•Estimate for $\mathcal{T}_{0}$ : The analysis is similar to the one for
$\mathcal{R}$. Indeed, let us fix $u^{(1)},...,u^{(6)}\in
H^{\sigma}(\mathbb{T})$. Then, decomposing the sum in (4.1) dyadically and
using Lemma 9.4 (with $\Omega(\vec{k})=0$), leads to:

$\begin{split}\sum_{\begin{subarray}{c}k_{1}-k_{2}+...-k_{6}=0\\\
\Omega(\vec{k})=0\end{subarray}}|\psi_{2s}(\vec{k})||u^{(1)}_{k_{1}}u^{(2)}_{k_{2}}...u^{(6)}_{k_{6}}|&\lesssim\sum_{\begin{subarray}{c}k_{1}-k_{2}+...-k_{6}=0\\\
\Omega(\vec{k})=0\end{subarray}}|k_{(1)}|^{2s-2}|k_{(3)}|^{2}|u^{(1)}_{k_{1}}u^{(2)}_{k_{2}}...u^{(6)}_{k_{6}}|\\\
&\lesssim\sum_{N_{1},...,N_{6}}\mathfrak{T}_{0}(N_{1},...,N_{6})\end{split}$

where,

$\mathfrak{T}_{0}(N_{1},...,N_{6}):=\sum_{\begin{subarray}{c}k_{1}-k_{2}+...-k_{6}=0\\\
\Omega(\vec{k})=0\end{subarray}}N_{(1)}^{2s-2}N_{(3)}^{2}\prod_{j=1}^{6}\mathds{1}_{|k_{j}|\sim
N_{j}}|u^{(j)}_{k_{j}}|$

Next, from Lemma 9.6 (with $f^{j}_{k_{j}}=\mathds{1}_{|k_{j}|\sim
N_{j}}|u^{(j)}_{k_{j}}|$), we have:

$\mathfrak{T}_{0}(N_{1},...,N_{6})\lesssim_{\varepsilon}N_{(1)}^{2s-2+\varepsilon}N_{(3)}^{2}\prod_{j=1}^{6}\left\lVert
P_{N_{j}}u^{(j)}\right\rVert_{\mathbb{L}^{2}(\mathbb{T})}$

From that point the proof goes the same as the one for $\mathcal{R}$ (see the
estimate (10.2)).

•Estimate of $\mathcal{T}_{j}$ for $j=1,2$: We only prove the estimate (4.2)
since the analysis for the estimate (4.3) is similar. Here, the computations
are a little more delicate. For the sake of readability, we restrict ourselves
to prove the estimate:

$\sum_{\begin{subarray}{c}k_{1}-k_{2}+...-k_{6}=0\\\
p_{1}-p_{2}+...+p_{5}=k_{1}\\\ \Omega(\vec{k})\neq
0\end{subarray}}\big{|}\frac{\psi_{2s}(\vec{k})}{\Omega(\vec{k})}\big{|}|u_{p_{1}}\overline{u_{p_{2}}}...u_{p_{5}}\overline{u_{k_{2}}}...\overline{u_{k_{6}}}|\leq
C_{s}\left\lVert u\right\rVert_{H^{\sigma}}^{10}$

In other words, we just prove the estimate (4.2) where all the $u^{(j)}$ and
$v^{(l)}$ are equal to a single $u$. To do so, we will use the following
lemma:

######  Lemma 10.1.

There exists a constant $C>0$ such that for any dyadic integers
$P_{1},...,P_{5}$ and any sequences
$\left(g^{P_{1}}_{p_{1}}\right)_{p_{1}\in\mathbb{Z}},...,\left(g^{P_{5}}_{p_{5}}\right)_{p_{5}\in\mathbb{Z}}$
of complex numbers that satisfy for $j=1,...,5$,
$g^{P_{j}}_{p_{j}}=\mathds{1}_{|p_{j}|\sim P_{j}}g_{p_{j}}$, if we set for
$k_{1}\in\mathbb{Z}$ :

$f_{k_{1}}:=\sum_{p_{1}-p_{2}+...+p_{5}=k_{1}}g_{p_{1}}^{P_{1}}g_{p_{2}}^{P_{2}}g_{p_{3}}^{P_{3}}g_{p_{4}}^{P_{4}}g_{p_{5}}^{P_{5}}$

Then,

$\left\lVert f\right\rVert_{l^{2}(\mathbb{Z})}\leq
C\left(P_{(2)}P_{(3)}P_{(4)}P_{(5)}\right)^{\frac{1}{2}}\prod_{j=1}^{5}\left\lVert
g^{P_{j}}\right\rVert_{l^{2}(\mathbb{Z})}$

For clarity, we postpone the proof of this lemma for the end of this section.
Now we are ready to prove the estimate. Once again, we start by decomposing
the sum as:

$\sum_{\begin{subarray}{c}k_{1}-k_{2}+...-k_{6}=0\\\
p_{1}-p_{2}+...+p_{5}=k_{1}\\\ \Omega(\vec{k})\neq
0\end{subarray}}\big{|}\frac{\psi_{2s}(\vec{k})}{\Omega(\vec{k})}\big{|}|u_{p_{1}}\overline{u_{p_{2}}}...u_{p_{5}}\overline{u_{k_{2}}}...\overline{u_{k_{6}}}|=\sum_{\begin{subarray}{c}M_{1},...,M_{6}\\\
P_{1},...,P_{5}\end{subarray}}\mathfrak{T}_{1}(M_{1},...,M_{6},P_{1},...,P_{5})$

where $M_{1},...,M_{6},P_{1},...,P_{5}$ are dyadic-valued and

$\mathfrak{T}_{1}(M_{1},...,M_{6},P_{1},...,P_{5}):=\sum_{\begin{subarray}{c}k_{1}-k_{2}+...-k_{6}=0\\\
p_{1}-p_{2}+...+p_{5}=k_{1}\\\ \Omega(\vec{k})\neq
0\end{subarray}}\big{|}\frac{\psi_{2s}(\vec{k})}{\Omega(\vec{k})}\big{|}|u_{p_{1}}^{P_{1}}...u_{p_{5}}^{P_{5}}u_{k_{2}}^{M_{2}}...u_{k_{6}}^{M_{6}}|\cdot\mathds{1}_{|k_{1}|\sim
M_{1}}$ (10.4)

with the practical notation $u_{l}^{N}:=\mathds{1}_{|l|\sim N}u_{l}$. Next, we
rewrite $\mathfrak{T}_{1}$ as:

$\mathfrak{T}_{1}(M_{1},...,P_{5})=\sum_{\begin{subarray}{c}k_{1}-k_{2}+...-k_{6}=0\\\
\Omega(\vec{k})\neq
0\end{subarray}}\big{|}\frac{\psi_{2s}(\vec{k})}{\Omega(\vec{k})}\big{|}\prod_{j=1}^{6}|f_{k_{j}}^{(j)}|$
(10.5)

where, for $j=2,...,6$ we denote $f_{k_{j}}^{M_{j}}:=u_{k_{j}}^{M_{j}}$, and:

$f_{k_{1}}^{(1)}:=\mathds{1}_{|k_{1}|\sim
M_{1}}\cdot\sum_{p_{1}-p_{2}+...+p_{5}=k_{1}}|u_{p_{1}}^{P_{1}}...u_{p_{5}}^{P_{5}}|$

At this stage, let us recall that $M_{(1)}\geq...\geq M_{(6)}$ and
$P_{(1)}\geq...\geq P_{(5)}$ are respectively a non-increasing rearrangement
of $M_{1},...,M_{6}$ and $P_{1},...,P_{5}$. We also introduce:

$N_{(1)}\geq N_{(2)}\geq...\geq N_{(10)}$ a non-increasing rearrangement of
$M_{2},...,M_{6},P_{1},...,P_{5}$

Note that the constraints in the sum in $\mathfrak{T}_{1}$ imply that

$M_{(1)}\sim M_{(2)}$, and: $P_{(1)}\sim P_{(2)}$ or $P_{(1)}\sim M_{1}$, and:
$N_{(1)}\sim N_{(2)}$.

