{\beta}^2-\left\lvert k+m\right\rvert_{\beta}^2\right\rvert\leq L^{\theta^{-1}}\right\}.\end{align*} $$

Then there exist finitely many functions $f_1,\ldots ,f_A$ , where $A\leq L^{C\theta ^{-1}}$ , such that for any $k\in \mathbb {Z}_L^d$ there exists $1\leq j\leq A$ such that $\left \lvert f_{(k)}-f_j\right \rvert \leq L^{-\theta ^{-1}}$ on $\mathrm {Dom}\left (f_{(k)}\right )$ .

Now it is not hard to see that Claim 3.7 allows us to obtain a bound of the form proved that is uniform in $(k,k')$ , after removing at most $O\left (L^{C\theta ^{-1}}\right )$ exceptional sets, each with probability $\lesssim e^{-L^{\theta }}$ . This then implies

$$ \begin{align*}\left\lVert(\mathcal{G}\mathcal{G}^*)^{D}\right\rVert_{\ell^2\to\ell^2}\lesssim L^{d+\theta}\rho^{2D\left(n_1+n_2+1\right)},\end{align*} $$

hence

$$ \begin{align*}\left\lVert\mathcal{G}\right\rVert_{\ell^2\to\ell^2}\lesssim L^{\frac{d+\theta}{2D}}\rho^{n_1+n_2+1}.\end{align*} $$

By fixing D to be a sufficiently large positive integer, we deduce the correct operator bound for $\mathcal {G}$ , and hence for $\mathcal {X}$ and $\mathcal {P}_+$ . This completes the proof of Proposition 2.6.

Proof of Claim 3.7. We will prove the result for any linear function $g(m)=x\cdot m+X$ , where $x\in \mathbb {R}^d$ and $X\in \mathbb {R}$ are arbitrary. We may also assume $m\in \mathbb {Z}^d$ instead of $\mathbb {Z}_L^d$ ; the domain $\mathrm {Dom}(g)$ will then be the set E of m such that $\lvert m\rvert \leq L^{1+\theta }$ and $\lvert g(m)\rvert \leq L^{2+\theta ^{-1}}$ .

Let the affine dimension $\dim (E)=r\leq d$ ; then E contains a maximal affine independent set $\left \{q_j:0\leq j\leq r\right \}$ . The number of choices for these $q_j$ is at most $L^{d+1}$ , so we may fix them. Let $\mathcal L$ be the primitive lattice generated by $\left \{q_j-q_0:1\leq j\leq r\right \}$ , and fix a reduced basis $\left \{\ell _j:1\leq j\leq r\right \}$ of $\mathcal L$ . For any $m\in E$ there is a unique integer vector $k=(k_1,\ldots ,k_r)\in \mathbb {Z}^r$ such that $\lvert k\rvert \lesssim L^{1+\theta }$ , $m-q_0=k_1\ell _1+\cdots +k_r\ell _r$ , and as a linear function we can write $g(m)=y\cdot k+Y$ , where $y\in \mathbb {R}^r$ and $Y=g(q_0)\in \mathbb {R}$ .

Now let the $k\in \mathbb {Z}^r$ corresponding to $m=q_j$ be $k^{\left (j\right )}$ , where $1\leq j\leq r$ ; then since $q_0\in E$ and $q_j\in E$ , we conclude that $\left \lvert y\cdot k^{\left (j\right )}\right \rvert \leq L^{3+\theta ^{-1}}$ . As the $k^{\left (j\right )}$ are linear independent integer vectors in $\mathbb {Z}^r$ with norm bounded by $L^{1+\theta }$ , we conclude that $\lvert y\rvert \leq L^{C+\theta ^{-1}}$ , and consequently $\lvert Y\rvert \leq L^{C+\theta ^{-1}}$ . We may then approximate $g(m)$ for $m\in E$ by $y_j\cdot k+Y_j$ , where $y_j$ and $Y_j$ are one of the $L^{C\theta ^{-1}}$ choices that approximate y and Y up to error $L^{-\theta ^{-1}}$ , and choose $g_{\left (j\right )}=y_j\cdot k+Y_j$ .