Proof. It suffices to consider the incidences on a half sphere, since \(\mathbb{S}^2\) can be covered by eight half spheres. We define the map \(\Psi\colon \{(x_1,x_2,x_3)\in \mathbb{S}^2 \mid x_3>0\} \to \mathbb{R}^2\), \(\Psi(x_1,x_2,x_3)=(x_1/x_3,x_2/x_3)\). It’s easy to see that \(\Psi\) is a bijection, hence it preserves the number of incidences. Besides, \(\Psi\) maps the intersection of great circles and half sphere into lines on the plane, the conclusion follows from Theorem 6. ◻

Proof. Suppose \((x_1,x_2,x_3) \in \mathrm{Cone}_M^{\text{irr}}\), i.e. \(x_1^2=x_2^2+x_3^2\) and \(\gcd(x_1,x_2,x_3)=1\). It’s clear that \(x_2,x_3\) cannot be both even, we may assume \(x_3\) is odd, and \(x_3 = \pm p_1^{\alpha_1}\dots p_r^{\alpha_r}\) is the prime factorization. We note that \[p_i^{2\alpha_i} | x_3^2 = (x_1-x_2)(x_1+x_2),\] so there exists some \(\gamma_i\in\mathbb{N}\) such that \(p^{\gamma_i}_i | (x_1-x_2)\) and \(p_i^{2\alpha_i-\gamma_i} | (x_1+x_2)\). If \(\gamma_i \neq 0,2\alpha_i\), then \(p_i\) divides both \(x_1-x_2\) and \(x_1+x_2\) and hence \(p_i\) divides both \(2x_1\) and \(2x_2\). Consequently \(p_i|\gcd(x_1,x_2,x_3)\), which is a contradiction. Hence we have exactly one of \(p_i^{2\alpha_i}|(x_1-x_2)\) and \(p_i^{2\alpha_i}|(x_1+x_2)\) holds. Denote \(I = \{ 1\leq i\leq r\mid p_i^{2\alpha_i} \text{ divides } x_1-x_2 \}\) and \[m = \prod_{i\in I} p_i^{\alpha_i},\quad n = \prod_{i\notin I} p_i^{\alpha_i} = |x_3|/m,\] where the product is defined to be \(1\) if the index set is empty. Then \(\gcd(m,n)=1\) and we have \((x_1-x_2 ,x_1+x_2 )= \pm (m^2,n^2)\), or equivalently \[(x_1, x_2) = \pm \left( \frac{n^2+m^2}{2}, \frac{n^2-m^2}{2} \right).\] Therefore, each point \((x_1,x_2,x_3) \in \mathrm{Cone}_M^{\text{irr}}\) can be represented by a pair \((m,n) \in \mathbb{Z}^2\) satisfying \(m^2+n^2 \lesssim M\), and hence \(\#\mathrm{Cone}_M^{\text{irr}} \lesssim M\). ◻

Proof. It suffices to consider the case \(\xi_1-\xi_2 \in \mathrm{Cone}\) and \(\xi_1\neq \xi_4\). For given \((\xi_1,\xi_4)\),

from previous discussion we know there exists some plane \(H\) contains both \(\xi_1,\xi_4\) and its normal vector belongs to \(\mathrm{Cone}\). It’s not hard to check that there exist at most two such planes. For each such plane \(H\), \(A(\xi_1-\xi_2)\) is a multiple of its normal vector and hence \(\xi_2\) lies on a line \(\ell\) passing through \(\xi_1\) with direction determined by \(H\).

We may write \(\xi_2 \in \ell \cap S \subset \mathbb{Z}^3\) as \(\xi_1+r\xi\) with \(\xi \in \mathbb{Z}^3\setminus\{0\}\), thus \(r\xi \in \mathbb{Z}^3\) and \(r\) belongs to an interval of length \(|\xi|^{-1} \operatorname{diam}(S)\). From Bézout’s identity, we know \(\gcd(\xi)\) can be written as linear combination of coordinates of \(\xi\) with integer coefficients, we have \(r \gcd(\xi) \in \mathbb{Z}\). Thus \[\label{number32of32points32on32intersection32of32line32and32S} \#(\ell \cap S) = \# \Bigl\{ r\in \frac{1}{\gcd(\xi)} \mathbb{Z} \Bigm| \xi_1+r\xi \in \ell\cap S \Bigr\} \leq \frac{\gcd(\xi)}{|\xi|} \operatorname{diam}(S),\tag{10}\] which implies for each pair \((\xi_1,\xi_4)\), there exists at most \(O(\operatorname{diam}(S))\) many choices of \((\xi_2,\xi_3)\) such that \({(\xi_1,\xi_2,\xi_3,\xi_4)} \in \mathcal{Q}_2\). We denote all the possible choices as \({(\xi_2,\xi_3)\in \mathcal{R}(\xi_1,\xi_4)}\).

On the other hand, for given \((\xi_2,\xi_3)\), we can also apply the same argument to \((\xi_1,\xi_4)\), and hence for each pair \((\xi_2,\xi_3)\), there exists at most \(O(\operatorname{diam}(S))\) many choices of \((\xi_1,\xi_4)\) such that \({(\xi_1,\xi_2,\xi_3,\xi_4)} \in \mathcal{Q}_2\). As a result, \[\begin{align} \Omega_2(f) &= \sum_{\xi_1,\xi_4} \Big( f(\xi_1)f(\xi_4) \sum_{(\xi_2,\xi_3)\in \mathcal{R}(\xi_1,\xi_4)} f(\xi_2)f(\xi_3) \Big) \\& \leq \Big( \sum_{\xi_1,\xi_4} |f(\xi_1)f(\xi_4)|^2 \Big)^{1/2} \Big( \sum_{\xi_1,\xi_4} \Big( \sum_{(\xi_2,\xi_3)\in \mathcal{R}(\xi_1,\xi_4)} f(\xi_2)f(\xi_3) \Big)^2 \Big)^{1/2} \\& \lesssim \operatorname{diam}(S)^{1/2} \|f\|^2_{\ell^2(\mathbb{Z}^3)} \Big( \sum_{\xi_1,\xi_4} \sum_{(\xi_2,\xi_3)\in \mathcal{R}(\xi_1,\xi_4)} | f(\xi_2)f(\xi_3) |^2 \Big)^{1/2} \\& =\operatorname{diam}(S)^{1/2} \|f\|^2_{\ell^2(\mathbb{Z}^3)} \Big( \sum_{\xi_2,\xi_3} \sum_{(\xi_1,\xi_4)\in\mathcal{R}(\xi_2,\xi_3)} | f(\xi_2)f(\xi_3) |^2 \Big)^{1/2} \\& \lesssim \operatorname{diam}(S) \|f\|_{\ell^2(\mathbb{Z}^3)}^4. \end{align}\] ◻

