Proof. First, we consider the case $|A| \lesssim N^{\frac {\alpha }{2}}$ , in which case we find $|\eta | \lesssim N^{\frac {\alpha }{2}+1}$ . Let $Z=[0,1] \times \mathbb {R} \times \mathbb {T}$ . We carry out the change of variables

(3.2) $$ \begin{align} x' = Nx, \; y' = N^{\frac{\alpha}{2}+1} y, \; t' = N^{\alpha+1} t, \; \xi' = N^{-1} \xi, \; \eta' = N^{-(\frac{\alpha}{2}+1)} \eta, \end{align} $$

and noting that $\omega _{\alpha }(N\xi ',N^{\frac {\alpha }{2}+1} \eta ') = N^{\alpha +1} \omega _\alpha (\xi ',\eta ')$ , to find with $Z'=[0,N^{\alpha +1}] \times \mathbb {R} \times 2 \pi N^{\frac {\alpha }{2}+1}$

$$ \begin{align*} \begin{aligned} &\, \| P_N S_\alpha(t) u_0 \|^4_{L^4_{t,x,y}([0,1], \mathbb{R} \times \mathbb{T}))} = \big\| \int_{\mathbb{R} \times \mathbb{Z}} e^{i(x \cdot \xi + y \cdot \eta + t \omega_\alpha(\xi,\eta))} \hat{u}_0(\xi,\eta) d\xi (d\eta)_1 \big\|^4_{L^4_{t,x,y}(Z)} \\ &= N^{-(\alpha+1)} N^{4(1-\frac{1}{4})} N^{-(\frac{\alpha}{2}+1)} N^{4 (\frac{\alpha}{2}+1)} \\ &\quad \times\big\| \int_{\mathbb{R}} N^{-(\frac{\alpha}{2}+1)} \sum_{\eta' \in \mathbb{Z} / N^{\frac{\alpha}{2}+1}} e^{i(x' \cdot \xi' + y' \eta' + t' \omega_\alpha(\xi',\eta'))} \hat{u}_0(N \xi' , N^{\frac{\alpha}{2}+1} \eta') d\xi' \big\|^4_{L^4_{t',x',y'}(Z')}. \end{aligned} \end{align*} $$

Note that by the support properties of $\hat {u}_0$ we can suppose that $|\xi '| \sim 1$ and $|\eta '| \lesssim 1$ . Note that $\eta ' \in \mathbb {Z}/N^{\frac {\alpha }{2}+1}$ .

Now we enlarge the y-domain of integration $\mathbb {R} / (2 \pi N^{\frac {\alpha }{2}+1} \mathbb {Z})$ perceived as an interval in $\mathbb {R}$ to an interval of size $N^{\alpha +1}$ by periodicity. To ease notation, we denote the integral over $\xi '$ and the summation over $\eta ' \in \mathbb {Z}/N^{\frac {\alpha }{2}+1}$ as integral $\int _*$ with measure given as product of Lebesgue and normalized counting measure $(d \eta ')_{N^{-(\frac {\alpha }{2}+1})}$ .

By spatial periodicity this amounts to a factor of $(N^{\alpha +1} / N^{\frac {\alpha }{2}+1} )^{-1} = N^{- \alpha /2}$ , and we continue the above, letting $Z = [0,N^{\alpha +1}] \times \mathbb {R} \times [0,N^{\alpha +1}]$ as

(3.3) $$ \begin{align} \begin{aligned} &= N^{-(\alpha+1)} N^{3} N^{-\big( \frac{\alpha}{2}+1 \big)} N^{4 (\frac{\alpha}{2}+1)} N^{- \frac{\alpha}{2}} \\ &\times \Big\| \int_* e^{i(x' \xi'+ y' \eta' + t'(|\xi'|^\alpha \xi' + (\eta')^2/\xi'))} \hat{u}_0(N \xi', N^{\frac{\alpha}{2}+1} \eta') d\xi' (d\eta')_{N^{-\frac{\alpha}{2}-1}} \Big\|^4_{L^4_{t',x',y'}(Z)}. \end{aligned} \end{align} $$

We invoke $\ell ^2$ -decoupling for elliptic surfaces in two dimensions. Here we note that the Hessian is for $\xi ' \neq 0$ given by

$$ \begin{align*} \partial^2 \omega_\alpha(\xi',\eta') = \begin{pmatrix} \alpha (\alpha + 1) \text{sgn}(\xi') |\xi'|^{\alpha-1} + 2 \frac{(\eta')^2}{(\xi')^3} & - \frac{2 \eta'}{(\xi')^2} \\ - \frac{2 \eta'}{(\xi')^2} & \frac{2}{\xi'} \end{pmatrix}. \end{align*} $$

$\partial ^2 \omega _\alpha $ is uniformly elliptic for $|\xi '| \sim 1$ and $|\eta '| \lesssim 1$ . Indeed, we compute

$$ \begin{align*} \det \partial^2 \omega_\alpha(\xi',\eta') = \alpha (\alpha-1) |\xi'|^{\alpha-2}, \quad |\text{tr} (\partial^2 \omega_\alpha)| \sim 1. \end{align*} $$

We infer for $\xi ' \sim 1$ the existence of two positive eigenvalues of size comparable to $1$ , and for $\xi ' \sim -1$ two negative eigenvalues of size comparable to $-1$ . The latter case can be reduced to the elliptic case by time reversal.

A few remarks about decoupling are in order: To apply decoupling as formulated in Theorem 3.2, we break up the domain of integration $[0,N^{\alpha +1}] \times \mathbb {R} \times [0,N^{\alpha +1}]$ into finitely overlapping balls of size $N^{\alpha +1}$ . The shift in x is admissible by translation invariance. Secondly, decoupling requires a continuous approximation, that is, approximating the exponential sum with a Fourier extension operator by mollifying the Dirac comb.

We give the details: For brevity let $\varphi (x',y',t',\xi ',\eta ') = x' \xi ' + y' \eta ' + t' \omega _\alpha (\xi ',\eta ')$ . Firstly, we break the x-integration into balls of size $N^{\alpha +1}$ .

$$ \begin{align*} \begin{aligned} &\quad \big\| \int e^{i \varphi(x',y',t',\xi',\eta')} \hat{u}_0(N \xi', N^{\frac{\alpha}{2}+1} \eta') d\xi' (d \eta')_{N^{-\frac{\alpha}{2}-1}} \big\|_{L^4_{t',x',y'}([0,N^{\alpha+1}] \times \mathbb{R} \times [0,N^{\alpha+1}])}^4 \\ &\lesssim \sum_{B_{N^{\alpha+1}}} \big\| \int e^{i\varphi(x',y',t',\xi',\eta')} \hat{u}_{0N} d\xi' (d \eta')_{N^{-\frac{\alpha}{2}-1}} \big\|_{L^4_{t',x',y'}([0,N^{\alpha+1}] \times B_{N^{\alpha+1}} \times [0,N^{\alpha+1}])}^4. \end{aligned} \end{align*} $$

Next, by a linear change of variables $x' \to x' + c(B_{N^{\frac {\alpha }{2}+1}})$ with $c(B)$ denoting the center of the ball, we can suppose that the ball is centered at the origin. The phase factor is absorbed into $\hat {u}_{0N}$ .

For the continuous approximation, we consider a sequence of smooth functions $\hat {f}_\lambda $ such that for $|x'|,|t'|,|y'| \lesssim N^{\alpha +1}$ we have

$$ \begin{align*} \int e^{i(x' \xi' + y' \eta' + t' \omega_\alpha(\xi',\eta'))} \hat{f}_\lambda(\xi',\eta') d\xi' d\eta' \to \int e^{i(x' \xi' + y' \eta' + t' \omega_\alpha(\xi',\eta'))} \hat{u}_{0N} d\xi' (d\eta')_{N^{-(\frac{\alpha}{2}+1)}}. \end{align*} $$

More specifically, we can choose $\text {supp}(\hat {f}_\lambda ) \subseteq \mathcal {N}_{\lambda ^{-1}}(\text {supp}(\hat {u}_{0N}))$ in a $\lambda ^{-1}$ -neighbourhood. Then, choosing $\lambda $ large enough, we can estimate by the theorem of dominated convergence:

$$ \begin{align*} \begin{aligned} &\quad \big\| \int e^{i\varphi(x',y',t',\xi',\eta')} \hat{u}_{0N} d\xi' (d \eta')_{N^{-\frac{\alpha}{2}-1}} \big\|_{L^4_{t',x',y'}([0,N^{\alpha+1}] \times B_{N^{\alpha+1}} \times N^{\alpha+1} \mathbb{T})} \\ &\lesssim \big\| \int e^{i\varphi(x',y',t',\xi',\eta')} \hat{f}_\lambda(\xi',\eta') d\xi' d\eta' \big\|_{L^4_{t',x',y'}(B_{N^{\alpha+1}})}. \end{aligned} \end{align*} $$

The final expression is amenable to decoupling as recorded and after reversing the continuous approximation by letting $\lambda \to \infty $ , we obtain an $\ell ^2$ -sum into balls on the frequency side of size $N^{- \frac {\alpha +1}{2}}$ . On this scale the phase function can be trivialized by Taylor expansion and we obtain from integration in time after reversing the continuous approximation:

$$ \begin{align*} \begin{aligned} &\quad \big\| \int_* e^{i\varphi(x',y',t',\xi',\eta')} \hat{u}_{0 \theta}(N \xi', N^{\frac{\alpha}{2}+1} \eta') d\xi' (d \eta')_{N^{-\frac{\alpha}{2}-1}} \big\|^2_{L_{t'}^4([0,N^{\alpha+1}],L^4_{x' y'}(B_{N^{\alpha+1}})} \\ &\lesssim N^{\frac{\alpha+1}{2}} \big\| \int_* e^{i(x' \xi' + y' \eta')} \hat{u}_{0 \theta}(N \xi', N^{\frac{\alpha}{2}+1} \eta') d\xi' (d \eta')_{N^{-\frac{\alpha}{2}-1}} \big\|^2_{L^4_{x' y'}(B_{{N^{\alpha+1}}})}. \end{aligned} \end{align*} $$

Now we reverse the scaling in x and y, which compensates the scaling factors $N^{3}$ and $N^{-(\frac {\alpha }{2}+1)} N^{4(\frac {\alpha }{2}+1)}$ in (3.3).

