# Improved global well-posedness for mass-critical nonlinear Schrödinger
equations on tori

Robert Schippa  Korea Institute of Advanced Study, Hoegi-ro 85, Dongdaemun-gu,
02455 Seoul<EMAIL_ADDRESS>
###### Abstract.

We show new global well-posedness results for mass-critical nonlinear
Schrödinger equations on tori in one and two dimensions. For the quintic
nonlinear Schrödinger equation on the circle we show global well-posedness for
initial data in $H^{s}(\mathbb{T})$ for $s>\frac{1}{3}$ and
$\|u_{0}\|_{L^{2}(\mathbb{T})}\ll 1$. In two dimensions we show global well-
posedness on possibly irrational tori for $s>\frac{3}{5}$. In the focusing
case we need to consider small $L^{2}$-norm, whereas in the defocusing case
large mass is covered.

##  1\. Introduction

This note is concerned with the global well-posedness of the mass-critical
nonlinear Schrödinger equation (NLS) on tori. The one-dimensional torus with
period $2\pi\beta$, $\beta>1/2$ is denoted by
$\mathbb{T}_{\beta}=\mathbb{R}/(2\pi\beta\mathbb{Z})$. For $d=2$ we denote
possibly irrational tori with $\underline{\gamma}\in(1/2,1]^{d-1}$ by
$\mathbb{T}^{d}_{\underline{\gamma}}=\mathbb{T}\times\mathbb{T}_{\gamma_{1}}\times\ldots\times\mathbb{T}_{\gamma_{d-1}}$.
The square torus is denoted by $\mathbb{T}^{d}$. To simplify notations, for
$d=1$ let $\underline{\gamma}=1$.

The Cauchy problems for the mass-critical NLS read

(1) $\left\\{\begin{array}[]{cl}i\partial_{t}u+\Delta
u&=\pm|u|^{\frac{4}{d}}u,\quad(t,x)\in\mathbb{R}\times\mathbb{T}^{d}_{\underline{\gamma}},\\\
u(0)&=u_{0}\in H^{s}(\mathbb{T}^{d}_{\underline{\gamma}}).\end{array}\right.$

The energy is given by

$E(u)(t)=\int_{\mathbb{T}^{d}_{\gamma}}\frac{|\nabla_{x}u|^{2}}{2}\pm\frac{d}{4+2d}|u|^{2+\frac{4}{d}}dx$

with signs matching the nonlinearity. The equation with $+$-sign on the right
hand-side is referred to as defocusing due to coercivity of the energy,
whereas the equation with $-$-sign is referred to as focusing.

The low regularity local well-posedness of nonlinear Schrödinger equations on
tori was thoroughly investigated for the first time by Bourgain [1], using
discrete Fourier restriction and introducing Fourier restriction norm spaces.

By using the $L^{{2+\frac{4}{d}}}_{t,x}$-Strichartz estimate

(2)
$\|e^{it\Delta}P_{N}f\|_{L^{{2+\frac{4}{d}}}_{t,x}([0,1]\times\mathbb{T}^{d})}\lesssim_{\varepsilon}N^{\varepsilon}\|f\|_{L^{2}(\mathbb{T})}$

and multilinear refinements, which result from Galilean invariance, Picard
iteration becomes possible in Fourier restriction norm spaces. In this way
Bourgain [1] proved local well-posedness of (1) for $d=1$ and $d=2$ on
$\mathbb{T}^{2}$ for $s>0$.

It is known since [1] that the derivative loss in (2) cannot be removed for
$d=1$, and Kishimoto [16] proved that (1) fails to be locally well-posed in
$L^{2}(\mathbb{T}^{d})$ with $C^{5}$-data-to-solution mapping in $d=1$ (resp.
$C^{3}$-data-to-solution mapping in $d=2$); see also [22].

The first global result on $\mathbb{T}$ below $H^{1}(\mathbb{T})$ was proved
by Bourgain [2] using normal form transformations. Using trilinear estimates
on frequency dependent times, global well-posedness for
$s>\frac{1}{2}-\varepsilon$ with some small $\varepsilon>0$ was established.
The key ingredient is a trilinear refinement of the $L^{6}_{t,x}$-Strichartz
estimate for separated frequencies on frequency-dependent time intervals.
Whereas Bourgain’s estimate was qualitative, the smoothing was recently
quantified in [21].

In [8] De Silva _et al._ reported global well-posedness in the defocusing case
in $H^{s}(\mathbb{T})$ for $s$ below $\frac{1}{2}$ in a quantitative
sense111An issue with the analysis of the resonant part was pointed out by
Tzirakis [24].. The authors used the $I$-method (see [5, 6]) to show energy
decay. With the method having become a canonical tool to show low regularity
well-posedness of dispersive equations (see [23, Chapter 3]), we presently
give only a very brief account for context:

Let $0<s<1$. The $I$-method revolves around a smoothing operator $I$, which is
smoothing of order $1-s$ for frequencies larger than $N$. For frequencies
lower than $N$, $I$ acts like the identity mapping. To introduce more
dispersion, we rescale to a large torus $\mathbb{T}^{d}_{\lambda}$. The effect
becomes more favorable for higher regularities. The relation between $\lambda$
and $N$ is given by

$\lambda=N^{\frac{1-s}{s}},$

which ensures that

$\|If_{\lambda}\|_{H^{1}(\mathbb{T}^{d}_{\lambda})}\lesssim\|f\|_{H^{s}(\mathbb{T}^{d})}.$

Although the energy cannot be used as conserved quantity for $f_{\lambda}\in
H^{s}(\mathbb{T}^{d}_{\lambda})$, $s<1$, it will be finite for $If_{\lambda}$.
We analyze the $I$-energy $E(Iu_{\lambda})$, were $u_{\lambda}$ denotes the
solution to (1) with $u_{\lambda}(0)=f_{\lambda}$, which serves as an almost
conserved quantity. The slow $I$-energy growth allows us to show well-
posedness of the rescaled system on a time interval
$N^{\beta}\lambda^{\alpha}$. However, this existence time depends on the very
large period $\lambda=\lambda(N)$. Scaling back to the unit torus, the time of
existence is given by $N^{\beta}\lambda^{\alpha-2}$. Provided that this
translates to $N^{\kappa}$ for some $\kappa>0$, we can conclude the global
well-posedness result by taking $N\to\infty$.

The result in [8] was further improved by Li–Wu–Xu [19] who showed global
well-posedness of the defocusing equation for $s>\frac{2}{5}$. They combined
the $I$-method argument with resonant decompositions. In one dimension we use
the same resonant decomposition and rescaling to large tori to increase
dispersive effects. Using suitable linear and bilinear Strichartz estimates on
time intervals much longer than the unit time scale, we can improve the result
due to Li–Wu–Xu [19] in case of small data $\|u_{0}\|_{L^{2}(\mathbb{T})}$.
Note that the $L^{2}$-norm remains invariant under rescaling, and our
arguments in one dimension currently do not allow for large data. It is
currently not clear how to bring the trilinear refinements from [21] into
play.

Precisely, we show the following small mass global well-posedness result:

######  Theorem 1.1.

Let $d=1$. For $s>\frac{1}{3}$ (1) is globally well-posed provided that
$\|u_{0}\|_{L^{2}(\mathbb{T})}\ll 1$.

Furthermore, we improve the global well-posedness in two dimensions on
arbitrary tori
$\mathbb{T}^{2}_{\gamma}=\mathbb{T}\times\mathbb{R}/(2\pi\gamma\mathbb{Z})$
for $\gamma\in(\frac{1}{2},1]$. In two dimensions the nonlinearity is cubic.
On the square torus $\mathbb{T}^{2}$, Bourgain proved the
$L^{4}_{t,x}$-Strichartz estimate in [1]

(3)
$\|P_{N}e^{it\Delta}u_{0}\|_{L^{4}_{t,x}(\mathbb{T}\times\mathbb{T}^{2})}\lesssim\exp\big{(}\frac{c\log
N}{\log\log N}\big{)}\|u_{0}\|_{L^{2}(\mathbb{T}^{2})}.$

Global well-posedness in the defocusing case was proved by De Silva _et al._
[8] for $s>\frac{2}{3}$ using the $I$-method, which confirmed an assertion of
Bourgain in [2]. De Silva–Pavlović–Staffilani–Tzirakis [8] relied on the
following bilinear Strichartz estimate on the square torus
$\mathbb{T}^{2}_{\lambda}$ with large period $\lambda\geq 1$ and separated
frequencies $N_{2}\ll N_{1}$:

(4)
$\|e^{it\Delta}P_{N_{1}}u_{1}e^{it\Delta}P_{N_{2}}u_{2}\|_{L^{2}_{t,x}([0,1]\times\mathbb{T}^{2}_{\lambda})}\lesssim\big{(}\frac{1}{\lambda}+\frac{N_{2}}{N_{1}}\big{)}^{\frac{1}{2}}\|u_{1}\|_{L^{2}(\mathbb{T}^{2}_{\lambda})}\|u_{2}\|_{L^{2}(\mathbb{T}^{2}_{\lambda})}.$

Using decoupling, Bourgain–Demeter [3] extended the $L^{4}_{t,x}$-Strichartz
estimate (3) to arbitrary tori. Moreover, using decoupling arguments, Fan _et
al._ [9] similarly extended the bilinear Strichartz estimate to arbitrary
tori, by which the global well-posedness results for the square torus could be
extended to the irrational torus. Trilinear Strichartz estimates in
$L_{t,x}^{4}$ were recently proved in [21].

Very recently, Herr–Kwak [14] proved the sharp $L^{4}_{t,x}$-Strichartz
estimate

$\|P_{S}e^{it\Delta}u_{0}\|_{L^{4}_{t,x}([0,(\log\\#S)^{-1}]\times\mathbb{T}^{2})}\lesssim\|u_{0}\|_{L^{2}(\mathbb{T}^{2})},$

where $P_{S}$ denotes the Fourier projection to a finite set
$S\subseteq\mathbb{Z}^{2}$ with $\\#S\geq 2$. This implied global well-
posedness on $\mathbb{T}^{2}$ for $s>0$ under the small mass constraint
$\|u_{0}\|_{L^{2}(\mathbb{T}^{2})}\ll 1$.

This raises two questions:

  * •

Do the arguments extend to irrational tori?

  * •

When does global well-posedness hold as well in the case of large mass?

Herr–Kwak use a Szemerédi–Trotter argument to show the $L^{4}_{t,x}$-estimate
and extension to irrational tori is currently unclear. Here instead we refine
the $I$-method argument originally due to De Silva _et al._ [8] and introduce
a correction term for the $I$-energy by resonant decomposition. We use linear
and bilinear Strichartz estimates, which hold regardless of the ratio of
periods. We show the following:

######  Theorem 1.2.

Let $d=2$. For $s>\frac{3}{5}$ (1) is globally well-posed in the defocusing
case. In the focusing case (1) is globally well-posed provided that
$\|u_{0}\|_{L^{2}(\mathbb{T}_{\gamma}^{2})}\ll 1$.

_Outline of the paper._ In Section 2 we fix notations and introduce the
function spaces on arbitrary tori. In one dimension we show linear and
bilinear Strichartz estimates on the time-scale $\lambda/N$ on large tori in
adapted function spaces. In two dimensions we stick to the unit time scale on
possibly large tori. In Section 3 we show local well-posedness on
$\mathbb{T}_{\lambda}$ on a time-scale comparable to $\lambda/N$ for $s>0$
with small mass constraint. On $\mathbb{T}^{2}_{\lambda}$ we show large data
well-posedness for $s>0$ on a time scale of length $\lambda^{-\delta}$.

In Section 4 we explain how we modify the $I$-energy by resonant decomposition
depending on the dimension. In Section 5 slow growth of the modified energy on
$\mathbb{T}_{\lambda}$ is established, and in Section 6 the corresponding
analysis on $\mathbb{T}^{2}_{\lambda}$ is carried out. The proofs of Theorems
1.1 and 1.2 are concluded in Section 7.

##  2\. Preliminaries

###  2.1. Fourier transform conventions

Recall that for $\underline{\gamma}\in(1/2,1]^{d-1}$ we set

$\mathbb{T}^{d}_{\underline{\gamma}}=\mathbb{T}\times\mathbb{T}_{\gamma_{1}}\times\ldots\times\mathbb{T}_{\gamma_{d-1}}.$

For fixed $\underline{\gamma}\in(1/2,1]^{d-1}$, we consider rescaled tori
$\mathbb{T}^{d}_{\lambda}:=\mathbb{T}_{\lambda}\times\mathbb{T}_{\gamma_{1}\lambda}\times\ldots\mathbb{T}_{\gamma_{d-1}\lambda}$
with large period $\lambda\geq 1$. We suppress the notation of
$\underline{\gamma}$ for the rescaled tori because it will be fixed at the
outset, and the estimates will be independent of
$\underline{\gamma}\in(1/2,1]^{d-1}$.

Let $f:\mathbb{T}^{d}_{\lambda}\to\mathbb{C}$ be measurable. The Lebesgue
norms are given by

$\|f\|_{L^{p}(\mathbb{T}^{d}_{\lambda})}=\big{(}\int_{\mathbb{T}^{d}_{\lambda}}|f(x)|^{p}dx\big{)}^{\frac{1}{p}}$

for $1\leq p<\infty$ with the usual modification for $p=\infty$.

The Fourier coefficients of $f\in L^{2}(\mathbb{T}^{d}_{\lambda})$ satisfy
$k\in\mathbb{Z}^{d}_{\lambda}:=\mathbb{Z}/\lambda\times\mathbb{Z}/(\lambda\gamma_{1})\times\ldots\mathbb{Z}/(\lambda\gamma_{d-1})$222To
be consistent, we suppress dependence on $\gamma$ likewise on the frequency
side. are given by

$\hat{f}(k)=\int_{\mathbb{T}^{d}_{\lambda}}e^{-ik\cdot x}f(x)dx,$

and the Fourier inversion formula reads

$f(x)=\frac{1}{(\gamma_{1}\ldots\gamma_{d-1})\lambda^{d}}\sum_{k\in\mathbb{Z}^{d}/\lambda}e^{ik\cdot
x}\hat{f}(k).$

The dual $(\mathbb{T}^{d}_{\lambda})^{*}=\mathbb{Z}^{d}_{\lambda}$ is endowed
with normalized counting measure. Let
$\tilde{f}:\mathbb{Z}^{d}_{\lambda}\to\mathbb{C}$ and
$A\subseteq\mathbb{Z}^{d}/\lambda$. We define

$\int_{A}\tilde{f}(k)(dk)_{\lambda}=\frac{1}{\lambda^{d}(\gamma_{1}\ldots\gamma_{d-1})}\sum_{k\in
A}\tilde{f}(k).$

The Fourier transform conventions are supposed to ensure that Plancherel’s
theorem holds true without constant depending on $\lambda$:

$\|f\|^{2}_{L^{2}(\mathbb{T}^{d}_{\lambda})}=\frac{1}{\lambda^{d}(\gamma_{1}\ldots\gamma_{d-1})}\sum_{k\in\mathbb{Z}^{d}_{\lambda}}|\hat{f}(k)|^{2}.$

Sobolev norms on $\mathbb{T}^{d}_{\lambda}$ are defined by

$\|f\|^{2}_{H^{s}(\mathbb{T}^{d}_{\lambda})}=\frac{1}{\lambda^{d}(\gamma_{1}\ldots\gamma_{d-1})}\sum_{k\in\mathbb{Z}^{d}_{\lambda}}\langle
k\rangle^{2s}|\hat{f}(k)|^{2}.$

For $A\subseteq\mathbb{Z}^{d}_{\lambda}$ we denote with
$P_{A}:L^{2}(\mathbb{T}^{d}_{\lambda})\to L^{2}(\mathbb{T}^{d}_{\lambda})$ the
Fourier projection to $A$. For $N\in 2^{\mathbb{N}_{0}}$ we denote with
$P_{N}$ the smooth frequency projection to frequencies of size $N$. For $N=1$,
we consider frequencies of size $\lesssim 1$.

###  2.2. Function spaces

In the following we define adapted function spaces, which behave well with
sharp time cutoff. This will be very useful in the analysis on “long” time-
intervals. $V^{p}$-spaces were firstly considered by Wiener [25]. The predual
space $U^{p}$ has an atomic structure and comprises a logarithmic refinement
of the Fourier restriction space; see [17, 18] for first applications.

For an introduction to $U^{p}$-/$V^{p}$-spaces we refer to Hadac–Herr–Koch
[11] (see also [12]). The $Y^{s}$-spaces introduced below were firstly used in
[15] and the following summary can also be found in [14].

######  Definition 2.1.

Let $\mathcal{Z}$ be the collection of finite non-decreasing sequences
$\\{t_{k}\\}_{k=0}^{K}$ in $\mathbb{R}$. We define $V^{2}$ as the space of all
right-continuous functions $u:\mathbb{R}\to\mathbb{C}$ with
$\lim_{t\to-\infty}u(t)=0$ and

$\|u\|_{V^{2}}=\big{(}\sup_{\\{t_{k}\\}_{k=0}^{K}\in\mathcal{Z}}\sum_{k=1}^{K}|u(t_{k})-u(t_{k-1})|^{2}\big{)}^{\frac{1}{2}}<\infty.$

The following $U^{p}_{\Delta}$-/$V^{p}_{\Delta}$-spaces will be useful for
intermediate steps. The function spaces, in which we formulate the well-
posedness results, are the $Y_{\pm}^{s}$-spaces defined in Definition 2.5.

######  Definition 2.2.

Let $1\leq p<\infty$. We define

$V^{p}_{\Delta}=\\{u:\mathbb{R}\times\mathbb{T}^{d}_{\lambda}\to\mathbb{C}:e^{-it\Delta}u\in
V^{p}L^{2}(\mathbb{T}^{d}_{\lambda})\\}$

with norm given by

$\|u\|_{V_{\Delta}^{p}}=\big{(}\sup_{\\{t_{k}\\}_{k=0}^{K}\in\mathcal{Z}}\sum_{k=1}^{K}\|e^{-it_{k}\Delta}u(t_{k})-e^{-it_{k-1}\Delta}u(t_{k-1})\|_{L^{2}}^{p}\big{)}^{\frac{1}{p}}<\infty.$

The predual $U^{p}$-spaces [11] admit an atomic decomposition:

######  Definition 2.3.

A $U^{p}_{\Delta}$-atom is a function
$f:\mathbb{R}\times\mathbb{T}^{d}_{\lambda}\to\mathbb{C}$, which admits the
representation

$f=\sum_{k=1}^{N}1_{[t_{k-1},t_{k})}a_{k},\quad\sum_{k=1}^{N}\|a_{k}\|_{L^{2}}^{p}=1.$

The $U^{p}$-norm of $u:\mathbb{R}\times\mathbb{T}^{d}_{\lambda}\to\mathbb{C}$
is given by

$\|u\|_{U^{p}_{\Delta}}=\inf\big{(}\\{\sum_{k=1}^{\infty}|\lambda_{k}|:\,u=\sum_{k=1}^{\infty}\lambda_{k}f_{k}\text{
with }\big{(}\lambda_{k}\big{)}\in\ell^{1},\quad f_{k}\text{ is a
}U^{p}_{\Delta}-\text{atom}\\}\big{)}.$

The convergence is in $L^{\infty}(\mathbb{R};L^{2})$.

######  Remark 2.4 (Transfer principle for $U^{p}_{\Delta}$-spaces).

The atomic structure allows one to transfer estimates like linear or bilinear
Strichartz estimates for free solutions to $U^{p}_{\Delta}$-functions, e.g.,
the linear Strichartz estimates yield

$\|P_{N}e^{it\Delta}f\|_{L^{p}_{t,x}(\mathbb{T}\times\mathbb{T}^{d}_{\lambda})}\lesssim_{\varepsilon}(\lambda
N)^{\varepsilon}\|f\|_{U^{p}_{\Delta}}$

for $p=\frac{2(d+2)}{d}$. Likewise, the bilinear Strichartz estimates transfer
to functions in $U^{2}_{\Delta}$. We refer to [11, Proposition 2.19] for
details.

######  Definition 2.5.

For $s\in\mathbb{R}$, we define $Y_{\pm}^{s}$ as the space
$u:\mathbb{R}\times\mathbb{T}^{d}_{\lambda}\to\mathbb{C}$ such that
$e^{it|\xi|^{2}}\widehat{u(t)}(\xi)$ lies in $V^{2}$ for each
$\xi\in\mathbb{Z}^{d}_{\lambda}$ and

$\|u\|_{Y_{\pm}^{s}}=\big{(}\frac{1}{\lambda^{d}}\sum_{\xi\in\mathbb{Z}^{d}_{\lambda}}(1+|\xi|^{2})^{s}\|e^{\pm
it|\xi|^{2}}\widehat{u(t)}(\xi)\|^{2}_{V^{2}}\big{)}^{\frac{1}{2}}<\infty.$

For $T>0$ and $u:[0,T]\times\mathbb{T}^{d}_{\lambda}\to\mathbb{C}$ we define

$\|u\|_{Y^{s}_{\pm,T}}=\big{(}\frac{1}{\lambda^{d}}\sum_{\xi\in\mathbb{Z}^{d}_{\lambda}}(1+|\xi|^{2})^{s}\|e^{\pm
it|\xi|^{2}}\widehat{u(t)}(\xi)\|^{2}_{V^{2}(0,T)}\big{)}^{\frac{1}{2}}.$

Note that $\|u\|_{Y^{s}_{+,T}}=\|\overline{u}\|_{Y^{s}_{-,T}}$. We write
$Y^{s}_{T}=Y^{s}_{+,T}$ for brevity.