Now, we use Lemma 9.7 to estimate $\mathfrak{T}_{1}$ in (10.5); we obtain that
for any $\varepsilon>0$:

$\mathfrak{T}_{1}(M_{1},...,P_{5})\lesssim_{\varepsilon}M_{(1)}^{2s-2+\varepsilon}M_{(3)}^{2}\prod_{j=1}^{6}\left\lVert
f^{(j)}\right\rVert_{l^{2}}\lesssim_{\varepsilon}M_{(1)}^{2s-2+\varepsilon}M_{(3)}^{2}\left\lVert
f^{(1)}\right\rVert_{l^{2}}\prod_{j=2}^{6}\left\lVert
u^{M_{j}}\right\rVert_{\mathbb{L}^{2}}$

Applying the well-suited estimate of $\left\lVert f^{(1)}\right\rVert_{l^{2}}$
from Lemma 10.1 leads to

$\mathfrak{T}_{1}(M_{1},...,P_{5})\lesssim_{\varepsilon}M_{(1)}^{2s-2+\varepsilon}M_{(3)}^{2}(P_{(2)}P_{(3)}P_{(4)}P_{(5)})^{\frac{1}{2}}\prod_{j=1}^{5}\left\lVert
u^{P_{j}}\right\rVert_{\mathbb{L}^{2}}\prod_{j=2}^{6}\left\lVert
u^{M_{j}}\right\rVert_{\mathbb{L}^{2}}$ (10.6)

and it follows that

$\mathfrak{T}_{1}(M_{1},...,P_{5})\lesssim_{\varepsilon}M_{(1)}^{2s-2+\varepsilon}M_{(3)}^{2}\left(M_{2}M_{3}M_{4}M_{5}M_{6}\right)^{-\sigma}P_{(1)}^{-\sigma}\left(P_{(2)}P_{(3)}P_{(4)}P_{(5)}\right)^{\frac{1}{2}-\sigma}\left\lVert
u\right\rVert_{H^{\sigma}}^{10}$

In the remaining part of the proof, we show that it implies that:

$\mathfrak{T}_{1}(M_{1},...,P_{5})\lesssim_{\varepsilon}N_{(1)}^{-}\left\lVert
u\right\rVert_{H^{\sigma}}^{10}$ (10.7)

for $\sigma<s-\frac{1}{2}$ close enough to $s-\frac{1}{2}$, which will
complete the estimate for $\mathcal{T}_{1}$. We start from the inequality
above and we use the fact that $M_{1}\lesssim P_{(1)}$ due to the constraint
$p_{1}-p_{2}+...+p_{5}=k_{1}$, along with the rough estimates
$(M_{(4)}M_{(5)}M_{(6)})^{-\sigma}\leq 1$ and
$(P_{(3)}P_{(4)}P_{(5)})^{\frac{1}{2}-\sigma}\leq 1$, in order to obtain:

$\begin{split}\mathfrak{T}_{1}(M_{1},...,P_{5})&\lesssim_{\varepsilon}M_{(1)}^{2s-2+\varepsilon}M_{(3)}^{2}(M_{1}M_{2}M_{3}M_{4}M_{5}M_{6})^{-\sigma}(M_{1}^{\sigma}P_{(1)}^{-\sigma})(P_{(2)}P_{(3)}P_{(4)}P_{(5)})^{\frac{1}{2}-\sigma}\left\lVert
u\right\rVert_{H^{\sigma}}^{10}\\\
&\lesssim_{\varepsilon}M_{(1)}^{2(s-1-\sigma)+\varepsilon}M_{(3)}^{2-\sigma}P_{(2)}^{\frac{1}{2}-\sigma}\left\lVert
u\right\rVert_{H^{\sigma}}^{10}\end{split}$ (10.8)

Besides, when $\sigma<s-\frac{1}{2}$ and $\varepsilon>0$ are respectively
arbitrarily close to $s-\frac{1}{2}$ and $0$, we have that
$2(s-1-\sigma)+\varepsilon$, $2-\sigma$ and $\frac{1}{2}-\sigma$ are
respectively arbitrarily close to

$\displaystyle 2(s-1-(s-\frac{1}{2}))$ $\displaystyle=-1,$ $\displaystyle
2-(s-\frac{1}{2})$ $\displaystyle=\frac{5}{2}-s,$
$\displaystyle\frac{1}{2}-(s-\frac{1}{2})=1-s$

Hence, $M_{(1)}$ is accompanied by a negative exponent, $P_{(2)}$ is
accompanied by a negative exponent (since $s>\frac{3}{2}>1$), and $M_{(3)}$ is
accompanied by a negative exponent when $s\geq\frac{5}{2}$ and by a non-
negative one when $s<\frac{5}{2}$.

–In the situation where $M_{(1)}\sim N_{(1)}$, we use the rough estimate
$P_{(2)}^{\frac{1}{2}-\sigma}\leq 1$ in (10.8), and we obtain:

$\mathfrak{T}_{1}(M_{1},...,P_{5})\lesssim_{\varepsilon}N_{(1)}^{2(s-1-\sigma)+\varepsilon}M_{(3)}^{2-\sigma}\left\lVert
u\right\rVert_{H^{\sigma}}^{10}$

which is conclusive as far as

$2(s-1-(s-\frac{1}{2}))+2-(s-\frac{1}{2})<0$

that is, as far as $s>\frac{3}{2}$.
–In the other situation, where $M_{(1)}\ll N_{(1)}$, we necessarily have
$P_{(1)}\sim P_{(2)}\sim N_{(1)}$, so we deduce from (10.8) that:

$\mathfrak{T}_{1}(M_{1},...,P_{5})\lesssim_{\varepsilon}N_{(1)}^{\frac{1}{2}-\sigma}M_{(3)}^{2(s-1-\sigma)+2-\sigma+\varepsilon}\left\lVert
u\right\rVert_{H^{\sigma}}^{10}$

Then, for $\sigma<s-\frac{1}{2}$ and $\varepsilon>0$ respectively arbitrarily
close to $s-\frac{1}{2}$ and $0$, we have:

$\mathfrak{T}_{1}(M_{1},...,P_{5})\lesssim
N_{(1)}^{(1-s)+}M_{(3)}^{(\frac{3}{2}-s)+}\left\lVert
u\right\rVert_{H^{\sigma}}^{10}\lesssim N_{(1)}^{-}\left\lVert
u\right\rVert_{H^{\sigma}}^{10}$

since $s>\frac{3}{2}$.
In conclusion, (10.7) is proven and the proof of Proposition 4.1 is completed.
∎

We finish this section by a proof of Lemma 10.1.

###### Proof of Lemma 10.1.

We have to estimate :

$\left\lVert
f\right\rVert_{l^{2}(\mathbb{Z})}^{2}=\sum_{k_{1}\in\mathbb{Z}}\bigg{|}\sum_{p_{1}-p_{2}+...+p_{5}=k_{1}}g_{p_{1}}^{P_{1}}g_{p_{2}}^{P_{2}}g_{p_{3}}^{P_{3}}g_{p_{4}}^{P_{4}}g_{p_{5}}^{P_{5}}\bigg{|}^{2}$

Using the Cauchy-Schwarz inequality we obtain:

$\left\lVert
f\right\rVert_{l^{2}(\mathbb{Z})}^{2}\leq\sum_{k_{1}\in\mathbb{Z}}\bigl{(}\sum_{p_{1},...,p_{5}}\mathds{1}_{p_{1}-p_{2}+...+p_{5}=k_{1}}\prod_{j=1}^{5}\mathds{1}_{|p_{j}|\sim
P_{j}}\bigr{)}\bigl{(}\sum_{p_{1},...,p_{5}}\mathds{1}_{p_{1}-p_{2}+...+p_{5}=k_{1}}|g_{p_{1}}^{P_{1}}g_{p_{2}}^{P_{2}}g_{p_{3}}^{P_{3}}g_{p_{4}}^{P_{4}}g_{p_{5}}^{P_{5}}|^{2}\bigr{)}$

Plugging the counting bound (9.1) into this formula yields

$\left\lVert f\right\rVert_{l^{2}(\mathbb{Z})}^{2}\lesssim
P_{(2)}P_{(3)}P_{(4)}P_{(5)}\sum_{p_{1},...,p_{5}}|g_{p_{1}}^{P_{1}}g_{p_{2}}^{P_{2}}g_{p_{3}}^{P_{3}}g_{p_{4}}^{P_{4}}g_{p_{5}}^{P_{5}}|^{2}\underbrace{\big{(}\sum_{k_{1}\in\mathbb{Z}}\mathds{1}_{p_{1}-p_{2}+...+p_{5}=k_{1}}\big{)}}_{\leq
1}$

Thus,

$\left\lVert f\right\rVert_{l^{2}(\mathbb{Z})}^{2}\lesssim
P_{(2)}P_{(3)}P_{(4)}P_{(5)}\prod_{j=1}^{5}\left\lVert
g^{P_{j}}\right\rVert_{l^{2}(\mathbb{Z})}^{2}$

which is the desired estimate. ∎

##  11\. Estimates for the weight of the weighted Gaussian measures

This section is dedicated to estimates on the weight
$\mathds{1}_{\\{\mathcal{C}(u)\leq R\\}}e^{-R_{s,N}(u)}$ of the weighted
Gaussian measure $\rho_{s,R,N}$ (defined in (7.1)). In particular, we show
that :

$\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}e^{-R_{s,N}(u)}\in\mathbb{L}^{1}(d\mu_{s})$

which ensures that $\rho_{s,R,N}$ is a finite measure on
$H^{\sigma}(\mathbb{T})$, $\sigma<s-\frac{1}{2}$. More precisely, we prove
Proposition 7.4. Yet, before doing so, we will need the two following lemmas :

######  Lemma 11.1.

Let $s>\frac{3}{2}$. Then, there exists $\beta\in(0,1)$ such that for every
$R>0$, there exists $C(s,R)>0$, such that for any $p\in[2,+\infty)$,

$\left\lVert\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}R_{s,N}\right\rVert_{\mathbb{L}^{p}(d\mu_{s})}\leq C(s,R)p^{\beta}$

uniformly in $N\in\mathbb{N}\cup\\{\infty\\}$.