It remains to estimate

$$ \begin{align*} \big\| \int_* e^{i(x \xi + y \eta)} \hat{u}_{0 \theta}(\xi,\eta) d\xi d\eta \big\|^2_{L^4_{xy}} \end{align*} $$

with $\theta $ denoting a rectangle of size $N^{\frac {1-\alpha }{2}} \times N^{\frac {1}{2}}$ . We can use Bernstein’s inequality to find

(3.4) $$ \begin{align} \big\| \int_{\mathbb{R} \times \mathbb{Z}} e^{i(x \xi + y \eta)} \hat{u}_{0 \theta}(\xi,\eta) d\xi (d\eta)_1 \big\|_{L^4_{xy}} \lesssim N^{\frac{1-\alpha}{8}} N^{\frac{1}{8}} \| u_{0 \theta} \|_{L^2_{xy}}. \end{align} $$

We summarize the estimates as follows: Scaling to unit frequencies and enlarging the spatial domain to $N^{\alpha +1}$ incurs a factor of

$$ \begin{align*} N^{-(\alpha+1)} N^{3} N^{-(\frac{\alpha}{2} + 1)} N^{4(\frac{\alpha}{2}+1)} N^{- \frac{\alpha}{2}}. \end{align*} $$

Invoking $\ell ^2$ -decoupling incurs a factor of $N^\varepsilon $ and decouples the integral into balls of size $N^{-\frac {\alpha +1}{2}}$ . Carrying out the time integral yields a factor $N^{\alpha +1}$ and reversing the scaling in x and $\xi $ , and y and $\eta $ gives factors $N^3$ and $N^{\frac {\alpha }{2}+1}$ , respectively. Moreover, we use periodicity to decrease the spatial domain to $\mathbb {T}$ again after inverting the scaling. This incurs a factor $N^{\frac {\alpha }{2}}$ .

We have proved so far

(3.5) $$ \begin{align} \| P_N S_\alpha(t) u_0 \|^4_{L_{t,x,y}^4([0,1], \mathbb{R} \times \mathbb{T})} \lesssim_\varepsilon N^{\varepsilon} \big( \sum_{\theta} \| u_{0 \theta} \|^2_{L^4_{xy}} \big)^{2} \end{align} $$

with the sum over $\theta $ being over essentially disjoint rectangles of size $N^{\frac {1-\alpha }{2}} \times N^{\frac {1}{2}}$ . Plugging (3.4) into (3.5) yields

$$ \begin{align*} \| P_N S_\alpha(t) u_0 \|^4_{L^4_{t,x,y}} \lesssim_\varepsilon N^{\varepsilon + \frac{2-\alpha}{2}} \| u_0 \|_{L^2_{xy}}^4, \end{align*} $$

which completes the proof in case $|A| \lesssim N^{\frac {\alpha }{2}}$ .

Now we turn to the case $|A| \gg N^{\frac {\alpha }{2}}$ : In the first step we apply the scaling (3.2), which maps the Fourier support to the set

$$ \begin{align*} |\xi'| \sim 1, \quad \big| \frac{\eta'}{\xi'} - A' \big| \lesssim 1 \end{align*} $$

for $A' = A/N^{\frac {\alpha }{2}}$ . Like above, by periodic extension in $y'$ , we reduce to estimate the expression up to scaling factors

$$ \begin{align*} \big\| \int e^{i(x'\xi' + y' \eta' + t' \omega_{\alpha}(\xi',\eta'))} \hat{u}_0(N \xi', N^{\frac{\alpha}{2}+1} \eta') d\xi' (d\eta')_{N^{-\frac{\alpha}{2}-1}} \big\|_{L^4_{t'}([0,N^{\alpha+1}], L^4_{x' y'}(\mathbb{R} \times B_{N^{\alpha+1}}))}. \end{align*} $$

We can apply the Galilean invariance

$$ \begin{align*} (\xi",\eta") = F'(\xi',\eta') = (\xi',\eta' - A' \xi') \end{align*} $$

to obtain

$$ \begin{align*} |\xi"| \sim 1, \quad \big| \frac{\eta"}{\xi"} \big| \lesssim 1. \end{align*} $$

Note that this does not change the domain of integration $\mathbb {R} \times B_{N^{\alpha +1}}$ , but it is important to note that in general $(\xi ",\eta ") \notin \mathbb {R} \times \mathbb {Z}$ . Now we again break the domain of integration $[0,N^{\alpha +1}] \times \mathbb {R} \times B_{N^{\alpha +1}}$ into balls of size $B_{N^{\alpha +1}}$ and use translation invariance.

We have by continuous approximation

$$ \begin{align*} \begin{aligned} &\quad \big\| \int d \xi" \sum_{\eta" \in \mathbb{Z} - A' \xi"} e^{i(x' \xi"+y' \eta" + t' \omega_\alpha(\xi",\eta"))} \hat{u}_{0N}(\xi",\eta"+ A' \xi') \big\|_{L^4_{t'}([0,N^{\alpha+1}], L^4_{x' y'}(B_{N^{\alpha+1}}))} \\ &\lesssim \big\| \int_{|\xi"| \sim 1, \, |\eta"| \lesssim 1} e^{i(x' \xi" + y' \eta" + t' \omega_\alpha(\xi",\eta"))} \hat{f}_\lambda(\xi",\eta") d\xi" d\eta" \big\|_{L^4_{t',x',y'}(B_{N^{\alpha+1}})}. \end{aligned} \end{align*} $$

Now we can apply $\ell ^2$ -decoupling and reverse the continuous approximation, which gives the estimate

$$ \begin{align*} \begin{aligned} &\quad \big\| \int d \xi" \sum_{\eta" \in \mathbb{Z} - A' \xi"} e^{i(x' \xi" + y' \eta" + t' \omega_\alpha(\xi",\eta")} \hat{u}_{0,N}(\xi",\eta" + A' \xi") \big\|_{L^4_{t',x',y'}(B_{N^{\alpha+1}})} \\ &\lesssim_\varepsilon N^{\varepsilon} \big( \sum_{\theta \in \mathcal{B}_{N^{-\frac{\alpha+1}{2}}}} \big\| \int d\xi" \sum_{\eta" \in \mathbb{Z} - A' \xi"} e^{i(x' \xi" + y' \eta" + t' \omega_\alpha(\xi",\eta"))}\\ &\quad \quad \times \chi_{\theta}(\xi",\eta") \hat{u}_{0N}(\xi",\eta"+A'\xi") \big\|_{L^4_{t',x',y'}(w_{B_{N^{\alpha+1}}})}^2 \big)^{\frac{1}{2}}. \end{aligned} \end{align*} $$

The sum over $\theta $ is over essentially disjoint $N^{-\frac {\alpha +1}{2}}$ -balls, which cover the set $\{|\xi "| \sim 1, |\eta "| \lesssim 1 \}$ .

Summing over the $B_{N^{\alpha +1}}$ -balls in the x-coordinate and reversing all linear transformations and the scalings, we arrive at the expression

$$ \begin{align*} \begin{aligned} &\quad \big\| \int d\xi \sum_{\eta \in \mathbb{Z}} e^{i(x\xi + y \eta + t \omega_\alpha(\xi,\eta))} \hat{u}_0(\xi,\eta) \big\|_{L_t^4([0,1], L^4_{xy}(\mathbb{R} \times \mathbb{T}))} \\ &\lesssim_\varepsilon N^\varepsilon \big( \sum_{\theta \in \mathcal{B}_{N^{-\frac{\alpha+1}{2}}}} \big\| \int d\xi \sum_{\eta \in \mathbb{Z}} e^{i(x \xi + y \eta + t \omega_\alpha(\xi,\eta))} \\ &\quad \quad \quad \times \chi_{\theta}(N^{-1} \xi, N^{-(\frac{\alpha}{2}+1)}(\eta - A \xi)) \hat{u}_0(\xi,\eta) \big\|^2_{L_t^4(w_1,L^4_{xy}(\mathbb{R} \times \mathbb{T}))}. \end{aligned} \end{align*} $$

Since there are no oscillations of $\omega _\alpha (\xi ,\eta )$ for $(N^{-1} \xi , N^{-(\frac {\alpha }{2}+1)}(\eta -A\xi ))$ contained in $\theta $ , we can carry out the integration in time without loss.

Let $\bar {\theta } = I_{N^{\frac {1-\alpha }{2}}} \times I_{N^{\frac {1}{2}}}$ be the anisotropic dilation of $\theta $ . Let

$$ \begin{align*} (\bar{\xi},\bar{\eta}) = F(\xi,\eta) = (\xi,\eta - A \xi) \end{align*} $$

denote the Galilean transform without anisotropic dilation.