######  Remark 2.6 (Transfer principle for $Y^{s}$-spaces).

The following embedding is immediate from the definition:

$Y^{0}_{T}\hookrightarrow V^{2}_{\Delta}.$

Moreover, it holds the following [11] for $p>2$:

$V^{2}_{\Delta}\hookrightarrow U^{p}_{\Delta}.$

######  Proposition 2.7 ([14]).

$Y^{s}$-norms have the following properties:

  * •

Let $A$, $B$ be disjoint subsets of $\mathbb{Z}^{d}_{\lambda}$. For
$s\in\mathbb{R}$, it holds

(5) $\|P_{A\cup
B}u\|^{2}_{Y^{s}}=\|P_{A}u\|^{2}_{Y^{s}}+\|P_{B}u\|^{2}_{Y^{s}}.$

  * •

For $s\in\mathbb{R}$, time $T>0$, and a function $f\in L^{1}H^{s}$ we have

(6)
$\|\chi_{[0,T)}\cdot\int_{0}^{t}e^{i(t-t^{\prime})\Delta}f(t^{\prime})dt^{\prime}\|_{Y^{s}}\lesssim\sup_{\begin{subarray}{c}v\in
Y^{-s}:\\\ \|v\|_{Y^{-s}}\leq
1\end{subarray}}\big{|}\int_{0}^{T}\int_{\mathbb{T}_{\lambda}}\overline{v}fdxdt\big{|}.$

###  2.3. Linear and bilinear Strichartz estimates

In the following we consider linear and bilinear Strichartz estimates on
$\mathbb{T}^{d}_{\lambda}$.

In one dimension, we consider times up to $\lambda/N$, which is the time-scale
on which we analyze (1) on $\mathbb{T}_{\lambda}$. We recall the bilinear
Strichartz estimate on $\mathbb{T}$ due to Moyua–Vega [20] and Hani [13] (see
[21, Appendix] for a different proof):

######  Theorem 2.8.

Let $N\in 2^{\mathbb{N}_{0}}$ and $I_{1}$, $I_{2}$ be intervals contained in
$[-10N,10N]$ with

$\inf_{k_{i}\in I_{i}}|k_{1}-k_{2}|\gtrsim N.$

Then the following estimate holds:

$\|P_{I_{1}}e^{it\partial_{xx}}f_{1}P_{I_{2}}e^{it\partial_{xx}}f_{2}\|_{L^{2}_{t}([0,N^{-1}],L_{x}^{2}(\mathbb{T}))}\lesssim
N^{-\frac{1}{2}}\|f_{1}\|_{L^{2}(\mathbb{T})}\|f_{2}\|_{L^{2}(\mathbb{T})}.$

Moreover, we have the following linear Strichartz estimates due to
Burq–Gérard–Tzvetkov [4] and Guo–Li–Yung [10].

######  Theorem 2.9.

Let $N\in 2^{\mathbb{N}_{0}}$, and $\varepsilon>0$. Then the following
estimates hold:

(7)
$\displaystyle\|P_{N}e^{it\partial_{xx}}f\|_{L^{6}_{t}([0,N^{-1}],L_{x}^{6}(\mathbb{T}))}$
$\displaystyle\lesssim\|f\|_{L^{2}(\mathbb{T})},$ (8)
$\displaystyle\|P_{N}e^{it\partial_{xx}}f\|_{L^{6}_{t,x}(\mathbb{T}^{2})}$
$\displaystyle\lesssim_{\varepsilon}\log(N)^{2+\varepsilon}\|f\|_{L^{2}(\mathbb{T})}.$

We have the following rescaled bilinear Strichartz estimates:

######  Proposition 2.10.

Let $M\in 2^{\mathbb{N}_{0}}$, and $I_{1}$, $I_{2}$ be intervals contained in
$[-10M,10M]$ with

$\inf_{k_{i}\in I_{i}}|k_{1}-k_{2}|\gtrsim M.$

Then the following estimate holds:

(9)
$\|P_{I_{1}}u_{1}P_{I_{2}}u_{2}\|_{L^{2}_{t,x}([0,\frac{\lambda}{N}]\times\mathbb{T}_{\lambda})}\lesssim(\log(\lambda)+1)^{2}\big{(}\frac{1}{M}+\frac{1}{N}\big{)}^{\frac{1}{2}}\|P_{I_{1}}u_{1}\|_{Y^{0}_{+}}\|P_{I_{2}}u_{2}\|_{Y^{0}_{+}}.$

###### Proof.

It suffices to show the following estimate for free solutions
$u_{i}=e^{it\partial_{x}^{2}}f_{i}$:

(10)
$\|P_{I_{1}}u_{1}P_{I_{2}}u_{2}\|_{L^{2}_{t,x}([0,\frac{\lambda}{N}]\times\mathbb{T}_{\lambda})}\lesssim\big{(}\frac{1}{M}+\frac{1}{N}\big{)}^{\frac{1}{2}}\|P_{I_{1}}f_{1}\|_{L^{2}(\mathbb{T}_{\lambda})}\|P_{I_{2}}f_{2}\|_{L^{2}(\mathbb{T}_{\lambda})}.$

Indeed, the above display implies by the transfer principle333This is a
consequence of the atomic representation. in $U^{p}$-spaces:

$\|P_{I_{1}}u_{1}P_{I_{2}}u_{2}\|_{L^{2}_{t,x}([0,\frac{\lambda}{N}]\times\mathbb{T}_{\lambda})}\lesssim\big{(}\frac{1}{M}+\frac{1}{N}\big{)}^{\frac{1}{2}}\|P_{I_{1}}u_{1}\|_{U^{2}_{\Delta}}\|P_{I_{2}}u_{2}\|_{U^{2}_{\Delta}}.$

This we can interpolate (see [11, Proposition 2.20] and [11, Corollary 2.21])
twice with the estimate

$\begin{split}\|P_{I_{1}}u_{1}P_{I_{2}}u_{2}\|_{L^{2}_{t,x}([0,\frac{\lambda}{N}]\times\mathbb{T}_{\lambda})}&\leq\|P_{I_{1}}u_{1}\|_{L^{4}_{t,x}([0,\frac{\lambda}{N}]\times\mathbb{T}_{\lambda})}\|P_{I_{2}}u_{2}\|_{L^{4}_{t,x}([0,\frac{\lambda}{N}]\times\mathbb{T}_{\lambda})}\\\
&\lesssim\lambda^{c}\|P_{I_{1}}u_{1}\|_{U^{4}_{\Delta}}\|P_{I_{2}}u_{2}\|_{U^{4}_{\Delta}},\end{split}$

which incurs a squared logarithmic factor in $\lambda$. Lastly, the estimate
(10) follows from rescaling to the unit torus:

$\|P_{I_{1}}u_{1}P_{I_{2}}u_{2}\|_{L^{2}_{t,x}([0,\frac{\lambda}{N}]\times\mathbb{T}_{\lambda})}=\lambda^{\frac{3}{2}}\|P_{\lambda
I_{1}}u_{1\lambda}P_{\lambda
I_{2}}u_{2\lambda}\|_{L^{2}_{t,x}([0,\frac{1}{\lambda N}]\times\mathbb{T})}.$

Note that $\lambda I_{j}$ are contained in $[-10\lambda M,10\lambda M]$. If
$M\gtrsim N$, we partition the time interval $[0,\frac{1}{\lambda N}]$ into
intervals $J$ of size $\frac{1}{\lambda M}$ on which we can use the bilinear
Strichartz estimate:

$\begin{split}\|P_{\lambda I_{1}}u_{1\lambda}P_{\lambda I_{2}}u_{\lambda
I_{2}}\|^{2}_{L^{2}_{t,x}([0,\frac{1}{\lambda
N}]\times\mathbb{T})}&=\sum_{J}\|P_{\lambda I_{1}}u_{1\lambda}P_{\lambda
I_{2}}u_{2\lambda}\|^{2}_{L^{2}_{t,x}(J\times\mathbb{T})}\\\
&\lesssim\frac{1}{N\lambda}\|u_{1\lambda}(0)\|^{2}_{L^{2}(\mathbb{T})}\|u_{2\lambda}(0)\|^{2}_{L^{2}(\mathbb{T})}.\end{split}$

The claim follows then from reversing the scaling.

If $M\lesssim N$, we can use the bilinear estimate directly:

$\|P_{\lambda I_{1}}u_{1\lambda}P_{\lambda
I_{2}}u_{2\lambda}\|^{2}_{L^{2}_{t,x}([0,\frac{1}{\lambda
N}]\times\mathbb{T})}\lesssim\frac{1}{M\lambda}\|u_{1\lambda}(0)\|^{2}_{L^{2}(\mathbb{T})}\|u_{2\lambda}(0)\|^{2}_{L^{2}(\mathbb{T})}.$

∎

######  Remark 2.11.

We note the following variant, which is immediate from complex conjugation:
Let $I_{1}$, $I_{2}$ be intervals contained in $[-10M,10M]$ with

$\inf_{k_{i}\in I_{i}}|k_{1}+k_{2}|\gtrsim M.$

Then the following estimate holds:

$\|P_{I_{1}}u_{1}P_{I_{2}}u_{2}\|_{L^{2}_{t,x}([0,\frac{\lambda}{N}]\times\mathbb{T}_{\lambda})}\lesssim(\log(\lambda)+1)^{2}\big{(}\frac{1}{M}+\frac{1}{N}\big{)}^{\frac{1}{2}}\|P_{I_{1}}u_{1}\|_{Y^{0}_{+,T}}\|P_{I_{2}}u_{2}\|_{Y^{0}_{-,T}}.$

######  Proposition 2.12.

Let $M\in 2^{\mathbb{N}_{0}}$. If $M\lesssim N$, the following estimate holds:

(11)
$\|P_{M}u\|_{L^{6}_{t,x}([0,\frac{\lambda}{N}]\times\mathbb{T}_{\lambda})}\lesssim\|u\|_{Y^{0}_{T}}.$

If $M\gg N$, the following estimate holds:

(12)
$\|P_{M}u\|_{L^{6}_{t,x}([0,\frac{\lambda}{N}]\times\mathbb{T}_{\lambda})}\lesssim\log(M)^{3}\|u\|_{Y^{0}_{T}}.$

###### Proof.

For estimate (11) it suffices to prove

$\|P_{M}u\|_{L^{6}_{t,x}([0,\frac{\lambda}{N}]\times\mathbb{T}_{\lambda})}\lesssim\|u\|_{U^{6}_{\Delta}}$

by embedding properties of $V^{2}_{\Delta}\hookrightarrow U^{6}_{\Delta}$. The
above display is immediate from the estimates in [4] and rescaling. The second
estimate (12) follows from the $L^{6}_{t,x}$-estimate on the torus due to
Guo–Li–Yung [10]. ∎

We recall the following linear and bilinear Strichartz estimates in adapted
function spaces (see [3, 9]):

######  Proposition 2.13.

Let $\lambda\geq 1$ and $N\in 2^{\mathbb{N}_{0}}$. The following estimate
holds:

(13)
$\|P_{N}u\|_{L^{4}_{t,x}(\mathbb{T}\times\mathbb{T}^{2}_{\lambda})}\lesssim_{\varepsilon}\lambda^{\varepsilon}\|u\|_{Y^{0}}.$

If $\lambda\geq N$, then the estimate holds without loss of
$\lambda^{\varepsilon}$.

Let $N_{1}\ll N_{2}$. Then the estimate

(14)
$\|P_{N_{1}}u_{1}P_{N_{2}}u_{2}\|_{L^{2}_{t,x}(\mathbb{T}\times\mathbb{T}^{2}_{\lambda})}\lesssim_{\varepsilon}(\lambda
N_{2})^{\varepsilon}\big{(}\frac{1}{\lambda}+\frac{N_{1}}{N_{2}}\big{)}^{\frac{1}{2}}\|u_{1}\|_{Y^{0}}\|u_{2}\|_{Y^{0}}$

holds.

###### Proof.

The linear estimate with $\lambda\geq N$ is a consequence of the scale-
invariant Strichartz estimates on frequency-dependent times due to
Burq–Gérard–Tzvetkov [4] and rescaling. For $N\geq\lambda$ we can use the
$L^{4}_{t,x}$-Strichartz estimates on the (irrational) torus

$\|P_{N}e^{it\Delta}f\|_{L^{4}_{t,x}(\mathbb{T}\times\mathbb{T}^{2}_{\underline{\gamma}})}\lesssim_{\varepsilon}N^{\varepsilon}\|f\|_{L^{2}(\mathbb{T}^{2}_{\underline{\gamma}})}$

and rescale. This extends to $Y^{0}$ following Remark 2.6. The bilinear
estimate (14) follows from the bilinear estimate for free solutions:

$\|P_{N_{1}}u_{1}P_{N_{2}}u_{2}\|_{L^{2}_{t,x}(\mathbb{T}\times\mathbb{T}^{2}_{\lambda})}\lesssim_{\varepsilon}(\lambda
N_{2})^{\varepsilon}\big{(}\frac{1}{\lambda}+\frac{N_{1}}{N_{2}}\big{)}^{\frac{1}{2}}\|u_{1}(0)\|_{L^{2}(\mathbb{T}^{d}_{\lambda})}\|u_{2}(0)\|_{L^{2}(\mathbb{T}^{d}_{\lambda})}.$

Interpolation with the $L^{4}_{t,x}$-Strichartz estimate and an application of
the transfer principle conclude the proof. ∎

##  3\. Auxiliary local well-posedness results

###  3.1. Local well-posedness on the torus on an intermediate time-scale

In the following we solve the quintic NLS on the rescaled torus

(15)
$\left\\{\begin{array}[]{cl}i\partial_{t}u+\partial_{xx}u&=\pm|u|^{4}u,\quad(t,x)\in\mathbb{R}\times\mathbb{T}_{\lambda},\\\
u(0)&=u_{0}\in H^{s}(\mathbb{T}_{\lambda}).\end{array}\right.$

With the quintic NLS on the unit torus being locally well-posed in
$H^{s}(\mathbb{T})$ for times $T\lesssim 1$, the quintic NLS on the rescaled
torus is expected to be locally well-posed up to times $\lambda^{2}$. We solve
the Cauchy problem in the following on the time-scale $\lambda/N$. Recall
$\lambda\sim N^{\frac{1-s}{s}+\varepsilon}$. Thus, for $s<\frac{1}{2}$, we
have $\lambda\gg N$, choosing $\varepsilon>0$ small enough.

With the $L^{6}$-Strichartz estimates in adapted function spaces at hand, we
can show the following auxiliary well-posedness result:

######  Proposition 3.1.

For $0<s<\frac{1}{2}$, the Cauchy problem (15) is locally well-posed in
$Y^{s}$ on $[0,T]\times\mathbb{T}_{\lambda}$ for $T\lesssim\frac{\lambda}{N}$
provided that $\|u_{0}\|_{H^{s}(\mathbb{T}_{\lambda})}\ll 1$. The solution
satisfies

$\|u\|_{Y_{T}^{s}}\lesssim\|u_{0}\|_{H^{s}}+\|u\|_{Y_{T}^{s}}^{5}.$

###### Proof.

We use the property (6) from Proposition 2.7 and Littlewood-Paley theory:

$\big{\|}\chi_{[0,T)}\int_{0}^{t}e^{i(t-t^{\prime})\Delta}(|u|^{4}u)(t^{\prime})dt^{\prime}\big{\|}_{Y^{s}}\lesssim\sup_{v}\big{|}\int_{0}^{T}\int_{\mathbb{T}_{\lambda}}\sum_{M\geq
1}P_{M}(|u|^{4}u)\overline{P_{M}v}dxdt\big{|}.$

We use an additional Littlewood-Paley decomposition of the factors to write

$\begin{split}&\quad\sum_{M\geq
1}\iint_{[0,T]\times\mathbb{T}_{\lambda}}P_{M}(|u|^{4}u)\overline{P_{M}v}dxdt\\\
&=\sum_{M\geq 1}\sum_{N_{1}\gtrsim N_{2}\ldots\gtrsim
N_{5}}\iint_{[0,T]\times\mathbb{T}_{\lambda}}P_{M}vP_{N_{1}}u_{1}\ldots
P_{N_{5}}u_{5}dxdt.\end{split}$

We omit the notation of complex conjugates above and in the following because
the estimates do not depend on these. We use Littlewood-Paley dichotomy to
argue

$\begin{split}&\quad\sum_{M\geq 1}\sum_{N_{1}\gtrsim N_{2}\ldots\gtrsim
N_{5}}\iint_{[0,T]\times\mathbb{T}_{\lambda}}P_{M}vP_{N_{1}}u_{1}\ldots
P_{N_{5}}u_{5}dxdt\\\ &=\sum_{M\sim N_{1}\gtrsim N_{2}\ldots\gtrsim
N_{5}}\iint_{[0,T]\times\mathbb{T}_{\lambda}}P_{M}vP_{N_{1}}u_{1}\ldots
P_{N_{5}}u_{5}dxdt\\\ &\quad+\sum_{M\ll N_{1}\sim N_{2}\gtrsim
N_{3}\ldots\gtrsim N_{5}}\iint P_{M}vP_{N_{1}}u_{1}\ldots
P_{N_{5}}u_{5}dxdt\\\ &=(I)+(II).\end{split}$

We turn to the estimate of $(I)$. We use almost orthogonality and decompose
$P_{M}v$ and $P_{N_{1}}u_{1}$ into frequency intervals of size $N_{2}$:

$\begin{split}&\quad\iint_{[0,T]\times\mathbb{T}_{\lambda}}P_{M}vP_{N_{1}}u_{1}\ldots
P_{N_{5}}u_{5}dxdt\\\
&=\sum_{I_{1},I_{2}}\iint_{[0,T]\times\mathbb{T}_{\lambda}}P_{M,I_{1}}vP_{N,I_{2}}u_{1}P_{N_{2}}u_{2}\ldots
P_{N_{5}}u_{5}dxdt.\end{split}$

After estimating the above expression, the sum over $I_{1}$ and $I_{2}$ can be
removed by an application of the Cauchy-Schwarz inequality and property (5) of
the $Y^{s}$-spaces.

Case $N_{2}\lesssim N$. We use the $L^{6}$-estimate from (11) to find

$\begin{split}&\quad\iint_{[0,T]\times\mathbb{T}_{\lambda}}P_{M,I_{1}}vP_{N_{1},I_{2}}u_{1}\ldots
P_{N_{5}}u_{5}dxdt\\\
&\lesssim\|P_{M,I_{1}}v\|_{L^{6}_{t,x}}\|P_{N_{1},I_{2}}u_{1}\|_{L^{6}_{t,x}}\ldots\|P_{N_{5}}u_{5}\|_{L^{6}_{t,x}}\\\
&\lesssim\|P_{M,I_{1}}v\|_{Y_{T}^{0}}\|P_{N_{1},I_{2}}u_{1}\|_{Y_{T}^{0}}\ldots\|P_{N_{5}}u_{5}\|_{Y_{T}^{0}}.\end{split}$

Now we can use almost orthogonality to sum over $I_{1}$ and $I_{2}$.

Case $N_{2}\gtrsim N$. In this case we can still use the
$L^{6}_{t,x}$-estimates from (12). We conclude

$\lesssim\log(\lambda
N_{2})^{18}\|P_{M,I_{1}}v\|_{Y_{T}^{0}}\|P_{N_{1}}u_{1}\|_{Y_{T}^{0}}\ldots\|P_{N_{5}}u_{5}\|_{Y_{T}^{0}}.$

Recalling that $\lambda\sim N^{\frac{1-s}{s}+\varepsilon}\lesssim
N_{2}^{\frac{1-s}{s}+\varepsilon}$, we have

$\lesssim\big{(}\frac{1}{s}\log(N_{2})\big{)}^{18}\|P_{M,I_{1}}v\|_{Y_{T}^{0}}\|P_{N_{1},I_{2}}u_{1}\|_{Y_{T}^{0}}\|P_{N_{2}}u_{2}\|_{Y_{T}^{0}}\ldots\|P_{N_{5}}u_{5}\|_{Y_{T}^{0}}.$

In this case we have easy summation to $\lesssim
N^{-\delta}\|u\|^{5}_{Y_{T}^{s}}$.