Recall that we denote $R_{s,\infty}=R_{s}$ for convenience.

######  Lemma 11.2.

Let $(X,\mathcal{A},d\nu)$ be a probability space and let
$F:X\longrightarrow\mathbb{C}$ be a mesurable function. Assume that there
exist constants $C_{0}>0$ and $\beta\in(0,1)$ such that for every
$p\in[2,+\infty)$,

$\left\lVert F\right\rVert_{\mathbb{L}^{p}(d\nu)}\leq C_{0}p^{\beta}$

Then, there exist $\delta>0$ and $C_{1}>0$ only depending on $\beta$ and
$C_{0}$ such that :

$\int_{X}e^{\delta|F(v)|^{\frac{1}{\beta}}}d\nu(v)\leq C_{1}$

We will only prove Lemma 11.1, Lemma 11.2 being just a slightly different
version of Lemma 4.5 from [32] (where a proof is given).

Before doing the proof of Lemma 11.1, let us briefly show how it implies
Proposition 7.4 when combining with Lemma 11.2.

###### Proof of Proposition 7.4.

Here, we assume the statements in Lemma 11.1 and Lemma 11.2.
–We start by proving the first statement in Proposition 7.4. Lemma 11.1 shows
that $F:=\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}R_{s,N}(u):H^{\sigma}(\mathbb{T})\longrightarrow\mathbb{C}$ satisfies the
assumptions of Lemma 11.2 (with $\nu=\mu_{s}$). Then, applying Lemma 11.2, we
obtain that there exist $\delta=\delta(s,R)>0$ and $C_{1}(s,R)>0$ such that :

$\int_{H^{\sigma}}e^{\delta\left|\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}R_{s,N}(u)\right|^{\frac{1}{\beta}}}d\mu_{s}\leq C_{1}(s,R)$

A fortiori we have,

$\int_{H^{\sigma}}\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}e^{\delta\left|R_{s,N}(u)\right|^{\frac{1}{\beta}}}d\mu_{s}\leq
C_{1}(s,R)$

Thanks to the fact that $\frac{1}{\beta}>1$, we deduce from the above
inequality that for every $p\in[1,+\infty)$ :

$\left\lVert\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}e^{|R_{s,N}(u)|}\right\rVert^{p}_{\mathbb{L}^{p}(d\mu_{s})}=\int_{H^{\sigma}}\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}e^{p|R_{s,N}(u)|}d\mu_{s}\leq C(s,R,p)<+\infty$

for a certain constant $C(s,R,p)>0$.
–As a consequence, we are now able to prove the second statement of
Proposition 7.4, that is:

$\left\lVert\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}e^{-R_{s}(u)}-\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}e^{-R_{s,N}(u)}\right\rVert_{\mathbb{L}^{p}(d\mu_{s})}\underset{N\rightarrow\infty}{\longrightarrow}0$

Firstly, we know from Proposition 3.3 that we have the pointwise convergence:

$\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}R_{s}(u)=\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}\cdot\frac{1}{6}\textnormal{Re}\sum_{\begin{subarray}{c}k_{1}-k_{2}+...-k_{6}=0\\\
\Omega(\vec{k})\neq
0\end{subarray}}\frac{\psi_{2s}(\vec{k})}{\Omega(\vec{k})}u_{k_{1}}\overline{u_{k_{2}}}...\overline{u_{k_{6}}}=\lim_{N\rightarrow\infty}\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}R_{s,N}(u)$

Consequently, from the continuity of the exponential, we also have the
pointwise convergence:

$\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}e^{-R_{s}(u)}=\lim_{N\rightarrow\infty}\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}e^{-R_{s,N}(u)}$

In particular, $\mathds{1}_{\\{\mathcal{C}(u)\leq R\\}}e^{-R_{s,N}(u)}$
converges to $\mathds{1}_{\\{\mathcal{C}(u)\leq R\\}}e^{-R_{s}(u)}$ in
measure. In addition, for a fixed $q\in(1,+\infty)$, we just proved that the
functions $\mathds{1}_{\\{\mathcal{C}(u)\leq R\\}}e^{-R_{s,N}(u)}$ are
uniformly bounded in $\mathbb{L}^{q}(d\mu_{s})$ (with respect to
$N\in\mathbb{N}$). Then, (using the same argument as the one from the proof of
Proposition 8.9) we can conclude that $\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}e^{-R_{s,N}(u)}$ converges to $\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}e^{-R_{s}(u)}$ in $\mathbb{L}^{p}(d\mu_{s})$ for any $p\in[1,q)$; and
since $q\in(1,+\infty)$ is arbitrary, the convergence holds for any
$p\in[1,+\infty)$. This completes the proof of Proposition 7.4. ∎

However, it remains to prove Lemma 11.1. Our analysis will rely on a
decomposition into two parts of $R_{s,N}$. We will be able to treat the first
part deterministically thanks to suitable ”exchanges of derivatives”. For the
second part, those ”exchanges of derivatives” will fail because we will be in
a high-high-high-low-low-low regime of frequency (where the three highest
frequencies in the sum (3.2) defining $R_{s,N}$ are in fact much more higher
than the three lowest frequencies). Instead, we will handle the second part
using the independence between the high frequency Gaussians and the low
frequency Gaussians, using Wiener chaos estimate.

###  11.1. Decomposition

Recall that we have (see Section 3, Proposition 3.1):

$R_{s,N}(u)=\frac{1}{6}\textnormal{Re}\hskip 2.84544pt\mathcal{R}(w)$

where $w:=\Pi_{N}u$ (with the convention $\Pi_{\infty}=id$), and

$\mathcal{R}(w):=\sum_{\begin{subarray}{c}k_{1}-k_{2}+...-k_{6}=0\\\
\Omega(\vec{k})\neq
0\end{subarray}}\frac{\psi_{2s}(\vec{k})}{\Omega(\vec{k})}w_{k_{1}}\overline{w_{k_{2}}}...\overline{w_{k_{6}}}$

It suffices to show the estimate of Lemma 11.1 for
$\mathds{1}_{\\{\mathcal{C}(u)\leq R\\}}\mathcal{R}(w)$ because
$|R_{s,N}(u)|\leq\frac{1}{6}|\mathcal{R}(w)|$.

Next, we split the set of indices over which we sum. We invoke the following
sets of indices :

$\begin{split}\Lambda_{D}:=\\{(k_{1},...,k_{6})\in\mathbb{Z}^{6}:\hskip
2.84544pt&\sum_{j=1}^{6}(-1)^{j-1}k_{j}=0,\hskip
2.84544pt\sum_{j=1}^{6}(-1)^{j-1}k_{j}^{2}\neq 0,\\\ &\hskip
2.84544pt|k_{(3)}|<|k_{(1)}|^{1-\delta_{0}}\hskip 2.84544pt\textbf{or}\hskip
2.84544pt|k_{(4)}|\geq|k_{(3)}|^{\delta_{0}}\\}\end{split}$ (11.1)

and,

$\begin{split}\Lambda_{W}:=\\{(k_{1},...,k_{6})\in\mathbb{Z}^{6}:\hskip
2.84544pt&\sum_{j=1}^{6}(-1)^{j-1}k_{j}=0,\hskip
2.84544pt\sum_{j=1}^{6}(-1)^{j-1}k_{j}^{2}\neq 0,\\\ &\hskip
2.84544pt|k_{(3)}|\geq|k_{(1)}|^{1-\delta_{0}}\hskip
2.84544pt\textbf{and}\hskip
2.84544pt|k_{(4)}|<|k_{(3)}|^{\delta_{0}}\\}\end{split}$ (11.2)

where $\delta_{0}\in(0,1)$111111We will see in the forthcoming proof that
there is no constraint on $\delta_{0}$, so we can chose it as any number in
$(0,1)$. Then, we split $\mathcal{R}$ as :

$\mathcal{R}(w)=\mathcal{R}^{(D)}(w)+\mathcal{R}^{(W)}(w)$

where,

$\displaystyle\mathcal{R}^{(D)}(w)$
$\displaystyle:=\sum_{\Lambda_{D}}\frac{\psi_{2s}(\vec{k})}{\Omega(\vec{k})}w_{k_{1}}\overline{w_{k_{2}}}...\overline{w_{k_{6}}},$
$\displaystyle\mathcal{R}^{(W)}(w):=\sum_{\Lambda_{W}}\frac{\psi_{2s}(\vec{k})}{\Omega(\vec{k})}w_{k_{1}}\overline{w_{k_{2}}}...\overline{w_{k_{6}}}$

To estimate $\mathcal{R}^{(D)}$ we will use the deterministic tools from
Section 9. To estimate $\mathcal{R}^{(W)}$, we note that in the sum we are in
a high-high-high-low-low-low regime because
$|k_{(1)}|,|k_{(2)}|,|k_{(3)}|\gg|k_{(4)}|,|k_{(5)}|,|k_{(6)}|$. It will then
be possible to make use of the independence between the Gaussians
$g_{k_{(1)}},g_{k_{(2)}},g_{k_{(3)}}$ and
$g_{k_{(4)}},g_{k_{(5)}},g_{k_{(6)}}$ via Wiener chaos.