Proof. We set \(\{H_i\}\) to be the set of all planes \(H\) with normal vector in \(\mathrm{Cone}_M^{\mathrm{irr}}\) and satisfying \(\| f\chi_H \|_{\ell^2(\mathbb{Z}^3)}^2 \geq M^{-2} \|f\|_{\ell^2(\mathbb{Z}^3)}^2\). For each \(n\in \mathrm{Cone}_M^{\mathrm{irr}}\), the planes with normal vector \(n\) are parallel with each other, which implies \[\# \{ H \in \mathcal{H}(\{n\}) \mid \| f\chi_H \|_{\ell^2(\mathbb{Z}^3)}^2 \geq M^{-2} \|f\|_{\ell^2(\mathbb{Z}^3)}^2 \} \leq M^2,\] thus \(\#\{H_i\} \leq M^2 \# \mathrm{Cone}_M^{\mathrm{irr}} \lesssim M^3\).

It remains to verify ?? . By the decomposition 9 and the facts that \(0\leq f^{\mathrm{error}} \leq f\), \(f^{\mathrm{error}} \chi_{H_i}=0\) and \(\mathcal{H}(\mathrm{Cone}_M^{\mathrm{irr}})= \mathcal{H}(\mathrm{Cone}_M)\), we get \[\begin{align} \Omega_2(f^{\mathrm{error}}) &\leq \sum_{ H \in \mathcal{H}(\mathrm{Cone}) } \Omega_2(f^{\mathrm{error}} \chi_H) \\& \leq \sum_{ H \in \mathcal{H}(\mathrm{Cone}\setminus \mathrm{Cone}_M) } \Omega_2(f \chi_H) + \sum_{ H \in \mathcal{H}(\mathrm{Cone}_M^{\mathrm{irr}} ) \setminus \{H_i\} } \Omega_2(f \chi_H). \end{align}\] Recall the estimate 10 , by using arguments similar to that in proof of Proposition 11, we have \[\sum_{ H \in \mathcal{H}(\mathrm{Cone}\setminus \mathrm{Cone}_M) } \Omega_2(f \chi_H) \lesssim M^{-1}\operatorname{diam}(S) \| f\|_{\ell^2(\mathbb{Z}^3)}^4.\] On the other hand, for each \(H \in \mathcal{H}(\mathrm{Cone}_M^{\mathrm{irr}}) \setminus \{H_i\},\) we have \(\| f\chi_H \|_{\ell^2(\mathbb{Z}^3)}^2 \leq M^{-2} \| f \|_{\ell^2(\mathbb{Z}^3)}^2\), and \[\begin{align} \sum_{ H \in \mathcal{H}(\mathrm{Cone}_M^{\mathrm{irr}} ) \setminus \{H_i\} } \Omega_2(f \chi_H) & ={} \sum_{n\in \mathrm{Cone}_M^{\mathrm{irr}}} \sum_{\substack{H \notin \{H_i\} \\ H \perp n}} \Omega_2(f\chi_H) \\& \lesssim{} \sum_{n\in \mathrm{Cone}_M^{\mathrm{irr}}} \sum_{\substack{H \notin \{H_i\} \\ H \perp n}}\operatorname{diam}(S) \| f\chi_H \|_{\ell^2(\mathbb{Z}^3)}^4 \\& \leq{} \frac{\operatorname{diam}(S) \|f \|_{\ell^2(\mathbb{Z}^3)}^2 }{M^2}\sum_{n\in \mathrm{Cone}_M^{\mathrm{irr}}} \sum_{\substack{H \notin \{H_i\} \\ H \perp n}} \|f\chi_H\|_{\ell^2(\mathbb{Z}^3)}^2 \\& \leq{} M^{-2}\#\mathrm{Cone}_M^{\mathrm{irr}} \operatorname{diam}(S) \|f\|_{\ell^2(\mathbb{Z}^3)}^4 \\& \lesssim M^{-1} \operatorname{diam}(S) \| f ||_{\ell^2(\mathbb{Z}^3)}^4 . \end{align}\] Here we used Proposition 11 for the first inequality and Lemma 10 for the last inequality. Hence \[\Omega_2(f^{\mathrm{error}}) \lesssim M^{-1}\operatorname{diam}(S) \| f\|_{\ell^2(\mathbb{Z}^3)}^4.\] ◻

Proof. For each line \(\ell\), we denote its direction vector as \(v_\ell \in \mathbb{S}^2\), and \(c_\ell = \{ v\in \mathbb{S}^2 \mid v \cdot Av_{\ell} =0 \}\). Then \(\mathfrak{c}(\xi,\ell,\ell') = 1\) is equivalent to \(v_\ell \in c_{\ell'}\). Set \(\mathcal{P} = \{ v_\ell \mid \ell \in \mathcal{L} \}\) and \(\mathcal{C} = \{ c_{\ell'} \mid \ell'\in \mathcal{L}' \}\), then the summation of \(\mathfrak{c}(\xi,\ell,\ell')\) is bounded by the number of incidences between points in \(\mathcal{P}\) and circles in \(\mathcal{C}\), see Theorem 8. ◻