We turn to the estimate of $(II)$: Let $N^{*}=\max(M,N_{3})$. We can again use
an almost orthogonal decomposition for $u_{N_{1}}$ and $u_{N_{2}}$ to
frequency intervals of size $N^{*}$. Again the $L^{6}_{t,x}$-estimates will be
the cornerstone of the analysis:

$\begin{split}&\quad\iint\sum_{M\ll N_{1}\sim N_{2}\gtrsim
N_{3}\gtrsim\ldots\gtrsim N_{5}=(*)}P_{M}vP_{N_{1},I_{1}}u_{1}\ldots
P_{N_{5}}u_{5}\\\
&\leq\sum_{(*)}\|P_{M}v\|_{L^{6}_{t,x}}\|P_{N_{1},I_{1}}u_{1}\|_{L^{6}_{t,x}}\|P_{N_{2},I_{2}}u_{2}\|_{L^{6}_{t,x}}\ldots\|P_{N_{5}}u_{5}\|_{L^{6}_{t,x}}.\end{split}$

If $N^{*}\lesssim N$, then we can use the estimate (11) to find

$\lesssim\sum_{(*)}\|P_{M}v\|_{Y_{T}^{0}}\|P_{N_{1},I_{1}}u_{1}\|_{Y_{T}^{0}}\ldots\|P_{N_{5}}u_{5}\|_{Y_{T}^{0}}.$

For $M\gtrsim N^{\varepsilon}$ we find from straight-forward summation the
estimate

$\lesssim N^{-\delta}\|u\|^{5}_{Y_{T}^{s}}.$

If $N^{*}\gg N$, then we can use the $L^{6}_{t,x}$-estimate (12) to close the
estimate. ∎

To carry out the $I$-method, we introduce notations: Let $m(\xi)$ be a smooth
non-negative symbol on $\mathbb{R}^{d}$, which equals $1$ for $|\xi|\leq 1$
and equals $|\xi|^{-1}$ for $|\xi|\geq 2$. For any $N\geq 1$ and
$\alpha\in\mathbb{R}$, let $I_{N}^{\alpha}$ be defined by

$(I_{N}^{\alpha}f)\widehat{(}\xi)=m\big{(}\frac{\xi}{N}\big{)}^{\alpha}\hat{f}(\xi).$

We have the following invariance lemma [7, Lemma 12.1]:

######  Lemma 3.2.

Let $\alpha_{0}>0$ and $n\geq 1$. Suppose that $Z$, $X_{1}$,$\ldots$,$X_{n}$
are translation invariant Banach spaces and $T$ is a translation invariant
$n$-linear operator such that we have the estimate

$\|I_{1}^{\alpha}T(u_{1},\ldots,u_{n})\|_{Z}\lesssim\prod_{i=1}^{n}\|I_{1}^{\alpha}u_{i}\|_{X_{i}}$

for all $u_{1}$,$\ldots$,$u_{n}$ and all $0\leq\alpha\leq\alpha_{0}$. Then the
following estimate holds:

$\|I_{N}^{\alpha}T(u_{1},\ldots,u_{n})\|_{Z}\lesssim\prod_{i=1}^{n}\|I_{N}^{\alpha}u_{i}\|_{X_{i}}$

with implicit constant independent of $\alpha$.

Proposition 3.1 and Lemma 3.2 yield the following result on the $I$-Cauchy
problem:

######  Proposition 3.3.

Let $0<s<1$ and $0<T\sim\lambda/N$. Let $I$ be a smooth radially decreasing
interpolant of

$(If)\widehat{(}\xi)=\begin{cases}\hat{f}(\xi),\quad&|\xi|\leq N,\\\
\big{(}\frac{N}{|\xi|}\big{)}^{1-s}\hat{f}(\xi),\quad&|\xi|>2N.\end{cases}$

Then the derived Cauchy problem

(16) $\left\\{\begin{array}[]{cl}i\partial_{t}Iu+\partial_{xx}Iu&=\pm
I(|u|^{4}u),\quad(t,x)\in[0,T]\times\mathbb{T}_{\lambda},\\\
Iu(x,0)&=Iu_{0}\in H^{1}(\mathbb{T}_{\lambda})\end{array}\right.$

is locally well-posed in $Y_{T}^{1}$ provided that
$\|u_{0}\|_{H^{s}(\mathbb{T}_{\lambda})}\ll 1$.

###  3.2. Auxiliary local well-posedness in two dimensions

Next, we show a local well-posedness result on the unit time scale in two
dimensions.

######  Proposition 3.4.

Let $0<s<\frac{2}{3}$ and $\lambda=N^{\frac{1-s}{s}}$. Then the $I$-system

$\left\\{\begin{array}[]{cl}i\partial_{t}Iu+\Delta Iu&=\pm
I(|u|^{2}u),\quad(t,x)\in\mathbb{R}\times\mathbb{T}^{2}_{\lambda},\\\
Iu(0)&=Iu_{0}\in H^{1}(\mathbb{T}^{2}_{\lambda})\end{array}\right.$

is locally well-posed on a time-scale $T\sim\lambda^{-\delta}$ for any
$\delta>0$. This means for any $C>0$ and
$\|Iu_{0}\|_{H^{1}(\mathbb{T}^{2}_{\lambda})}\leq C$ there exists a solution
$Iu\in Y^{1}_{T}$ which satisfies

$\|Iu\|_{Y^{1}_{T}}\leq\bar{C}(C,\delta)$

independently of $N\gg 1$ and which depends Lipschitz-continuously on the
initial data.

###### Proof.

We invoke the invariance lemma 3.2 to reduce to show local well-posedness

$\left\\{\begin{array}[]{cl}i\partial_{t}u+\Delta
u&=\pm|u|^{2}u,\quad(t,x)\in\mathbb{R}\times\mathbb{T}^{2}_{\lambda},\\\
u(0)&=u_{0}\in H^{s}(\mathbb{T}^{2}_{\lambda})\end{array}\right.$

for $0<s<1$. By the property (5) of $Y^{s}$-spaces and Littlewood-Paley
decomposition it suffices to show estimates for ordered dyadic frequency
ranges $N_{1}\geq N_{2}\geq N_{3}$:

(17)
$\big{|}\iint_{[0,T]\times\mathbb{T}^{2}_{\lambda}}P_{N}vP_{N_{1}}u_{1}P_{N_{2}}u_{2}P_{N_{3}}u_{3}dxdt\big{|}\lesssim_{\varepsilon}\lambda^{0-}(N_{3}^{*})^{\varepsilon}(N_{4}^{*})^{\varepsilon}\|P_{N}v\|_{Y^{0}}\prod_{i=1}^{3}\|P_{N_{i}}u_{i}\|_{Y^{0}}.$

for $N\sim N_{1}\gtrsim N_{2}\gtrsim N_{3}\gtrsim N_{4}$ or $N\ll N_{1}\sim
N_{2}$. For the proof, decompose the highest and second to highest frequency
ranges $N_{1}^{*}$, $N_{2}^{*}$ into balls of size $N_{3}^{*}$ by almost
orthogonality. Now, for a function frequency localized in a ball
$B=B(x,N_{3}^{*})$ we have the following Strichartz estimates as a consequence
of Galilean invariance for $4\leq q\leq\infty$:

$\|P_{B}u\|_{L^{q}_{t,x}([0,T]\times\mathbb{T}^{2}_{\gamma})}\lesssim_{\varepsilon^{\prime}}(\lambda
N_{3}^{*})^{\varepsilon^{\prime}}(N_{3}^{*})^{4\big{(}\frac{1}{4}-\frac{1}{q}\big{)}}\|u\|_{Y^{0}}.$

This follows from interpolating the estimates at $q=4$ and $q=\infty$.

Consequently, an application of Hölder’s inequality yields for $0\leq
T\lesssim\lambda^{-\delta}$:

$\|P_{B}u\|_{L^{4}_{t,x}([0,T]\times\mathbb{T}^{2}_{\gamma})}\lesssim_{\varepsilon^{\prime}}\lambda^{-\delta\big{(}\frac{1}{4}-\frac{1}{q}\big{)}+\varepsilon^{\prime}}(N_{3}^{*})^{4\big{(}\frac{1}{4}-\frac{1}{q}\big{)}+\varepsilon^{\prime}}\|P_{B}u\|_{Y^{0}}.$

For given $\varepsilon>0$, choose $q>4$ small enough and
$\varepsilon^{\prime}$ small enough, such that by an application of Hölder’s
inequality, we find

$\big{|}\iint_{[0,T]\times\mathbb{T}^{2}_{\lambda}}P_{N}vP_{N_{1}}u_{1}P_{N_{2}}u_{2}P_{N_{3}}u_{3}dxdt\big{|}\lesssim_{\varepsilon}\lambda^{0-}(N_{3}^{*})^{\varepsilon}\|P_{N}v\|_{Y^{0}}\prod_{i=1}^{3}\|P_{N_{i}}u_{i}\|_{Y^{0}}.$

The estimate suffices to show local well-posedness for $s=2\varepsilon$ by a
standard application of the contraction mapping principle. The details are
omitted. ∎

##  4\. Modified energies

###  4.1. Modified energies in one dimension

With the local well-posedness on the time-scale $\lambda/N$ ensured, we next
analyze the growth of the modified energy. We use the resonant decomposition
due to Li–Wu–Xu [19] and recall the (necessary) notations. Let $n\in
2\mathbb{N}$ and we define $n$-multipliers $M_{n}(k_{1},\ldots,k_{n})$ on

$\Gamma_{n}=\\{\underline{k}=(k_{1},\ldots,k_{n})\in\mathbb{R}^{n}:k_{1}+\ldots+k_{n}=0\\}.$

We define

$\Lambda_{n}(M_{n};f_{1},\ldots,f_{n})=\int_{\Gamma_{n}}M_{n}(k_{1},\ldots,k_{n})\prod_{j=1}^{n}\hat{f}_{j}(t,k_{j})(d\Gamma_{n})_{\lambda}\text{
with }(d\Gamma_{n})_{\lambda}=\prod_{j=1}^{n-1}(dk_{j})_{\lambda}$

and let
$\Lambda_{n}(M_{n})=\Lambda_{n}(M_{n};u,\overline{u},\ldots,u,\overline{u})$.
We find by substituting the equation (1) for $d=1$:

$\frac{d}{dt}\Lambda_{n}(M_{n})=\Lambda_{n}(M_{n}\alpha_{n})+i\Lambda_{n+4}\big{(}\sum_{j=1}^{n}(-1)^{j}X_{j}(M_{n})\big{)},$

where

$\alpha_{n}=i\sum_{j=1}^{n}(-1)^{j}k_{j}^{2};\quad
X_{j}(M_{n})=M_{n}(k_{1},\ldots,k_{j-1},k_{j}+\ldots+k_{j+4},k_{j+5},\ldots,k_{n+4}).$

We define the $I$-energy by

$E_{I}^{1}(u(t))=E(Iu)=\frac{1}{2}\|\partial_{x}Iu\|^{2}_{2}\pm\frac{1}{6}\|Iu(t)\|^{6}_{L^{6}}=\Lambda_{2}(\sigma_{2})\pm\Lambda_{6}(\sigma_{6}),$

where

$\sigma_{2}=-\frac{1}{2}m(k_{1})k_{1}m(k_{2})k_{2};\quad\sigma_{6}=\frac{1}{6}m(k_{1})\ldots
m(k_{6}).$

We compute by $\alpha_{2}=0$ due to convolution constraint:

(18)
$\frac{d}{dt}\Lambda_{2}(\sigma_{2})=i\Lambda_{6}\big{(}(-1)X_{1}(\sigma_{2})+X_{2}(\sigma_{2})\big{)}=\frac{i}{6}\sum_{j=1}^{6}(-1)^{j+1}m^{2}(k_{j})k_{j}^{2}.$

Secondly,

(19)
$\frac{d}{dt}\Lambda_{6}(\sigma_{6})=\Lambda_{6}(\sigma_{6}\alpha_{6})\pm\underbrace{i\Lambda_{10}\big{(}\sum_{j=1}^{6}(-1)^{j}X_{j}(\sigma_{6})\big{)}}_{M^{1}_{10}}.$

We denote the sum of (18) and the first term from (19) as $M_{6}$. In the
following we focus on the defocusing case to simplify notations. Since we
establish estimates of the modulus of the involved terms, the same estimates
work in the focusing case.

The keypoint is to show that the estimates from Li–Wu–Xu hold on the much
longer time interval $T\sim\lambda/N$ compared to $T\lesssim 1$. With the
energy decay being the same, we can improve on the regularity from Li–Wu–Xu
[19] under small mass assumption.

Let

$\Upsilon=\\{(k_{1},\ldots,k_{6})\in\Gamma_{6}:|k_{1}^{*}|\sim|k_{2}^{*}|\gtrsim
N\\}$

and denote with $\chi_{X}$ the characteristic function for some set $X$.

We decompose the multiplier $M_{6}^{1}$ in resonant $\tilde{A}$ and non-
resonant $\tilde{A}^{c}$ region, which will be defined in Subsection 5.1,
after using the cancellation by conservation of energy:

$\begin{split}M_{6}(k_{1},\ldots,k_{6})&=\frac{i}{6}\sum_{j=1}^{6}(-1)^{j+1}m^{2}(k_{j})k_{j}^{2}+\sigma_{6}\alpha_{6}\\\
&=(M_{6}^{1}+M_{6}^{2})(\chi_{\Upsilon}+\chi_{\Upsilon^{c}})=(M_{6}^{1}+M_{6}^{2})\chi_{\Upsilon}\\\
&=\underbrace{M_{6}^{2}\chi_{\Upsilon}+M_{6}^{1}\chi_{\tilde{A}^{c}}\chi_{\Upsilon}}_{\tilde{M}_{6}}+\underbrace{M_{6}^{1}\chi_{\tilde{A}}\chi_{\Upsilon}}_{\overline{M}_{6}}.\end{split}$

For $\tilde{M}_{6}$ we have a favorable estimate after integration by parts.
But this cannot be carried out for $\overline{M}_{6}$. We define the modified
$I$-energy $E_{I}^{2}(u(t))$ by

$E_{I}^{2}(u(t))=E_{I}^{1}(u(t))+\Lambda_{6}(\tilde{\sigma}_{6}),\quad\tilde{\sigma}_{6}=-\tilde{M}_{6}/\alpha_{6}.$

Now it holds

$\frac{d}{dt}E_{I}^{2}(u(t))=\Lambda_{6}(\overline{M}_{6})+\Lambda_{10}(\overline{M}_{10}),$

where

$\overline{M}_{10}=i\sum_{j=1}^{6}(-1)^{j}(X_{j}(\sigma_{6})+X_{j}(\tilde{\sigma}_{6})).$

Below we use the following formula:

(20)
$E_{I}^{1}(u(t))=E_{I}^{1}(u(0))-\big{[}\Lambda_{6}(\tilde{\sigma}_{6})\big{]}_{0}^{t}+\int_{0}^{t}\Lambda_{6}(\overline{M}_{6})+\Lambda_{10}(\overline{M}_{10})ds.$

In the following we focus on the proof of the following estimates for
$0<t\lesssim\frac{\lambda}{N}$ and $\frac{1}{3}<s<\frac{1}{2}$:

(21)
$\displaystyle\big{|}\int_{0}^{t}\int\overline{M}_{6}(k_{1},\ldots,k_{6})\hat{u}(s,k_{1})\ldots\hat{\overline{u}}(s,k_{6})d\Gamma_{6}ds\big{|}$
$\displaystyle\lesssim N^{-3+}\|Iu\|^{6}_{Y_{T}^{1}},$ (22)
$\displaystyle\big{|}\int_{0}^{t}\int\overline{M}_{10}(k_{1},\ldots,k_{10})\hat{u}(s,k_{1})\ldots\hat{\overline{u}}(s,k_{10})d\Gamma_{10}ds\big{|}$
$\displaystyle\lesssim N^{-3+}\|Iu\|^{10}_{Y_{T}^{1}},$ (23)
$\displaystyle\big{|}\Lambda_{6}(\tilde{\sigma}_{6})(t)\big{|}$
$\displaystyle\lesssim
N^{-\delta}\|Iu(t)\|^{6}_{H^{1}(\mathbb{T}_{\lambda})}.$

The first and second estimate translate to favorable energy estimates on the
extended time-interval of length $\frac{\lambda}{N}$. Since
$|\sigma_{6}|,|\tilde{\sigma}_{6}|\lesssim 1$, the proof of the second and
third estimate is more straight-forward. The bulk of the work is to estimate
the resonant part. Moreover, we need to establish size and regularity
estimates of the multilinear Fourier multipliers. The size estimates are
essentially due to Li–Wu–Xu [19]. However, the regularity was not analyzed in
[19], which will be done through Fourier series argument in the forthcoming
sections. For this reason also the analysis of the size is repeated.

###  4.2. Modified energies in two dimensions

We turn to the analysis in two dimensions. The energy is given by

$E^{1}(u)(t)=\int_{\mathbb{T}^{2}_{\lambda}}\frac{|\nabla_{x}u|^{2}}{2}dx\pm\int_{\mathbb{T}^{2}_{\lambda}}\frac{|u|^{4}}{4}dx,$

and we decompose the $I$-energy
$E_{I}=E^{1}(Iu)=\Lambda_{2}(\sigma_{2})\pm\Lambda_{4}(\sigma_{4})$ with

$\sigma_{2}=-\frac{1}{2}k_{1}m(k_{1})\cdot
k_{2}m(k_{2}),\quad\sigma_{4}=\frac{1}{4}m(k_{1})\ldots m(k_{4}).$

Then we compute from substituting (1) and symmetrization

$\frac{d}{dt}\Lambda_{2}(\sigma_{2})=\Lambda_{4}(M_{4}),\quad
M_{4}=\frac{i}{4}(-k_{1}^{2}m^{2}(k_{1})+k_{2}^{2}m^{2}(k_{2})-k_{3}^{2}m^{2}(k_{3})+k_{4}^{2}m^{2}(k_{4})).$

Secondly,

$\frac{d}{dt}\Lambda_{4}(\sigma_{4})=\frac{i}{4}m(k_{1})\ldots
m(k_{4})(-k_{1}^{2}+k_{2}^{2}-k_{3}^{2}+k_{4}^{2})\pm
i\sum_{j=1}^{4}X_{j}(\sigma_{4}).$

We carry out a very simple resonant decomposition, using a correction term for
the first term of the above display and some part of $M_{4}$. The aim is to
remain with the estimate $|\overline{M}_{4}|\lesssim
m(N_{1}^{*})N_{1}^{*}m(N_{3}^{*})N_{3}^{*}$ for the remainder term
$\overline{M}_{4}$ of $M_{4}$.

To this end, we suppose that $|k_{i}|\sim N_{i}$. If $N_{2},N_{4}\ll N_{1}\sim
N_{3}$ or $N_{1},N_{3}\ll N_{2}\sim N_{4}$, then clearly the resonance
function

$\Omega_{4}(k_{1},k_{2},k_{3},k_{4})=k_{1}^{2}-k_{2}^{2}+k_{3}^{2}-k_{4}^{2},\quad(k_{1},\ldots,k_{4})\in\Gamma_{4}$

satisfies the size estimate
$|\Omega_{4}(k_{1},\ldots,k_{4})|\gtrsim(N_{1}^{*})^{2}$. We define the
relevant region (since the multipliers become trivial in the complement due to
energy conservation and convolution constraint):

$\Upsilon=\\{(k_{1},\ldots,k_{4})\in\Gamma_{4}:\,|k_{i}|\sim
N_{i},\;N_{1}^{*}\sim N_{2}^{*}\gtrsim N\\}.$

We define the non-resonant region by

$\tilde{A}=\\{(k_{1},\ldots,k_{4})\in\Upsilon:\,|k_{1}|\sim|k_{3}|\gg
N_{3}^{*}\text{ or }|k_{2}|\sim|k_{4}|\gg N_{3}^{*}\\}.$

With the non-resonant region given by $\tilde{A}$ we define the correction
term

$\tilde{\sigma}_{4}=\mp\sigma_{4}-\underbrace{i\frac{k_{1}^{2}m^{2}(k_{1})-m^{2}(k_{2})k_{2}^{2}+m^{2}(k_{3})k_{3}^{2}-m^{2}(k_{4})k_{4}^{2}}{\Omega_{4}(k_{1},\ldots,k_{4})}\chi_{\tilde{A}}}_{\sigma_{4,2}}.$

We define the modified $I$-energy by

$E_{I}^{2}=E_{I}^{1}+\Lambda_{4}(\tilde{\sigma}_{4}),$

and it follows

$\frac{d}{dt}E_{I}^{2}=\Lambda_{4}(\overline{M}_{4})+\Lambda_{6}(\overline{M}_{6})$

with

$\overline{M}_{4}=M_{4}\cdot\chi_{\Upsilon\cap\tilde{A}^{c}}\quad\text{ and
}\quad\overline{M}_{6}=\pm
i\sum_{j=1}^{4}(X_{j}(\sigma_{4})+X_{j}(\sigma_{4,2})).$

We rewrite the formula for $E_{I}^{1}$ by the fundamental theorem of calculus
as

$E_{I}^{1}(t)=E_{I}^{1}(0)+\int_{0}^{t}\Lambda_{4}(\overline{M}_{4})ds+\int_{0}^{t}\Lambda_{6}(\overline{M}_{6})ds-[\Lambda_{4}(\tilde{\sigma}_{4})(t)-\Lambda_{4}(\tilde{\sigma}_{4})(0)].$

We shall establish the following estimates for $\frac{1}{2}<s\leq\frac{2}{3}$
and $0\leq T\lesssim\lambda^{-\delta}$:

(24) $\displaystyle\big{|}\int_{0}^{T}\Lambda_{4}(\overline{M}_{4})ds\big{|}$
$\displaystyle\lesssim N^{-1+}\lambda^{-\frac{1}{2}}\|Iu\|^{4}_{Y^{1}_{T}},$
(25) $\displaystyle\big{|}\int_{0}^{T}\Lambda_{6}(\overline{M}_{6})ds\big{|}$
$\displaystyle\lesssim N^{-2+}\|Iu\|^{6}_{Y^{1}_{T}},$ (26)
$\displaystyle\big{|}\Lambda_{4}(\tilde{\sigma}_{4})(t)\big{|}$
$\displaystyle\lesssim
N^{-\delta}\|Iu(t)\|^{4}_{H^{1}(\mathbb{T}^{2}_{\lambda})}.$

The main part is to show estimates for $\Lambda_{4}(\overline{M}_{4})$. The
estimates for $\overline{M}_{6}$ and the boundary term are simpler because the
multiplier is bounded.