####  11.1.1. Absence of pairing:

######  Definition 11.3 (Pairing).

Consider a constraint under the form :

$\varepsilon_{1}k_{1}+\varepsilon_{2}k_{2}+...+\varepsilon_{m}k_{m}=0$

where $k_{j}\in\mathbb{Z}$ and $\varepsilon_{j}\in\\{-,+\\}$. We say that
$k_{j}$ and $k_{l}$ are paired if

$\varepsilon_{j}k_{j}+\varepsilon_{l}k_{l}=0$

######  Remark 11.4.

In $\Lambda_{W}$ (see (11.2) above), there is (for $k_{(1)}$ large enough) no
pairing within the three highest frequencies (relatively to the constraint
$k_{1}-k_{2}+...-k_{6}=0$). Indeed, suppose there is a pairing between two of
the three highest frequencies. Then, the constraint would take the form:

the remaining high frequency = sum of three low frequencies

which is impossible because in $\Lambda_{W}$ we have:

$|$high frequencies$|$ $\gg$ $|$low frequencies$|$

###  11.2. Proof of the estimate

We are now ready to prove Lemma 11.1.

###### Proof of Lemma 11.1.

We organize the proof in two steps. In the first step, we provide a
(deterministic) estimate for $\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}\mathcal{R}$. In the second step, we show that this estimate is only
conclusive for the contribution $\mathcal{R}^{(D)}$, and that for the
remaining contribution $\mathcal{R}^{(W)}$, further analysis using Wiener
chaos is required.

Step 1, Deterministic estimate : We rely on the estimate (10.2) we obtained in
Section 10 (which we do not reprove here). Thus, we start our analysis from
the following estimates:

$\displaystyle\left|\mathcal{R}(w)\right|$
$\displaystyle\lesssim\sum_{N_{1},...,N_{6}}\mathfrak{R}(N_{1},...,N_{6})$
and: $\displaystyle\mathfrak{R}(N_{1},...,N_{6})$
$\displaystyle\lesssim_{\varepsilon}N_{(1)}^{2s-2+\varepsilon}N_{(3)}^{2}\prod_{j=1}^{6}\left\lVert
P_{N_{j}}w\right\rVert_{\mathbb{L}^{2}(\mathbb{T})}$ (11.3)

where $N_{1},...,N_{6}$ are dyadic valued, $P_{N}$ is the projector onto
frequencies $|k|\sim N$, and:

$\mathfrak{R}(N_{1},...,N_{6}):=\sum_{\begin{subarray}{c}k_{1}-k_{2}+...-k_{6}=0\\\
\Omega(\vec{k})\neq
0\end{subarray}}\Big{|}\frac{\psi_{2s}(\vec{k})}{\Omega(\vec{k})}\Big{|}\prod_{j=1}^{6}\mathds{1}_{|k_{j}|\sim
N_{j}}|w_{k_{j}}|$

We decompose $\mathfrak{R}$ as
$\mathfrak{R}=\mathfrak{R}^{(D)}+\mathfrak{R}^{(W)}$, according to the
decomposition of $\mathcal{R}$.

Now, let $1<\sigma^{\prime}<\sigma<s-\frac{1}{2}$. We have:

$\prod_{j=1}^{6}\left\lVert
P_{N_{j}}w\right\rVert_{\mathbb{L}^{2}(\mathbb{T})}\lesssim\prod_{j=1}^{6}\left\lVert
P_{N_{j}}u\right\rVert_{\mathbb{L}^{2}(\mathbb{T})}\lesssim(N_{(1)}^{-\sigma}\left\lVert
u\right\rVert_{H^{\sigma}})(N_{(2)}^{-\sigma^{\prime}}\left\lVert
u\right\rVert_{H^{\sigma^{\prime}}})(N_{(3)}N_{(4)}N_{(5)}N_{(6)})^{-1}\left\lVert
u\right\rVert_{H^{1}}^{4}$ (11.4)

Recall that the constraints in the sum in $\mathfrak{R}$ above implies that
$N_{(1)}\sim N_{(2)}$. Moreover, if we write:

$\sigma^{\prime}=\alpha 1+(1-\alpha)\sigma$

for $\alpha\in(0,1)$, then, by interpolating $H^{\sigma^{\prime}}$ between
$H^{1}$ and $H^{\sigma}$, we obtain from (11.4) that:

$\begin{split}\prod_{j=1}^{6}\left\lVert
P_{N_{j}}u\right\rVert_{\mathbb{L}^{2}(\mathbb{T})}&\lesssim
N_{(1)}^{-2\sigma+\alpha(\sigma-1)}(N_{(3)}N_{(4)}N_{(5)}N_{(6)})^{-1}\left\lVert
u\right\rVert_{H^{1}}^{4+\alpha}\left\lVert
u\right\rVert^{2-\alpha}_{H^{\sigma}}\\\ &\lesssim
N_{(1)}^{-2\sigma+\alpha(s-\frac{3}{2})}(N_{(3)}N_{(4)}N_{(5)}N_{(6)})^{-1}\left\lVert
u\right\rVert_{H^{1}}^{4+\alpha}\left\lVert
u\right\rVert^{2-\alpha}_{H^{\sigma}},\hskip 8.5359pt(\textnormal{because
$\alpha(\sigma-1)<\alpha(s-3/2)$ })\end{split}$

Plugging this into (11.3), we obtain

$\mathfrak{R}(N_{1},...,N_{6})\lesssim_{\varepsilon}N_{(1)}^{2(s-\sigma-1)+\varepsilon+\alpha(s-\frac{3}{2})}N_{(3)}(N_{(4)}N_{(5)}N_{(6)})^{-1}\left\lVert
u\right\rVert_{H^{1}}^{4+\alpha}\left\lVert
u\right\rVert^{2-\alpha}_{H^{\sigma}}$

In particular, thanks to Remark 7.2,

$\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}\mathfrak{R}(N_{1},...,N_{6})\lesssim_{\varepsilon}N_{(1)}^{2(s-\sigma-1)+\varepsilon+\alpha(s-\frac{3}{2})}N_{(3)}(N_{(4)}N_{(5)}N_{(6)})^{-1}R^{4+\alpha}\left\lVert
u\right\rVert^{2-\alpha}_{H^{\sigma}}$

And since we will not need the smallness provided by $(N_{(5)}N_{(6)})^{-1}$,
we simply write :

$\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}\mathfrak{R}(N_{1},...,N_{6})\lesssim_{\varepsilon,R}N_{(1)}^{2(s-\sigma-1)+\varepsilon+\alpha(s-\frac{3}{2})}N_{(3)}N_{(4)}^{-1}\left\lVert
u\right\rVert^{2-\alpha}_{H^{\sigma}}$

Using now (9.3), we conclude that

$\left\lVert\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}\mathfrak{R}(N_{1},...,N_{6})\right\rVert_{\mathbb{L}^{p}(d\mu_{s})}\leq
CN_{(1)}^{2(s-\sigma-1)+\varepsilon+\alpha(s-\frac{3}{2})}N_{(3)}N_{(4)}^{-1}p^{\frac{2-\alpha}{2}}$
(11.5)

Where the constant $C>0$ depends on $R$, $s$, $\sigma$ and $\varepsilon$. This
is the estimate we will work with later in the proof. We stress the fact that
(11.5) is true for $\mathfrak{R}$ and also for $\mathfrak{R}^{(D)}$ and
$\mathfrak{R}^{(W)}$ (with the exact same proof).

Step 2, Estimates for $\mathcal{R}^{(D)}$ and $\mathcal{R}^{(W)}$:
•We start with $\mathcal{R}^{(D)}$. Recalling that $\sigma<s-\frac{1}{2}$, we
see that the exponent of $N_{(1)}$ in (11.5) satisfies:

$\begin{cases}2(s-\sigma-1)+\varepsilon+\alpha(s-\frac{3}{2})>-1\\\
2(s-\sigma-1)+\varepsilon+\alpha(s-\frac{3}{2})\longrightarrow-1\hskip
5.69046pt\text{as}\begin{cases}\varepsilon\rightarrow 0\\\ \alpha\rightarrow
0\\\ \sigma\rightarrow s-\frac{1}{2}\end{cases}\end{cases}$ (11.6)

Moreover, for a fixed $\delta_{0}\in(0,1)$, the conditions on the indices
lying in $\Lambda_{D}$ imply that $N_{(3)}\lesssim N_{(1)}^{1-\delta_{0}}$ or
$N_{(4)}\gtrsim N_{(3)}^{\delta_{0}}$. In both cases, we deduce from (11.5)
that for $\varepsilon$ and $\alpha$ close enough to $0$ and $\sigma$ close
enough to $s-\frac{1}{2}$, we have121212The notation $N^{-}$ means that the
power of $N$ is $-\gamma$ for a certain $\gamma>0$.