To conclude the argument, we apply Bernstein’s inequality for which we need to understand $F^{-1}(\bar {\theta })$ . Since $\xi \in I_{N^{\frac {1-\alpha }{2}}}$ the condition $\eta - A \xi \in I_{N^{\frac {1}{2}}}$ yields the bound

$$ \begin{align*} \# \{ \eta \in \mathbb{Z} \; | \; \exists \xi \in I_{N^{\frac{1}{2}}}: \eta - A \xi \in I_{N^{\frac{1}{2}}} \} \lesssim N^{\frac{1}{2}} + A N^{\frac{1-\alpha}{2}} \sim A N^{\frac{1-\alpha}{2}}. \end{align*} $$

The final estimate is due to the size assumption on A. For fixed $\eta $ , we estimate the measure of $\xi $ such that $F(\xi ,\eta ) \in \bar {\theta }$ as

$$ \begin{align*} \text{meas}( \{ \xi \in \mathbb{R} : F(\xi,\eta) \in \bar{\theta} \}) \lesssim N^{\frac{1}{2}} / A. \end{align*} $$

Then it is a consequence of Bernstein’s inequality that

$$ \begin{align*} \begin{aligned} &\quad \big\| \int_{F(\xi,\eta) \in \bar{\theta}} e^{i(x \xi + y \eta)} \hat{f}(\xi,\eta) d\xi d\eta \big\|_{L^4_{xy}(\mathbb{R} \times \mathbb{T})} \\ &\lesssim (A N^{\frac{1-\alpha}{2}})^{\frac{1}{4}} (N^{\frac{1}{2}} / A)^{\frac{1}{4}} \| \int_{(\xi,\eta) \in F^{-1}(\bar{\theta})} e^{i(x\xi + y \eta)} \hat{u}_0(\xi,\eta) \|_{L^2_{xy}} \\ &\lesssim N^{\frac{2-\alpha}{8}} \| \int_{(\xi,\eta) \in F^{-1}(\bar{\theta})} e^{i(x\xi + y \eta)} \hat{u}_0(\xi,\eta) \|_{L^2_{xy}}. \end{aligned} \end{align*} $$

The proof is then concluded by the essential disjointness of $F^{-1}(\bar {\theta })$ .

Proof. We can follow along the arguments of the proof of Proposition 3.1, that is, employ the anisotropic scaling (3.2), use a continuous approximation to invoke $\ell ^2$ -decoupling, and finally reverse the continuous approximation and scaling. After carrying out the integration in time, we arrive at the expression:

$$ \begin{align*} \| S_\alpha(t) u_0 \|_{L_t^4([0,1],L^4_{xy}(\mathbb{T}^2_\gamma))} \lesssim_\varepsilon N^\varepsilon \big( \sum_{\theta \in \mathcal{R}_{N^{\frac{1-\alpha}{2}} \times N^{\frac{1}{2}}}} \| u_{0 \theta} \|^2_{L^4_{xy}(\mathbb{T}^2_\gamma)} \big)^{\frac{1}{2}}, \end{align*} $$

where the sum is carried out over essentially disjoint rectangles $\theta $ of size $N^{\frac {1-\alpha }{2}} \times N^{\frac {1}{2}}$ , which cover $A_{N,N^{\frac {\alpha }{2}+1}}$ . An application of Bernstein’s inequality yields

$$ \begin{align*} \| u_{0 \theta} \|_{L^4_{xy}(\mathbb{T}^2_\gamma)} \lesssim N^{\frac{1}{8}} \| u_{0 \theta} \|_{L^2 (\mathbb{T}^2_\gamma)}, \end{align*} $$

since presently the estimate is carried out with counting measure compared to the previous section. The proof is concluded by essential disjointness of $\theta $ .

Proof. Suppose that $L_1 = \min (L_1,L_2)$ and $L_2 = \max (L_1,L_2)$ by symmetry. We obtain by the Cauchy-Schwarz inequality:

$$ \begin{align*} \begin{aligned} &\quad \| (f_{1,N_1,L_1} * f_{2,N_2,L_2}) \|_{L^2_{\tau,\xi,\eta}} \\ &\leqslant \sup_{(\tau,\xi,\eta) \in \mathbb{R}^3} \text{meas}( S= \{ (\tau_1,\xi_1,\eta_1) \in \text{supp}(f_{1,N_1,L_1}), \; \xi_1 \in I_1 : \\ &\quad \quad (\tau-\tau_1,\xi-\xi_1,\eta-\eta_1) \in \text{supp}(f_{2,N_2,L_2}), \; \xi - \xi_1 \in I_2 \} )^{\frac{1}{2}} \prod_{i=1}^2 \|f_{i,N_i,L_i} \|_{L_{\tau,\xi,\eta}^2}. \end{aligned} \end{align*} $$

The assumption (4.4) yields

$$ \begin{align*} \big| \frac{\eta_1}{\xi_1} - \frac{\eta - \eta_1}{\xi - \xi_1} \big| \gtrsim N_{\max}^{\frac{\alpha}{2}}. \end{align*} $$

Note moreover that

(4.6) $$ \begin{align} \big| \big( \tau_1 - \omega_\alpha(\xi_1,\eta_1) \big) + \big( (\tau - \tau_1) - \omega_\alpha(\xi - \xi_1,\eta- \eta_1) \big) \big| \lesssim L_2. \end{align} $$

We estimate the measure of S by fixing $\xi _1$ , which amounts to a factor $|I_1|$ , and counting $\tau _1$ and $\eta _1$ . From (4.4) and (4.6) follows that for fixed $\tau $ , $\xi _1$ , $\xi $ , $\eta $ that on $\mathbb {R}^2$ the measure of $\eta _1$ is estimated by $\big ( \frac {L_2}{ N_{\max }^{\frac {\alpha }{2}}} \big )^{\frac {1}{2}}$ . On $\mathbb {R} \times \mathbb {T}$ there are at most $\langle \frac {L_2}{ N_{\max }^{\frac {\alpha }{2}}} \rangle ^{\frac {1}{2}}$ values of $\eta _1$ such that $(\xi _1,\eta _1,\tau _1) \in S$ . This is a consequence of the mean-value theorem. Finally, with $(\xi _1,\eta _1)$ fixed, we trivially estimate

$$ \begin{align*} \text{meas} ( \{ \tau_1 : |\tau_1 - \omega_\alpha(\xi_1,\eta_1)| \leqslant L_1 \} ) \leqslant L_1. \end{align*} $$

In conclusion we obtain

$$ \begin{align*} \begin{aligned} \text{meas} (S) &\leqslant |I_1| \cdot \# \{ \eta_1 : (4.4) \text{ and } (4.6) \text{ holds } \} \cdot \big| \{ \tau_1 : |\tau_1 - \omega_\alpha(\xi_1,\eta_1) \big| \leqslant L_1 \} \big| \\ &\lesssim N_{\min} \cdot C_{1,\mathbb{D}}(L_2,N_{\max}) \cdot L_1. \end{aligned} \end{align*} $$

with $C_{1,\mathbb {D}}$ defined in (4.3). This finishes the proof.

Proof. Since

$$ \begin{align*} \begin{aligned} &\quad |\Omega_\alpha(\xi_1,\eta_1,\xi-\xi_1,\eta-\eta_1)| \\ &= |\tau - \omega_\alpha(\xi,\eta) - (\tau_1 - \omega_\alpha(\xi_1,\eta_1)) - ((\tau-\tau_1) - \omega_\alpha(\xi-\xi_1,\eta-\eta_1)) | \\ &\ll N_{\max}^\alpha N_{\min}, \end{aligned} \end{align*} $$

we are in the resonant case (4.1) and we have by (4.2)

(4.8) $$ \begin{align} \big| \frac{\eta_1}{\xi_1} - \frac{\eta - \eta_1}{\xi - \xi_1} \big| \sim N_{\max}^{\frac{\alpha}{2}}. \end{align} $$

The assumptions of Proposition 4.1 are satisfied, which yields the claim.