Proof of Proposition 13. For each dyadic integer \(s\), we put \(\mathcal{L}_s\) to be the set of lines \(\ell\) such that \(s\leq \#(\ell\cap S) < 2s\) and the direction vector of \(\ell\) doesn’t belong to \(\mathrm{Cone}\). Then \[\begin{align} \Omega_1(f) &= \sum_{s,t\text{ dyadic}} \#\{ {(\xi_1,\xi_2,\xi_3,\xi_4)}\in \mathcal{Q}_1 \mid \ell_{\xi_1\xi_2} \in \mathcal{L}_s,\ell_{\xi_1,\xi_4}\in\mathcal{L}_t \} \\& \leq \sum_{s,t\text{ dyadic}} \sum_{\ell \in \mathcal{L}_s} \sum_{\ell'\in \mathcal{L}_t} \sum_{\xi_1\in S} \mathfrak{c}(\xi_1,\ell,\ell')\#(\ell\cap S)\#(\ell'\cap S). \label{bound:Omega1-inci} \end{align}\tag{11}\] For \(s,t\leq (\#S)^{1/3}\), we use Lemma 14 to estimate \[\begin{align} \sum_{s,t\text{ dyadic}} \sum_{\xi_1\in S}st \Big( \sum_{\ell \in \mathcal{L}_s} \sum_{\ell'\in \mathcal{L}_t} \mathfrak{c}(\xi_1,\ell,\ell') \Big) \lesssim \sum_{s,t\text{ dyadic}} st \sum_{\xi_1\in S}( (\mathcal{J}_{\xi_1}^s\mathcal{J}_{\xi_1}^t)^{2/3} + \mathcal{J}_{\xi_1}^s+\mathcal{J}_{\xi_1}^t ), \end{align}\] where \(\mathcal{J}_{\xi_1}^s,\mathcal{J}_{\xi_1}^t\) denote the number of lines in \(\mathcal{L}_s,\mathcal{L}_t\) passing through \(\xi_1\). By Corollary 7, we have \[\sum_{\xi_1\in S} \mathcal{J}_{\xi_1}^s \lesssim \frac{(\#S)^2}{s^2},\quad \sum_{\xi_1\in S} \mathcal{J}_{\xi_1}^t \lesssim \frac{(\#S)^2}{t^2},\] hence \[\sum_{s,t \leq (\#S)^{1/3}\text{ dyadic}} st \sum_{\xi_1\in S}(\mathcal{J}_{\xi_1}^s+\mathcal{J}_{\xi_1}^t) \lesssim (\#S)^{7/3}.\] On the other hand, since all the lines passing through \(\xi_1\) are pairwise disjoint (excluding the common point \(\xi_1\)), we have \(\mathcal{J}_{\xi_1}^s \lesssim \#S/s\) and \(\mathcal{J}_{\xi_1}^t \lesssim \#S/t\). Therefore, \[\begin{align} \sum_{s,t\text{ dyadic}} st \sum_{\xi_1\in S} (\mathcal{J}_{\xi_1}^s\mathcal{J}_{\xi_1}^t)^{2/3} &\lesssim \sum_{s,t\text{ dyadic}} st \sum_{\xi_1 \in S} \left( \frac{(\#S)^2}{st} \right)^{1/6} (\mathcal{J}_{\xi_1}^s\mathcal{J}_{\xi_1}^t)^{1/2} \\&\leq \sum_{s,t\text{ dyadic}} {(\#S)^{1/3}}{(st)^{5/6}} \Big( \sum_{\xi_1\in S} \mathcal{J}_{\xi_1}^s \Big)^{1/2}\Big( \sum_{\xi_1\in S} \mathcal{J}_{\xi_1}^t\Big)^{1/2} \\& \lesssim \sum_{s,t\text{ dyadic}} {(\#S)^{1/3}}{(st)^{5/6}}\frac{\#S}{s}\frac{\#S}{t} \lesssim (\#S)^{7/3}. \end{align}\] This proves ?? when \(s,t\leq (\#S)^{1/3}\).

For \(s\geq(\#S)^{1/3}\) or \(t\geq (\#S)^{1/3}\), we assume \(s\geq (\#S)^{1/3}\) without loss of generality. We notice that for any fixed \(\ell\) and \(\xi_1\in\ell\), the lines \(\ell'\) such that \(\mathfrak{c}(\xi_1,\ell,\ell')\neq 0\) belong to some plane determined by \(\ell\) and \(\xi_1\). Furthermore, for different choices of \(\xi_1\), these planes are pairwise disjoint due to the fact that the direction vector of \(\ell\) doesn’t belong to \(\mathrm{Cone}\). Thus all these lines \(\ell'\) are pairwise disjoint (excluding the possible common points on \(\ell\)), and \[\sum_t \sum_{\ell'\in\mathcal{L}_t} \sum_{\xi_1\in S} \mathfrak{c}(\xi_1,\ell,\ell') \#(\ell'\cap S) \lesssim \#S. \label{est:case2-1}\tag{12}\] On the other hand, by Corollary 7 we have \[\sum_{s\geq (\#S)^{1/3}} \sum_{\ell \in \mathcal{L}_s} \#(\ell \cap S) \lesssim (\#S)^{4/3}.\label{est:case2-2}\tag{13}\] 11 , 12 and 13 together imply the bound ?? . ◻