$\left\lVert\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}\mathfrak{R}^{(D)}(N_{1},...,N_{6})\right\rVert_{\mathbb{L}^{p}(d\mu_{s})}\leq
CN_{(1)}^{-}p^{\frac{2-\alpha}{2}}$

Consequently,

$\left\lVert\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}\mathcal{R}^{(D)}\right\rVert_{\mathbb{L}^{p}(d\mu_{s})}\leq
Cp^{\frac{2-\alpha}{2}}$

•We continue our analysis with the term $\mathcal{R}^{(W)}$. Once again, we
start by decomposing $\mathcal{R}^{(W)}$ dyadically as
$\mathcal{R}^{(W)}(w)=\sum_{N_{1},...,N_{6}}\mathcal{R}_{N_{1},...,N_{6}}^{(W)}(w)$,
where131313For the analysis below, we need to keep the complex conjugation
bars, so here we don’t use $\mathfrak{R}^{(W)}$:

$\mathcal{R}_{N_{1},...,N_{6}}^{(W)}(w):=\sum_{\begin{subarray}{c}k_{1}-k_{2}+...-k_{6}=0\\\
\Omega(\vec{k})\neq 0\\\
|k_{(4)}|\leq|k_{(3)}|^{\delta_{0}}\end{subarray}}\frac{\psi_{2s}(\vec{k})}{\Omega(\vec{k})}w_{k_{1}}\overline{w_{k_{2}}}...\overline{w_{k_{6}}}\prod_{j=1}^{6}\mathds{1}_{|k_{j}|\sim
N_{j}}$

Henceforth, we denote $w_{k_{j}}^{N_{j}}:=\mathds{1}_{|k_{j}|\sim
N_{j}}w_{k_{j}}$ for better readability.

Without loss of generality, we assume that
$\\{N_{1},N_{2},N_{3}\\}=\\{N_{(1)},N_{(2)},N_{(3)}\\}$, meaning that the
three highest frequencies are $k_{1},k_{2},k_{3}$. The other cases are similar
or simpler.
We denote $\mathcal{B}_{\ll N_{(3)}}$ the $\sigma$-algebra generated by the
Gaussians $\left(g_{k}\right)_{|k|\leq N_{(3)}/100}$. We only need to consider
the contribution when $N_{(1)}$ is large because when $N_{(1)}$ is small, all
the $N_{j}$’s are small, and we can use (11.5) without fearing issues of
summability in the $N_{j}$’s. In particular, we assume that $N_{(1)}$ is
sufficiently large so that $N_{(3)}$ (which is $\gtrsim
N_{(1)}^{1-\delta_{0}}$) is large enough to ensure that $N_{(4)}$ (which is
$\lesssim N_{(3)}^{\delta_{0}}$) is $\leq N_{(3)}/100$. As a consequence, we
have that the random variables :

$\begin{split}&w_{k_{4}}^{N_{4}},w_{k_{4}}^{N_{4}},w_{k_{5}}^{N_{6}}\hskip
5.69046pt\textnormal{are $\mathcal{B}_{\ll N_{(3)}}$mesurable}\\\
&w_{k_{1}}^{N_{1}},w_{k_{2}}^{N_{2}},w_{k_{3}}^{N_{3}}\hskip
5.69046pt\textnormal{are independent of $\mathcal{B}_{\ll
N_{(3)}}$}\end{split}$

With this set up, we can now begin the analysis. Thanks to Remark 7.2, we have

$\begin{split}\left\lVert\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}\mathcal{R}_{N_{1},...,N_{6}}^{(W)}\right\rVert_{\mathbb{L}^{p}(d\mu_{s})}&\leq\left\lVert\mathds{1}_{B_{R}^{H^{1}}}(u)\cdot\mathcal{R}_{N_{1},...,N_{6}}^{(W)}\right\rVert_{\mathbb{L}^{p}(d\mu_{s})}\\\
&\leq\left\lVert\mathds{1}_{B_{R}^{H^{1}}}(P_{N_{(3)/100}}u)\cdot\mathcal{R}_{N_{1},...,N_{6}}^{(W)}\right\rVert_{\mathbb{L}^{p}(d\mu_{s})}\end{split}$

And, $\mathds{1}_{B_{R}^{H^{1}}}(P_{N_{(3)/100}}u)$ is $\mathcal{B}_{\ll
N_{(3)}}$ mesurable. So, using the $\mathbb{L}^{p}$-norm conditioned to the
$\sigma$-algebra $\mathcal{B}_{\ll N_{(3)}}$, denoted
$\mathbb{L}^{p}(d\mu_{s}|\mathcal{B}_{\ll N_{(3)}})$, we obtain :

$\begin{split}\left\lVert\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}\mathcal{R}_{N_{1},...,N_{6}}^{(W)}\right\rVert_{\mathbb{L}^{p}(d\mu_{s})}&\leq\Big{\|}\big{\|}\mathds{1}_{B_{R}^{H^{1}}}(P_{N_{(3)/100}}u)\cdot\mathcal{R}_{N_{1},...,N_{6}}^{(W)}\big{\|}_{\mathbb{L}^{p}(d\mu_{s}|\mathcal{B}_{\ll
N_{(3)}})}\Big{\|}_{\mathbb{L}^{\infty}(d\mu_{s})}\\\
&\leq\Big{\|}\big{\|}\mathcal{R}_{N_{1},...,N_{6}}^{(W)}\big{\|}_{\mathbb{L}^{p}(d\mu_{s}|\mathcal{B}_{\ll
N_{(3)}})}\cdot\mathds{1}_{B_{R}^{H^{1}}}(P_{N_{(3)/100}}u)\Big{\|}_{\mathbb{L}^{\infty}(d\mu_{s})}\end{split}$
(11.7)

Now, the conditional Wiener-chaos estimate from Lemma 9.9 (with $m=3$),
followed by Lemma 9.10 combined with the absence of paring (see Remark 11.4),
allows us to obtain :

$\begin{split}\left\lVert\mathcal{R}_{N_{1},...,N_{6}}^{(W)}\right\rVert_{\mathbb{L}^{p}(d\mu_{s}|\mathcal{B}_{\ll
N_{(3)}})}&\lesssim
p^{\frac{3}{2}}\left\lVert\mathcal{R}_{N_{1},...,N_{6}}^{(W)}\right\rVert_{\mathbb{L}^{2}(d\mu_{s}|\mathcal{B}_{\ll
N_{(3)}})}\\\ &\lesssim
p^{\frac{3}{2}}(N_{(1)}N_{(2)}N_{(3)})^{-s}\bigg{(}\sum_{k_{1},k_{2},k_{3}}\Big{|}\sum_{k_{4},k_{5},k_{6}}C(\vec{k})\cdot\frac{\psi_{2s}(\vec{k})}{\Omega(\vec{k})}\overline{w_{k_{4}}^{N_{4}}}w_{k_{5}}^{N_{5}}\overline{w_{k_{6}}^{N_{6}}}\Big{|}^{2}\bigg{)}^{\frac{1}{2}}\end{split}$
(11.8)

where we gathered all the constraints into the term :

$C(\vec{k}):=\mathds{1}_{k_{1}-k_{2}+...-k_{6}=0}\cdot\mathds{1}_{\Omega(\vec{k})\neq
0}\cdot\prod_{j=1}^{6}\mathds{1}_{|k_{j}|\sim N_{j}}$

Next, by Cauchy-Schwarz,

$\sum_{k_{1},k_{2},k_{3}}\Big{|}\sum_{k_{4},k_{5},k_{6}}C(\vec{k})\cdot\frac{\psi_{2s}(\vec{k})}{\Omega(\vec{k})}\overline{w_{k_{4}}^{N_{4}}}w_{k_{5}}^{N_{5}}\overline{w_{k_{6}}^{N_{6}}}\Big{|}^{2}\lesssim\sum_{k_{1},k_{2},k_{3}}\Big{(}\sum_{k_{4},k_{5},k_{6}}C(\vec{k})\cdot\Big{|}\frac{\psi_{2s}(\vec{k})}{\Omega(\vec{k})}\Big{|}^{2}|w_{k_{4}}^{N_{4}}w_{k_{5}}^{N_{5}}|^{2}\Big{)}\Big{(}\sum_{k_{4},k_{5},k_{6}}C(\vec{k})|w_{k_{6}}^{N_{6}}|^{2}\Big{)}$

From Lemma 9.4 we get
$\left|\frac{\psi_{2s}(\vec{k})}{\Omega(\vec{k})}\right|\lesssim|k_{(1)}|^{2s-2}(1+\frac{|k_{(3)}|^{2}}{|\Omega(\vec{k})|})\lesssim|k_{(1)}|^{2s-2}|k_{(3)}|^{2}\lesssim
N_{(1)}^{2s-2}N_{(3)}^{2}$. Then, on the one hand :