Proof. If $L_{\max } = \max (L_1,L_2) \gtrsim N_1^\alpha N_2$ , this is a consequence of Lemma 4.3:

$$ \begin{align*} \begin{aligned} \| f_{1,N_1,L_1} * f_{2,N_2,L_2} \|_{L^2_{\tau,\xi,\eta}} &\lesssim N_2^{\frac{3}{4}} L_{\min}^{\frac{1}{2}} L_{\max}^{\frac{1}{4}} \prod_{i=1}^2 \| f_{i,N_i,L_i} \|_2 \\ &\lesssim \frac{N_2^{\frac{1}{2}}}{N_1^{\frac{\alpha}{4}}} \prod_{i=1}^2 L_i^{\frac{1}{2}} \| f_{i,N_i,L_i} \|_2. \end{aligned} \end{align*} $$

In the following we suppose that $L_{\max } \ll N_1^\alpha N_2$ . We carry out a dyadic decomposition (Whitney) in the transversality parameter:

$$ \begin{align*} D \sim \big| \frac{\eta_1}{\xi_1} - \frac{\eta_2}{\xi_2} \big| \end{align*} $$

with the range $D \in [\big ( \frac {L_{\max }}{N_2} \big )^{\frac {1}{2}}, \infty )$ :

$$ \begin{align*} \| f_{1,N_1,L_1} * f_{2,N_2,L_2} \|_{L^2_{\tau,\xi,\eta}} \leqslant \sum_{D \geqslant \big( \frac{L_{\max}}{N_2} \big)^{\frac{1}{2}}} \sum_{I_1^D \sim I_2^D} \| f_{1,N_1,L_1}^{I_1^D} * f_{2,N_2,L_2}^{I_2^D} \|_{L^2_{\tau,\xi,\eta}}. \end{align*} $$

Here we have broken the support of $f_{i,N_i,L_i}$ into intervals $I_i^D$ of $(\eta _i/\xi _i)$ of length $\sim D$ with

$$ \begin{align*} \big| \frac{\eta_1}{\xi_1} - \frac{\eta_2}{\xi_2} \big| \sim D \text{ for } \frac{\eta_i}{\xi_i} \in I_i^D. \end{align*} $$

The contribution of $D \gtrsim N_1^{\frac {\alpha }{2}}$ is handled by Proposition 4.1:

$$ \begin{align*} \begin{aligned} \sum_{D \gtrsim N_1^{\frac{\alpha}{2}}} \sum_{I_1^D \sim I_2^D} \| f_{1,N_1,L_1}^{I_1^D} * f_{2,N_2,L_2}^{I_2^D} \|_{L^2_{\tau,\xi,\eta}} &\lesssim \sum_{D \gtrsim N_1^{\frac{\alpha}{2}}} \sum_{I_1^D \sim I_2^D} \frac{N_2^{\frac{1}{2}}}{D^{\frac{1}{2}}} \prod_{i=1}^2 L_i^{\frac{1}{2}} \| f_{i,N_i,L_i}^{I_i^D} \|_{L^2_{\tau,\xi,\eta}} \\ &\lesssim \frac{N_2^{\frac{1}{2}}}{N_1^{\frac{\alpha}{4}}} \prod_{i=1}^2 L_i^{\frac{1}{2}} \| f_{i,N_i,L_i} \|_{L^2_{\tau,\xi,\eta}}. \end{aligned} \end{align*} $$

For $D \ll N_1^{\frac {\alpha }{2}}$ we can use an almost orthogonal decomposition to effectively reduce the $\xi $ -support of $f_{i,N_i,L_i}$ . Without loss of generality we can suppose that $\pi _{\xi }(f_{i,N_i,L_i}) \subseteq \mathbb {R}_{>0}$ by complex conjugation and symmetry of the dispersion relation.

The convolution constraint reads

$$ \begin{align*} \left\{ \begin{array}{cl} \xi_1 + \xi_2 &= \xi_3 + \xi_4, \\ \eta_1 + \eta_2 &= \eta_3 + \eta_4, \\ \xi_1^{\alpha+1} + \frac{\eta_1^2}{\xi_1} + \xi_2^{\alpha+1} + \frac{\eta_2^2}{\xi_2} &= \xi_3^{\alpha+1} + \frac{\eta_3^2}{\xi_3} + \xi_4^{\alpha+1} + \frac{\eta_4^2}{\xi_4} + \mathcal{O}(L_{\max}). \end{array} \right. \end{align*} $$

We rescale to unit $\xi $ -frequencies by $\xi \to \frac {\xi }{N_1}$ , $\eta \to \frac {\eta }{N_1^{\frac {\alpha }{2} + 1}}$ to find the following for the renormalized frequencies:

$$ \begin{align*} \left\{ \begin{array}{cl} \xi^{\prime}_1 + \xi^{\prime}_2 &= \xi^{\prime}_3 + \xi^{\prime}_4, \\ \eta^{\prime}_1 + \eta^{\prime}_2 &= \eta^{\prime}_3 + \eta^{\prime}_4, \\ (\xi^{\prime}_1)^{\alpha+1} + \frac{(\eta^{\prime}_1)^2}{\xi^{\prime}_1} + (\xi^{\prime}_2)^{\alpha+1} + \frac{(\eta^{\prime}_2)^2}{\xi^{\prime}_2} &= (\xi^{\prime}_3)^{\alpha+1} + \frac{(\eta^{\prime}_3)^2}{\xi^{\prime}_3} \\ &\quad +\ (\xi^{\prime}_4)^{\alpha+1} + \frac{(\eta^{\prime}_4)^2}{\xi^{\prime}_4} + \mathcal{O}(L_{\max}/N_1^{\alpha+1}). \end{array} \right. \end{align*} $$

We subtract

$$ \begin{align*} \frac{(\eta_1' + \eta_2')^2}{\xi_1'+\xi_2'} = \frac{(\eta_3'+\eta_4')^2}{\xi_3'+\xi_4'} \end{align*} $$

from the third equation to find

$$ \begin{align*} \left\{ \begin{array}{cl} \xi_1' + \xi_2' &= \xi_3' + \xi_4', \\ (\xi_1')^{\alpha+1} + (\xi_2')^{\alpha+1}- \frac{(\eta_1' \xi_2' - \eta_2' \xi_1')^2}{\xi_1' \xi_2' (\xi_1'+\xi_2')} &= (\xi_3')^{\alpha+1} + (\xi_4')^{\alpha+1} \\ &\quad \quad - \frac{(\eta_3' \xi_4' - \eta_4' \xi_3')^2}{\xi_3' \xi_4' (\xi_3'+\xi_4')} + \mathcal{O}(L_{\max}/N_1^{\alpha+1}). \end{array} \right. \end{align*} $$

This yields

$$ \begin{align*} \left\{ \begin{array}{cl} \xi_1' + \xi_2' &= \xi_3' + \xi_4', \\ (\xi_1')^{\alpha+1} + (\xi_2')^{\alpha+1} &= (\xi_3')^{\alpha+1} + (\xi_4')^{\alpha+1} + \mathcal{O}(\frac{D^2 N_2}{N_1^{\alpha+1}} + \frac{L_{\max}}{N_1^{\alpha+1}}). \end{array} \right. \end{align*} $$

Note that for our Whitney decomposition we always have $D^2 N_2 \gtrsim L_{\max }$ . Note that the curve $\xi ' \mapsto (\xi ')^{\alpha +1}$ degenerates at the origin for $\alpha> -1$ . To still find an almost orthogonality resembling the Córdoba–Fefferman square function estimate, we rewrite the second line as

$$ \begin{align*} f'(\xi_{1*}) (\xi_1' - \xi_3') + f'(\xi_{2*})(\xi_2'-\xi_4') = \mathcal{O}\big( \frac{D^2 N_2}{N_1^\alpha N_1} \big). \end{align*} $$

We have $f'(\xi _{2*}) = f'(\xi _{1*}) + \Delta $ with $|\Delta | \gtrsim 1$ , for which reason

$$ \begin{align*} f'(\xi_{1*}) (\xi_1' - \xi_3' + \xi_2' - \xi_4') + \Delta (\xi_2'-\xi_4') = \mathcal{O}\big( \frac{D^2 N_2}{N_1^\alpha N_1} \big). \end{align*} $$

Consequently, $\xi _i' = \xi _{i+2}' + \mathcal {O}\big ( \frac {D^2 N_2}{N_1^\alpha N_1} \big )$ for $i \in \{1,2\}$ , and we obtain an almost orthogonal decomposition of

$$ \begin{align*} \| f^{\prime}_{1,N_1,L_1} * f^{\prime}_{2,N_2,L_2} \|^2_{L^2_{\tau,\xi,\eta}} \lesssim \sum_{I_1', I^{\prime}_2} \| f^{\prime}_{1,N_1,L_1,I^{\prime}_1} * f^{\prime}_{2,N_2,L_2,I^{\prime}_2} \|^2_{L^2} \end{align*} $$

with $\xi '$ -intervals of length $\mathcal {O}(\frac {D^2 N_2}{N_1^\alpha N_1})$ . After rescaling we obtain a decomposition into intervals of length $D^2 N_2 / N_1^\alpha $ .

First we handle the contribution $D \gg L_{\max }/N_2$ . Applying the bilinear Strichartz estimate Proposition 4.1 gives

$$ \begin{align*} \| f^{I_1}_{1,N_1,L_1} * f^{I_2}_{2,N_2,L_2} \|_{L^2_{\tau,\xi,\eta}} \lesssim \frac{D N_2^{\frac{1}{2}}}{N_1^{\frac{\alpha}{2}}} D^{-\frac{1}{2}} \prod_{i=1}^2 L_i^{\frac{1}{2}} \| f^{I_i}_{i,N_i,L_i} \|_2. \end{align*} $$

Summation in the intervals can be carried out by almost orthogonality. Summation in $L_{\max } / N_2 \ll D \lesssim N_1^{\alpha /2}$ gives

$$ \begin{align*} \sum_{L_{\max} / N_2 \ll D \lesssim N_1^{\alpha/2}} \frac{D^{\frac{1}{2}} N_2^{\frac{1}{2}}}{N_1^{\frac{\alpha}{2}}} \lesssim \frac{N_2^{\frac{1}{2}}}{N_1^{\frac{\alpha}{4}}}. \end{align*} $$

For the contribution $D \sim \big ( \frac {L_{\max }}{N_2} \big )^{\frac {1}{2}}$ we use Lemma 4.3 after almost orthgonal decomposition into intervals of length $\big ( \frac {L_{\max }}{N_1^{\alpha }} \big )^{\frac {1}{2}}$ :

$$ \begin{align*} \| f_{1,N_1,L_1}^{I_1} * f^{I_2}_{2,N_2,L_2} \|_{L^2_{\tau,\xi,\eta}} \lesssim L_{\min}^{\frac{1}{2}} \big( \frac{L_{\max}}{N_1^\alpha} \big)^{\frac{1}{2}} N_2^{\frac{1}{4}} L_{\max}^{\frac{1}{4}} \prod_{i=1}^2 \| f_{i,N_i,L_i}^{I_i} \|_2. \end{align*} $$

By the upper bound for $L_{\max }$ this gives

$$ \begin{align*} \| f_{1,N_1,L_1}^{I_1} * f^{I_2}_{2,N_2,L_2} \|_{L^2_{\tau,\xi,\eta}} \lesssim \frac{N_2^{\frac{1}{2}}}{N_1^{\frac{\alpha}{4}}} \prod_{i=1}^2 \| f_{i,N_i,L_i}^{I_i} \|_2. \end{align*} $$

This handles all possible transversalities $D \in [\big ( \frac {L_{\max }}{N_2} \big )^{\frac {1}{2}}, N_1^\alpha ]$ . The proof is complete.