Proof of Theorem 1. We set \(f = |\hat{\phi}| \chi_{[-N,N]^3}\) and enumerate \(\mathbb{Z}^3 \cap [-N,N]^3\) as \(\xi_1,\xi_2,\dots\) such that \[f(\xi_1) \geq f(\xi_2) \geq \dots.\] Let \(S_j = \{ \xi_{2^{j}},\dots,\xi_{2^{j+1}-1} \}\), \(f_j = f(\xi_{2^j}) \chi_{S_j}\) and \(\lambda_j = 2^{j/2} f(\xi_{2^j})\) for \(0\leq j \leq {j_{\text{max}}}\) with \(2^{j_{\text{max}}}\lesssim N^3\). We have \(\# S_j \leq 2^j\), \(f \leq \sum_{j} f_j\), \(|\lambda_j|=\|f_j\|_{\ell^2(\mathbb{Z}^3)}\) and \[\| \lambda_j\|_{\ell^2_{j\leq {j_{\text{max}}}}}^2 = \sum_{j=0}^{{j_{\text{max}}}} 2^j|f(\xi_{2^j})|^2 \leq |f(\xi_1)|^2 + \sum_j \| f\chi_{S_{j-1} }\|_{\ell^2(\mathbb{Z}^3)}^2 \lesssim \|f\|_{\ell^2(\mathbb{Z}^3)}^2. \label{est:lambda-f}\tag{14}\] Let \(\delta>0\) sufficiently small. Given such a decomposition of \(f\), we say \(j\) is good if \(\Omega_2(f_j) \lesssim N^{1-\delta} \| f_j \|_{\ell^2(\mathbb{Z}^3)}^4\), otherwise we say \(j\) is bad. For each bad \(j\), by taking \(M=N^\delta\) in Proposition 12, we can find at most \(O(N^{3\delta})\) planes \(\{H_{j}^{i_j}\} \subset \mathcal{H}(\mathrm{Cone}_M^{\mathrm{irr}})\), such that \(f_j^{\mathrm{error}}:=f\chi_{S_j \setminus \cup_{i_j} H_{j}^{i_j}}\) satisfies \[\Omega_2(f_j^{\mathrm{error}}) \lesssim N^{1-\delta} \| f_j\|_{\ell^2(\mathbb{Z}^3)}^4. \label{def:fgood}\tag{15}\] We denote \(f_j^{\mathrm{good}}=f_j^{}\) if \(j\) is good and \(f_j^{\mathrm{good}}=f_j^{\mathrm{error}}\) if \(j\) is bad, and \[f_{\text{bad}} = \sum_{j \text{ bad} } (f_j^{}-f_j^{\mathrm{good}}).\] Then from Proposition 13 and 15 , we get bounds for each \(f_j^{\mathrm{good}}\) \[\begin{align} \| \mathrm{e}^{\mathrm{i}t\Box} \check{f}_j^\text{good} \|^4_{{L_{t,x}^4([0,1]\times \mathbb{T}^3)}} &= \Omega_1(f_j^\text{good}) + \Omega_2(f_j^\text{good})\\ & \lesssim (2^{{j}/{3}}+N^{1-\delta})\|f_j^\text{good}\|^4_{\ell^2(\mathbb{Z}^3)}. \label{est:f-good} \end{align}\tag{16}\] Now we control the contribution of \(f_{\text{bad}}\), which can be written as \[\begin{align} &\| \mathrm{e}^{\mathrm{i}t\Box} \check f_{\text{bad}}\|_{{L_{t,x}^4([0,1]\times \mathbb{T}^3)}}^4 = \Omega_1(f_{\text{bad}}) +\Omega_2(f_{\text{bad}}) . \end{align}\] Proposition 11 indicates that \(\Omega_2(f_{\text{bad}}) \lesssim N \| f_{\text{bad}} \|_{{\ell^2(\mathbb{Z}^3)}}^4\). Meanwhile, we have the estimate \[\begin{align} \Omega_1(f_{\text{bad}}) &\lesssim \sum_{\substack{j_1,j_2,j_3,j_4 \text{ bad} \\ i_{j_1},i_{j_2},i_{j_3},i_{j_4} \lesssim N^{3\delta}}} \Omega_1(f_{j_1}\chi_{H_{j_1}^{i_{j_1}}},f_{j_2}\chi_{H_{j_2}^{i_{j_2}}}, f_{j_3}\chi_{H_{j_3}^{i_{j_3}}}, f_{j_4}\chi_{H_{j_4}^{i_{j_4}}}) \\ &\lesssim \sum_{j_1,j_2,j_3,j_4} N^{3/4+O(\delta)} \prod_{k=1}^4 \| f_{j_k} \|_{\ell^2(\mathbb{Z}^3)} \lesssim N \| f_{\text{bad}}\|_{\ell^2(\mathbb{Z}^3)}^4, \end{align}\] which is a consequence of applying the following Proposition 15 to each quadruple \((i_{j_1},i_{j_2},i_{j_3},i_{j_4})\) with \(M=N^\delta\).

Combining all the estimates, we get \[\begin{align} \| \mathrm{e}^{\mathrm{i}t\Box}\check{f} \|_{{L_{t,x}^4([0,1]\times \mathbb{T}^3)}} &\lesssim \sum_{j} \| \mathrm{e}^{\mathrm{i}t\Box} \check{f}_j^\text{good} \|_{{L_{t,x}^4([0,1]\times \mathbb{T}^3)}} + \| \mathrm{e}^{\mathrm{i}t\Box} \check f_{\text{bad}}\|_{{L_{t,x}^4([0,1]\times \mathbb{T}^3)}}\\& \lesssim \sum_{j \leq {j_{\text{max}}}} (2^{j/12} + N^{(1-\delta)/4})\lambda_j + \| \mathrm{e}^{\mathrm{i}t\Box} \check f_{\text{bad}}\|_{{L_{t,x}^4([0,1]\times \mathbb{T}^3)}}\\& \lesssim N^{1/4} \| \lambda_j \|_{\ell_{j\leq {j_{\text{max}}}}^2} + \| \mathrm{e}^{\mathrm{i}t\Box} \check f_{\text{bad}}\|_{{L_{t,x}^4([0,1]\times \mathbb{T}^3)}}\\ &\lesssim N^{1/4} \| f\|_{\ell^2(\mathbb{Z}^3)}. \end{align}\] From 6 , we see \(\| \mathrm{e}^{\mathrm{i}t\Box}\phi \|_{{L_{t,x}^4([0,1]\times \mathbb{T}^3)}}\leq \| \mathrm{e}^{\mathrm{i}t\Box}\check{f} \|_{{L_{t,x}^4([0,1]\times \mathbb{T}^3)}}\). Thus we finish the proof for the \(L^4\) estimate ?? , and the general case ?? follows from interpolation of \(L^4\) with \(L^2\) and \(L^\infty\) estimates. ◻

Proof. We decompose \(\mathcal{Q}_1\) into \[\bigcup_{(a,b) \in \mathbb{Z}^3\times \mathbb{Z} } \Gamma_{a,b} :=\bigcup \Bigg\{ {(\xi_1,\xi_2,\xi_3,\xi_4)} \in \mathcal{Q}_1 \Biggm| \begin{array}{cc} \xi_1+\xi_3=a=\xi_2+\xi_4 \\ h(\xi_1)+h(\xi_3) = b = h(\xi_2)+h(\xi_4)\end{array}\Bigg\}.\] Also we denote \(\mathcal{A}_{a,b} = \{ \xi \in \mathbb{Z}^3 \mid h(\xi)+h(a-\xi) = b \}\). Then \({(\xi_1,\xi_2,\xi_3,\xi_4)} \in \Gamma_{a,b}\) implies \(\xi_j \in \mathcal{A}_{a,b}\) for \(1\leq j\leq 4\). It’s easy to see that \[\begin{align} \Omega_1{(f_1,f_2,f_3,f_4)} &= \sum_{a,b} \sum_{\Gamma_{a,b}} f_1(\xi_1)f_2(\xi_2)f_3(\xi_3)f_4(\xi_4) \tag{17} \\& = \sum_{a,b} \Big( \sum_{\substack{\xi_1,\xi_3\in \mathcal{A}_{a,b}\\\xi_1+\xi_3=a}} f_1(\xi_1)f_3(\xi_3) \Big)\Big( \sum_{\substack{\xi_2,\xi_4\in \mathcal{A}_{a,b}\\\xi_2+\xi_4=a}} f_2(\xi_2)f_4(\xi_4) \Big) . \\\tag{18} \end{align}\] By Cauchy-Schwarz, it’s further bounded by \[\bigg( \sum_{a,b} \Big( \sum_{\substack{\xi_1 \in \mathcal{A}_{a,b} }} f_1(\xi_1)f_3(a-\xi_1) \Big)^2 \bigg)^{1/2} \bigg( \sum_{a,b} \Big( \sum_{\substack{\xi_2 \in \mathcal{A}_{a,b} }} f_2(\xi_2)f_4(a-\xi_2) \Big)^2 \bigg)^{1/2}.\label{est:4f-Aab-CS}\tag{19}\]