$\begin{split}\sum_{k_{4},k_{5},k_{6}}C(\vec{k})\cdot\Big{|}\frac{\psi_{2s}(\vec{k})}{\Omega(\vec{k})}\Big{|}^{2}|w_{k_{4}}^{N_{4}}w_{k_{5}}^{N_{5}}|^{2}&\lesssim(N_{(1)}^{2s-2}N_{(3)}^{2})^{2}\sum_{k_{4},k_{5},k_{6}}C(\vec{k})\cdot|w_{k_{4}}^{N_{4}}w_{k_{5}}^{N_{5}}|^{2}\\\
&\lesssim(N_{(1)}^{2s-2}N_{(3)}^{2})^{2}\sum_{k_{4},k_{5}}|w_{k_{4}}^{N_{4}}w_{k_{5}}^{N_{5}}|^{2}\cdot\underbrace{\sum_{k_{6}}\mathds{1}_{k_{6}=k_{1}-k_{2}...+k_{5}}}_{\leq
1}\\\ &\lesssim(N_{(1)}^{2s-2}N_{(3)}^{2}\left\lVert
w^{N_{4}}\right\rVert_{\mathbb{L}^{2}}\left\lVert
w^{N_{5}}\right\rVert_{\mathbb{L}^{2}})^{2}\end{split}$

and on the other hand,

$\begin{split}\sum_{k_{1},k_{2},k_{3}}\Big{(}\sum_{k_{4},k_{5},k_{6}}C(\vec{k})|w_{k_{6}}^{N_{6}}|^{2}\Big{)}=\sum_{k_{6}}|w_{k_{6}}^{N_{6}}|^{2}\sum_{k_{1},k_{2},k_{3},k_{4},k_{5}}C(\vec{k})\lesssim
N_{(2)}N_{(3)}N_{(4)}N_{(5)}\cdot\left\lVert
w^{N_{6}}\right\rVert_{\mathbb{L}^{2}}^{2}\end{split}$

where we used the counting bound from Lemma 9.3. We deduce from the two
inequalities above that

$\begin{split}\bigg{(}\sum_{k_{1},k_{2},k_{3}}\Big{|}\sum_{k_{4},k_{5},k_{6}}C(\vec{k})\cdot\frac{\psi_{2s}(\vec{k})}{\Omega(\vec{k})}\overline{w_{k_{4}}^{N_{4}}}w_{k_{5}}^{N_{5}}\overline{w_{k_{6}}^{N_{6}}}\Big{|}^{2}\bigg{)}^{\frac{1}{2}}&\lesssim
N_{(1)}^{2s-2}N_{(3)}^{2}N_{(1)}^{\frac{1}{2}}N_{(3)}^{\frac{1}{2}}(N_{(4)}N_{(5})^{\frac{1}{2}}\left\lVert
w^{N_{4}}\right\rVert_{\mathbb{L}^{2}}\left\lVert
w^{N_{5}}\right\rVert_{\mathbb{L}^{2}}\left\lVert
w^{N_{6}}\right\rVert_{\mathbb{L}^{2}}\\\ &\lesssim
N_{(1)}^{2s-2+\frac{1}{2}}N_{(3)}^{\frac{5}{2}}(N_{(4)}N_{(5})^{-\frac{1}{2}}N_{(6)}^{-1}\left\lVert
w^{N_{4}}\right\rVert_{H^{1}}\left\lVert
w^{N_{5}}\right\rVert_{H^{1}}\left\lVert w^{N_{6}}\right\rVert_{H^{1}}\\\
&\lesssim N_{(1)}^{2s-\frac{3}{2}}N_{(3)}^{\frac{5}{2}}\left\lVert
w^{N_{4}}\right\rVert_{H^{1}}\left\lVert
w^{N_{5}}\right\rVert_{H^{1}}\left\lVert
w^{N_{6}}\right\rVert_{H^{1}}\end{split}$

Coming back to (11.8), we deduce that

$\left\lVert\mathcal{R}_{N_{1},...,N_{6}}^{(W)}\right\rVert_{\mathbb{L}^{p}(d\mu_{s}|\mathcal{B}_{\ll
N_{(3)}})}\lesssim
p^{\frac{3}{2}}N_{(1)}^{-\frac{3}{2}}N_{(3)}^{\frac{5}{2}-s}\left\lVert
w^{N_{4}}\right\rVert_{H^{1}}\left\lVert
w^{N_{5}}\right\rVert_{H^{1}}\left\lVert w^{N_{6}}\right\rVert_{H^{1}}$

And plugging this into (11.7), we obtain:

$\left\lVert\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}\mathcal{R}_{N_{1},...,N_{6}}^{(W)}\right\rVert_{\mathbb{L}^{p}(d\mu_{s})}\lesssim_{R}p^{\frac{3}{2}}N_{(1)}^{-\frac{3}{2}}N_{(3)}^{\frac{5}{2}-s}$

Interpolating the above inequality with (11.5) (more precisely (11.5) with
$\mathfrak{R}^{(W)}$ instead of $\mathfrak{R}$), we have that for any
$\theta\in(0,1)$ :

$\left\lVert\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}\mathcal{R}_{N_{1},...,N_{6}}^{(W)}\right\rVert_{\mathbb{L}^{p}(d\mu_{s})}\lesssim
p^{\frac{3}{2}\theta+\frac{2-\alpha}{2}(1-\theta)}N_{(1)}^{-\frac{3}{2}\theta+\left(2(s-1-\sigma)+\varepsilon+\alpha(s-\frac{3}{2})\right)(1-\theta)}N_{(3)}^{(\frac{5}{2}-s)\theta+(1-\theta)}$
(11.9)

Finally, we use the following lemma to finish the proof of Lemma 11.1:

######  Lemma 11.5.

There exist $\sigma<s-\frac{1}{2}$ close enough to $s-\frac{1}{2}$,
$\varepsilon>0$ close enough to $0$, $\theta\in(0,1)$ and $\alpha\in(0,1)$,
such that

$\displaystyle\frac{3}{2}\theta+\frac{2-\alpha}{2}(1-\theta)$
$\displaystyle<1,$ $\displaystyle
N_{(1)}^{-\frac{3}{2}\theta+\left(2(s-1-\sigma)+\varepsilon+\alpha(s-\frac{3}{2})\right)(1-\theta)}N_{(3)}^{(\frac{5}{2}-s)\theta+(1-\theta)}$
$\displaystyle\lesssim N_{(1)}^{-}$

Let us provide a proof of this lemma:

###### Proof of Lemma 11.5.

Firstly, we have:

$\begin{split}\frac{3}{2}\theta+\frac{2-\alpha}{2}(1-\theta)<1&\iff\theta\left(\frac{1+\alpha}{2}\right)<1+\frac{\alpha-2}{2}=\frac{\alpha}{2}\\\
&\iff\theta<\frac{\alpha}{1+\alpha}\end{split}$ (11.10)

Secondly, regarding the exponent of $N_{(3)}$, we have:

$(\frac{5}{2}-s)\theta+(1-\theta)\geq 0\iff\theta\leq\frac{1}{s-\frac{3}{2}}$
(11.11)

Let us take $\theta\leq\frac{1}{s-\frac{3}{2}}$ (always true when
$s\leq\frac{5}{2}$). Then,

$N_{(1)}^{-\frac{3}{2}\theta+\left(2(s-1-\sigma)+\varepsilon+\alpha(s-\frac{3}{2})\right)(1-\theta)}N_{(3)}^{(\frac{5}{2}-s)\theta+(1-\theta)}\lesssim
N_{(1)}^{-\frac{3}{2}\theta+\left(2(s-1-\sigma)+\varepsilon+\alpha(s-\frac{3}{2})\right)(1-\theta)}N_{(1)}^{(\frac{5}{2}-s)\theta+(1-\theta)}$

We want to have appropriate parameters such that the conditions on $\theta$
given in (11.10) and (11.11) are satisfied along with the following one:

$-\frac{3}{2}\theta+\left(2(s-1-\sigma)+\varepsilon+\alpha(s-\frac{3}{2})\right)(1-\theta)+(\frac{5}{2}-s)\theta+(1-\theta)<0$

Since $2(s-1-\sigma)+\varepsilon\longrightarrow-1$ as $\sigma\rightarrow
s-\frac{1}{2}$ and $\varepsilon\rightarrow 0$, the previous condition will be
true for $\sigma$ and $\varepsilon$ respectively close enough to
$s-\frac{3}{2}$ and $0$ if:

$-\frac{3}{2}\theta+\left(-1+\alpha(s-\frac{3}{2})\right)(1-\theta)+(\frac{5}{2}-s)\theta+(1-\theta)<0$

And this condition is equivalent to the following one:

$\begin{split}&\theta\left(1-s-\alpha(s-\frac{3}{2})\right)<-\alpha(s-\frac{3}{2})\\\
\iff&\theta>\frac{\alpha(s-\frac{3}{2})}{s-1+\alpha(s-\frac{3}{2})}=\frac{\alpha}{\frac{s-1}{s-\frac{3}{2}}+\alpha}=\frac{\alpha}{1+\frac{1}{2(s-\frac{3}{2})}+\alpha}\end{split}$
(11.12)

To conclude, if we first take $\alpha$ small enough such that
$\frac{\alpha}{1+\alpha}<\frac{1}{s-\frac{3}{2}}$, and then $\theta$ such that

$\frac{\alpha}{1+\frac{1}{2(s-3/2)}+\alpha}<\theta<\frac{\alpha}{1+\alpha},$

the conditions given in (11.10), (11.11) and (11.12) are satisfied. We
illustrate this with the following drawing :