Proof. First, we suppose that $L_{\max } \gtrsim N_1^\alpha N_2$ . In this case we apply Lemma 4.3 to find

$$ \begin{align*} \| f_{1,N_1,L_1} * f_{2,N_2,L_2} \|_{L^2} \lesssim N_2^{\frac{1}{2}} L_{\min}^{\frac{1}{2}} \langle L_{\max} N_2 \rangle^{\frac{1}{4}} \prod_{i=1}^2 \| f_{i,N_i,L_i} \|_2, \end{align*} $$

which implies (4.11).

Next, suppose that $L_{\max } \ll N_1^\alpha N_2$ . We carry out a Whitney decomposition in the transversality parameter

$$ \begin{align*} D \sim \big| \frac{\eta_1}{\xi_1} - \frac{\eta_2}{\xi_2} \big| \end{align*} $$

in the range $D \in [ \big ( \frac {L_{\max }}{N_2} \big )^{\frac {1}{2}}, D^*]$ :

$$ \begin{align*} \| f_{1,N_1,L_1} * f_{2,N_2,L_2} \|_{L^2_{\tau,\xi,\eta}} \leqslant \sum_{D \in \big[ \big( \frac{L_{\max}}{N_2}\big)^{\frac{1}{2}}, D^* ]} \sum_{I_1^D \sim I_2^D} \| f_{1,N_1,L_1}^{I_{1}^D} * f_{2,N_2,L_2}^{I_{2}^D} \|_{L^2_{\tau,\xi,\eta}}. \end{align*} $$

By the argument from the proof of Lemma 4.4 we can effectively localize the $\xi $ -support to intervals of length $\mathcal {O}(\frac {D^2 N_2}{N_1^\alpha } \big )$ . We obtain for $D= \big ( L_{\max } / N_2 \big )^{\frac {1}{2}}$ by applying Lemma 4.3

(4.12) $$ \begin{align} \| f^{I_{1}^D}_{1,N_1,L_1} * f^{I_{2}^D}_{2,N_2,L_2} \|_{L^2_{\tau,\xi,\eta}} \lesssim L_{\min}^{\frac{1}{2}} \frac{L_{\max}^{\frac{1}{2}}}{N_1^{\frac{\alpha}{2}}} \langle L_{\max} N_2 \rangle^{\frac{1}{4}} \prod_{i=1}^2 \| f_{i,N_i,L_i}^{I_{i}^D} \|_{L^2}. \end{align} $$

This is sufficient for $L_{\max } \ll N_1^\alpha N_2$ .

Next, we handle the contribution $D \in \big ( \big ( \frac {L_{\max }}{N_2} \big )^{\frac {1}{2}}, N_1^{\alpha /2} \big ]$ . After localizing the $\xi $ -support to intervals of length $\mathcal {O}\big ( D^2 N_2 / N_1^\alpha \big )$ by almost orthogonality, we can apply Proposition 4.1 to find

(4.13) $$ \begin{align} \begin{aligned} &\quad \sum_{D \in \big( \big( \frac{L_{\max}}{N_2}\big)^{\frac{1}{2}}, N_1^{\frac{\alpha}{2}} \big) ]} \sum_{I_1^D \sim I_2^D} \| f_{1,N_1,L_1}^{I_{1}^D} * f_{2,N_2,L_2}^{I_{2}^D} \|_{L^2_{\tau,\xi,\eta}} \\ &\lesssim L_{\min}^{\frac{1}{2}} \sum_{D \in \big( \big( \frac{L_{\max}}{N_2}\big)^{\frac{1}{2}}, N_1^{\frac{\alpha}{2}} \big) ]} \frac{D N_2^{\frac{1}{2}}}{N_1^{\frac{\alpha}{2}}} \langle L_{\max} / D \rangle^{\frac{1}{2}} \prod_{i=1}^2 \| f_{i,N_i,L_i} \|_2 \\ &\lesssim L_{\min}^{\frac{1}{2}} N_2^{\frac{1}{2}} \langle L_{\max} / N_1^{\alpha/2} \rangle^{\frac{1}{2}} \prod_{i=1}^2 \| f_{i,N_i,L_i} \|_2. \end{aligned} \end{align} $$

Finally, we turn to the contribution with large transversality. Here we apply Proposition 4.1 directly to find

(4.14) $$ \begin{align} \begin{aligned} &\quad \sum_{D \in [ N_1^{\frac{\alpha}{2}} , D^* ]} \sum_{I_1^D \sim I_2^D} \| f_{1,N_1,L_1}^{I_{1}^D} * f_{2,N_2,L_2}^{I_{2}^D} \|_{L^2_{\tau,\xi,\eta}} \\ &\lesssim N_2^{\frac{1}{2}} L_{\min}^{\frac{1}{2}} \sum_{D \in [ N_1^{\frac{\alpha}{2}} , D^* ]} \langle L_{\max} / D \rangle^{\frac{1}{2}} \prod_{i=1}^2 \| f_{i,N_i,L_i} \|_2 \\ &\lesssim \log(D^*) N_2^{\frac{1}{2}} \langle L_{\max} / N_1^{\alpha/2} \rangle^{\frac{1}{2}} \prod_{i=1}^2 \| f_{i,N_i,L_i} \|_2. \end{aligned} \end{align} $$

Collecting contributions (4.12)-(4.14) the proof is complete.

Assumption 4.7. For $i=1,2,3$ there exist $0 < \beta \leqslant 1$ , $b>0$ , $A \geqslant 1$ , $F_i \in C^{1,\beta }(\mathcal {U}_i)$ , where the $\mathcal {U}_i$ denote open and convex sets in $\mathbb {R}^2$ and $G_i \in O(3)$ such that

1. (i) the oriented surfaces $S_i$ are given by

   $$ \begin{align*} S_i = G_i \text{gr}(F_i), \quad \text{gr}(F_i) = \{ (x,y,z) \in \mathbb{R}^3 : z = F_i(x,y), \; (x,y) \in \mathcal{U}_i \}; \end{align*} $$

2. (ii) the vector field $\mathfrak {n}_i : S_i \to \mathbb {S}^2$ of outward unit normals on $S_i$ satisfies the Hölder condition:

   $$ \begin{align*} \sup_{\sigma,\tilde{\sigma}} \frac{|\mathfrak{n}_i(\sigma) - \mathfrak{n}_i(\tilde{\sigma})|}{|\sigma - \tilde{\sigma}|^\beta} + \frac{|\mathfrak{n}_i(\sigma).(\sigma - \tilde{\sigma})|}{|\sigma - \tilde{\sigma}|^{1+\beta}} \leqslant b; \end{align*} $$

3. (iii) there is $A \geqslant 1$ such that for all $\sigma _i \in S_i$ , $i=1,2,3$ the following estimate holds:

   $$ \begin{align*} A^{-1} \leqslant |\mathfrak{n}_1(\sigma_1) \wedge \mathfrak{n}_2(\sigma_2) \wedge \mathfrak{n}_3(\sigma_3) | \leqslant 1. \end{align*} $$

Proof. We follow the argument in [Reference Kinoshita and Schippa30]. By duality it suffices to show the trilinear expression

$$ \begin{align*} \int_{S_3(\varepsilon)} (f_1 * f_2) f_3 d \nu_\gamma \lesssim A^{\frac{1}{2}} \varepsilon^{\frac{\gamma}{2}} \prod_{i=1}^3 \| f_i \|_{L^2(S_i(\varepsilon),d \nu_\gamma)}. \end{align*} $$

To this end, let $(B_{\varepsilon ,j})_{j \in \mathbb {N}}$ be an essentially disjoint cover of $\mathbb {R}^3$ with balls of size $\varepsilon $ . Set

$$ \begin{align*} J_{3,\varepsilon} = \{ j \in \mathbb{N} : B_{\varepsilon,j} \cap S_3(\varepsilon) \neq \emptyset \} \text{ and } S_{3,j}(\varepsilon) = B_{\varepsilon,j} \cap S_3(\varepsilon). \end{align*} $$

By Minkowski’s inequality we find

$$ \begin{align*} \big| \int_{\mathbb{R}^3} (f_1 * f_2)(\lambda) f_3(\lambda) d\nu_\gamma(\lambda) \big| \leqslant \sum_{j \in J_{3,\varepsilon}} \big| \int_{\mathbb{R}^3} (f_1 * f_2)(\lambda) f_3 \big\vert_{S_{3,j}(\varepsilon)}(\lambda) d\nu_\gamma(\lambda) \big|. \end{align*} $$