Next we estimate the size of \(\mathcal{A}_{a,b}\cap H_1\cap[-N,N]^3\), the same argument holds for \(j=2,3,4\). Since \(n_1\in \mathrm{Cone}_M^{\mathrm{irr}}\) is the normal vector of \(H_1\) and \(n_1 \perp An_1\), hence \(\{n_1,An_1,n_1\times An_1\}\) is an orthogonal basis of \(\mathbb{R}^3\). We decompose \(\xi\) with respect to this orthogonal basis and denote \[\xi^\perp = \xi - \frac{ \xi\cdot n_1 }{|n_1|^2} n_1 - \frac{\xi\cdot An_1}{|n_1|^2}An_1 \in \frac{1}{|n_1|^2} \mathbb{Z}^3. \label{definition32of32perp32part}\tag{20}\] Clearly \(\xi^\perp\) belongs to the one-dimensional linear subspace spanned by \(n_1\times An_1\). For \(\xi\in H_1\), the inner product \(\xi\cdot n_1 = c_1\) is constant, and we compute \(h(\xi)\) by using the decomposition 20 \[\begin{align} h(\xi) &= \xi \cdot A\xi = \xi^\perp \cdot A\xi^\perp + (\xi - \xi^\perp)\cdot A(\xi-\xi^\perp) + 2\xi^\perp \cdot A(\xi-\xi^\perp) \\ &= h(\xi^\perp) + 2c_1 \frac{\xi\cdot An_1}{|n_1|^2} . \end{align}\] As a result, it holds that for \(\xi \in \mathcal{A}_{a,b} \cap H_1\) \[\begin{align} b &= h(\xi) + h(a-\xi) = 2h(\xi)+h(a)-2\xi\cdot Aa \\& = 2h(\xi^\perp) + 4c_1 \frac{ \xi \cdot An_1 }{|n_1|^2} + h(a) -2\xi^\perp\cdot Aa - \frac{2( c_1 a\cdot An_1 + (\xi_1\cdot An_1)(a\cdot n_1) )}{|n_1|^2} \end{align}\] i.e. \[\frac{4c_1 - 2a\cdot n_1}{|n_1|^2} \xi\cdot An_1 = -2h(\xi^\perp)+2\xi^\perp\cdot Aa + \Big[ b - h(a) + \frac{2c_1}{|n_1|^2} a\cdot An_1 \Big]. \label{equation32of32xi44a44b}\tag{21}\]

The following lines are dedicated to further simplify 21 , which gives us the desired estimate of the size of \(\mathcal{A}_{a,b}\cap H_1\cap[-N,N]^3\). As in 20 , we denote \[a^\perp = a - \frac{ a \cdot n_1 }{|n_1|^2} n_1 - \frac{a \cdot An_1}{|n_1|^2}An_1 \in \frac{1}{|n_1|^2} \mathbb{Z}^3. \label{a32perp32part}\tag{22}\] There exists some \(\eta \in \mathbb{Z}^3\) such that \(|\eta| \leq |n_1\times An_1| \leq M^2\) and \(2\xi^\perp - a^\perp = \frac{\lambda}{|n_1|^2} \eta\) for some \(\lambda \in\mathbb{Z}\). Since the matrix \(A\) is non-singular, \(h(\eta)\) must be non-zero. Now 21 can be written as \[\begin{align} \frac{\lambda^2}{|n_1|^4} h(\eta) &= h(2\xi^\perp-a^\perp) \\ &= 2\left[ b+h(a^\perp)-h(a)+\frac{2c_1}{|n_1|^2}a\cdot n_1 \right] - \frac{8c_1-4a\cdot n_1}{|n_1|^2}\xi\cdot An_1. \end{align}\] Set \(z = \lambda h(\eta)\), \(y = |n_1|^2 \xi \cdot An_1\) and \[\begin{align} q&= h(\eta )( 8c_1-4a_1\cdot n_1) , \\ \omega&=2|n_1|^4h(\eta )\left[ b+h(a^\perp)-h(a)+\frac{2c_1}{|n_1|^2}a\cdot n_1 \right], \end{align}\] then \((y,z,q,\omega)\) is an integer solution of \[z^2=qy+\omega, \quad |q|,|y|,|z| \lesssim M^8N. \label{eq:claim}\tag{23}\]

For given \(a,b\), if \(a \cdot n_1 \neq 2c_1\), which is equivalent to \(a/2 \notin H_1\), then from 21 , \(\xi\cdot An_1\) is determined by \(\xi^\perp\). From 20 , \(\xi\in \mathcal{A}_{a,b} \cap H_1\) is determined by \(\xi^\perp\) since \(\xi\cdot n_1=c_1\). Let us state the following claim, postponing its proof for later.

Claim: For given \(N,q\in\mathbb{Z}_+\) and \(\omega \in \mathbb{Z}\), \[\#\Big( \{ (y,z)\in \mathbb{Z}^2 \mid z^2= qy+\omega \} \cap [-N,N]^2\Big)\lesssim \sqrt{N}+\sqrt{q}.\] Applying the claim to 23 , we see that there are \(O(M^{4}N^{1/2})\) many choices of \(\xi^\perp\), hence \[\#(\mathcal{A}_{a,b}\cap H_1\cap[-N,N]^3) \lesssim M^{4} N^{1/2}.\label{Aab-bound:curve}\tag{24}\]

On the other hand, if \(a\cdot n_1 = 2c_1\), i.e. \(a/2\in H_1\), the left hand side of 21 is \(0\) and we have at most two choices of \(\xi^\perp\), hence from 20 we know \(\xi\) belongs to the union of two lines contained in \(H_1\), \[\#(\mathcal{A}_{a,b} \cap H_1\cap[-N,N]^3) \lesssim N. \label{Aab-bound:line}\tag{25}\]

Now we prove the main estimate ?? .