$0$$\frac{1}{s-3/2}$$\frac{\alpha}{1+\frac{1}{2(s-3/2)}+\alpha}$$\frac{\alpha}{1+\alpha}$$\theta$

With those parameters, Lemma 11.5 is proven. ∎

Hence, coming back to (11.9), the proof of Lemma 11.1 is completed. ∎

##  12\. Estimates for the modified energy derivative at 0

This section is dedicated to $L^{p}(d\mu_{s})$ estimates on $Q_{s,N}$ (defined
in (4.4), see also Definition 2.3). More precisely, we prove Proposition 8.2.
The strategy of our proof is the same as the one of Lemma 11.1. We first
obtain a deterministic estimate that will be conclusive except for a frequency
regime high-high-high-low-…-low (where the three highest frequencies are in
fact much more higher than the others). This will lead us to decompose
$Q_{s,N}$ into two parts, one that captures the high-high-high-low-…-low
regime, and one that captures the other regime of frequencies. We will handle
the high-high-high-low-…-low regime using the independence of Gaussians via
Wiener-chaos estimate. It will be slightly more complicated than the proof of
Lemma 11.1 because of the presence of more indices. However, in our situation,
we will not encounter the possibility of a ”pairing between
generations”141414For example, considering the constraint
$k_{1}-k_{2}+...-k_{6}=0\hskip 2.84544pt\&\hskip
2.84544ptp_{1}-p_{2}+...+p_{5}=k_{1}$, then according to Definition 11.3, a
”pairing between generations” corresponds to the situation when one of the
$k_{j}$ is paired with one of the $p_{l}$. which could have required the
”remarkable cancellation” that has been presented in [29] (sections 5 and 7).
The reason why we do not encounter such a pairing stems from the fact that we
perform the Wiener chaos estimate with respect to three high-frequency
Gaussians, that is with $m=3$ in Lemma 9.9. In doing so, we prevent a pairing
between generations from occurring (see Paragraph 12.1.1). In [29], the Wiener
chaos estimates are performed with respect to two high-frequency Gaussians,
that is with $m=2$ in Lemma 9.9, and in this situation, a pairing between
generation may occur.

##### Preliminaries

Fix $N\in\mathbb{N}\cup\\{\infty\\}$. Recall that from (4.4) we have:

$Q_{s,N}(u)=Q_{s,N}(w)=\textnormal{Im}(-\frac{1}{6}\mathcal{Q}_{0}(w)+\frac{1}{2}\mathcal{Q}_{1}(w)-\frac{1}{2}\mathcal{Q}_{2}(w))$

where $w:=\Pi_{N}u$ (with the convention $\Pi_{\infty}=id$), and:

$\mathcal{Q}_{0}(w):=\sum_{\begin{subarray}{c}k_{1}-k_{2}+...-k_{6}=0\\\
\Omega(\vec{k})=0\end{subarray}}\psi_{2s}(\vec{k})w_{k_{1}}\overline{w_{k_{2}}}...\overline{w_{k_{6}}}$

and,

$\mathcal{Q}_{1}(w):=\sum_{\begin{subarray}{c}k_{1}-k_{2}+...-k_{6}=0\\\
p_{1}-p_{2}+...+p_{5}=k_{1}\\\ \Omega(\vec{k})\neq
0\end{subarray}}\frac{\psi_{2s}(\vec{k})}{\Omega(\vec{k})}w_{p_{1}}\overline{w_{p_{2}}}...w_{p_{5}}\overline{w_{k_{2}}}...\overline{w_{k_{6}}}$

and,

$\mathcal{Q}_{2}(w):=\sum_{\begin{subarray}{c}k_{1}-k_{2}+...-k_{6}=0\\\
q_{1}-q_{2}+...+q_{5}=k_{2}\\\ \Omega(\vec{k})\neq
0\end{subarray}}\frac{\psi_{2s}(\vec{k})}{\Omega(\vec{k})}w_{k_{1}}\overline{w_{q_{1}}}w_{q_{2}}...\overline{w_{q_{5}}}w_{k_{3}}...\overline{w_{k_{6}}}$

It suffices to show the estimates for $\mathcal{Q}_{0},\mathcal{Q}_{1}$ and
$\mathcal{Q}_{2}$ because
$|Q_{s,N}(u)|\leq|\mathcal{Q}_{0}(w)|+|\mathcal{Q}_{1}(w)|+|\mathcal{Q}_{2}(w)|$.

Estimate for $\mathcal{Q}_{0}$ : Actually, the estimate :

$\left\lVert\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}\mathcal{Q}_{0}\right\rVert_{\mathbb{L}^{p}(d\mu_{s})}\leq
C(s,R)p^{\beta}$ (12.1)

has somehow already been proven in the proof of Lemma 11.1. Indeed, the proof
is very similar, and we only sketch the beginning of it.

###### Sketch of the proof of (12.1).

We decompose $\mathcal{Q}_{0}$ dyadically and we use Lemma 9.4 (with
$\Omega(\vec{k})=0$), and we obtain :

$\left|\mathcal{Q}_{0}(w)\right|\lesssim\sum_{N_{1},...,N_{6}}\mathfrak{Q}_{0}(N_{1},...,N_{6})$

where,

$\mathfrak{Q}_{0}(N_{1},...,N_{6}):=\sum_{\begin{subarray}{c}k_{1}-k_{2}+...-k_{6}=0\\\
\Omega(\vec{k})=0\end{subarray}}N_{(1)}^{2s-2}N_{(3)}^{2}\prod_{j=1}^{6}\mathds{1}_{|k_{j}|\sim
N_{j}}|w_{k_{j}}|$

To estimate $\mathfrak{Q}_{0}(N_{1},...,N_{6})$, we use Lemma 9.6 (with
$f^{j}_{k_{j}}=\mathds{1}_{|k_{j}|\sim N_{j}}|w_{k_{j}}|$), which yields:

$\mathfrak{Q}_{0}(N_{1},...,N_{6})\lesssim_{\varepsilon}N_{(1)}^{2s-2+\varepsilon}N_{(3)}^{2}\prod_{j=1}^{6}\left\lVert
P_{N_{j}}w\right\rVert_{\mathbb{L}^{2}(\mathbb{T})}$

with $P_{N}$ the projector onto frequencies $|k|\sim N$. This estimate is the
analogue of (11.3). And from this point, the proof of (12.1) goes exactly the
same as the proof of Lemma 11.1. ∎

To sum up, in order to prove Proposition 8.2, it remains to establish the
estimates in the following lemma :

######  Lemma 12.1.

Let $s>\frac{3}{2}$. Then, there exists $\beta\in(0,1)$ such that for every
$R>0$, there exists a constant $C(s,R)>0$, such that for any
$p\in[2,+\infty)$,

$\left\lVert\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}\mathcal{Q}_{j}\right\rVert_{\mathbb{L}^{p}(d\mu_{s})}\leq
C(s,R)p^{\beta}$ (12.2)

for $j=1,2$.

Since the analysis for $\mathcal{Q}_{1}$ and $\mathcal{Q}_{2}$ is the same, we
only prove the estimate for $\mathcal{Q}_{1}$.

####  12.0.1. Notations and remarks on set of indices:

Before decomposing $\mathcal{Q}$, let us first introduce some notations :

######  Notations 12.2.

Given a set of frequency
$k_{1},k_{2},...,k_{6},p_{1},p_{2},...,p_{5}\in\mathbb{Z}$, we denote :

  * •

$k_{(1)},...,k_{(6)}$ a rearrangement of the $k_{j}$ such that

$|k_{(1)}|\geq|k_{(2)}|\geq...\geq|k_{(6)}|$

  * •

$p_{(1)},...,p_{(6)}$ a rearrangement of the $p_{j}$ such that

$|p_{(1)}|\geq|p_{(2)}|\geq...\geq|p_{(5)}|$

  * •

$n_{(1)},n_{(2)},...,n_{(10)}$ a rearrangement of
$k_{2},...,k_{6},p_{1},...,p_{5}$ such that

$|n_{(1)}|\geq|n_{(2)}|\geq...\geq|n_{(10)}|$

Also, in the sequel we will use :

  * •

the letter $M_{j}$ for the localization of the frequency $k_{j}$,

  * •

the letter $P_{j}$ for the localization of the frequency $p_{j}$,

Finally, $N_{(1)}\geq N_{(2)}\geq...\geq N_{(10)}$ will be a non-increasing
rearrangement of $P_{1},...,P_{5},M_{2},...,M_{6}$.

######  Remark 12.3.

– when $k_{1}-k_{2}+...-k_{6}=0$, we have $|k_{(1)}|\sim|k_{(2)}|$.
– when $k_{1}=p_{1}-p_{2}+...+p_{5}$, we have $|p_{(1)}|\sim|p_{(2)}|$ or
$|p_{(1)}|\sim|k_{1}|$.
– when $k_{1}-k_{2}+...-k_{6}=0$ and $k_{1}=p_{1}-p_{2}+...+p_{5}$, we have
$|n_{(1)}|\sim|n_{(2)}|$.