##  5\. Energy growth bounds in one dimension

###  5.1. Estimate of the resonant contribution

####  5.1.1. Resonant decomposition

In the following we shall estimate the multiplier

$M_{6}(k_{1},\ldots,k_{6})=\sum_{i=1}^{6}(-1)^{i+1}m^{2}(k_{i})k_{i}^{2}$

on $\Gamma_{6}$ after dyadic localization $|k_{i}|\sim N_{i}\in 2^{N_{0}}$
against the resonance function

$\Omega_{6}(k_{1},\ldots,k_{6})=\sum_{i=1}^{6}(-1)^{i+1}k_{i}^{2}.$

By symmetry we suppose in the following

(27) $N_{1}\geq N_{3}\geq N_{5},\,N_{2}\geq N_{4}\geq N_{6},\text{ and
}N_{1}\geq N_{2}.$

By Littlewood-Paley dichotomy we have $N_{1}^{*}\sim N_{2}^{*}$. In the
following we carry out possibly finer decompositions than on the dyadic scale.
We shall only consider additional decompositions into intervals with tractable
summation. Consider

$A=\\{\underline{k}\in\Gamma_{6}:|\Omega_{6}|\lesssim|M_{6}|\\}.$

We shall find $\tilde{A}\supseteq A$ with $\tilde{A}$ given as union of
products of intervals and let

$\overline{M}_{6}=M_{6}\chi_{\tilde{A}}=\sum_{I_{1},\ldots,I_{6}}M_{6}\chi_{I_{1}}(k_{1})\ldots\chi_{I_{6}}(k_{6}).$

Here $\chi_{I}$ denotes the indicator function for the interval $I$. The
contribution of $\overline{M}_{6}$ will then be estimated with size and
regularity of $M_{6}$ on $I_{1},\ldots,I_{6}$ (see Proposition 5.6) and linear
and bilinear Strichartz estimates.

It is easy to see that the following cases are non-resonant:

######  Lemma 5.1.

Suppose that $|k_{i}|\sim N_{i}$, $i=1,\ldots,6$, and (27) holds. If

(28) $N_{2}\ll N_{1},$

or

(29) $N_{1}^{*}\sim N_{2}^{*}\gtrsim N_{3}^{*}\gg N_{4}^{*},$

then we are in the non-resonant case

(30) $|M_{6}(k_{1},\ldots,k_{6})|\lesssim|\Omega_{6}(k_{1},\ldots,k_{6})|.$

###### Proof.

In case of (28), it is immediate that

$|\Omega_{6}(k_{1},\ldots,k_{6})|=|k_{1}^{2}+k_{3}^{2}+k_{5}^{2}-(k_{2}^{2}+k_{4}^{2}+k_{6}^{2})|\gtrsim
N_{1}^{2}-N_{2}^{2}\gtrsim N_{1}^{2}.$

So, when we consider (29), we suppose that
$\\{N_{1},N_{2},N_{3}\\}=\\{N_{1}^{*},N_{2}^{*},N_{3}^{*}\\}$ or
$\\{N_{1}^{*},N_{2}^{*},N_{3}^{*}\\}=\\{N_{1},N_{2},N_{4}\\}$. We obtain from
convolution constraint in the first scenario

$\begin{split}\Omega_{6}(k_{1},\ldots,k_{6})&=k_{1}^{2}-k_{2}^{2}+k_{3}^{2}+O((N_{4}^{*})^{2})=(k_{1}-k_{2})(k_{1}+k_{2})+k_{3}^{2}+O((N_{4}^{*})^{2})\\\
&=(k_{2}-k_{1}-k_{3})k_{3}+O(N_{1}^{*}N_{4}^{*})\\\
&=-2k_{1}k_{3}+O(N_{1}^{*}N_{4}^{*}).\end{split}$

Hence, the lower bound $|\Omega_{6}(k_{1},\ldots,k_{6})|\gtrsim
N_{1}^{*}N_{3}^{*}$ is immediate. In the second scenario
$\\{N_{1}^{*},N_{2}^{*},N_{3}^{*}\\}=\\{N_{1},N_{2},N_{4}\\}$ we can argue
symmetrically.

If $N_{1}^{*}\sim N_{3}^{*}$, the size estimate $|M_{6}|\lesssim
N_{1}^{*}N_{3}^{*}$ is trivial. If $N_{1}^{*}\sim N_{2}^{*}\gg N_{3}^{*}$ we
have $|M_{6}|\lesssim m(N_{1}^{*})N_{1}^{*}N_{3}^{*}$ by invoking the mean
value theorem to estimate
$|m^{2}(k_{1})k_{1}^{2}-m^{2}(k_{2})k_{2}^{2}|\lesssim
m(N_{1}^{*})N_{1}^{*}N_{3}^{*}$ (see Proposition 5.6 (i)). ∎

The following lemma yields our criterion for non-resonance in case the third
highest frequency is much lower than the highest frequency.

######  Lemma 5.2.

Suppose that $|k_{i}|\sim N_{i}$, $i=1,\ldots,6$, and (27) holds. Moreover,
suppose that $N_{1}^{*}\sim N_{2}^{*}\gg N_{3}^{*}\sim N_{4}^{*}$, and
$\\{N_{1}^{*},N_{2}^{*}\\}=\\{N_{1},N_{2}\\}$, and $k_{1}$ and $k_{2}$ are of
different signs. If $|k_{1}+k_{2}|\sim
N_{12}\gg\frac{(N_{3}^{*})^{2}}{N_{1}^{*}}$, then we have $|\Omega_{6}|\gtrsim
N_{1}^{*}N_{12}\gtrsim|M_{6}(k_{1},\ldots,k_{6})|$.

###### Proof.

We write

$\Omega_{6}=(k_{1}-k_{2})(k_{1}+k_{2})+k_{3}^{2}-k_{4}^{2}+k_{5}^{2}-k_{6}^{2}=(k_{1}-k_{2})(k_{1}+k_{2})+O((N_{3}^{*})^{2}).$

With $|k_{1}-k_{2}|\sim N_{1}^{*}$, we find that under our assumption
$|k_{1}+k_{2}|\sim N_{12}\gg\frac{(N_{3}^{*})^{2}}{N_{1}^{*}}$, we have

$|\Omega_{6}|\sim N_{12}N_{1}^{*}.$

For an upper bound on $M_{6}$ we write (supposing that $k_{1}>-k_{2}$, the
other case is symmetric):

$\begin{split}M_{6}(k_{1},\ldots,k_{6})&=m^{2}(k_{1})k_{1}^{2}-m^{2}(k_{2})k_{2}^{2}+m^{2}(k_{3})k_{3}^{2}-\ldots-m^{2}(k_{6})k_{6}^{2}\\\
&=\int_{-k_{2}}^{k_{1}}f^{\prime}(\xi)d\xi+O(m^{2}(N_{3}^{*})(N_{3}^{*})^{2})\\\
&=(k_{1}+k_{2})f^{\prime}(k_{*})+O(m^{2}(N_{3}^{*})(N_{3}^{*})^{2})\end{split}$

with $f(\xi)=m^{2}(\xi)\xi^{2}$. This yields the claimed estimate by our
assumptions on $|k_{1}+k_{2}|$. ∎

Note that the above lemma amounts to introducing a bilinear Fourier
localization to describe the non- resonant set. On the other hand, for a
superset of the resonant set, where
$|k_{1}+k_{2}|\lesssim\frac{(N_{3}^{*})^{2}}{N_{1}^{*}}$, we can localize
$k_{1}$ and $k_{2}$ to intervals $I_{1}$ and $I_{2}$ of size
$\frac{(N_{3}^{*})^{2}}{N_{1}^{*}}$ which summation can be carried out via
almost orthogonality.

Lastly, we consider the case with four comparable high frequencies and the
fifth frequency being significantly lower.

######  Lemma 5.3.

Suppose that $N_{1}^{*}\sim N_{4}^{*}\gg N_{5}^{*}$. Then one of the following
holds:

(31) $\displaystyle\\{N_{1}^{*},\ldots,N_{4}^{*}\\}$
$\displaystyle=\\{N_{1},N_{2},N_{3},N_{4}\\},$ (32)
$\displaystyle\\{N_{1}^{*},\ldots,N_{4}^{*}\\}$
$\displaystyle=\\{N_{1},N_{2},N_{4},N_{6}\\},$ (33)
$\displaystyle\\{N_{1}^{*},\ldots,N_{4}^{*}\\}$
$\displaystyle=\\{N_{1},N_{2},N_{3},N_{5}\\}.$

  * (i)

If $k_{2},k_{4},k_{6}$ in Case (32) or $k_{1},k_{3},k_{5}$ in Case (33) are of
the same sign, then it holds $|\Omega_{6}|\gtrsim(N_{1}^{*})^{2}$.

  * (ii)

Suppose that we are in Case (32) and $k_{2},k_{4},k_{6}$ are not of the same
sign. If $k_{i}$, $i\in\\{2,4,6\\}$ and $k_{1}$ are of the same sign, and
$|k_{i}-k_{1}|\ll N_{1}^{*}$, then $|\Omega_{6}|\gtrsim(N_{1}^{*})^{2}$. If
$k_{i}$ and $k_{1}$ are of different signs, and $|k_{i}+k_{1}|\ll N_{1}^{*}$,
then $|\Omega_{6}|\gtrsim(N_{1}^{*})^{2}$.

  * (iii)

Suppose that we are in Case (33) and $k_{1}$,$k_{3}$,$k_{5}$ are not of the
same sign. If $k_{i}$, $i\in\\{1,2,3\\}$ and $k_{2}$ are of the same sign and
$|k_{i}-k_{2}|\ll N_{1}^{*}$, then $|\Omega_{6}|\gtrsim(N_{1}^{*})^{2}$. If
$k_{i}$ and $k_{2}$ are of different signs and $|k_{i}+k_{2}|\ll N_{1}^{*}$,
then $|\Omega_{6}|\gtrsim(N_{1}^{*})^{2}$.

###### Proof.

Ad (i): Suppose that we are in Case (31) and $k_{2},\,k_{4},\,k_{6}$ are of
the same sign. Then we find

(34)
$\begin{split}\Omega_{6}&=k_{1}^{2}-k_{2}^{2}-k_{4}^{2}-k_{6}^{2}+O((N_{5}^{*})^{2})\\\
&=(k_{2}+k_{4}+k_{6}+O(N_{5}^{*}))^{2}-k_{2}^{2}-k_{4}^{2}-k_{6}^{2}+O((N_{5}^{*})^{2})\\\
&=2k_{2}k_{4}+2k_{2}k_{6}+2k_{4}k_{6}+O(N_{1}^{*}N_{5}^{*}),\end{split}$

and hence $|\Omega_{6}|\gtrsim(N_{1}^{*})^{2}$. The second statement in (i)
follows from the same argument.

Ad (ii) and (iii): Suppose that we are in Case (32). It holds

$\begin{split}\Omega_{6}&=k_{1}^{2}-k_{2}^{2}-k_{4}^{2}-k_{6}^{2}+O((N_{5}^{*})^{2})=(k_{1}-k_{2})(k_{1}+k_{2})-k_{4}^{2}-k_{6}^{2}+O((N_{5}^{*})^{2})\\\
&=(k_{1}-k_{2})(k_{1}+k_{2})-k_{4}^{2}-k_{6}^{2}+O((N_{5}^{*})^{2}).\end{split}$

If $k_{1}$ and $k_{2}$ are of the same sign, we have $|k_{1}+k_{2}|\sim
N_{1}^{*}$. If now $|k_{1}-k_{2}|\ll N_{1}^{*}$, then the lower bound
$|\Omega_{6}|\gtrsim(N_{1}^{*})^{2}$ is immediate. Similarly, if $k_{1}$ and
$k_{2}$ are of different signs, we have $|k_{1}-k_{2}|\sim N_{1}^{*}$. If now
$|k_{1}+k_{2}|\ll N_{1}^{*}$, then again $|\Omega_{6}|\gtrsim(N_{1}^{*})^{2}$.
The other cases described in (ii) and (iii) follow from obvious modifications
of the above argument. ∎

####  5.1.2. Description of the resonant set

We summarize the regions, which we treat as resonant: Recall that we have
dyadically localized $|k_{i}|\sim N_{i}$. Write
$A_{N_{i}}=[-2N_{i},-N_{i}/2]\cup[N_{i}/2,2N_{i}]$ for $N_{i}\in
2^{\mathbb{N}}$ and $A_{1}=[-2,2]$.

(*): _We have that_

(35) $N_{1}\sim N_{2},\;N_{1}^{*}\sim N_{2}^{*},\;N_{3}^{*}\sim N_{4}^{*}$

_and one of the following cases:_

  * (i)

_if $N_{1}^{*}\sim N_{2}^{*}\gg N_{3}^{*}\sim N_{4}^{*}$, then $k_{1}$ and
$k_{2}$ are of different signs and
$|k_{1}+k_{2}|\lesssim\frac{(N_{3}^{*})^{2}}{N_{1}^{*}},$ _

  * (ii)

_if $N_{1}^{*}\sim N_{4}^{*}\gg N_{5}^{*}$, then it holds one of (31)-(33). _

_In case of ( 32) $k_{2}$, $k_{4}$, $k_{6}$ are not of the same sign. If
$k_{i}$, $i\in\\{2,4,6\\}$ and $k_{1}$ are of the same sign, then
$|k_{i}-k_{1}|\sim N_{1}^{*}$. If $k_{i}$ and $k_{1}$ are of different signs,
then $|k_{i}+k_{1}|\sim N_{1}^{*}$._

_In case of ( 33) $k_{1}$, $k_{3}$, $k_{5}$ are not of the same sign. If
$k_{i}$, $i\in\\{1,3,5\\}$ and $k_{2}$ are of the same sign, then
$|k_{i}-k_{2}|\sim N_{1}^{*}$. If $k_{i}$ and $k_{2}$ are of different signs,
then $|k_{i}+k_{2}|\sim N_{1}^{*}$_,

  * (iii)

$N_{1}^{*}\sim N_{5}^{*}$.

We clarify that we include the case $N_{1}^{*}\sim N_{5}^{*}$ because it can
be estimated without taking advantage of possibly large resonance function.
However, we recall the example by V. Sohinger [24], which states that for any
$K\in\mathbb{N}$,

$(k_{1},\ldots,k_{6})=K(5,-3,6,-2,1,-7)\text{ satisfies
}\sum_{i=1}^{6}k_{i}=\sum_{i=1}^{6}(-1)^{i}k_{i}^{2}=0.$

Hence, in case $N_{1}^{*}\sim N_{5}^{*}$, the resonance function can also
vanish.

We define

$\overline{M}_{6}(k_{1},\ldots,k_{6})=\sum_{\begin{subarray}{c}I_{1},\ldots,I_{6}:\\\
I_{i}\subseteq A_{N_{i}},\\\ (*)\text{
holds}\end{subarray}}M_{6}(k_{1},\ldots,k_{6})\chi_{I_{1}}(k_{1})\ldots\chi_{I_{6}}(k_{6}).$

####  5.1.3. Estimates in the resonant region

In the following we establish size and regularity estimates on the multiplier

(36)
$\overline{M}_{6}(k_{1},\ldots,k_{6})=\sum_{i=1}^{6}(-1)^{i+1}m^{2}(k_{i})k_{i}^{2}$

by extending the symbol from $\Gamma_{6}$ to $\mathbb{R}^{6}$. The extended
symbol will be denoted by $\overline{M}_{6}^{\mathbb{R}}$. We take advantage
of additional frequency localization such that we only have to show regularity
of the derivatives in the sense, which takes into account the localization of
the frequencies to intervals and not the absolute size of the frequencies. By
symmetry we can assume

$|k_{1}|\geq|k_{3}|\geq|k_{5}|\text{ and }|k_{2}|\geq|k_{4}|\geq|k_{6}|.$

Hence, we have $\\{k_{1}^{*},k_{2}^{*}\\}=\\{k_{1},k_{2}\\}$.

We use a Fourier series argument on

$\overline{M}_{6}(k_{1},\ldots,k_{6})\chi_{I_{1}}(k_{1})\ldots\chi_{I_{6}}(k_{6})$

with the frequencies $k_{i}$ being localized to intervals of length $L_{i}\in
2^{\mathbb{Z}}$ with absolute size $N_{i}\in 2^{\mathbb{N}_{0}}$. The
following is well-known. We state a precise version for future reference.

######  Proposition 5.4.

Suppose that there is
$\overline{M}^{\mathbb{R}}_{6}:\mathbb{R}^{6}\to\mathbb{C}$ with

$\overline{M}^{\mathbb{R}}_{6}(k_{1},\ldots,k_{6})\chi_{I_{1}}(k_{1})\ldots\chi_{I_{6}}(k_{6})=\overline{M}_{6}(k_{1},\ldots,k_{6})\chi_{I_{1}}(k_{1})\ldots\chi_{I_{6}}(k_{6})\text{
on }\Gamma_{6},$

and there is $\overline{M}_{6}(I_{1},\ldots,I_{6})>0$ such that

(37)
$|\partial^{\alpha}_{\ell_{j}}\overline{M}^{\mathbb{R}}_{6}(\ell_{1},\ldots,\ell_{6})|\lesssim\frac{\overline{M}_{6}(I_{1},\ldots,I_{6})}{L_{j}^{\alpha}}\text{
for }0\leq|\alpha|\leq 15.$

Then we have the Fourier series expansion:

(38) $\overline{M}^{\mathbb{R}}_{6}(k_{1},\ldots,k_{6})=\frac{1}{L_{1}\ldots
L_{6}}\sum_{\xi_{1},\ldots,\xi_{6}\in\mathbb{Z}}m_{6}(\frac{\xi_{1}}{L_{1}},\ldots,\frac{\xi_{6}}{L_{6}})e^{i\frac{k_{1}\xi_{1}}{L_{1}}}\ldots
e^{i\frac{k_{6}\xi_{6}}{L_{6}}}$

and the Fourier coefficients are given by

(39)
$m_{6}(\frac{\xi_{1}}{L_{1}},\ldots,\frac{\xi_{6}}{L_{6}})=\int_{\ell_{i}\in
L_{i}}e^{-i\frac{\xi_{1}\ell_{1}}{L_{1}}}\ldots
e^{-\frac{\xi_{6}\ell_{6}}{L_{6}}}\overline{M}^{\mathbb{R}}_{6}(\ell_{1},\ldots,\ell_{6})d\ell_{1}\ldots
d\ell_{6}.$

The coefficients satisfy the estimate

(40)
$\big{|}m_{6}(\frac{\xi_{1}}{L_{1}},\ldots,\frac{\xi_{6}}{L_{6}})\big{|}\lesssim
L_{1}\ldots
L_{6}\overline{M}_{6}(I_{1},\ldots,I_{6})\langle\xi_{\max}\rangle^{-15}.$

###### Proof.

For $\xi=0$, (40) is immediate from (37) for $\alpha=0$. For $\xi\neq 0$ we
have through integration by parts

$\begin{split}&\quad\int\big{(}\frac{L_{i}}{-i\xi_{j}}\big{)}\frac{\partial}{\partial\ell_{j}}\big{(}e^{-i\frac{\xi_{j}\ell_{j}}{L_{j}}}\big{)}\big{(}\prod_{i\neq
j}e^{-i\frac{\xi_{i}\ell_{i}}{L_{i}}}\big{)}\overline{M}_{6}(\ell_{1},\ldots,\ell_{6})d\ell_{1}\ldots
d\ell_{6}\\\
&=\frac{L_{j}}{(-i\xi_{j})}\int\big{(}e^{-i\frac{\xi_{1}\ell_{1}}{L_{1}}}\ldots
e^{-i\frac{\xi_{6}\ell_{6}}{L_{6}}}\big{)}\big{(}\frac{\partial}{\partial\ell_{j}}\overline{M}_{6}(\ell_{1},\ldots,\ell_{6})\big{)}d\ell_{1}\ldots
d\ell_{6}.\end{split}$

Hence, (40) is immediate from repeated integration by parts.

∎

######  Remark 5.5.

We use the Fourier series expansion in the following way: We want to estimate

$\int_{\Gamma_{6}\times[0,T]}\overline{M}_{6}(k_{1},\ldots,k_{6})\chi_{I_{1}}(k_{1})\ldots\chi_{I_{6}}(k_{6})\hat{u}_{1}(k_{1},t)\ldots\hat{u}_{6}(k_{6},t)d\Gamma_{6}dt.$

We use (38) to find

$\begin{split}&\quad\int_{\Gamma_{6}\times[0,T]}\overline{M}_{6}(k_{1},\ldots,k_{6})\chi_{I_{1}}(k_{1})\ldots\chi_{I_{6}}(k_{6})\hat{u}_{1}(k_{1},t)\ldots\hat{u}_{6}(k_{6},t)d\Gamma_{6}dt\\\
&=\int_{\Gamma_{6}\times[0,T]}\frac{1}{L_{1}\ldots
L_{6}}\sum_{\xi_{1},\ldots,\xi_{6}\in\mathbb{Z}}m_{6}\big{(}\frac{\xi_{1}}{L_{1}},\ldots,\frac{\xi_{6}}{L_{6}}\big{)}e^{i\frac{k_{1}\xi_{1}}{L_{1}}}\ldots
e^{i\frac{k_{6}\xi_{6}}{L_{6}}}\prod_{i=1}^{6}\hat{u}_{i}(k_{i},t)d\Gamma_{6}dt\\\
&=\frac{1}{L_{1}\ldots
L_{6}}\sum_{\xi_{1},\ldots,\xi_{6}\in\mathbb{Z}}m_{6}(\frac{\xi_{1}}{L_{1}},\ldots,\frac{\xi_{6}}{L_{6}})\int_{\mathbb{T}_{\lambda}\times[0,T]}u_{1}(x+\frac{\xi_{1}}{L_{1}},t)\ldots
u_{6}(x+\frac{\xi_{6}}{L_{6}},t)dxdt.\end{split}$

At this point we use Strichartz estimates on the product $u_{1}\ldots u_{6}$.
Since Strichartz estimates are translation invariant, it suffices to show an
estimate

$\big{|}\int_{\mathbb{T}_{\lambda}\times[0,T]}u_{1}(x,t)\ldots
u_{6}(x,t)dxdt\big{|}\lesssim
C(I_{1},\ldots,I_{6})\prod_{i=1}^{6}\|u_{i}\|_{Y^{1}_{T}}$

to conclude by the estimate (40):

$\begin{split}&\quad\big{|}\int_{\Gamma_{6}\times[0,T]}\overline{M}_{6}(k_{1},\ldots,k_{6})\chi_{I_{1}}(k_{1})\ldots\chi_{I_{6}}(k_{6})\hat{u}_{1}(k_{1},t)\ldots\hat{u}_{6}(k_{6},t)d\Gamma_{6}dt\big{|}\\\
&\lesssim\overline{M}_{6}(I_{1},\ldots,I_{6})C(I_{1},\ldots,I_{6})\prod_{i=1}^{6}\|u_{i}\|_{Y^{1}_{T}}.\end{split}$

In the following we establish size and regularity estimates for
$\overline{M}_{6}$ by extending the symbol to $\mathbb{R}^{6}$. We take
advantage of additional frequency localization such that we only have to show
regularity of the derivatives in the above sense: the required regularity
hinges on the possibly smaller frequency localization, compared to the
possibly large dyadic frequency. The size estimates are like in [19].