Case 1: \(a\cdot n_{j_0} \neq 2c_{j_0}\), i.e. \(a/2\notin H_{j_0}\), for some \(1\leq j_0\leq 4\). Without loss of generality, we assume \(j_0=1\). By Cauchy-Schwarz and 24 we have \[\begin{align} \Big( \sum_{\xi \in \mathcal{A}_{a,b} } f_{1}(\xi) f_{3}(a-\xi) \Big)^2 \lesssim M^{4} N^{1/2} \sum_{\xi\in\mathcal{A}_{a,b}} |f_{1}(\xi)|^2 | f_{3}(a-\xi)|^2, \end{align}\] summing over \(a,b\) gives that \[\sum_{\frac{a}{2} \notin H_1} \sum_b \Big( \sum_{\xi \in \mathcal{A}_{a,b} } f_{1}(\xi) f_{3}(a-\xi) \Big)^2 \lesssim M^{4}N^{1/2} \| f_1 \|_{\ell^2(\mathbb{Z}^3)}^2 \| f_3 \|_{\ell^2(\mathbb{Z}^3)}^2.\] On the other hand \[\begin{align} \sum_{a,b} \Big( \sum_{\xi \in \mathcal{A}_{a,b} } f_{2}(\xi) f_{4}(a-\xi) \Big)^2 &\lesssim \max\{ M^4N^{1/2}, N\} \sum_{a,b} \sum_{\xi\in\mathcal{A}_{a,b}} |f_{2}(\xi)|^2 | f_{4}(a-\xi)|^2 \\& \lesssim \max\{M^4N^{1/2},N\} \| f_2 \|_{\ell^2(\mathbb{Z}^3)}^2 \| f_4 \|_{\ell^2(\mathbb{Z}^3)}^2, \end{align}\] as a result \[\begin{align} \sum_{\frac{a}{2} \notin H_1} \sum_b \sum_{\Gamma_{a,b}} f_1(\xi_1)f_2(\xi_2)f_3(\xi_3)f_4(\xi_4) \lesssim M^{2}N^{1/2}(M^2+N^{1/4}) \prod_{j=1}^4 \| f_j \|_{\ell^2(\mathbb{Z}^3)}. \end{align}\]

Case 2: \(a/2 \in H_j\) for all \(j\).

•If \(\dim \operatorname{span}_\mathbb{R} \{n_1,n_2,n_3,n_4\} = 3\). In this case there are at most one such \(a\), and hence from 18 \[\begin{align} & \sum_{\frac{a}{2} \in \cap_j H_j} \sum_b \sum_{\Gamma_{a,b}} f_1(\xi_1)f_2(\xi_2)f_3(\xi_3)f_4(\xi_4)\\ \lesssim & \Big(\sum_{\xi_1} f_1(\xi_1)f_3(a-\xi_1)\Big)\Big( \sum_{\xi_2} f_2(\xi_2)f_4(a-\xi_2) \Big) \leq \prod_{j=1}^4 \| f_j \|_{\ell^2(\mathbb{Z}^3)}. \end{align}\]

•If \(\dim \operatorname{span}_\mathbb{R} \{n_1,n_2,n_3,n_4\} = 2\). Let \(2\leq j \leq 4\) be such that \[\operatorname{span}_\mathbb{R} \{ n_1,n_j\} = \operatorname{span}_{\mathbb{R}} \{ n_1,n_2,n_3,n_4 \},\] we will prove \[\sum_{\frac{a}{2}\in H_1\cap H_j} \sum_{b} \Big( \sum_{\substack{\xi \in \mathcal{A}_{a,b} }} f_1(\xi)f_3(a-\xi) \Big)^2 \lesssim \| f_1\|_{\ell^2(\mathbb{Z}^3)}^2 \| f_3 \|_{\ell^2(\mathbb{Z}^3)}^2.\] By symmetry, the same estimate holds for \(f_2,f_4\).

Recall the definition of \(a^\perp\) in 22 , then we can write that \[2c_j = a\cdot n_j = \left[ a^\perp \cdot n_j + \frac{2c_1}{|n_1|^2}{ n_1\cdot n_j} \right] + \frac{a\cdot An_1}{|n_1|^2} An_1\cdot n_j.\] We claim that \(An_1 \cdot n_j = n_1\cdot An_j\neq 0\). Otherwise, since \(n_1,n_j \in \mathrm{Cone}\), it holds that \[\operatorname{span}_{\mathbb{R}}\{An_1,An_j\} \perp \operatorname{span}_{\mathbb{R}}\{n_1,n_j\},\] but their dimensions are both \(2\) , which is a contradiction. As a result, we can solve \(a\cdot An_1\) in terms of \(a^\perp\), and hence \(a\) is determined by \(a^\perp\).

We set \(\mathcal{A}_{a,b}^\perp = \{ \xi^\perp \mid \xi \in \mathcal{A}_{a,b} \}\), where \(\xi^\perp\) is defined as 20 . Then \[\mathcal{A}_{a,b} =\bigcup_{\beta\in \mathcal{A}_{a,b}^\perp}\{\xi \in \mathcal{A}_{a,b} \mid \xi^\perp=\beta\} .\] From 21 we see that \(\# \mathcal{A}_{a,b}^\perp \leq 2\) for each \(a,b\).

By Cauchy-Schwarz inequality, \[\begin{align} \Big( \sum_{\substack{\xi \in \mathcal{A}_{a,b} }} f_1(\xi)f_3(a-\xi) \Big)^2 &\lesssim \sum_{{\beta} \in \mathcal{A}_{a,b}^\perp} \Big(\sum_{\substack{\xi\in [-N,N]^3, \xi^\perp = {\beta}}} f_1(\xi)f_3(a-\xi) \Big)^2 \\&\leq \sum_{{\beta} \in \mathcal{A}_{a,b}^\perp } {\Big( \sum_{\xi^\perp={\beta}} |f_1(\xi)|^2 \Big)} {\Big( \sum_{\xi^\perp = a^\perp - {\beta}} |f_3(\xi)|^2\Big)}. \end{align}\] Therefore, \[\begin{align} & \mathrel{\phantom{\lesssim}} \sum_{\frac{a}{2}\in H_1\cap H_j} \sum_{b} \Big( \sum_{\substack{\xi \in \mathcal{A}_{a,b} }} f_1(\xi)f_3(a-\xi) \Big)^2 \\ &\lesssim \sum_{\frac{a}{2}\in H_1\cap H_j} \sum_{b} \sum_{{\beta} \in \mathcal{A}_{a,b}^\perp } {\Big( \sum_{\xi^\perp={\beta}} |f_1(\xi)|^2 \Big)} {\Big( \sum_{\xi^\perp = a^\perp - {\beta}} |f_3(\xi)|^2\Big)}\\& = \sum_{\frac{a}{2}\in H_1\cap H_j} \sum_{{\beta}} {\Big( \sum_{\xi^\perp={\beta}} |f_1(\xi)|^2 \Big)} {\Big( \sum_{\xi^\perp = a^\perp - {\beta}} |f_3(\xi)|^2\Big)} \\& = \| f_1\|_{\ell^2(\mathbb{Z}^3)}^2 \| f_3 \|_{\ell^2(\mathbb{Z}^3)}^2. \end{align}\]