By Hölder’s inequality and (1.7), we find

$$ \begin{align*} \big| \int_{\mathbb{R}^3} (f_1 * f_2)(\lambda) f_3 \big\vert_{S_{3,j_3}(\varepsilon)}(\lambda) d\nu_\gamma (\lambda) \big| \lesssim \varepsilon^{\frac{\gamma}{2}} \prod_{i=1}^3 \| f_i \|_{L^2(S_{i,j,\varepsilon})}. \end{align*} $$

Above we denote

$$ \begin{align*} S_{1,j,\varepsilon} &= \{ \lambda_1 \in S_1(\varepsilon) : \exists \lambda' \in B_{\varepsilon,j} : \lambda' - \lambda_1 \in S_2(\varepsilon) \}, \\ S_{2,j,\varepsilon} &= \{ \lambda_2 \in S_2(\varepsilon) : \exists \lambda' \in B_{\varepsilon,j} : \lambda' - \lambda_2 \in S_1(\varepsilon)\}. \end{align*} $$

The following estimate was proved in [Reference Kinoshita and Schippa30, Eq. (16), p. 13]:

(4.15) $$ \begin{align} \sum_{j \in J_{3,\varepsilon}} \chi_{S_{1,j,\varepsilon} \times S_{2,j,\varepsilon}}(\lambda_1,\lambda_2) \lesssim A. \end{align} $$

With this at hand, we can conclude as follows:

$$ \begin{align*} \begin{aligned} &\quad \sum_{j \in J_{3,\varepsilon}} \big| \int_{\mathbb{R}^3} (f_1 * f_2)(\lambda) f_3 \big\vert_{S_{3,j}(\varepsilon)}(\lambda) d \nu_\gamma(\lambda) \big| \\ &\lesssim \varepsilon^{\frac{\gamma}{2}} \sum_{j \in J_{3,\varepsilon}} \| f_1 \|_{L^2(S_{1,j,\varepsilon}, d\nu_\gamma)} \| f_2 \|_{L^2(S_{2,j,\varepsilon},d \nu_\gamma)} \| f_3\|_{L^2(S_{3,j}(\varepsilon),d\nu_\gamma)} \\ &\lesssim \varepsilon^{\frac{\gamma}{2}} \big( \sum_{ j \in J_{3,\varepsilon}} \| f_3 \|_{L^2(S_{3,j}(\varepsilon),d\nu_\gamma)}^2 \big)^{\frac{1}{2}} \big( \sum_{j \in J_{3,\varepsilon}} \| f_1 \|^2_{L^2(S_{1,j,\varepsilon}, d\nu_\gamma)} \| f_2 \|_{L^2(S_{2,j,\varepsilon},d\nu_\gamma)}^2 \big)^{\frac{1}{2}} \\ &\lesssim \varepsilon^{\frac{\gamma}{2}} \| f_3 \|_{L^2(S_3(\varepsilon),d\nu_\gamma)} \big( \int_{\mathbb{R}^3 \times \mathbb{R}^3} \sum_{j \in J_{3,\varepsilon}} \chi_{S_{1,j,\varepsilon} \times S_{2,j,\varepsilon}}(\lambda_1,\lambda_2) \\ &\quad \qquad \quad \times |f_1(\lambda_1)|^2 |f_2(\lambda_2)|^2 d\nu_\gamma(\lambda_1) d\nu_\gamma(\lambda_2) \big)^{\frac{1}{2}} \\ &\lesssim \varepsilon^{\frac{\gamma}{2}} A^{\frac{1}{2}} \prod_{i=1}^3 \| f_i \|_{L^2(S_i(\varepsilon)}. \end{aligned}\\[-46pt] \end{align*} $$

Proof. We start with the proof of the estimate in Euclidean space. To ease notation, let $N = N_1$ . Firstly, we decompose $f_{i,j_i,k_i} = \sum _c f^c_{i,j_i,k_i}$ into $\sim \langle L_i \rangle $ functions, which are in the $1$ -neighbourhood of a translation of the characteristic surface. Write below $g_{i,N_i,L_i} = f^c_{i,N_i,L_i}$ to ease notation. We use the anisotropic rescaling to find

$$ \begin{align*} \begin{aligned} &\quad \int (g_{1,N_1,L_1} * g_{2,N_2,L_2} ) g_{3,N_3,L_3} d \tau d \xi d \eta \\ &= N^{2(\alpha+1)} N^2 N^{\alpha + 2} \int (g^{\prime}_{1,N_1,L_1} * g^{\prime}_{2,N_2,L_2} ) g^{\prime}_{3,N_3,L_3} d \tau' d \xi' d \eta'. \end{aligned} \end{align*} $$

This makes the rescaled functions $f^c_{i,N_i,L_i}$ (or a translate thereof) supported in the $N^{-(\alpha +1)}$ -neighbourhood of the characteristic surface. After the almost orthogonal decomposition from above, we can suppose that $|\eta _i'| \lesssim 1$ . We have computed for the transversality:

$$ \begin{align*} |\mathfrak{n}(\xi_1',\eta_1') \wedge \mathfrak{n}(\xi_2',\eta_2') \wedge \mathfrak{n}(\xi_3',\eta_3') | \sim \frac{N_2}{N_1}. \end{align*} $$

In the above $(\xi _i',\eta _i')$ are contained in balls of size $c(N_2 / N_1)$ , which constitute an almost orthogonal decomposition of the spatial frequency support.

Hence, we can apply Theorem 1.1 with Lebesgue measure and $\varepsilon = N^{-(\alpha +1)}$ to find

$$ \begin{align*} | \int (g^{\prime}_{1,N_1,L_1} * g^{\prime}_{2,N_2,L_2} ) g^{\prime}_{3,N_3,L_3} d \tau' d \xi' d \eta' | \lesssim \big( \frac{N_1}{N_2} \big)^{\frac{1}{2}} N^{-\frac{3(\alpha+1)}{2}} \prod_{i=1}^3 \| g^{\prime}_{i,N_i,L_i} \|_{L^2}. \end{align*} $$

The claim follows from reversing the scaling (4.16), which yields a factor of $N^{-\frac {3(\alpha +1)}{2}} N^{-\frac {3}{2}} N^{-\frac {3(\alpha +2)}{4}}$ , and summing over $c_i$ with $\# c_i \sim L_i$ with Cauchy-Schwarz. This finishes the proof in the Euclidean case.

We turn to the cylinder case. First we suppose that $L_{\text {med}} \ll N_1^{\alpha /2}$ . If $L_{\max } \leqslant N_1^{\alpha /2}$ , we do not decompose. If $L_{\max } \geqslant N_1^{\alpha /2}$ , we decompose $f_{i,N_i,L_i}$ with $L_i = L_{\max }$ into layers having modulation size $N_1^{\alpha /2}$ such that in the following we suppose $L_{\max } \lesssim N_1^{\alpha /2}$ . We use the anisotropic scaling

$$ \begin{align*} \tau \to \tau' = \frac{\tau}{N_1^{\alpha+1}}, \quad \xi \to \xi' = \frac{\xi}{ N}, \quad \eta \to \eta'=\frac{\eta}{N_1^{\frac{\alpha}{2}+1}}, \end{align*} $$

which leads to

$$ \begin{align*} \begin{aligned} &\quad \big| \int_{\mathbb{R} \times \mathbb{R} \times \mathbb{Z}/\gamma} (f_{1,N_1,L_1} * f_{2,N_2,L_2} ) f_{3,N_3,L_3} d \xi (d \eta)_\gamma d\tau \big| = N_1^2 N_1^{2(\alpha+1)} \\ &\times (\gamma N_1^{\frac{\alpha}{2}+1}) \big| \int_{\mathbb{R} \times \mathbb{R} \times \mathbb{Z}/(\gamma N_1^{\frac{\alpha}{2}+1})} (f^{\prime}_{1,N_1,L_1} * f^{\prime}_{2,N_2,L_2} ) f_{3,N_3,L_3}' d\xi' (d\eta')_{\gamma N_1^{\frac{\alpha}{2}+1}} d\tau' \big|. \end{aligned} \end{align*} $$

Write $L_i' = L_i / N_1^{\alpha +1}$ . We carry out the Galilean transform

$$ \begin{align*} (\xi",\eta") = (\xi',\eta' - A' \xi') \end{align*} $$

with $A' = N_1^{-\frac {\alpha }{2}} A$ such that given $\xi "$ we have $\eta " + A' \xi " \in \mathbb {Z} / N_1^{\frac {\alpha }{2}+1}$ . Moreover, we have $|\eta "| \lesssim 1$ and we can carry out like above an almost orthogonal decomposition of $(\xi ",\eta ")$ into balls of size $c(N_2/N_1)$ .