######  Proposition 5.6.

Let $I_{1},\ldots,I_{6}$ denote intervals of length $L_{i}\in 2^{\mathbb{Z}}$.
The multiplier
$\overline{M}_{6}(k_{1},\ldots,k_{6})\chi_{I_{1}}(k_{1})\ldots\chi_{I_{6}}(k_{6})$
defined in (36) on $\Gamma_{6}$ extends trivially to $\mathbb{R}^{6}$ and
satisfies size and regularity assumptions under the following conditions on
$I_{1},\ldots,I_{6}$.

  * (i)

For $|k_{i}|\sim N_{i}\in 2^{\mathbb{N}_{0}}$ and $N_{1}^{*}\sim N_{3}^{*}$ we
have the size estimate and regularity estimates (37) with

(41) $|\overline{M}_{6}(I_{1},\ldots,I_{6})|\lesssim
m(N_{1}^{*})N_{1}^{*}m(N_{3}^{*})N_{3}^{*}.$

If $N_{1}^{*}\gg N_{3}^{*}$ and $|I_{1}|\sim|I_{2}|\sim N_{3}^{*}$, we have
the size and regularity estimate (37) with

(42) $|\overline{M}_{6}(I_{1},\ldots,I_{6})|\lesssim
m(N_{1}^{*})N_{1}^{*}m(N_{3}^{*})N_{3}^{*}.$

  * (ii)

If $N_{1}^{*}\sim|k_{1}|\sim|k_{2}|\gtrsim N\gg N_{3}^{*}\sim N_{4}^{*}$,
$|I_{1}|\sim|I_{2}|\lesssim\frac{(N_{3}^{*})^{2}}{N_{1}^{*}}$, and
$|k_{1}+k_{2}|\lesssim\frac{(N_{3}^{*})^{2}}{N_{1}^{*}}$ for $k_{i}\in I_{i}$,
then the trivial extension satisfies the size and regularity estimate (37)
with

$|\overline{M}_{6}(I_{1},\ldots,I_{6})|\lesssim(N_{3}^{*})^{2}.$

  * (iii)

If $|I_{j}|\lesssim N_{5}^{*}$ for $j=1,2,3,4$, and
$\max(|k_{1}+k_{2}|,|k_{3}+k_{4}|)\lesssim N_{5}^{*}$, then we have the size
and regularity estimate

$|\overline{M}_{6}(I_{1},\ldots,I_{6})|\lesssim
m(N_{1}^{*})N_{1}^{*}N_{5}^{*}.$

  * (iv)

If $|I_{j}|\lesssim N_{12}$ for $j=1,2,3,4$, and $N_{1}\gtrsim
N_{12}\sim|k_{1}+k_{2}|\sim|k_{3}+k_{4}|\gg|k_{5}^{*}|$, then the size and
regularity estimates (37) hold with
$|\overline{M}_{6}(I_{1},\ldots,I_{6})|\lesssim m(N^{*}_{1})N_{1}^{*}N_{12}$.

###### Proof.

Case (i): The size and regularity estimates (41) in case $N_{1}^{*}\sim
N_{3}^{*}$ are straight-forward. If $N_{1}^{*}\gg N_{3}^{*}$, the size
estimate (42) follows from the mean-value theorem and convolution constraint,
which yields

$|m^{2}(k_{1})k_{1}^{2}-m^{2}(k_{2})k_{2}^{2}|\lesssim
m^{2}(N_{1}^{*})N_{1}^{*}N_{3}^{*}.$

Lastly,

$m^{2}(N_{1}^{*})N_{1}^{*}N_{3}^{*}+m^{2}(N_{3}^{*})(N_{3}^{*})^{2}\lesssim
m(N_{1}^{*})m(N_{3}^{*})N_{1}^{*}N_{3}^{*}$

because $m(k)|k|$ is increasing in $|k|$.

Case (ii): The size estimate is given by

$\begin{split}|m^{2}(k_{1})k_{1}^{2}-m^{2}(k_{2})k_{2}^{2}|&\lesssim
m^{2}(N_{1}^{*})|k_{1}-k_{2}||k_{1}+k_{2}|\\\ &\lesssim
m^{2}(N_{1}^{*})N_{1}^{*}\frac{(N_{3}^{*})^{2}}{N_{1}^{*}}\lesssim(N_{3}^{*})^{2}.\end{split}$

and $m^{2}(k_{i})k_{i}^{2}\lesssim(N_{3}^{*})^{2}$ for $i=3,\ldots,6$. To show
the regularity estimate for $k_{1}$ and $k_{2}$, we need to take into account
the additional frequency localization.

Case (iii): In this case we find like in [19, Lemma 3.4(iii)]

$\begin{split}&\quad|m^{2}(k_{1})k_{1}^{2}-m^{2}(k_{2})k_{2}^{2}+\ldots+m^{2}(k_{5})k_{5}^{2}-m^{2}(k_{6})k_{6}^{2}|\\\
&\leq|m^{2}(k_{1})k_{1}^{2}-m^{2}(k_{2})k_{2}^{2}|+|m^{2}(k_{3})k_{3}^{2}-m^{2}(k_{4})k_{4}^{2}|+|m^{2}(k_{5})k_{5}^{2}-m^{2}(k_{6})k_{6}^{2}|\\\
&\lesssim
m^{2}(N_{1}^{*})N_{1}^{*}N_{5}^{*}+m^{2}(N_{3}^{*})N_{3}^{*}N_{5}^{*}+m^{2}(N_{5}^{*})(N_{5}^{*})^{2}\\\
&\lesssim
m(N_{1}^{*})N_{1}^{*}N_{5}^{*}+m(N_{1}^{*})N_{1}^{*}m(N_{3}^{*})N_{5}^{*}+m(N_{1}^{*})N_{1}^{*}m(N_{5}^{*})N_{5}^{*}\\\
&\lesssim m(N_{1}^{*})N_{1}^{*}N_{5}^{*}.\end{split}$

The central estimate follows from the mean-value theorem, the pen-ultimate
estimate from $m(|k|)|k|$ being increasing in $|k|$, and the ultimate estimate
from uniform boundedness of $m$.

Case (iv): In this case we can argue like in (iii). ∎

With the Fourier series expansion at hand, we can show the following estimate
for the resonant part:

######  Proposition 5.7.

For any $\frac{1}{3}<s<\frac{1}{2}$, $0<T\lesssim\frac{\lambda}{N}$, we have

$\big{|}\int_{0}^{T}\Lambda_{6}(\overline{M}_{6})dt\big{|}\lesssim
N^{-3+}\|Iu\|^{6}_{Y^{1}_{T}}.$

###### Proof.

We carry out a dyadic frequency localization such that $|k_{i}|\sim N_{i}\in
2^{\mathbb{N}_{0}}$. A decreasing rearrangement of $N_{1},\ldots,N_{6}$ is
denoted by $N_{1}^{*}\geq N_{2}^{*}\ldots\geq N_{6}^{*}$. Note by LP-dichotomy
that $N_{1}^{*}\sim N_{2}^{*}$. In the resonant case we have $N_{1}^{*}\sim
N_{2}^{*}\gtrsim N_{3}^{*}\sim N_{4}^{*}$ and by symmetry considerations we
can suppose that

$N_{1}\geq N_{3}\geq N_{5}\text{ and }N_{2}\geq N_{4}\geq N_{6}\text{ and
}N_{1}\geq N_{2}.$

Moreover, in the resonant case we have
$\\{N_{1},N_{2}\\}=\\{N_{1}^{*},N_{2}^{*}\\}$, and we have further
restrictions on the signs and separation of $k_{i}$ if $N_{1}^{*}\sim
N_{4}^{*}$, which we shall recall below.

Since $\overline{M}_{6}=0$ for $|k_{1}|,\ldots,|k_{6}|<N$ we suppose that
$N_{1}^{*}\sim N_{2}^{*}\gtrsim N$. Denote $I_{T}=[0,T]$. It suffices to show
the following estimate:

(43)
$\big{|}\int_{\Gamma_{6}}\int_{I_{T}}\frac{\overline{M}_{6}(k_{1},\ldots,k_{6})\hat{u}_{1}(k_{1},t)\ldots\hat{u}_{6}(k_{6},t)}{\langle
k_{1}\rangle m(k_{1})\ldots\langle k_{6}\rangle
m(k_{6})}d\Gamma_{6}dt\big{|}\lesssim
N^{-3+}(N_{1}^{*})^{0-}\prod_{i=1}^{6}\|u_{i}\|_{Y^{0}_{i,T}}$

with $\|u_{i}\|_{Y^{0}_{i,T}}=\|u_{i}\|_{Y^{0}_{(-1)^{i},T}}$.

We consider the following regions, which take into account the size of $m$:

$\begin{split}A_{1}&=\\{\underline{k}\in\Gamma_{6}:N_{2}^{*}\gtrsim N\gg
N_{3}^{*}\sim N_{4}^{*}\\},\\\
A_{2}&=\\{\underline{k}\in\Gamma_{6}:N_{3}^{*}\gtrsim N\gg N_{4}^{*}\\},\\\
A_{3}&=\\{\underline{k}\in\Gamma_{6}:N_{4}^{*}\gtrsim N\gg N_{5}^{*}\\},\\\
A_{4}&=\\{\underline{k}\in\Gamma_{6}:N_{5}^{*}\gtrsim N\\}.\end{split}$

Since $A_{2}$ is a non-resonant region, we do not estimate this here.

$\bullet$ Estimate in $A_{1}$: We can use almost orthogonality to decompose
the support of $k_{1}$ and $k_{2}$ into intervals of length $N_{3}^{*}$. By
Proposition 5.6 we have
$\overline{M}_{6}(I_{1},\ldots,I_{6})\lesssim(N_{3}^{*})^{2}$, which implies
by the Fourier series argument (see Remark 5.5), Hölder’s inequality, and two
bilinear Strichartz estimates from Proposition 2.10:

$\begin{split}\eqref{eq:KeyEstimateResonantPart}&\lesssim
N^{2s-2}\big{|}\int_{A_{1}\times
I_{T}}\frac{\hat{u}_{1}(k_{1},t_{1})\ldots\hat{u}_{6}(k_{6},t)}{|k_{1}^{*}|^{s}|k_{2}^{*}|^{s}\langle
k_{5}^{*}\rangle\langle k_{6}^{*}\rangle}d\Gamma_{6}dt\big{|}\\\ &\lesssim
N^{-2}N^{2s}(N_{1}^{*})^{-2s}\big{|}\int_{\mathbb{T}_{\lambda}\times
I_{T}}u^{*}_{1}u^{*}_{3}\,u_{2}^{*}u_{4}^{*}\,J_{x}^{-1}u_{5}^{*}J_{x}^{-1}u_{6}^{*}dxdt\big{|}\\\
&\lesssim
N^{-2}N^{2s}(N_{1}^{*})^{-2s}\|u_{1}^{*}u_{3}^{*}\|_{L^{2}_{t,x}(I_{T}\times\mathbb{T}_{\lambda})}\|u_{2}^{*}u_{4}^{*}\|_{L^{2}_{t,x}(I_{T}\times\mathbb{T}_{\lambda})}\|J_{x}^{-1}u_{5}^{*}\|_{L^{\infty}_{t,x}}\|J_{x}^{-1}u_{6}^{*}\|_{L^{\infty}_{t,x}}\\\
&\lesssim
N^{-3+}(N_{1}^{*})^{0-}\prod_{i=1}^{6}\|u_{i}\|_{Y^{0}_{i,T}}.\end{split}$

Hence, the proof is complete in $A_{1}$.

$\bullet$ Estimate in $A_{3}$: we have

$lhs\eqref{eq:KeyEstimateResonantPart}\lesssim
N^{4s-4}\big{|}\int_{A_{3}\times
I_{T}}\frac{\overline{M}_{6}(k_{1},\ldots,k_{6})\hat{u}_{1}(k_{1},t)\ldots\hat{u}_{6}(k_{6},t)}{|k_{1}^{*}|^{s}|k_{2}^{*}|^{s}|k_{3}^{*}|^{s}|k_{4}^{*}|^{s}\langle
k_{5}^{*}\rangle\langle k_{6}^{*}\rangle}d\Gamma_{6}dt\big{|}.$

We split $A_{3}$ into two regions:

$\begin{split}A_{31}&=\\{\underline{k}\in A_{3}:N_{1}^{*}\gg N_{3}^{*}\\},\\\
A_{32}&=\\{\underline{k}\in A_{3}:N_{1}^{*}\sim N_{3}^{*}\\}.\end{split}$

In the first case we can use the same argument like for the estimate of
$A_{1}$, using the size and regularity of $\overline{M}_{6}$ as stated in
Proposition 5.6(i), and two bilinear Strichartz estimates by Proposition 2.10.
For the estimate of $A_{32}$ we recall that $N_{3}^{*}\sim N_{4}^{*}$ in the
resonant case, so that

$N_{1}^{*}\sim N_{4}^{*}\text{ in }A_{32}.$

We recall notations from [19]: Split $A_{32}$ into three regions

$\begin{split}A_{321}&=\\{\underline{k}\in
A_{32}:\\{N_{1}^{*},\ldots,N_{4}^{*}\\}=\\{N_{1},\ldots,N_{4}\\}\\},\\\
A_{322}&=\\{\underline{k}\in
A_{32}:\\{N_{1}^{*},\ldots,N_{4}^{*}\\}=\\{N_{1},N_{2},N_{3},N_{5}\\}\\},\\\
A_{323}&=\\{\underline{k}\in
A_{32}:\\{N_{1}^{*},\ldots,N_{4}^{*}\\}=\\{N_{1},N_{2},N_{4},N_{6}\\}\\}.\end{split}$

Case 2(a): Estimate in $A_{321}$. We may assume that $k_{1}>0$ by symmetry.
Then we have to consider the following four cases of signs distributed on
$k_{1},\ldots,k_{4}$:

$\begin{split}A_{3211}&=\\{\underline{k}\in
A_{321}:k_{1}>0,k_{2}>0,k_{3}<0,k_{4}<0\\},\\\ A_{3212}&=\\{\underline{k}\in
A_{321}:k_{1}>0,k_{2}<0,k_{3}<0,k_{4}>0\\},\\\ A_{3213}&=\\{\underline{k}\in
A_{321}:k_{1}>0,k_{2}<0,k_{3}<0,k_{4}<0\\},\\\ A_{3214}&=\\{\underline{k}\in
A_{321}:k_{1}>0,k_{2}<0,k_{3}>0,k_{4}<0\\}.\end{split}$

For the estimates of $A_{3211}$,$A_{3212}$,$A_{3213}$ we use bilinear
Strichartz estimates, taking advantage of the different directions of
propagation. For $A_{3211}$ and $A_{3212}$ we have by Proposition 5.6(i) that
$\overline{M}_{6}(I_{1},\ldots,I_{6})\lesssim
m(N_{1}^{*})m(N_{3}^{*})N_{1}^{*}N_{3}^{*}$ and can use two bilinear
Strichartz estimates on $u_{1}u_{3}$ and $u_{2}u_{4}$. Here we use that

(44) $|k_{1}-k_{3}|\sim|k_{2}-k_{4}|\gtrsim N_{1}^{*}$

and hence, by Proposition 5.6:

$\begin{split}lhs\eqref{eq:KeyEstimateResonantPart}&\lesssim
N^{2s-2}(N_{1}^{*})^{-2s}\big{|}\int_{\mathbb{T}_{\lambda}\times
I_{T}}u_{1}u_{3}\cdot u_{2}u_{4}\cdot
J_{x}^{-1}u_{5}J_{x}^{-1}u_{6}dxdt\big{|}\\\ &\lesssim
N^{2s-2}(N_{1}^{*})^{-2s}\|u_{1}u_{3}\|_{L^{2}_{t,x}(I_{T}\times\mathbb{T}_{\lambda})}\|u_{2}u_{4}\|_{L^{2}_{t,x}(I_{T}\times\mathbb{T}_{\lambda})}\|J_{x}^{-1}u_{5}\|_{L^{\infty}_{t,x}}\|J_{x}^{-1}u_{6}\|_{L^{\infty}_{t,x}}\\\
&\lesssim
N^{2s-2}(N_{1}^{*})^{-2s}\frac{1}{N}\prod_{i=1}^{6}\|u_{i}\|_{Y^{0}_{i,T}}.\end{split}$

For $A_{3213}$ we observe that

(45) $|k_{1}+k_{4}|\sim|k_{2}+k_{3}|=|k_{2}+k_{3}|\gtrsim N_{1}^{*}\gtrsim N.$

Hence, we can use the size and regularity estimate $\overline{M}_{6}\lesssim
m(N_{1}^{*})m(N_{3}^{*})N_{1}^{*}N_{3}^{*}$, and we can use two bilinear
Strichartz estimates on $u_{1}u_{2}$, and $u_{2}u_{3}$. Note that in this case
we use Strichartz estimates on pairs of functions, which are estimated in
$Y$-spaces of different signs. Hence, we need to check (45) in contrast to
(44). We find

$\begin{split}lhs\eqref{eq:KeyEstimateResonantPart}&\lesssim
N^{-2+2s}(N_{1}^{*})^{-2s}\big{|}\int_{\mathbb{T}_{\lambda}\times
I_{T}}u_{1}u_{4}\cdot u_{2}u_{3}\cdot
J_{x}^{-1}u_{5}J_{x}^{-1}u_{6}dxdt\big{|}\\\ &\lesssim
N^{-2+2s}(N_{1}^{*})^{-2s}\|u_{1}u_{4}\|_{L^{2}_{t,x}(I_{T}\times\mathbb{T}_{\lambda})}\|u_{2}u_{3}\|_{L^{2}_{t,x}(I_{T}\times\mathbb{T}_{\lambda})}\prod_{i=5}^{6}\|J_{x}^{-1}u_{i}\|_{L^{\infty}_{t,x}}\\\
&\lesssim
N^{-3+}(N_{1}^{*})^{0-}\prod_{i=1}^{6}\|u_{i}\|_{Y^{0}_{i,T}}.\end{split}$

Finally, we turn to $A_{3214}$ and distinguish two cases:

$\begin{split}A_{3214a}&=\\{\underline{k}\in
A_{3214}:|k_{1}+k_{2}|\lesssim|k_{5}^{*}|\\},\\\
A_{3214b}&=\\{\underline{k}\in
A_{3214}:|k_{1}+k_{2}|\gg|k_{5}+k_{6}\\}.\end{split}$

For the estimate in $A_{3214a}$, we observe that

$|k_{1}+k_{2}|\sim|k_{3}+k_{4}|\lesssim N_{5}^{*}.$

We can localize $k_{1}$, $k_{2}$, $k_{3}$, $k_{4}$ to intervals of size
$N_{5}^{*}$ without summation loss by almost orthogonality. Then, we can use
the size and regularity estimate from Proposition 5.6(iii) to infer that
$\overline{M}_{6}(I_{1},\ldots,I_{6})\lesssim m(N_{1}^{*})N_{1}^{*}N_{5}^{*}$
and use one bilinear and three $L^{6}_{t,x}$-Strichartz estimates to infer:

$\begin{split}lhs\eqref{eq:KeyEstimateResonantPart}&\lesssim
N^{3s-3}(N_{1}^{*})^{-3s}\big{|}\int_{\Gamma_{6}\times
I_{T}}\hat{u}_{1}(k_{1},t)\chi_{I_{1}}(k_{1})\ldots\hat{u_{6}}(k_{6},t)\chi_{I_{6}}(k_{6})d\Gamma_{6}dt\big{|}\\\
&\lesssim N^{3s-3}(N_{1}^{*})^{-3s}\big{|}\int_{\mathbb{T}_{\lambda}\times
I_{T}}u_{1}u_{5}^{*}\cdot u_{2}\cdot u_{3}\cdot u_{4}\cdot
J_{x}^{-1}u_{6}^{*}dxdt\big{|}\\\ &\lesssim
N^{3s-3}(N_{1}^{*})^{-3s}\|u_{1}u_{5}^{*}\|_{L^{2}_{t,x}(I_{T}\times\mathbb{T}_{\lambda})}\prod_{i=2}^{4}\|u_{i}\|_{L^{6}_{t,x}}\|J_{x}^{-1}u_{6}^{*}\|_{L^{\infty}_{t,x}}\\\
&\lesssim
N^{-3+}(N_{1}^{*})^{0-}N^{-\frac{1}{2}}\prod_{i=1}^{6}\|u_{i}\|_{Y^{0}_{i,T}}.\end{split}$