• If \(\dim\operatorname{span}_\mathbb{R}\{ n_1,n_2,n_3,n_4\}=1\). In this case \(n_1=n_2=n_3=n_4\) and \(H_1=H_2=H_3=H_4\), otherwise the intersection is empty. Suppose \({(\xi_1,\xi_2,\xi_3,\xi_4)} \in \mathcal{Q}\) with \(\xi_1-\xi_2\) is not a multiple of \(An_1\), we have \({(\xi_1-\xi_2)\cdot A(\xi_1-\xi_4)=0}\) and \(An_1\cdot A(\xi_1-\xi_4)=0\) since \(n_1\) is the normal vector of \(H_1\), which implies \(A(\xi_1-\xi_4)\) is also a normal vector of \(H_1\) and hence \(A(\xi_1-\xi_4)\) is a multiple of \(n_1\) and \(\xi_1-\xi_4\in \mathrm{Cone}\). As a result, we must have \({(\xi_1,\xi_2,\xi_3,\xi_4)}\in\mathcal{Q}_2\) and \(\Omega_1{(f_1,f_2,f_3,f_4)}=0\). ◻

Proof of the claim. Denote \(\wp_{q,\omega} = \{ (y,z)\in \mathbb{Z}^2 \mid z^2= qy+\omega \} \cap [-N,N]^2\) for fixed \(N,q\in\mathbb{Z}_+\) and \(\omega \in \mathbb{Z}\). Suppose \(q = p_1^{\alpha_1}\dots p_r^{\alpha_r}\) is the prime factorization, we denote \(\theta_i(z)\) the minimal residue of \(z\pmod {p_i^{\alpha_i}}\) for \(1\leq i\leq r\) and \(\theta_0(z) = \lfloor z/q\rfloor\). Then the map \[\mathbb{Z} \to \mathbb{Z}^{1+r},\quad z\mapsto (\theta_0(z),\theta_1(z),\dots,\theta_r(z))\] is an injection. In fact, if \(\theta_i(z) = \theta_i(\tilde{z} )\) for all \(1\leq i\leq r\), then \(z-\tilde{z}\) is divided by all \(p_i^{\alpha_i}\) and hence \(z -\tilde{z}\) is divided by \(q\). But \(\theta_0(z)=\theta_0(\tilde{z})\) implies \(0\leq |z-\tilde{z}|\leq q-1\), which forces that \(z=\tilde{z}\). As a result, \[\# \wp_{q,\omega} \leq \prod_{i=0}^r \#\{ \theta_i(z) \mid (y,z) \in \wp_{q,\omega} \}.\] Without loss of generality, we only consider the case \(z>0\) and fix some \((y_0,z_0) \in \wp_{q,\omega}\).

We note that \(\omega-qN \leq z^2 \leq \omega+qN\) for all \((y,z) \in \wp_{q,\omega}\). If \(\omega>2qN\), then \(|z|,|z_0|>\sqrt{\omega/2}\) and hence \[\left|\theta_0(z) - \theta_0(z_0) \right| \leq 1+\frac{ |z^2- z_0^2| }{ q |z+z_0| }= 1+\frac{|y-y_0|}{|z+z_0|} \lesssim 1+ \frac{N}{\sqrt \omega} \lesssim 1+ \sqrt {N/q}.\] If \(\omega \leq 2qN\), then \(|z| \lesssim \sqrt{qN}\) and hence \(|\theta_0(z)| \lesssim \sqrt{ N/q}\). Therefore, we know \(\theta_0(z)\) belongs to some interval of length \(O(\sqrt{N/q}+1)\) for all \((y,z) \in \wp_{q,\omega}\).

On the other hand, for any \((y,z) \in \wp_{q,\omega}\) we have \(q|(z^2-z_0^2)\), and hence \(p_i^{\alpha_i} | \theta_i(z-z_0)\theta_i(z+z_0)\) for \(1\leq i\leq r\). Consequently, we can find some \(\gamma_i\in\mathbb{N}\) which depends on \(z\), such that \(p^{\gamma_i}_i | \theta_i (z-z_0)\) and \(p_i^{\alpha_i-\gamma_i} | \theta_i(z+z_0)\). This further implies \[\theta_i(z) \in \big( \theta_i(z_0) + p_i^{\gamma_i} \mathbb{Z}\big) \cap \big(-\theta_i(z_0)+p_i^{\alpha_i-\gamma_i}\mathbb{Z}\big).\] Let us put \(\tilde{\gamma}_i =\max_{z} \min\{\gamma_i, \alpha_i-\gamma_i\}\). Notice \(p_i^{\tilde{\gamma}_i} | 2\theta_i(z_0)\) and \(p_i^{\max\{\gamma_i, \alpha_i-\gamma_i\}} \mathbb{Z}\subset p_i^{\alpha_i-\tilde{\gamma}_i}\mathbb{Z}\) , we see that at least one of \[\theta_i(z)\in \big( \theta_i(z_0) + p_i^{\alpha_i-\tilde{\gamma}_i} \mathbb{Z}\big) ,\qquad \theta_i(z)\in \big(-\theta_i(z_0)+p_i^{\alpha_i-\tilde{\gamma}_i}\mathbb{Z}\big)\] holds true. Thus \[\{ \theta_i(z) \mid (y,z) \in \wp_{q,\omega} \} \subset \big( \theta_i(z_0) + p_i^{\alpha_i-\tilde{\gamma}_i} \mathbb{Z}\big) \cup \big(-\theta_i(z_0)+p_i^{\alpha_i-\tilde{\gamma}_i}\mathbb{Z}\big),\] and \[\begin{align} \#\{ \theta_i(z) \mid (y,z) \in \wp_{q,\omega} \} &\leq \begin{cases} 2 p_i^{\tilde{\gamma}_i} , & \tilde{\gamma}_i\neq \alpha_i/2, \\ p_i^{\alpha_i/2}, & \tilde{\gamma}_i=\alpha_i/2, \end{cases} \\ &\leq p_i^{\alpha_i/2} \max \{1, 2p_i^{-1/2}\} . \end{align}\] As a result \(\# \wp_{q,\omega} \lesssim \sqrt N+\sqrt{\vphantom{N} q}\). ◻