Next, we decompose the supports of $f^{\prime }_{i,N_i,L_i}$ into balls $\theta _i$ of size $L_{\max }' = \frac {L_{\max }}{N_1^{\alpha +1}} \lesssim N_1^{-\frac {\alpha }{2} - 1}$ . Then, we obtain

(4.19) $$ \begin{align} \begin{aligned} &\quad \sum_{\theta_3} \big| \int \sum_{\theta_1 \sim_{\theta_3} \theta_2} \big( f^{\prime}_{1,\theta_1} * f^{\prime}_{2,\theta_2} \big) f^{\prime}_{3,\theta_3} \big| \\ &\leqslant \sum_{\theta_3} \sum_{\theta_1 \sim_{\theta_3} \theta_2} \| f^{\prime}_{1,\theta_1} * f^{\prime}_{2,\theta_2} \|_{L^2_{(\tau,\xi,\eta)'}} \| f^{\prime}_{3,\theta_3} \|_{L^2_{(\tau,\xi,\eta)'}}. \end{aligned} \end{align} $$

By symmetry we can suppose that $L_1' = L_{\min }'$ . We estimate the convolution with the Cauchy-Schwarz inequality. Let $I_{L_{\max }'}$ denote the interval of length $L^{\prime }_{\max } = L_{\max }/(N_1^{\alpha +1})$ , which contains $\pi _{\eta "}(\text {supp}(f^{\prime }_{1,\theta _1}))$ . We let $J_{L_{\max }'}$ denote the interval which contains $\pi _{\xi "}(\text {supp}(f^{\prime }_{1,\theta _1})$ . We count the number of $\eta '$ such that $\eta ' - A' \xi ' \in I_{L_{\max }'}$ for $\xi " \in I^{\prime }_{L_{\max }'}$ . This is similar to the proof of Proposition 3.3:

$$ \begin{align*} \# \{ \eta' \in \mathbb{Z}/(N_1^{\frac{\alpha}{2}+1}) : (\xi",\eta") \in I_{L_{\max}'} \times I^{\prime}_{L_{\max}'} \} \lesssim \langle A' \cdot L^{\prime}_{\max} \cdot N_1^{\frac{\alpha}{2}+1} \rangle \sim \langle A' \cdot \frac{L_{\max}}{N_1^{\frac{\alpha}{2}}} \rangle. \end{align*} $$

Now, for fixed $\eta '$ from the above set, we estimate

$$ \begin{align*} \text{meas}( \{ \xi" \in I_{L_{\max}'} : (\xi",\eta") \in I^{\prime}_{L^{\prime}_{\max}} \times I_{L_{\max}} \} \lesssim \frac{L^{\prime}_{\max}}{A'}. \end{align*} $$

Given $(\xi ",\eta ")$ from above, we finally have

$$ \begin{align*} \text{meas} ( \{ \tau' \in \mathbb{R} : (\tau',\xi",\eta") \in \theta_1 \} ) \lesssim L^{\prime}_{\min}. \end{align*} $$

For this reason we find

$$ \begin{align*} \| f^{\prime}_{1,\theta_1} * f^{\prime}_{2,\theta_2} \|_{L^2_{\tau',\xi',\eta'}} \lesssim (L_{\min}' (L^{\prime}_{\max}/A') \langle A' \cdot \frac{L_{\max}}{N_1^{\frac{\alpha}{2}}} \rangle)^{\frac{1}{2}} \| f^{\prime}_{1,\theta_1} \|_{L^2_{\tau',\xi',\eta'}} \| f^{\prime}_{2,\theta_2} \|_{L^2_{\tau',\xi',\eta'}}. \end{align*} $$

By two more applications of the Cauchy-Schwarz inequality to carry out the summation over $\theta _i$ , we incur a factor of $A^{\frac {1}{2}} \sim \Big (\frac {N_1}{N_2}\Big )^{\frac {1}{2}}$ and obtain

$$ \begin{align*} \begin{aligned} &\quad \sum_{\theta_3} \sum_{\theta_1 \sim_{\theta_3} \theta_2} \| f^{\prime}_{1,\theta_1} * f^{\prime}_{2,\theta_2} \|_{L^2_{\tau',\xi',\eta'}} \| f^{\prime}_{3,\theta_3} \|_{L^2_{\tau',\xi',\eta'}} \\ &\lesssim (L^{\prime}_{\min} L^{\prime}_{\max})^{\frac{1}{2}} \sum_{\theta_3} \sum_{\theta_1 \sim_{\theta_3} \theta_2} \| f^{\prime}_{1,\theta_1} \|_{L^2_{\tau',\xi',\eta'}} \| f^{\prime}_{2,\theta_2} \|_{L^2_{\tau',\xi',\eta'}} \| f^{\prime}_{3,\theta_3} \|_{L^2_{\tau',\xi',\eta'}} \\ &\lesssim (L^{\prime}_{\min} L_{\max}')^{\frac{1}{2}} (N_1 /N_2)^{\frac{1}{2}} \prod_{i=1}^3 \| f^{\prime}_{i,N_i,L_i} \|_{L^2_{\tau',\xi',\eta'}}. \end{aligned} \end{align*} $$

Now we reverse the scaling which yields a factor $N_1^{-\frac {3}{2}} N_1^{-\frac {3}{2}(\alpha +1)}$ . Taking into account the scaling factor from above gives

$$ \begin{align*} N_1^2 N_1^{2(\alpha+1)} \big( \frac{N_1}{N_2} \big)^{\frac{1}{2}} N_1^{-\frac{3}{2}} N_1^{-\frac{3}{2}(\alpha+1)} N_1^{-(\alpha+1)} (L_{\min} L_{\max})^{\frac{1}{2}} = N_1^{\frac{1}{2}-\frac{\alpha}{2}} N_2^{-\frac{1}{2}} (L_{\min} L_{\max})^{\frac{1}{2}}. \end{align*} $$

This proves (4.18) in case $L_{\text {med}} \ll N_1^{\alpha /2}$ .

We turn to the case $L_{\text {med}} \gtrsim N_1^{\alpha /2}$ and suppose that $L_{\min } \ll N_1^{\alpha /2}$ . We decompose the modulation of $f_{i_j}$ with $L_{i_1} = L_{\text {med}}$ and $L_{i_2} = L_{\text {max}}$ into layers of thickness $N_1^{\frac {\alpha }{2}}$ , such that we can apply the previous result with $L_{\text {med}} \sim L_{\max } \sim N_1^{\alpha /2}$ . The claim follows then from the Cauchy-Schwarz inequality which incurs factors

$$ \begin{align*} (L_{\text{med}} / N_1^{\alpha/2})^{\frac{1}{2}} (L_{\text{max}} / N_1^{\alpha/2})^{\frac{1}{2}}. \end{align*} $$

Finally, we turn to the case $L_{\min } \gtrsim N_1^{\alpha /2}$ . In this case we decompose all functions into layers of modulation with size $N_1^{\alpha /2}$ . Then we can apply the previous arguments and finally, we can apply the Cauchy-Schwarz inequality, which incurs a factor of

$$ \begin{align*} \big( \prod_{i=1}^3 L_i \big)^{\frac{1}{2}} / N_1^{\frac{3 \alpha}{2}}. \end{align*} $$

This completes the proof of the Loomis-Whitney inequality on the cylinder.

Proof. We recall that the resonance function is given by

$$ \begin{align*} \Omega_{\alpha}(\xi_1,\xi_2,\eta_1,\eta_2) = \Omega_{\alpha}^1(\xi_1,\xi_2) - \Omega_{\alpha}^2(\xi_1,\xi_2,\eta_1,\eta_2), \end{align*} $$

where

$$ \begin{align*} \begin{aligned} \Omega_{\alpha}^1(\xi_1,\xi_2) &= |\xi_1+\xi_2|^{\alpha}(\xi_1+\xi_2) - |\xi_1|^{\alpha}\xi_1 - |\xi_2|^{\alpha}\xi_2,\\ \Omega_{\alpha}^2(\xi_1,\xi_2,\eta_1,\eta_2) &= \frac{(\eta_1\xi_2-\eta_2\xi_1)^2}{\xi_1\xi_2( \xi_1+\xi_2)}. \end{aligned} \end{align*} $$

We define functions via their Fourier transforms as follows:

(5.3) $$ \begin{align} \begin{aligned} \hat{\phi}_1(\xi_1,\eta_1) &= \frac{1_{D_1}(\xi_1,\eta_1)}{N^{-\frac{\alpha}{2}+\frac{1}{4}}}, \quad D_1 = [N^{-\frac{\alpha-1}{2}}, 2N^{-\frac{\alpha-1}{2}}] \times [-N^{-\frac{\alpha}{2}}, N^{-\frac{\alpha}{2}}],\\ \hat{\phi}_2(\xi_2,\eta_2) &= \frac{1_{D_2}(\xi_2,\eta_2)}{N^{s_1 +(1+\frac{\alpha}{2})s_2}N^{-\frac{\alpha}{2}+\frac{1}{4}}}, \\ D_2 &= [N,N+ N^{-\frac{\alpha-1}{2}}] \times [\sqrt{1+\alpha}N^{\frac{\alpha}{2}+1}, \sqrt{1+\alpha}N^{\frac{\alpha}{2}+1} +N^{-\frac{\alpha}{2}}]. \end{aligned} \end{align} $$

From [Reference Sanwal and Schippa41, Lemma 3.1], we estimate the size of the resonance function by

$$ \begin{align*} |\Omega_{\alpha}| \sim 1. \end{align*} $$

In Lemma 5.1 we computed

$$ \begin{align*} |\hat{v}(t,\xi,\eta)| \sim t \frac{N}{N^{s_1} N^{(1+\frac{\alpha}{2})s_2}} \end{align*} $$

for a significant measure of $(\xi ,\eta ) = (\xi _1,\eta _1) + (\xi _2,\eta _2)$ with $(\xi _i,\eta _i) \in D_i$ . For $0<t\leqslant c \ll 1$ , the Sobolev norm of $v(t)$ is given by

$$ \begin{align*} \|v(t)\|_{H^{s_1,s_2}(\mathbb{R}^2)} \sim tN^{\frac{5}{4}-\frac{\alpha}{2}}. \end{align*} $$

For $\Gamma _t$ to be $C^2$ -differentiable, it needs to hold

$$ \begin{align*} 1 \sim \|\phi_1\|_{H^{s_1,s_2}(\mathbb{R}^2)} \|\phi_2\|_{H^{s_1,s_2}(\mathbb{R}^2)} \gtrsim t N^{\frac{5}{4}-\frac{\alpha}{2}}, \end{align*} $$

which requires $\alpha \geqslant \frac {5}{2}$ for $N\gg 1$ . Clearly, $\phi _i$ are not real-valued, but letting $u^*_0 = \phi _1 + \phi _2$ and symmetrizing the Fourier transform we find real-valued initial data $u_0$ with comparable Sobolev norm. The above estimates remain unchanged, which completes the proof.