For the estimate in $A_{3214b}$ we make a dyadic decomposition in
$|k_{1}+k_{2}|\sim N_{12}$ with $N_{5}^{*}\ll N_{12}\lesssim N_{1}^{*}$ and by
almost orthogonality we localize $k_{1}$,$k_{2}$,$k_{3}$,$k_{4}$ to intervals
$I_{1},\ldots,I_{4}$ of length $N_{12}$. Now we use the size and regularity
estimate $\overline{M}_{6}(I_{1},\ldots,I_{6})\lesssim
m(N_{1}^{*})N_{1}^{*}N_{12}$ provided by Proposition 5.6 and use Hölder’s
inequality to find

(46) $\begin{split}lhs\eqref{eq:KeyEstimateResonantPart}&\lesssim
N^{3s-3}(N_{1}^{*})^{-3s}\big{|}\int u_{1}(x,t)\ldots u_{6}(x,t)dxdt\big{|}\\\
&\lesssim
N^{3s-3}N_{12}(N_{1}^{*})^{-3s}\|u_{1}u_{2}\|_{L^{2}_{t,x}}\|u_{3}u_{4}\|_{L^{2}_{t,x}}\|J_{x}^{-1}u_{5}\|_{L^{\infty}_{t,x}}\|J_{x}^{-1}u_{6}\|_{L^{\infty}_{t,x}}.\end{split}$

We want to apply two bilinear Strichartz estimates. Note the following: we can
write $\|u_{1}u_{2}\|_{L^{2}_{t,x}}=\|u_{1}\overline{u}_{2}\|_{L^{2}_{t,x}}$
and $u_{1}$, $\overline{u}_{2}$ have spatial frequency support in intervals of
length $N_{12}$ contained in the positive real numbers. Moreover, the
intervals are separated by distance $N_{12}$. By Galilean transform we can use
a bilinear Strichartz estimate at frequencies $N_{12}$ and time scale
$\frac{\lambda}{N}$ to find

$\begin{split}\|u_{1}\overline{u}_{2}\|_{L^{2}_{t,x}}&\lesssim\big{(}\frac{1}{N_{12}}+\frac{1}{N}\big{)}^{\frac{1}{2}}\|u_{1}\|_{Y^{0}_{+,T}}\|\overline{u}_{2}\|_{Y^{0}_{+,T}}\\\
&\lesssim\big{(}\frac{1}{N_{12}}+\frac{1}{N}\big{)}^{\frac{1}{2}}\|u_{1}\|_{Y^{0}_{+,T}}\|u_{2}\|_{Y^{0}_{-,T}}.\end{split}$

In a similar spirit we find

$\|u_{3}u_{4}\|_{L^{2}_{t,x}}\lesssim\big{(}\frac{1}{N_{12}}+\frac{1}{N}\big{)}^{\frac{1}{2}}\|u_{3}\|_{Y^{0}_{+,T}}\|u_{4}\|_{Y^{0}_{-,T}}.$

For $N_{12}\lesssim N$, we have from straight-forward summation

(47) $\eqref{eq:A3214bEstimate}\lesssim
N^{-3+}(N_{1}^{*})^{0-}\prod_{i=1}^{6}\|u_{i}\|_{Y^{0}_{i,T}}.$

If $N\ll N_{12}$, we find after summation over $N_{12}\lesssim N_{1}^{*}$
using $s>\frac{1}{3}$

$\begin{split}\eqref{eq:A3214bEstimateII}&\lesssim
N^{3s-3}(N_{1}^{*})^{-3s}N_{1}^{*}N^{-1}\prod_{i=1}^{6}\|u_{i}\|_{Y^{0}_{i,T}}\\\
&\lesssim
N^{-3+}(N_{1}^{*})^{0-}\prod_{i=1}^{6}\|u_{i}\|_{Y^{0}_{i,T}}.\end{split}$

Case 2(b): We turn to the estimate of

$A_{322}=\\{\underline{k}\in
A_{32}:\\{N_{1}^{*},\ldots,N_{4}^{*}\\}=\\{N_{1},N_{2},N_{3},N_{5}\\}\\}.$

In the resonant case, $k_{1}$, $k_{3}$, and $k_{5}$ must have different signs.
Suppose that $k_{1}>0$ and $k_{2}>0$. If $k_{3}>0$, $k_{5}<0$, then we can use
two bilinear Strichartz estimates on $u_{1}u_{2}$ and $u_{3}u_{5}$. If
$k_{1}>0$, $k_{2}>0$, $k_{3}<0$, $k_{5}<0$, then we can write

$k_{1}^{2}+k_{5}^{2}+k_{3}^{2}-k_{2}^{2}=k_{1}^{2}+k_{5}^{2}+(k_{3}-k_{2})(k_{3}+k_{2}).$

Hence, in the non-resonant case we have $|k_{3}+k_{2}|\gtrsim N_{1}^{*}$,
which implies that we can use two bilinear Stricharzt estimates on
$u_{1}u_{5}$ and $u_{2}u_{3}$.

Case 2(c): Estimate in $A_{323}$. We can apply two bilinear Strichartz
estimates for the same reasons like in Case 2(b).

Step 4: Estimate in $A_{4}$. We can apply the size and regularity estimate of
$\overline{M}_{6}$ provided by Proposition 5.6(i) and use six
$L^{6}$-estimates from Proposition 2.12:

$\begin{split}lhs\eqref{eq:KeyEstimateResonantPart}&\lesssim
N^{3s-3}\big{|}\int_{A_{4}\times
I_{T}}\frac{\hat{u}_{1}(k_{1},t)\ldots\hat{u}_{6}(k_{6},t)}{(N_{2}^{*})^{s}(N_{4}^{*})^{s}(N_{5}^{*})^{s}m(k_{6})\langle
k_{6}^{*}\rangle}d\Gamma_{6}\big{|}\\\ &\lesssim
N^{-3+}(N_{1}^{*})^{0-}\|J_{x}^{0-}u_{1}\|_{L^{6}_{x,t}}\ldots\|J_{x}^{0-}u_{6}\|_{L^{6}_{x,t}}\lesssim
N^{-3+}\prod_{i=1}^{6}\|u_{i}\|_{Y^{0}_{T}}.\end{split}$

∎

###  5.2. Estimate of the non-resonant contribution

Next we show the expected size and regularity estimate of $M_{6}/\Omega_{6}$
with $|k_{i}|\sim N_{i}$ and possibly finer decomposition dictated by the
resonance. Recall that we further suppose by symmetry

$N_{1}\geq N_{3}\geq N_{5},\quad N_{2}\geq N_{2}\geq N_{4}.$

We show the following:

######  Lemma 5.8.

In the non-resonant region
$M^{1}_{6}/\Omega_{6}:\Gamma_{6}\cap\tilde{A}^{c}\to\mathbb{C}$ satisfies the
size and regularity estimate:

$\big{|}\frac{M_{6}}{\Omega_{6}}(k_{1},\ldots,k_{6})\big{|}\lesssim 1.$

###### Proof.

This follows from Case-by-Case analysis.

Case: $N_{1}\gg N_{2}$. In this non-resonant case we do not have a finer
frequency decomposition than on the dyadic scale: Note that

$|\partial_{\xi_{i}}^{\alpha}M_{6}(\xi_{1},\ldots,\xi_{6})|\lesssim
m^{2}(N_{i})N_{i}^{2-\alpha}.$

Secondly, note the representation

(48)
$\partial_{\xi_{i}}^{\alpha}\Omega_{6}^{-1}=\sum_{j}\kappa_{j}\frac{\xi_{i}^{\gamma_{j}}}{\Omega_{6}^{\beta_{j}}}$

with $2\beta_{j}-\gamma_{j}\geq\alpha+2$. This follows from induction on
$\alpha$. Consequently,

$|\partial_{\xi_{i}}^{\alpha}\Omega_{6}^{-1}|\lesssim\Omega_{6}^{-1}/N_{i}^{\alpha}.$

Case: $N_{1}\sim N_{2}\gtrsim N_{3}^{*}\gg N_{4}^{*}$. We suppose by symmetry
that $N_{3}^{*}=N_{3}$. We carry out an additional almost orthogonal
decomposition of frequency ranges $|k_{i}|\sim N_{i}$, $i=1,2$ into intervals
of size $N_{3}^{*}$ (if $N_{3}^{*}\ll N_{1}^{*}$). Above we had shown

$|\Omega_{6}(k_{1},\ldots,k_{6})|\gtrsim N_{1}N_{3}\text{ for
}(k_{1},\ldots,k_{6})\in\Gamma_{6}.$

We choose the following extension from $\Gamma_{6}$ to $\mathbb{R}^{6}$:

$\begin{split}\Omega_{6}(k_{1},\ldots,k_{6})&=(k_{1}-k_{2})(k_{1}+k_{2})+k_{3}^{2}-k_{4}^{2}+k_{5}^{2}-k_{6}^{2}\\\
&=-(k_{1}-k_{2})(k_{3}+\ldots+k_{6})+k_{3}^{2}-k_{4}^{2}+k_{5}^{2}-k_{6}^{2}\\\
&=k_{3}(k_{3}+k_{2}-k_{1})-(k_{1}-k_{2})(k_{4}+\ldots+k_{6})-k_{4}^{2}+k_{5}^{2}-k_{6}^{2}\\\
&=-2k_{1}k_{3}-k_{3}(k_{4}+k_{5}+k_{6})-(k_{1}-k_{2})(k_{4}+\ldots+k_{6})-k_{4}^{2}+k_{5}^{2}-k_{6}^{2}.\end{split}$

This extension denoted by $\tilde{\Omega}_{6}(k_{1},\ldots,k_{6})$ satisfies
favorable bounds and also the estimates for the derivatives are satisfied.

Secondly, we recall the estimate

$|M_{6}|\lesssim m(N_{1}^{*})N_{1}^{*}m(N_{3}^{*})N_{3}^{*}.$

We have

$\begin{split}\big{|}\partial_{\xi_{1}}\big{(}\frac{M_{6}}{\tilde{\Omega_{6}}}\big{)}\big{|}&\lesssim\big{|}\frac{\partial_{\xi_{1}}M_{6}}{\tilde{\Omega}_{6}}\big{|}+\big{|}\frac{M_{6}}{\tilde{\Omega}_{6}^{2}}\partial_{\xi_{1}}\tilde{\Omega}_{6}\big{|}\\\
&\lesssim\frac{m^{2}(N_{1})N_{1}}{N_{1}N_{3}}+\frac{m(N_{1})m(N_{3})}{N_{1}N_{3}}N_{3}\lesssim
N_{3}^{-1}.\end{split}$

The general claim follows from (48). Note that in the worst case the
derivatives hit $\tilde{\Omega}_{6}^{-1}$ successively. This gives the bound

$\big{|}\partial_{\xi_{1}}^{\alpha}\frac{M_{6}}{\Omega_{6}}\big{|}\lesssim
N_{3}^{-\alpha}.$

Case: $N_{1}^{*}\sim N_{2}^{*}\gg N_{3}^{*}\sim N_{4}^{*}$. Recall that
$N_{i}^{*}=N_{i}$ for $i=1,\ldots,4$, $k_{1}$ and $k_{2}$ are of different
signs and $|k_{1}+k_{2}|\gg\frac{(N_{3}^{*})^{2}}{N_{1}^{*}}$. Let
$|k_{1}+k_{2}|\sim N_{12}$. This allows us an additional almost orthogonal
decomposition of frequency ranges $|k_{i}|\sim N_{i}$ into intervals of size
$N_{12}$. We have

$|\Omega_{6}|\gtrsim N_{1}N_{12},\quad|M_{6}|\lesssim
m^{2}(N_{1}^{*})N_{1}^{*}N_{12}+m^{2}(N_{3}^{*})(N_{3}^{*})^{2}.$

This shows the size estimate. For the regularity estimate for $\xi_{1}$ and
$\xi_{2}$, we use again (48) and note that the critical case is when the
derivatives hit $\Omega_{6}^{-k}$ in the denominator: The resulting term has
the form

$\big{|}\Omega_{6}^{-(\alpha+1)}\xi_{1}^{\alpha}\big{|}\lesssim\frac{1}{N_{12}^{\alpha+1}N_{1}},$

which is acceptable, since we have reduced the support of $k_{1}$ and $k_{2}$
to intervals of size $N_{12}$.

Case: $N_{1}^{*}\sim N_{4}^{*}\gg N_{5}^{*}$. In the non-resonant case we have
the negation of $(*)(ii)$. If $k_{2}$, $k_{4}$, $k_{6}$ are of the same sign
(or similarly, $k_{1}$, $k_{3}$, $k_{5}$), then we choose the extension
described by (34). For the remaining cases arising from the negation of
$(*)(ii)$ we choose the trivial extension of $\Omega_{6}$. Then we have
$|\Omega_{6}|\gtrsim(N_{1}^{*})^{2}$ and $|M_{6}|\lesssim
m^{2}(N_{1})(N_{1}^{*})^{2}$. With these at hand and the suitablve extension,
it becomes straight-forward to check the size and regularity. ∎

For the sake of completeness, we record the following trivial size and
regularity estimate for $\sigma_{6}=m(k_{1})\ldots m(k_{6})$:

######  Proposition 5.9.

We have the following size and regularity estimates for
$\sigma_{6}:\Gamma_{6}\to\mathbb{C}$. Suppose that $|k_{i}|\sim N_{i}$, then
the trivial extension of $\sigma_{6}$ to $\mathbb{R}^{6}$ satisfies the size
and regularity estimates with:

$|\sigma_{6}(\xi_{1},\ldots,\xi_{6})|\lesssim 1.$

We obtain in the non-resonant case for $k_{i}\in I_{i}$, denoting intervals
taking into account possibly finer than dyadic frequency localization:

(49)
$\begin{split}&\quad\frac{M_{6}}{\Omega_{6}}(k_{11}+\ldots+k_{15},k_{2},\ldots,k_{6})\\\
&=\frac{1}{L_{1}\ldots
L_{6}}\sum_{\xi_{1},\ldots,\xi_{6}\in\mathbb{Z}}m_{6}(\frac{\xi_{1}}{L_{1}},\ldots,\frac{\xi_{6}}{L_{6}})e^{i\frac{(k_{11}+\ldots+k_{15})\xi_{1}}{L_{1}}}\ldots
e^{i\frac{k_{6}\xi_{6}}{L_{6}}}.\end{split}$

We obtain for the modified energy (here we only consider integration by parts
with respect to the first function and disregard possible complex conjugation
since the following estimates are independent thereof):

(50)
$\begin{split}&\quad\int_{0}^{t}\sum_{\begin{subarray}{c}k_{1}+\ldots+k_{6}=0,\\\
k_{11}+k_{12}+\ldots+k_{15}=k_{1}\end{subarray}}\frac{M_{6}}{\Omega_{6}}(k_{11}+\ldots+k_{15},k_{2},\ldots,k_{6})\\\
&\times\chi_{I_{1}}(k_{11}+\ldots+k_{15})\hat{u}(t,k_{11})\ldots\hat{u}(t,k_{15})\chi_{I_{2}}(k_{2})\hat{u}(t,k_{2})\ldots\chi_{I_{6}}(k_{6})\hat{u}(t,k_{6})ds.\end{split}$

We carry out an additional dyadic decomposition $|k_{1j}|\sim N_{1j}$.
Redenoting

$N_{11},\ldots,N_{15},N_{2},\ldots,N_{6}\to K_{1},\ldots,K_{10}$

we have $K_{1}^{*}\sim K_{2}^{*}\gtrsim N$, where $K_{i}^{*}$ denotes a
decreasing rearrangement of $K_{1},\ldots,K_{10}$.

We plug in the Fourier series expansion (38) and Fourier inversion for
$\chi_{I_{1}}$:

(51) $\chi_{I_{1}}(k_{11}+\ldots+k_{15})=\int
e^{-i(k_{11}+\ldots+k_{15})\xi}a_{I_{1}}(\xi)d\xi$

with $\|a_{I_{1}}\|_{L^{1}}\lesssim 1$ independent of $I_{1}$.

By Fourier inversion in (50), the size and regularity estimates provided by
Lemma 5.8, and (51), which allows us to consider functions
$\hat{u}(k_{11},t),\ldots,\hat{u}(k_{15},t)$ separately, we see it suffices to
estimate

(52) $\big{|}\int_{0}^{T}\int_{\mathbb{T}_{\lambda}}u_{1}(t,x)\ldots
u_{10}(t,x)dxdt\big{|}\lesssim
N^{-3+}\prod_{i=1}^{10}\|m(D)u_{i}\|_{Y^{1}_{(i),T}}$

for any choice of $i\in\\{+,-\\}$ and where we can assume that

$\text{supp}(\hat{u}_{i}(t,\cdot))\subseteq A_{K_{i}}.$

Moreover, we suppose by Littlewood-Paley dichotomy that

(53) $K_{1}^{*}\sim K_{2}^{*}\gtrsim N.$

The key decay estimate is proved in the following proposition:

######  Proposition 5.10.

Suppose that $0<T\leq\frac{\lambda}{N}$. Then the following estimate holds:

$\big{|}\int_{0}^{T}\Lambda_{10}(\overline{M}_{10})ds\big{|}\lesssim
N^{-3+}\|Iu\|_{Y_{T}^{1}}^{10}.$

###### Proof.

With the above reductions, we can focus on the estimate

$\big{|}\int_{0}^{T}\int_{\mathbb{T}_{\lambda}}P_{K_{1}}u(t,x)\ldots
P_{K_{10}}u(t,x)dxds\big{|}\lesssim
N^{-3+}(K_{1}^{*})^{0-}\prod_{i=1}^{10}\|P_{K_{i}}Iu\|_{Y_{i,T}^{1}}.$

To ease notation, suppose that $K_{i}=K_{i}^{*}$.

Case  $K_{1}^{*}\sim K_{2}^{*}\gtrsim N$ and $K_{3}^{*}\ll N$. Then we can use
two bilinear Strichartz estimates on $P_{K_{1}}uP_{K_{3}}u$ and
$P_{K_{2}}uP_{K_{4}}u$ and estimate the remaining terms in $L^{\infty}_{t,x}$:

$\begin{split}&\quad\big{|}\int_{0}^{T}\int_{\mathbb{T}_{\lambda}}P_{K_{1}}u(t,x)\ldots
P_{K_{10}}u(t,x)dxds\big{|}\\\
&\lesssim\|P_{K_{1}}uP_{K_{3}}u\|_{L^{2}_{t,x}}\|P_{K_{2}}uP_{K_{4}}u\|_{L^{2}_{t,x}}\prod_{i=5}^{10}\|P_{K_{i}}u\|_{L^{\infty}_{t,x}}\\\
&\lesssim\frac{\log(N)^{2}}{N}\|P_{K_{1}}u\|_{Y^{0}_{1,T}}\|P_{K_{2}}u\|_{Y^{0}_{2,T}}\prod_{i=3}^{10}K_{i}^{\frac{1}{2}}\|P_{K_{i}}u\|_{Y^{0}_{i,T}}.\end{split}$

Taking the multiplier $m(K_{i})K_{i}$ into account, we find

$\lesssim\frac{\log(N)^{2}}{N^{3}}K_{1}^{0-}\prod_{i=1}^{10}\|P_{K_{i}}Iu\|_{Y^{1}_{i,T}}.$

Case $N_{3}^{*}\gtrsim N$ and $N_{4}^{*}\ll N$. In this case there are still
two bilinear Strichartz estimates applicable such that we find the same
estimate like above. This can be further improved properly taking into account
the weight $m(K_{i})K_{i}$.