Proof. We may assume that \(N_1\leq N_2\). We decompose \(\mathbb{Z}^3 = \bigcup_j C_j\) into almost disjoint cubes with side length \(N_1\) and write \[u_1u_2 = \sum_{C_j} u_1 P_{C_j} u_2.\] Their Fourier supports are finitely overlapped, hence we have the almost orthogonality \[\begin{align} \| u_1u_2\|_{{L_{t,x}^2([0,1]\times \mathbb{T}^3)}}^2 &\approx \sum_j \| u_1 P_{C_j} u_2\|_{{L_{t,x}^2([0,1]\times \mathbb{T}^3)}}^2 \\& \leq \sum_j \| u_1 \|_{{L_{t,x}^4([0,1]\times \mathbb{T}^3)}}^2 \| P_{C_j} u_2 \|_{{L_{t,x}^4([0,1]\times \mathbb{T}^3)}}^2. \end{align}\] By Corollary ?? , Remark 25 and the embedding properties in Remark 24, \[\begin{align} \| u_1u_2\|_{{L_{t,x}^2([0,1]\times \mathbb{T}^3)}}^2 &\lesssim{} \sum_j N_1 \| u_1\|_{Y^0([0,1])}^2 \| P_{C_j} u_2 \|_{Y^0([0,1])}^2 \\& ={} N_1\| u_1\|_{Y^0([0,1])}^2 \| u_2 \|_{Y^0([0,1])}^2. \end{align}\] ◻

Proof. It suffices to show that for any \(u_0 \in Y^{-s}([0,T))\), we have \[\left| \int_{[0,T)\times \mathbb{T}^3} u_0\prod_{i=1}^{2k+1}u_i \mathop{}\!\mathrm{d}x\mathop{}\!\mathrm{d}t \right| \lesssim \| u_0\|_{Y^{-s}([0,T))} \prod_{i=1}^{2k+1} \| u_i \|_{X^{s}([0,T))}.\] We apply Littlewood-Paley decomposition to each \(u_i\) to write \[u_i = \sum_{N_i\text{ dyadic}} P_{N_i}u_i = \sum_{N_i\text{ dyadic}} u_{N_{i}}^{(i)},\] hence it suffices to estimate \[\sum_{N_0,\dots,N_{2k+1}} \left| \int_{[0,T)\times \mathbb{T}^3} u_{N_{0}}^{(0)}\prod_{i=1}^{2k+1} u_{N_{i}}^{(i)} \mathop{}\!\mathrm{d}x\mathop{}\!\mathrm{d}t \right|.\] In order to make the integral non-zero, we must have that the two highest frequencies are comparable. Due to symmetry, it’s harmless to assume \(N_1\geq N_2\geq \dots\geq N_{2k+1}\). Following Proposition 28 we have that \[\begin{align} &\left| \int_{[0,T)\times \mathbb{T}^3} u_{N_{0}}^{(0)}\prod_{i=1}^{2k+1} u_{N_{i}}^{(i)} \mathop{}\!\mathrm{d}x\mathop{}\!\mathrm{d}t \right| \\\leq&{} \| u_{N_{0}}^{(0)}u_{N_{2}}^{(2)}\|_{{L_{t,x}^2([0,1]\times \mathbb{T}^3)}} \| u_{N_{1}}^{(1)} u_{N_{3}}^{(3)} \|_{{L_{t,x}^2([0,1]\times \mathbb{T}^3)}} \prod_{i\geq 4} \| u_{N_{i}}^{(i)} \|_{{L_{t,x}^\infty([0,1]\times \mathbb{T}^3)}} \\ \lesssim&{} \min\{ N_0,N_2\}^{\frac{1}{2}}_{\vphantom{3}} N_3^{\frac{1}{2}} \| u_{N_{0}}^{(0)}\|_{Y^0} \| u_{N_{1}}^{(1)}\|_{Y^0} \| u_{N_{2}}^{(2)}\|_{Y^0}\| u_{N_{3}}^{(3)}\|_{Y^0} \prod_{i\geq 4} N_i^{3/2} \| u_{N_{i}}^{(i)} \|_{Y^0} \\ \approx&{} \min\{ N_0,N_2\}^{\frac{1}{2}}_{\vphantom{3}} \frac{N_0^sN_3^{\frac{1}{2}-s}}{N_1^{s}N_2^s} \| u_{N_{0}}^{(0)}\|_{Y^{-s}} \| u_{N_{1}}^{(1)}\|_{Y^s} \| u_{N_{2}}^{(2)}\|_{Y^s}\| u_{N_{3}}^{(3)}\|_{Y^s} \prod_{i\geq 4} N_i^{\frac{3}{2}-s} \| u_{N_{i}}^{(i)} \|_{Y^s}. \end{align}\] For \(k=1\), since \(s>1/2\), we directly apply Cauchy-Schwarz to the summation over the two lower frequencies and the two highest frequencies respectively to the desired conclusion. For \(k\geq 2\), applying Cauchy-Schwarz to summation over \(N_i\) for \(i\geq 4\), we get that \[\begin{align} &\min\{ N_0,N_2\}^{\frac{1}{2}}_{\vphantom{3}} \frac{N_0^sN_3^{s-\frac{1}{2}}}{N_1^{s}N_2^s} \| u_{N_{0}}^{(0)}\|_{Y^{-s}} \| u_{N_{1}}^{(1)}\|_{Y^s} \| u_{N_{2}}^{(2)}\|_{Y^s}\| u_{N_{3}}^{(3)}\|_{Y^s} \prod_{i\geq 4} \| u_i\|_{{X^{s}([0,T))}} \\ \leq{}& \left( \frac{N_0}{N_1} \right)^{s} \left( \frac{ N_3 }{N_2} \right)^{s-\frac{1}{2}}\| u_{N_{0}}^{(0)}\|_{Y^{-s}} \| u_{N_{1}}^{(1)}\|_{Y^s} \| u_{N_{2}}^{(2)}\|_{Y^s}\| u_{N_{3}}^{(3)}\|_{Y^s} \prod_{i\geq 4} \| u_i\|_{{X^{s}([0,T))}}. \end{align}\] Then apply Cauchy-Schwarz to the summation over the two lower frequencies and the two highest frequencies respectively, we get the desired conclusion. ◻