Proof of Theorem 5.3

We define the input functions $\phi _1$ and $\phi _2$ via their spatial Fourier transforms as follows:

$$ \begin{align*} \begin{aligned} \hat{\phi}_1(\xi_1,\eta_1) &= q^{-\frac{1}{\alpha+2}} \gamma^{-\frac{1}{2}} 1_{D_1}(\xi_1,\eta_1), \text{ where } D_1 = [q^{\frac{2}{\alpha+2}}\gamma, 2q^{\frac{2}{\alpha+2}}\gamma] \times \{0\},\\ \hat{\phi}_2(\xi_2,\eta_2) &= q^{-\frac{1}{\alpha+2}} \gamma^{-\frac{1}{2}} (q^{\frac{2}{\alpha+2}}N)^{-s_1} p^{-s_2}1_{D_2}(\xi_2,\eta_2), \\ &\quad \text{ where } D_2 = [q^{\frac{2}{\alpha+2}} (N-\gamma), q^{\frac{2}{\alpha+2}}N] \times \{p\}. \end{aligned} \end{align*} $$

In the above definition we choose q such that $[q^{\frac {2}{\alpha +2}}\gamma , 2q^{\frac {2}{\alpha +2}}\gamma ] \subseteq [0,1]$ , that is, $ q^{\frac {2}{\alpha +2}} \gamma \lesssim 1$ which follows from (5.10) and $\alpha \geqslant 1$ .

It is straight-forward to check that $\|\phi _i\|_{H^{s_1,s_2}} \sim 1$ for $i=1,2$ . We consider the contribution to the second Picard iterate given by the Duhamel integral:

$$ \begin{align*} v(t) = \int_0^t S_{\alpha}(t-s)\partial_x\big(S_{\alpha}(s)\phi_1 S_{\alpha}(s)\phi_2\big)ds. \end{align*} $$

To show that the flow map $\Gamma _t$ is not $C^2$ differentiable at the origin, we show that

$$ \begin{align*} \|v(t)\|_{H^{s_1,s_2}} \to \infty \text{ as } N \to \infty. \end{align*} $$

From Lemma 5.1, we have

$$ \begin{align*} \hat{v}(t,\xi,\eta) = \xi e^{it\omega_{\alpha}(\xi,\eta)} \int_{*} \frac{1-e^{-it\Omega_{\alpha}(\xi_1,\xi_2,\eta_1,\eta_2)}}{\Omega_{\alpha}(\xi_1,\xi_2,\eta_1,\eta_2)} \hat{\phi}_1(\xi_1,\eta_1)\hat{\phi}_2(\xi_2,\eta_2) d\xi_1d\eta_1, \end{align*} $$

where $*$ denotes the convolution constraint and we have the counting measure in the $\eta $ variable. To compute the $H^{s_1,s_2}$ norm of v, we estimate the size of $\hat {v}$ for $0<t\leqslant c \ll 1$ :

$$ \begin{align*} |\hat{v}(t,\xi,\eta)| \sim t \frac{q^{\frac{2}{\alpha+2}}N}{(q^{\frac{2}{\alpha+2}}N)^{s_1}p^{s_2}}. \end{align*} $$

Thus,

$$ \begin{align*} \|v(t)\|_{H^{s_1,s_2}}^2 \sim \int \langle \xi\rangle^{2s_1}\langle \eta \rangle^{2s_2}|\hat{v}(t,\xi,\eta)|^2d\xi d\eta \sim t^2 q^{\frac{4}{\alpha+2}} N^2 q^{\frac{2}{\alpha+2}}\gamma. \end{align*} $$

For the flow map $\Gamma _t$ to be $C^2$ , we require

$$ \begin{align*} 1\sim \|\phi_1\|_{H^{s_1,s_2}} \|\phi_2\|_{H^{s_1,s_2}} \gtrsim \|v(t)\|_{H^{s_1,s_2}} \sim t q^{\frac{3}{\alpha+2}}N\gamma^{\frac{1}{2}} \sim t N^{-\frac{1}{4} \frac{5-\alpha}{\alpha+3}} N^{\frac{5-\alpha}{4}}, \end{align*} $$

where we used (5.9) and (5.10) to obtain the last term in the above display. Hence, we conclude that $\Gamma _t$ can be $C^2$ -differentiable only for $\alpha \geqslant 5$ .

Proof. Considering initial data, which do not depend on the second variable, the evolution becomes

$$ \begin{align*} \partial_t u + \partial_x D_x^\alpha u = u \partial_x u. \end{align*} $$

For this equation it is easy to see, choosing initial data

$$ \begin{align*} \hat{u}_0(\xi) = \delta_1(\xi) + \delta_N(\xi) \end{align*} $$

for $N \gg 1$ , the data-to-solution mapping fails to be $C^2$ for $\alpha < 1$ .

In the following we suppose that $\alpha \geqslant 1$ and consider the following functions:

(5.12) $$ \begin{align} \begin{aligned} \hat{\phi}_1(\xi_1,\eta_1) &= 1_{D_1}(\xi_1,\eta_1), \quad D_1 = \{ 1 \} \times \{ 0 \},\\ \hat{\phi}_2(\xi_2,\eta_2) &= N^{-s_1} p^{-s_2} 1_{D_2}(\xi_2,\eta_2), \quad D_2 = \{ N \} \times \{ p \}. \end{aligned} \end{align} $$

We choose p such that

$$ \begin{align*} \Omega_\alpha(1,0,N,p) = (N+1)^{\alpha+1} - 1 - N^{\alpha+1} - \frac{p^2}{(N+1)N} = 0. \end{align*} $$

Note that this requires a choice of $\gamma \in (\frac {1}{2},1]$ such that $p \in \gamma ^{-1} \mathbb {Z}$ , that is, fixing the ratio of the periods.

Clearly, for the Fourier transform of the Duhamel integral

$$ \begin{align*} |\hat{v}(t,\xi,\eta)| \sim \frac{N t}{N^{s_1} p^{s_2}}, \quad \xi = N+1, \; \eta = p. \end{align*} $$

Hence,

$$ \begin{align*} \| v(t) \|_{H^{s_1,s_2}} \gtrsim N t. \end{align*} $$

This implies that for $\alpha \geqslant 1$ , the data-to-solution mapping of (1.1) fails to be $C^2$ for some $\gamma = \gamma (\alpha )$ .

Proof of Theorem 6.1

The proof follows along the same lines as for [Reference Sanwal and Schippa41, Theorem 6.1], and we shall be brief. We use $X^{s,b}$ spaces as the auxiliary spaces to run a fixed point argument for the operator given by:

$$ \begin{align*} \Gamma(u)(t):= \eta(t) S_{\alpha}(t)u_0 + \eta(t) \int_0^t S_{\alpha}(t-s)(u \partial_x u)(s)ds. \end{align*} $$

Here, $\eta $ is a smooth compactly supported time cut-off. Using the linear estimate, the energy estimate for $X^{s,b}$ spaces [Reference Tao46, Section 2.6] and Proposition 6.2, we obtain

$$ \begin{align*} \|\Gamma(u)\|_{X^{s,b}_1} \leqslant C (\|u(0)\|_{H^{s,0}} + \|u\|_{X^{s,b}_1}^2). \end{align*} $$

Similarly,

$$ \begin{align*} \| \Gamma(u_1) -\Gamma(u_2)\|_{X^{s,b}_1} \leqslant 2\tilde{C}(\|u_0\|_{H^{s,0}}) \|u_1-u_2\|_{X^{s,b}_1}. \end{align*} $$

For small initial data (attributed in the constants $C,\tilde {C}$ ), we can prove local well-posedness for (5.1) on $X^{s,b}_1$ . Using scaling and subcriticality of the regularity, any large data can be scaled to be small and one obtains well-posedness on a time interval $[0,T]$ , with T depending on the size of the initial data.

Proof of Theorem 6.3

For small initial data and a fixed time interval $[0,1]$ , the proof of local well-posedness is the same as in the $\mathbb {R}^2$ case. In the large data case, we use the leeway in the modulation regularity in the proof of Proposition 6.4 to apply Lemma 6.9. This yields a modified bilinear estimate

$$ \begin{align*} \| \partial_x(u v) \|_{X^{s,b-1}_T} \lesssim T^{\delta} \| u \|_{X^{s,b}} \| v \|_{X^{s,b}}. \end{align*} $$

Thus, we see that the time of existence of the solution will depend on the size of the initial data and the well-posedness result follows similarly as in the $\mathbb {R}^2$ case.