Case $N_{4}^{*}\gtrsim N$. In this case it suffices to apply six
$L^{6}_{t,x}$-Strichartz estimates and four $L^{\infty}_{t,x}$-estimates to
find

(54)
$\begin{split}\big{|}\int_{0}^{T}\int_{\mathbb{T}_{\lambda}}\prod_{i=1}^{10}P_{K_{i}}u\big{|}&\lesssim\prod_{i=1}^{6}\|P_{K_{i}}u\|_{L^{6}_{t,x}}\prod_{i=7}^{10}\|P_{K_{i}}u\|_{L^{\infty}_{t,x}}\\\
&\lesssim\log(K_{1})^{20}(K_{7}\ldots
K_{10})^{\frac{1}{2}}\prod_{i=1}^{10}\|P_{K_{i}}u\|_{Y^{0}_{i,T}}.\end{split}$

The weight

$\frac{1}{m(K_{1})K_{1}}\ldots\frac{1}{m(K_{10})K_{10}}$

is least favorable when $K_{10}\gtrsim N$. In this case we have

$K_{i}m(K_{i})=\big{(}\frac{N}{K_{i}}\big{)}^{1-s}K_{i}=N^{1-s}K_{i}^{s}$

and for $\frac{1}{3}<s<1$:

$\frac{(K_{7}\ldots K_{10})^{\frac{1}{2}}}{m(K_{1})K_{1}\ldots
m(K_{10})K_{10}}=N^{-4(1-s)}(K_{1}\ldots
K_{4})^{-s}\underbrace{\big{(}\prod_{i=5}^{10}N^{-(1-s)}K_{5}^{-s}\big{)}(K_{7}\ldots
K_{10})^{\frac{1}{2}}}_{\lesssim 1}.$

Taking (54) together with the above display yields

$\lesssim K_{1}^{0-}N^{-4+}\prod_{i=1}^{10}\|P_{K_{i}}Iu\|_{Y^{1}_{i,T}},$

which is sufficient. ∎

Finally, we show the estimate for the boundary term:

######  Proposition 5.11.

Let $\frac{1}{3}<s<\frac{1}{2}$. Then there is $\delta(s)>0$ such that the
estimate holds:

$\big{|}\Lambda_{6}(\tilde{\sigma}_{6})(t)\big{|}\lesssim
N^{-\delta}\|Iu(t)\|_{H^{1}(\mathbb{T}_{\lambda})}^{6}.$

###### Proof.

The estimates will be very crude. By Fourier series expansion, which is
possible by Proposition 5.9, and Littlewood-Paley decomposition it suffices to
estimate

$\big{|}\int_{\mathbb{T}_{\lambda}}P_{N_{1}}u_{1}(t,x)\ldots
P_{N_{6}}u_{6}(t,x)dx\big{|}\lesssim
N_{1}^{-\delta}\prod_{i=1}^{6}N_{i}\|P_{N_{i}}m(D)u_{i}\|_{L^{2}(\mathbb{T}_{\lambda})}$

for $N_{1}\geq N_{2}\ldots\geq N_{6}$, $N_{1}\sim N_{2}\gtrsim N$. To show the
above display, we use Hölder’s inequality and obtain

(55)
$\begin{split}&\quad\big{|}\int_{\mathbb{T}_{\lambda}}P_{N_{1}}u_{1}(t,x)\ldots
P_{N_{6}}u_{6}(t,x)dx\big{|}\\\
&\lesssim\|P_{N_{1}}u_{1}\|_{L^{2}}\|P_{N_{2}}u_{2}\|_{L^{2}}\prod_{i=3}^{6}\|P_{N_{i}}u_{i}\|_{L^{\infty}}\\\
&\lesssim(N_{3}\ldots
N_{6})^{\frac{1}{2}}N^{-2(1-s)}N_{1}^{-2s}(m(N_{1})N_{1}\|P_{N_{1}}u_{1}\|_{L^{2}}m(N_{2})N_{2}\|P_{N_{2}}u_{2}\|_{L^{2}})\prod_{i=3}^{6}\|P_{N_{i}}u_{i}\|_{L^{2}}.\end{split}$

To conclude the proof, we need to take into account the weight
$m(N_{3})N_{3}\ldots m(N_{6})N_{6}$, which is least favorable when
$N_{6}\gtrsim N$. In this case we still obtain

$(m(N_{3})N_{3}\ldots m(N_{6})N_{6})^{-1}=N^{-4(1-s)}(N_{3}^{-s}\ldots
N_{6}^{-s}).$

Consequently,

(56) $\begin{split}&\quad(N_{3}\ldots
N_{6})^{\frac{1}{2}}\prod_{i=3}^{6}\|P_{N_{i}}u_{i}\|_{L^{2}}\\\
&\lesssim(N_{3}\ldots N_{6})^{\frac{1}{2}}N^{-4(1-s)}(N_{3}^{-s}\ldots
N_{6}^{-s})\prod_{i=3}^{6}N_{i}\|P_{N_{i}}m(D)u_{i}\|_{L^{2}}\\\ &\lesssim
N^{-4(1-s)}(N_{3}\ldots
N_{6})^{\frac{1}{6}}\prod_{i=3}^{6}N_{i}\|P_{N_{i}}m(D)u_{i}\|_{L^{2}}.\end{split}$

Since $s>\frac{1}{3}$, plugging (56) into (55) yields

$\big{|}\int_{\mathbb{T}_{\lambda}}P_{N_{1}}u_{1}(t,x)\ldots
P_{N_{6}}u_{6}(t,x)dx\big{|}\lesssim
N^{-6(1-s)}N_{1}^{-\delta}\prod_{i=1}^{6}N_{i}\|P_{N_{i}}m(D)u\|_{L^{2}(\mathbb{T}_{\lambda})}.$

∎

##  6\. Energy growth bounds in two dimensions

###  6.1. Estimate of the resonant contribution

The first step is to show size and regularity estimates for
$\overline{M_{4}}$. The following analysis remains fairly rough. To improve on
this, one should look into a refined resonant decomposition of $M_{4}$. In the
following we suppose dyadic frequency localization

$|k_{i}|\sim N_{i},\quad N_{1}\geq N_{3},\quad N_{2}\geq N_{4}.$

We show the following straight-forward size and regularity estimate:

######  Lemma 6.1.

Suppose the frequencies are in dyadic ranges like above and $N_{1}\sim N_{2}$.
Then the following size and regularity estimate holds:

(57) $|\overline{M_{4}}|\lesssim m(N_{1}^{*})N_{1}^{*}m(N_{3}^{*})N_{3}^{*}.$

###### Proof.

In case $N_{1}^{*}\sim N_{3}^{*}$ the claimed size and regularity estimate is
immediate since $m^{2}(N_{i})N_{i}^{2}\lesssim
m^{2}(N_{1}^{*})(N_{1}^{*})^{2}$ by monotonicity of the function $m(k)k$ for
$k\in[1,\infty)$. In case $N_{1}^{*}\gg N_{3}^{*}$ we firstly decompose the
frequency ranges of $k_{1}$ and $k_{2}$ (corresponding to $N_{1}^{*}$ and
$N_{2}^{*}$) into balls of size $N_{3}^{*}$, which causes no summation loss by
almost orthogonality. With $N_{1}^{*}\sim N_{2}^{*}\sim N_{1}\sim N_{2}$ by
our assumptions, we can use the mean-value theorem on
$m^{2}(k_{1})k_{1}^{2}-m^{2}(k_{2})(-k_{2})^{2}$ to estimate

$|m^{2}(k_{1})k_{1}^{2}-m^{2}(-k_{2})(-k_{2})^{2}|\lesssim|\nabla
f(\xi_{*})||k_{1}+k_{2}|$

with $f(\xi)=m^{2}(\xi)|\xi|^{2}$. With
$\xi_{*}\in\\{k_{1}+t(-k_{2}-k_{1}):t\in[0,1]\\}$ the estimate
$|\xi_{*}|\gtrsim N_{1}$ is evident (because by convolution constraint
$|k_{1}+k_{2}|\lesssim N_{3}^{*}\ll N_{1}$).

This gives $|\nabla f(\xi_{*})|\sim m^{2}(N_{1}^{*})N_{1}^{*}$ and, in
conclusion,

$|m^{2}(k_{1})k_{1}^{2}-m^{2}(-k_{2})(-k_{2})^{2}|\lesssim
m^{2}(N_{1}^{*})N_{1}^{*}N_{3}^{*}.$

Hence,

$|M_{4}|\lesssim
m^{2}(N_{1}^{*})N_{1}^{*}N_{3}^{*}+m^{2}(N_{3}^{*})(N_{3}^{*})^{2}\lesssim
m(N_{1}^{*})m(N_{3}^{*})N_{1}^{*}N_{3}^{*}.$

The regularity of the trivial extension

$M_{4}^{\mathbb{R}}(\xi_{1},\ldots,\xi_{4})=m^{2}(\xi_{1})\xi_{1}^{2}-m^{2}(\xi_{2})\xi_{2}^{2}+m^{2}(\xi_{3})\xi_{3}^{2}-m^{2}(\xi_{4})\xi_{4}^{2}$

is checked like in Proposition 5.6. For the estimates in $\xi_{1}$ and
$\xi_{2}$ we use the additional localization in Fourier space to argue

$|\partial_{\xi_{i}}^{\alpha}M_{4}|\lesssim
m(N_{1}^{*})N_{1}^{*}m(N_{3}^{*})N_{3}^{*}(N_{3}^{*})^{-\alpha}\text{ for
}i=1,2.$

∎

We are ready to prove the estimate for the energy growth of the resonant part:

######  Proposition 6.2.

Let $0\leq T\leq\lambda^{-\delta}$ and $\frac{1}{2}<s\leq\frac{2}{3}$. Let
$\lambda=N^{\frac{1-s}{s}}$. Then the following estimate holds for any
$\varepsilon>0$

(58)
$\big{|}\int_{0}^{T}\Lambda_{4}(\overline{M}_{4})ds\big{|}\lesssim_{\varepsilon}N^{-1+\varepsilon}\lambda^{-\frac{1}{2}}\|Iu\|^{4}_{Y^{1}_{T}}.$

###### Proof.

Since the multiplier is otherwise trivial we can suppose that $N_{1}^{*}\sim
N_{2}^{*}\gtrsim N$. By symmetry, we suppose like above $N_{1}\sim N_{2}\sim
N_{1}^{*}\sim N_{2}^{*}\gg N_{3}^{*}$.

Case A $N_{3}^{*}\ll N$. In this case the size and regularity estimate holds
with $|M_{4}|\lesssim m(N_{1})N_{1}m(N_{3}^{*})N_{3}^{*}$. In the following
suppose that $N_{3}^{*}=N_{3}$ to ease notation. After using the Fourier
series argument and changing to position space, we can use two bilinear
Strichartz estimates (due to large frequency separation):

$\begin{split}&\quad\big{|}\int_{0}^{T}\int_{\Gamma_{4}}M_{4}(k_{1},\ldots,k_{4})\prod_{i=1}^{4}\chi_{I_{i}}\hat{u}_{i}(k_{i})d\Gamma_{4}ds\big{|}\\\
&\lesssim
m(N_{1})N_{1}m(N_{3}^{*})N_{3}^{*}\big{|}\int_{0}^{T}\int_{\mathbb{T}^{2}_{\gamma}}u_{1}(t,x)\ldots
u_{4}(t,x)dxdt\big{|}\\\
&\lesssim_{\varepsilon}m(N_{1})N_{1}m(N_{3}^{*})N_{3}^{*}(\lambda
N_{1})^{\varepsilon}\big{(}\frac{1}{\lambda}+\frac{N_{3}}{N_{1}}\big{)}^{\frac{1}{2}}\big{(}\frac{1}{\lambda}+\frac{N_{4}}{N_{1}}\big{)}^{\frac{1}{2}}\prod_{i=1}^{4}\|P_{N_{i}}u\|_{Y^{0}_{T}}.\end{split}$

We analyze the expression depending on the size of $N_{1}$ relative to $N_{3}$
and $N_{4}$.

Case AI $N_{3}\gtrsim N_{1}^{\frac{1}{2}}$. Note that for
$s\in[\frac{1}{2},\frac{2}{3}]$, $\lambda^{-1}\lesssim N^{-\frac{1}{2}}$ and
therefore, by $N\lesssim N_{1}$ and $N_{3}\gtrsim N^{\frac{1}{2}}$, we have

$\frac{1}{\lambda}+\frac{N_{3}}{N_{1}}\lesssim\frac{1}{N^{\frac{1}{2}}}+\frac{N_{3}}{N}\lesssim\frac{N_{3}}{N}.$

_Subcase AI.i_ $N_{3}\sim N_{4}$. In this case we obtain

$\begin{split}&\quad
N_{1}^{\varepsilon}m(N_{1})N_{1}m(N_{2})N_{2}(N^{-(1-s)}N_{2}^{-s})N_{3}\frac{N_{4}}{N}\prod_{i=1}^{4}\|P_{N_{i}}u_{i}\|_{Y^{0}_{T}}\\\
&\lesssim
N_{1}^{-\varepsilon}N^{-2+}\prod_{i=1}^{4}m(N_{i})N_{i}\|P_{N_{i}}u\|_{Y^{0}_{T}}.\end{split}$

This is sufficient.

_Subcase AI.ii_ $N_{4}\ll N_{3}$. Then the above expression becomes

$\begin{split}&\quad
N_{1}^{\varepsilon}m(N_{1})N_{1}m(N_{2})N_{2}(N^{-(1-s)}N_{2}^{-s})N_{3}\big{(}\frac{N_{3}}{N}\big{)}^{\frac{1}{2}}\big{(}\lambda^{-1}+\frac{N_{4}}{N}\big{)}^{\frac{1}{2}}\\\
&\lesssim
N_{1}^{-\varepsilon}N^{-1+}\lambda^{-\frac{1}{2}}\prod_{i=1}^{4}N_{i}m(N_{i}).\end{split}$

The ultimate step is clear if $\lambda^{-1}\geq N_{4}/N$. If
$N_{4}/N\geq\lambda^{-1}$, we can verify that
$(N_{4}N)^{-\frac{1}{2}}\leq\lambda^{-\frac{1}{2}}$ for
$s\in[\frac{1}{2},\frac{2}{3}]$.

Case AII $N_{3}\lesssim N_{1}^{\frac{1}{2}}$. In this case we have

$\frac{1}{\lambda}+\frac{N_{i}}{N_{1}}\lesssim\frac{1}{N^{\frac{1}{2}}}\text{
for }i=3,4,$

and we find

$N_{1}^{\varepsilon}m(N_{1})N_{1}m(N_{2})N_{2}(N^{-(1-s)}N_{2}^{-s})N_{3}\frac{1}{N^{\frac{1}{4}}}\big{(}\frac{N_{4}}{N}+\frac{1}{\lambda}\big{)}^{\frac{1}{2}}\lesssim
N_{1}^{-\varepsilon}N^{-\frac{5}{4}+}\lambda^{-\frac{1}{2}}\prod_{i=1}^{4}m(N_{i})N_{i},$

which suffices.

Case B $N_{3}^{*}\gtrsim N$ and $N_{4}^{*}\ll N$. In this case we can use one
bilinear Strichartz estimate involving $u_{N_{1}^{*}}$ and $u_{N_{4}^{*}}$ and
two $L^{4}_{t,x}$-Strichartz estimates to find by the Fourier series argument:

$\begin{split}&\quad\big{|}\int_{0}^{T}\int_{\Gamma_{4}}\overline{M}_{4}(k_{1},\ldots,k_{4})\hat{u}_{1}(k_{1})\ldots\hat{\overline{u}}(k_{4})d\Gamma_{4}ds\big{|}\\\
&\lesssim
m(N_{1}^{*})N_{1}^{*}m(N_{3}^{*})N_{3}^{*}\big{|}\int_{0}^{T}\int_{\mathbb{T}^{2}_{\gamma}}u_{1}\ldots
u_{4}(s,x)dxds\big{|}\\\
&\lesssim_{\varepsilon}N_{1}^{\varepsilon}m(N_{1}^{*})N_{1}^{*}m(N_{3}^{*})N_{3}^{*}\big{(}\frac{1}{\lambda}+\frac{N_{4}}{N_{1}}\big{)}^{\frac{1}{2}}\prod_{i=1}^{4}\|u_{i}\|_{Y^{0}_{T}}\\\
&\lesssim
N_{1}^{\varepsilon}m(N_{1})N_{1}m(N_{2})N_{2}(N^{-(1-s)}N_{2}^{-s})m(N_{3})N_{3}\frac{m(N_{4})N_{4}}{\lambda^{\frac{1}{2}}}\prod_{i=1}^{4}\|u_{i}\|_{Y^{0}_{T}}\\\
&\lesssim
N_{1}^{-\varepsilon}N^{-1+}\lambda^{-\frac{1}{2}}\prod_{i=1}^{4}m(N_{i})N_{i}\|u_{i}\|_{Y^{0}_{T}}.\end{split}$

This suffices.

Case C  $N_{4}^{*}\gtrsim N$. In this case we infer from
$L^{4}_{t,x}$-Strichartz estimates and the estimate
$(m(N_{i})N_{i})^{-1}\lesssim N^{-(1-s)}N_{i}^{-s}$:

$\begin{split}&\quad\big{|}\int_{0}^{T}\int_{\Gamma_{4}}\overline{M}_{4}(k_{1},\ldots,k_{4})\hat{u}_{1}(k_{1})\ldots\hat{\overline{u}}(k_{4})d\Gamma_{4}ds\big{|}\\\
&\lesssim_{\varepsilon}N_{1}^{\varepsilon}m(N_{1}^{*})N_{1}^{*}m(N_{3}^{*})N_{3}^{*}\prod_{i=1}^{4}\|u_{i}\|_{Y^{0}_{T}}\\\
&\lesssim
N^{-2+}N_{1}^{0-}\prod_{i=1}^{4}m(N_{i})N_{i}\|u_{i}\|_{Y^{0}_{T}},\end{split}$

which suffices. ∎

###  6.2. Estimate of the non-resonant contribution

We turn to estimates for the boundary term. Under the dyadic frequency
localization $N_{1}\sim N_{3}\gg N_{3}^{*}$ or $N_{2}\sim N_{4}\gg N_{3}^{*}$
we can use almost orthogonality to additionally localize the range of $k_{1}$
and $k_{3}$ to balls of size $N_{3}^{*}$. Then we can argue like above to
obtain the following:

######  Lemma 6.3.

Let $N_{1}\sim N_{3}\gg N_{3}^{*}$ or $N_{2}\sim N_{4}\gg N_{3}^{*}$. Then it
holds the size and regularity estimates:

$\big{|}\frac{M_{4}}{\Omega_{4}}\big{|}\lesssim 1.$

We are ready to estimate the boundary term:

######  Lemma 6.4.

Let $\frac{1}{2}<s\leq\frac{2}{3}$. Then it holds for $\delta=\delta(s)$

(59) $|\Lambda_{4}(\tilde{\sigma}_{4})(t)|\lesssim
N^{-\delta}\|Iu(t)\|_{H^{1}(\mathbb{T}^{2}_{\lambda})}^{4}.$

###### Proof.

We denote $u_{i}^{*}=P_{N_{i}}u$. Using the Fourier series argument, it
suffices to show

$\big{|}\int_{\mathbb{T}^{2}_{\lambda}}u_{1}^{*}(x)\ldots
u_{4}^{*}(x)dx\big{|}\lesssim
N_{1}^{-\varepsilon}N^{-\delta}\prod_{i=1}^{4}m(N_{i})N_{i}\|P_{N_{i}}u_{i}\|_{L^{2}(\mathbb{T}^{2}_{\lambda})}.$

We apply Hölder’s and Bernstein’s inequality to find

$\begin{split}\big{|}\int_{\mathbb{T}^{2}_{\lambda}}u_{1}^{*}(x)\ldots
u_{4}^{*}(x)dx\big{|}&\lesssim\|P_{N_{1}^{*}}u_{1}^{*}\|_{L^{2}(\mathbb{T}^{2}_{\lambda})}\|P_{N_{2}^{*}}u_{2}^{*}\|_{L^{2}(\mathbb{T}^{2}_{\lambda})}\|P_{N_{3}^{*}}u_{3}^{*}\|_{L^{\infty}(\mathbb{T}^{2}_{\lambda})}\|P_{N_{4}^{*}}u_{4}^{*}\|_{L^{\infty}(\mathbb{T}^{2}_{\lambda})}\\\
&\lesssim
N_{3}^{*}N_{4}^{*}\prod_{i=1}^{4}\|P_{N_{i}}u_{i}\|_{L^{2}(\mathbb{T}^{2}_{\lambda})}.\end{split}$

Suppose that $N_{1}^{*}\sim N_{2}^{*}\gtrsim N$ and $N_{3}^{*}\lesssim N$.
Since $(m(N_{i}^{*})N_{i}^{*})^{-1}\lesssim N^{-(1-s)}(N_{1}^{*})^{-s}$ for
$i=1,2$, we find

$\big{|}\int_{\mathbb{T}^{2}_{\lambda}}u_{1}^{*}\ldots
u_{4}^{*}dx\big{|}\lesssim
N^{-2+}(N_{1}^{*})^{0-}\prod_{i=1}^{4}m(N_{i})N_{i}\|P_{N_{i}}u_{i}\|_{L^{2}(\mathbb{T}^{2}_{\lambda})}.$

If $N_{3}^{*}\gtrsim N$ and $N_{4}^{*}\ll N$, note that for
$\frac{1}{3}<s\leq\frac{2}{3}$, we have

$N_{3}^{*}\lesssim N^{0-}\prod_{i=1}^{3}m(N_{i}^{*})N_{i}^{*},$

which finishes this case.

Finally, suppose that $N_{4}^{*}\gtrsim N$. Since
$m(N_{i}^{*})N_{i}^{*}=\frac{N^{1-s}}{(N_{i}^{*})^{1-s}}N_{i}^{*}=N^{1-s}(N_{i}^{*})^{s}$,
we have for $s\in(\frac{1}{2},\frac{2}{3})$

$(N_{i}^{*})^{\frac{1}{2}}\lesssim_{s}m(N_{i}^{*})N_{i}^{*}N^{-\frac{1}{3}}(N_{i}^{*})^{0-}.$

Consequently,

$(N_{3}^{*})^{\frac{3}{2}}(N_{4}^{*})^{\frac{1}{2}}\lesssim
N^{-\frac{4}{3}}(N_{1}^{*})^{0-}\prod_{i=1}^{4}m(N_{i}^{*})N_{i}^{*}.$

The proof is complete. ∎

###  6.3. Estimate for the non-resonant part

In the following we show a growth bound for the non-resonant part:

######  Proposition 6.5.

For $\frac{1}{2}<s\leq\frac{2}{3}$ the following estimate holds:

(60) $\big{|}\int_{0}^{T}\Lambda_{6}(\overline{M}_{6})dt\big{|}\lesssim
N^{-2+}\prod_{i=1}^{6}m(N_{i})N_{i}\|u_{i}\|_{Y^{0}_{T}}.$

Recall that

$\overline{M}_{6}=\pm i(\sum_{j=1}^{4}X_{j}(\sigma_{4})+X_{j}(\sigma_{4,2})).$

Since we have already showed the size and regularity estimate for

$\sigma_{4,2}=\frac{M_{4}}{\Omega_{4}}\cdot\chi_{\tilde{A}},$

it remains to check the size and regularity estimate for

$\sigma_{4}=m(k_{1})\ldots m(k_{4}),$

which is straight-forward and recorded in the following:

######  Lemma 6.6.

$\sigma_{4}$ satisfies the size and regularity estimate:

$|\sigma_{4}|\lesssim 1.$

This way we can apply the Fourier series argument to $\sigma_{4}$ and
$\sigma_{4,2}$ to reduce the estimate (60) to the following:

######  Lemma 6.7.

Let $\frac{1}{2}<s<\frac{2}{3}$ and $T\lesssim 1$. Suppose that $u_{i}$ is
frequency localized at $N_{i}$ and $N_{1}\sim N_{2}\gtrsim N_{3}\ldots\gtrsim
N_{6}$ with $N_{1}\gtrsim N$. Then the following estimate holds:

$\big{|}\int_{0}^{T}\int_{\mathbb{T}^{2}_{\gamma}}u_{1}(t,x)\ldots
u_{6}(t,x)dxdt\big{|}\lesssim
N^{-2+}N_{1}^{0-}\prod_{i=1}^{6}m(N_{i})N_{i}\|u_{i}\|_{Y^{0}_{T}}.$

###### Proof.

Suppose that $N_{5}\ll N$. Then we can use four linear Strichartz estimates to
find

$\begin{split}\big{|}\int_{0}^{T}\int_{\mathbb{T}^{2}_{\gamma}}u_{1}(t,x)\ldots
u_{6}(t,x)dxdt\big{|}&\lesssim\prod_{i=1}^{4}\|u_{i}\|_{L^{4}_{t,x}}\|u_{5}\|_{L^{\infty}_{t,x}}\|u_{6}\|_{L^{\infty}_{t,x}}\\\
&\lesssim_{\varepsilon}(\lambda
N_{1})^{\varepsilon}N_{5}N_{6}\prod_{i=1}^{6}\|u_{i}\|_{Y^{0}_{T}}.\end{split}$

Since $N_{1}\sim N_{2}\gtrsim N$, we have

$(m(N_{1})N_{1}m(N_{2})N_{2})^{-1}\lesssim N^{-2(1-s)}N_{1}^{-2s}$

and thus,

$(\lambda
N_{1})^{\varepsilon}N_{5}N_{6}\prod_{i=1}^{6}\|u_{i}\|_{Y^{0}_{T}}\lesssim
N^{-2+}N_{1}^{0-}\prod_{i=1}^{6}m(N_{i})N_{i}\|u_{i}\|_{Y^{0}_{T}}.$

Suppose that $N_{5}\gtrsim N$ or $N_{6}\gtrsim N$. We still find

$N_{5}N_{6}\lesssim
m(N_{5})N_{5}m(N_{6})N_{6}\big{(}\prod_{i=1}^{4}m(N_{i})N_{i}\big{)}N_{1}^{0-}N^{-2+},$

which allows us to conclude like above. ∎

##  7\. Conclusion of the proofs of Theorems 1.1 and 1.2

###  7.1. Proof of Theorem 1.1

We give a sketch of the argument. The slow energy growth $N^{-3+}$ on the time
interval $\lambda/N$ tells us that we can iterate the local well-posedness
$N^{3-}$ times until the energy has grown significantly. This gives for the
rescaled solution $u_{\lambda}$ on $\mathbb{T}_{\lambda}$ a time interval of
existence of

$\frac{\lambda}{N}\cdot N^{3-}=\lambda N^{2-}.$

Scaling back we obtain that the solution $u$ exists on a time scale

$T=N^{2-}\lambda^{-1}=N^{-\frac{1}{s}+3-}.$

Consequently, for $s>\frac{1}{3}$, we obtain global well-posedness letting
$N\to\infty$.

We turn to the details. To make the continuity argument transparent, we
collect the previously established estimates and implicit constants. Firstly,
there is a universal constant $C_{1}\geq 1$ such that

$\|Iu_{\lambda}(t)\|_{H^{1}}^{2}\leq
C_{1}(E_{I}^{1}(u_{\lambda})(t)+\|u_{\lambda}(t)\|_{L^{2}}^{2}).$

In the defocusing case this is clear, whereas in the focusing case this
requires an application of the Gagliardo-Nirenberg inequality and the small
mass assumption. Secondly, there is a universal constant $C_{2}\geq 1$ such
that

$E_{I}^{1}(u_{0\lambda})\leq C_{2}\|Iu_{0\lambda}\|^{2}_{H^{1}}.$

By Proposition 3.1, there is $\bar{C}>0$ such that

$\|Iu_{\lambda}\|_{Y^{1}_{T}}\leq\bar{C}$

for $T=\frac{\lambda}{N}$ provided that

$\|Iu_{0\lambda}\|^{2}_{H^{1}(\mathbb{T}_{\lambda})}\leq c\ll 1.$

Let $c_{2}\leq c/(10C_{2}C_{3})$. The key bounds to control the energy growth,
which were proved in Propositions 5.7, 5.10, and 5.11, are presently recalled:

(61)
$\begin{split}\big{|}\int_{0}^{t}\int\overline{M}_{6}(k_{1},\ldots,k_{6})\hat{u}(s,k_{1})\ldots\hat{\overline{u}}(s,k_{6})d\Gamma_{6}ds\big{|}&\lesssim
N^{-3+}\|Iu\|^{6}_{Y_{T}^{1}},\\\
\big{|}\int_{0}^{t}\int\overline{M}_{10}(k_{1},\ldots,k_{10})\hat{u}(s,k_{1})\ldots\hat{\overline{u}}(s,k_{10})d\Gamma_{10}ds\big{|}&\lesssim
N^{-3+}\|Iu\|^{10}_{Y_{T}^{1}},\\\
\big{|}\Lambda_{6}(\tilde{\sigma}_{6})(t)\big{|}&\lesssim
N^{-\delta}\|Iu(t)\|^{6}_{H^{1}(\mathbb{T}_{\lambda})}.\end{split}$

We show the following:

######  Proposition 7.1.

Suppose that

$\|Iu_{0\lambda}\|^{2}_{H^{1}(\mathbb{T}_{\lambda})}\leq c_{2}.$

Choose $N$ large enough such that

$N^{-\varepsilon}C_{1}(\bar{C}^{6}+\bar{C}^{10})\leq C_{1}C_{2}c,\quad
C_{\delta}N^{-\delta}(2(10C_{1}C_{2}c)^{6})\leq C_{2}c.$

Then, for $T=\lambda N^{2-2\varepsilon}$, the following estimate holds:

$\|Iu_{\lambda}(t)\|_{H^{1}(\mathbb{T}_{\lambda})}^{2}\leq
10C_{2}C_{3}c_{2}\ll 1.$

###### Proof.

The proof is carried out via a continuity argument. For times $0\leq
t\leq\frac{\lambda}{N}$ this follows from the auxiliary local well-posedness
result. We turn to larger times: Write

$\begin{split}E_{I}^{1}(u_{\lambda}(t))&=E_{I}^{1}(u_{0\lambda})-\big{(}\Lambda_{6}(\tilde{\sigma}_{6}(t))-\Lambda_{6}(\tilde{\sigma}_{6})(0))\\\
&\quad+\int_{0}^{t}\Lambda_{6}(\overline{M}_{6})ds+\int_{0}^{t}\Lambda_{10}(\overline{M}_{10})ds.\end{split}$

Suppose that $\|Iu_{\lambda}(t)\|_{H^{1}(\mathbb{T}_{\lambda})}^{2}\leq
10C_{1}C_{2}c$ and $0\leq t\leq N^{2-}\lambda$, $M\in\mathbb{N}$ such that
$T\in[\frac{(M-1)\lambda}{N},\frac{M\lambda}{N}]$. By the above display and
the bounds (61), we obtain:

$\begin{split}\|Iu_{\lambda}(t)\|_{H^{1}}^{2}&\leq
C_{1}(E_{I}^{1}u_{\lambda}(t)+\|u_{0\lambda}\|_{L^{2}}^{2})\\\ &\leq
C_{1}(E_{I}^{1}(u_{0\lambda})-\big{(}\Lambda_{6}(\tilde{\sigma}_{6})(t)-\Lambda_{6}(\tilde{\sigma})(0)\big{)}\\\
&\quad+\sum_{k=0}^{M-2}\int_{[k\lambda/N,(k+1)\lambda/N]}\big{(}\Lambda_{6}(\overline{M}_{6})+\Lambda_{10}(\overline{M}_{10})\big{)}ds\\\
&\quad+\int_{[\frac{(M-1)\lambda}{N},T]}\Lambda_{6}(\overline{M}_{6})+\Lambda_{10}(\overline{M}_{10})ds\\\
&\leq
C_{1}(E_{I}^{1}(u_{0\lambda})+C_{\delta}N^{-\delta}\big{(}\|Iu_{\lambda}(t)\|_{H^{1}}^{6}+\|Iu_{\lambda}(0)\|_{H^{1}}^{6}\big{)}+MN^{-3+}(\bar{C}^{6}+\bar{C}^{10})\big{)}\\\
&\leq
C_{1}(C_{2}\|Iu_{0\lambda}\|_{H^{1}}^{2}+MN^{-3+}(\bar{C}^{6}+\bar{C}^{10})+C_{\delta}N^{-\delta}(2(10C_{1}C_{2}c)^{6})\\\
&\leq 5C_{1}C_{2}c.\end{split}$

Hence, we have small $H^{1}$-norms on a timescale $N^{2-}\lambda$. ∎

###### Proof of Theorem 1.1.

Suppose that $u_{0}\in H^{s}(\mathbb{T})$ for $s>\frac{1}{3}$ with
$\|u_{0}\|_{L^{2}(\mathbb{T})}\leq c\ll 1$. We rescale $u_{0\lambda}$ to the
large torus $\mathbb{T}_{\lambda}$ and obtain with the choice
$\lambda=N^{\frac{1-s}{s}+\varepsilon}$:

$\begin{split}\|Iu_{0\lambda}\|^{2}_{H^{1}(\mathbb{T}_{\lambda})}&\leq
2(\|u_{0\lambda}\|^{2}_{L^{2}(\mathbb{T}_{\lambda})}+N^{2(1-s)}\|u_{0\lambda}\|^{2}_{\dot{H}^{s}(\mathbb{T}_{\lambda})})\\\
&\leq
2(\|u_{0}\|^{2}_{L^{2}(\mathbb{T})}+N^{2(1-s)}\lambda^{-2s}\|u_{0}\|^{2}_{\dot{H}^{s}(\mathbb{T})})\\\
&\leq
2(\|u_{0}\|^{2}_{L^{2}(\mathbb{T})}+N^{-2s\varepsilon}\|u_{0}\|^{2}_{\dot{H}^{s}(\mathbb{T})}).\end{split}$

Now we can choose $N=N(\|u_{0}\|_{\dot{H}^{s}},\|u_{0}\|_{L^{2}(\mathbb{T})})$
large enough such that

$\|Iu_{0\lambda}\|^{2}_{H^{1}(\mathbb{T}_{\lambda})}\leq c_{2}\ll 1$

with $c_{2}$ like in Proposition 7.1. By Proposition 7.1 the $I$-Cauchy
problem (16) can be solved for times $T\sim\lambda N^{2-}$.

Scaling back, we find that the solution on $\mathbb{T}$ exists for times

$T^{*}\sim\lambda^{-1}N^{2-}=N^{3-\frac{1}{s}-}.$

Hence, for $s>\frac{1}{3}$ and letting $N\to\infty$, we obtain global well-
posedness.

∎

###  7.2. Proof of Theorem 1.2

In this section we finish the proof of Theorem 1.2. We collect the proved
estimates and formulae for convenience.

Starting with local solutions to

(62) $\left\\{\begin{array}[]{cl}i\partial_{t}u+\Delta
u&=\pm|u|^{2}u,\quad(t,x)\in\mathbb{R}\times\mathbb{T}^{2}_{\gamma},\\\
u(0)&=u_{0}\in H^{s}(\mathbb{T}_{\underline{\gamma}}^{2})\end{array}\right.$

for $\frac{1}{2}<s\leq\frac{2}{3}$, we rescale to tori
$\mathbb{T}^{2}_{\lambda}$ with large period and damp the high frequencies to
consider the $I$-system

(63) $\left\\{\begin{array}[]{cl}i\partial_{t}Iu_{\lambda}+\Delta
Iu_{\lambda}&=\pm
I(|u_{\lambda}|^{2}u_{\lambda}),\quad(t,x)\in\mathbb{R}\times\mathbb{T}^{2}_{\lambda},\\\
Iu_{\lambda}(0)&=Iu_{\lambda 0}\in
H^{1}(\mathbb{T}^{2}_{\lambda}).\end{array}\right.$

With our choice of $\lambda=N^{\frac{1-s}{s}}$ we accomplish

$\|Iu_{\lambda}(0)\|_{H^{1}(\mathbb{T}^{2}_{\lambda})}\lesssim\|u_{\lambda
0}\|_{H^{s}(\mathbb{T}^{2}_{\lambda})},$

and we have local well-posedness of (63) on times $T=\lambda^{-\delta}$.

We use the $I$-energy as an almost conserved quantity:

$E_{I}^{1}(u_{\lambda})(t)=E_{I}^{1}(u_{\lambda
0})-\big{[}\Lambda_{4}(\tilde{\sigma}_{4})(t)-\Lambda_{4}(\tilde{\sigma}_{4})(0)\big{]}+\int_{0}^{t}\big{(}\Lambda_{4}(\overline{M}_{4})+\Lambda_{6}(\overline{M}_{6})\big{)}ds.$

The relevant growth bounds for $\frac{1}{2}<s\leq\frac{2}{3}$ and $0\leq
T\leq\lambda^{-\delta}$ are given by Proposition 6.2 and 6.5:

$\big{|}\int_{0}^{T}\Lambda_{4}(\overline{M}_{4})ds\big{|}\lesssim
N^{-1+}\lambda^{-\frac{1}{2}}\|Iu_{\lambda}\|^{4}_{Y^{1}_{T}},\quad\big{|}\int_{0}^{T}\Lambda_{6}(\overline{M}_{6})ds\big{|}\lesssim
N^{-2+}\|Iu_{\lambda}\|^{6}_{Y^{1}_{T}}.$

Note that the estimate for the resonant part is clearly inferior to the
estimate for the non-resonant part. The estimate for the boundary term for
$\frac{1}{2}<s\leq\frac{2}{3}$ is given by some $C_{\delta}$ and $N^{-\delta}$

$\big{|}\Lambda_{4}(\tilde{\sigma}_{4})(t)\big{|}\leq
C_{\delta}N^{-\delta}\|Iu_{\lambda}(t)\|^{4}_{H^{1}(\mathbb{T}^{2}_{\lambda})}.$

In the defocusing case we have

(64) $\|Iu_{\lambda}\|^{2}_{\dot{H}^{1}(\mathbb{T}^{2}_{\lambda})}\leq
2E_{I}^{1}(u(t)),\quad\|Iu_{\lambda}(t)\|^{2}_{L^{2}(\mathbb{T}^{2}_{\lambda})}\leq\|u_{0\lambda}\|^{2}_{L^{2}(\mathbb{T}^{2}_{\lambda})}.$

For the first estimate to hold in the focusing case, we require a smallness
condition on
$\|u_{0\lambda}\|_{L^{2}(\mathbb{T}_{\lambda})}=\|u_{0\lambda}\|_{L^{2}(\mathbb{T}^{2}_{\gamma})}\ll
1$. Then follows from the Gagliardo–Nirenberg–Ladyzhenskaya inequality

$\|Iu_{\lambda}(t)\|_{L^{4}}^{4}\ll\|Iu_{\lambda}(t)\|_{\dot{H}^{1}}^{2}.$

Consequently, we obtain as well in the focusing case:

$\|Iu_{\lambda}(t)\|_{\dot{H}^{1}}^{2}\leq
2\big{(}\|Iu_{\lambda}(t)\|_{\dot{H}^{1}}^{2}-\int_{\mathbb{T}^{2}_{\gamma}}\frac{|Iu_{\lambda}|^{4}}{4}dx\big{)}.$

We choose $N$ large enough depending on the following. Firstly, note that
there is a universal constant $C_{2}$ such that

$E_{I}^{1}(u_{0\lambda})\leq C_{2}\|Iu_{0\lambda}\|_{H^{1}}^{2}.$

Given $u_{0}$, by the auxiliary local well-posedness we find $\bar{C}$ such
that

$\|Iv_{\lambda}\|_{Y^{1}_{T}}\leq\bar{C}\|Iv_{0\lambda}\|_{H^{1}}$

for $T\leq\lambda^{-\delta}$ as long as

$\|Iv_{0\lambda}\|^{2}_{H^{1}}\leq 10C_{2}\|Iu_{0\lambda}\|^{2}_{H^{1}}.$

Now we choose $N\gg 1$ such that

$N^{-\varepsilon}\big{(}\bar{C}^{4}(5C_{2})^{4}\|Iu_{0\lambda}\|_{H_{\lambda}^{1}}^{4}+\bar{C}^{6}(5C_{2})^{6}\|Iu_{0\lambda}\|_{H_{\lambda}^{1}}^{6})\leq\|Iu_{0}\|_{H_{\lambda}^{1}}^{2}.$

Secondly, we choose $N\gg 1$ possibly larger such that

$26C_{2}C_{\delta}N^{-\delta}\|Iu_{0\lambda}\|_{H_{\lambda}^{1}}^{4}\leq\|Iu_{0\lambda}\|_{H_{\lambda}^{1}}^{2}.$

With this choice of $N$, we show the following proposition:

######  Proposition 7.2.

Suppose that $\|u_{0}\|_{L^{2}(\mathbb{T}_{\gamma})}\ll 1$ in the focusing
case such that (64) holds. There is $N=N(\|u_{0}\|_{H^{s}})$ such that for
$0\leq t\leq T\lesssim N^{1-}\lambda^{\frac{1}{2}}$ the following estimate
holds for the solution to (63):

(65) $\|Iu_{\lambda}(t)\|^{2}_{H^{1}(\mathbb{T}^{2}_{\lambda})}\leq
5C_{2}\|Iu_{0\lambda}\|_{H^{1}}^{2}.$

###### Proof.

We use a continuity argument. The claim holds for $|t|\leq\lambda^{-\delta}$
by the auxiliary local well-posedness. Next suppose (65) holds up to $T\ll
N^{1-}\lambda^{\frac{1}{2}}$. Write $T=M\lambda^{-\delta}$ for
$M\in\mathbb{N}$. Then we estimate

$\begin{split}E_{I}^{1}(u_{\lambda}(T))&=E_{I}^{1}(u_{0\lambda})-\big{[}\Lambda_{4}(\tilde{\sigma}_{4})(t)-\Lambda_{4}(\tilde{\sigma}_{4})(0)\big{]}+\int_{0}^{\lambda^{-\delta}}\big{(}\Lambda_{4}(\overline{M}_{4})+\Lambda_{6}(\overline{M}_{6})\big{)}\\\
&\quad+\int_{\lambda^{-\delta}}^{2\lambda^{-\delta}}\big{(}\Lambda_{4}(\overline{M}_{4})+\Lambda_{6}(\overline{M}_{6})\big{)}+\ldots+\int_{(M-1)\lambda^{-\delta}}^{M\lambda^{-\delta}}\big{(}\Lambda_{4}(\overline{M}_{4})+\Lambda_{6}(\overline{M}_{6})\big{)}.\end{split}$

By the bootstrap assumption we have the following estimate for the boundary
term:

$C_{\delta}N^{-\delta}(\|Iu_{\lambda}(t)\|^{4}_{H^{1}}+\|Iu_{0\lambda}\|_{H^{1}}^{4})\leq
C_{\delta}N^{-\delta}26C_{2}\|Iu_{0\lambda}\|^{4}_{H^{1}}\leq\|Iu_{0\lambda}\|_{H^{1}}^{2}.$

Consequently,

$\begin{split}E_{I}^{1}(u_{\lambda}(T))&\leq
E_{I}^{1}(u_{0\lambda})+\|Iu_{0\lambda}\|_{H^{1}}^{2}\\\
&\quad+M\lambda^{-\delta}N^{-1+\varepsilon}\lambda^{-\frac{1}{2}}N^{-\varepsilon}\tilde{C}(\bar{C}^{4}(5C_{2})^{4}\|Iu_{0\lambda}\|_{H^{1}}^{4}+\bar{C}^{6}(5C_{2})^{6}\|Iu_{0\lambda}\|^{6}_{H^{1}}).\end{split}$

The final term can be bounded by $\|Iu_{0\lambda}\|_{H^{1}}^{2}$ and thus,

$\|Iu_{\lambda}(T)\|_{H^{1}}^{2}\leq
E_{I}^{1}(u_{\lambda}(T))+\|Iu_{\lambda}(T)\|_{L^{2}}^{2}\leq
4\|Iu_{0\lambda}\|_{H^{1}}^{2}.$

Hence, we can continue the bootstrap assumption until $T\sim
N^{1-}\lambda^{\frac{1}{2}}$. ∎

With the above proposition at hand, the conclusion is now a matter of
rescaling:

###### Proof of Theorem 1.2.

We choose $N$ large enough to have (65) from Proposition 7.2. Reversing the
scaling we find that the solution to (62) exists on a time

$T\sim\lambda^{-\frac{3}{2}}N^{1-}\sim N^{-\frac{3}{2s}+\frac{5}{2}-}.$

The exponent is positive provided that $s>\frac{3}{5}$. Hence, taking $N$ to
infinity for $s>\frac{3}{5}$ establishes global well-posedness. ∎

## Acknowledgement

Financial support by Korea Institute for Advanced Study, grant No. MG093901 is
gratefully acknowledged.

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