###  12.1. Decomposition

For convenience, we will denote $\mathcal{Q}_{1}$ simply as $\mathcal{Q}$.

Next, in the same spirit as the decomposition in Section 11, we decompose the
set of indices over which we sum in $\mathcal{Q}$. We invoke the following set
of indices :

$\begin{split}\mathcal{I}_{D}:=\\{(k_{1},...,k_{6},p_{1},...,p_{5})&\in\mathbb{Z}^{11}:\hskip
2.84544pt\sum_{j=1}^{6}(-1)^{j-1}k_{j}=0,\hskip
5.69046ptk_{1}=\sum_{j=1}^{5}(-1)^{j-1}p_{j},\hskip
5.69046pt\sum_{j=1}^{6}(-1)^{j-1}k_{j}^{2}\neq 0,\\\
&(n_{(1)},n_{(2)})\in\\{p_{1},...,p_{5}\\}^{2}\hskip
5.69046pt\textbf{or}\hskip 5.69046pt|n_{(3)}|<|n_{(1)}|^{1-\delta_{0}}\hskip
2.84544pt\textbf{or}\hskip
5.69046pt|n_{(4)}|\geq|n_{(3)}|^{\delta_{0}}\\}\end{split}$ (12.3)

and,

$\begin{split}\mathcal{I}_{W}:=\\{(k_{1},...,k_{6},p_{1},...&,p_{5})\in\mathbb{Z}^{11}:\hskip
2.84544pt\sum_{j=1}^{6}(-1)^{j-1}k_{j}=0,\hskip
5.69046ptk_{1}=\sum_{j=1}^{5}(-1)^{j-1}p_{j},\hskip
5.69046pt\sum_{j=1}^{6}(-1)^{j-1}k_{j}^{2}\neq 0,\\\
&(n_{(1)},n_{(2)})\notin\\{p_{1},...,p_{5}\\}^{2}\hskip
5.69046pt\textbf{and}\hskip
5.69046pt|n_{(3)}|\geq|n_{(1)}|^{1-\delta_{0}}\hskip
2.84544pt\textbf{and}\hskip
5.69046pt|n_{(4)}|<|n_{(3)}|^{\delta_{0}}\\}\end{split}$ (12.4)

for a fixed $\delta_{0}\in(0,1)$151515Once again, $\delta_{0}$ can be any
number in $(0,1)$. Note that what differs from the decomposition in Section 11
is the additional constraint
$(n_{(1)},n_{(2)})\notin\\{p_{1},...,p_{5}\\}^{2}$ in $\mathcal{I}_{W}$. This
is because if $(n_{(1)},n_{(2)})\in\\{p_{1},...,p_{5}\\}^{2}$, we will not
need to use Wiener-chaos.
In the same vein, we split $\mathcal{Q}$ as :

$\mathcal{Q}(w)=\mathcal{Q}^{(D)}(w)+\mathcal{Q}^{(W)}(w)$

where,

$\displaystyle\mathcal{Q}^{(D)}(w)$
$\displaystyle:=\sum_{\mathcal{I}_{D}}\frac{\psi_{2s}(\vec{k})}{\Omega(\vec{k})}w_{p_{1}}\overline{w_{p_{2}}}...w_{p_{5}}\overline{w_{k_{2}}}...\overline{w_{k_{6}}},$
$\displaystyle\mathcal{Q}^{(W)}(w):=\sum_{\mathcal{I}_{W}}\frac{\psi_{2s}(\vec{k})}{\Omega(\vec{k})}w_{p_{1}}\overline{w_{p_{2}}}...w_{p_{5}}\overline{w_{k_{2}}}...\overline{w_{k_{6}}}$
(12.5)

####  12.1.1. Absence of pairing

Recall that we have defined what a ”pairing” is in Definition 11.3.
In $\mathcal{I}_{W}$ (see (12.4) above), there is (for $n_{(1)}$ large enough)
no pairing within the three highest frequencies, relatively to the constraint:

$p_{1}-p_{2}...+p_{5}-k_{2}+k_{3}...-k_{6}=0$

Indeed, suppose there is a pairing between two of the three highest
frequencies. Then, the constraint would take the form:

the remaining high frequency = sum of seven low frequencies

which is impossible because in $\mathcal{I}_{W}$ we have:

$|$high frequencies$|$ $\gg$ $|$low frequencies$|$

###  12.2. Proof of the estimate

We are now prepared to prove Lemma 12.1 (for $j=1$). The forthcoming proof is
organized as follows. Firstly, we establish an estimate that will be
conclusive only for the contribution $\mathcal{Q}^{(D)}$. For the contribution
$\mathcal{Q}^{(W)}$, we will also exploit the independence between high
frequency Gaussians and low frequency Gaussians using Wiener chaos estimate.

###### Proof of Lemma 12.1.

Step 1, Deterministic estimate : Let $w\in H^{s-\frac{1}{2}-}(\mathbb{T})$.
For convenience, for any dyadic number $N$ we denote
$w_{k}^{N}:=\mathds{1}_{|k|\sim N}|w_{k}|$. We start by taking the absolute
value and summing over dyadic blocks

$\left|\mathcal{Q}(w)\right|\lesssim\sum_{M_{1},...,M_{6},P_{1},...,P_{5}}\mathfrak{Q}(M_{1},...,M_{6},P_{1},...,P_{5})$

where $M_{1},...,M_{6},P_{1},...,P_{5}$ are dyadic-valued and

$\mathfrak{Q}(M_{1},...,M_{6},P_{1},...,P_{5}):=\sum_{\begin{subarray}{c}k_{1}-k_{2}+...-k_{6}=0\\\
p_{1}-p_{2}+...+p_{5}=k_{1}\\\ \Omega(\vec{k})\neq
0\end{subarray}}\big{|}\frac{\psi_{2s}(\vec{k})}{\Omega(\vec{k})}\big{|}|w_{p_{1}}^{P_{1}}...w_{p_{5}}^{P_{5}}w_{k_{2}}^{M_{2}}...w_{k_{6}}^{M_{6}}|\cdot\mathds{1}_{|k_{1}|\sim
M_{1}}$

We decompose $\mathfrak{Q}$ as
$\mathfrak{Q}=\mathfrak{Q}^{(D)}+\mathfrak{Q}^{(W)}$, according to the
decomposition of $\mathcal{Q}$.

In regard to (10.4) and (10.6) from the proof of Proposition 4.1, we have:

$\mathfrak{Q}(M_{1},...,P_{5})\lesssim
M_{(1)}^{2s-2+\varepsilon}M_{(3)}^{2}\left(P_{(2)}P_{(3)}P_{(4)}P_{(5)}\right)^{\frac{1}{2}}\prod_{j=1}^{5}\left\lVert
w^{P_{j}}\right\rVert_{\mathbb{L}^{2}}\prod_{j=2}^{6}\left\lVert
w^{M_{j}}\right\rVert_{\mathbb{L}^{2}}$

Using Notations 12.2, it means that

$\mathfrak{Q}(M_{1},...,P_{5})\lesssim
M_{(1)}^{2s-2+\varepsilon}M_{(3)}^{2}\left(P_{(2)}P_{(3)}P_{(4)}P_{(5)}\right)^{\frac{1}{2}}\prod_{j=1}^{10}\left\lVert
w^{N_{j}}\right\rVert_{\mathbb{L}^{2}}$ (12.6)

Moreover, for $1<\sigma^{\prime}<\sigma<s-\frac{1}{2}$, we have :

$\prod_{j=1}^{10}\left\lVert
w^{N_{j}}\right\rVert_{\mathbb{L}^{2}}\lesssim(N_{(1)}^{-\sigma}\left\lVert
u\right\rVert_{H^{\sigma}})(N_{(2)}^{-\sigma^{\prime}}\left\lVert
u\right\rVert_{H^{\sigma^{\prime}}})(N_{(3)}N_{(4)}...N_{(10)})^{-1}\left\lVert
u\right\rVert_{H^{1}}^{8}$ (12.7)

Then, since $N_{(1)}\sim N_{(2)}$, and writing $\sigma^{\prime}$ as:

$\sigma^{\prime}=\alpha 1+(1-\alpha)\sigma$

for $\alpha\in(0,1)$, we deduce from (12.7), by interpolating
$H^{\sigma^{\prime}}$ between $H^{1}$ and $H^{\sigma}$, that:

$\begin{split}\prod_{j=1}^{10}\left\lVert
w^{N_{j}}\right\rVert_{\mathbb{L}^{2}}&\lesssim
N_{(1)}^{-2\sigma+\alpha(\sigma-1)}(N_{(3)}N_{(4)}...N_{(10)})^{-1}\left\lVert
u\right\rVert_{H^{1}}^{8+\alpha}\left\lVert
u\right\rVert^{2-\alpha}_{H^{\sigma}}\\\ &\lesssim
N_{(1)}^{-2\sigma+\alpha(s-\frac{3}{2})}(N_{(3)}N_{(4)}...N_{(10)})^{-1}\left\lVert
u\right\rVert_{H^{1}}^{8+\alpha}\left\lVert
u\right\rVert^{2-\alpha}_{H^{\sigma}},\hskip 8.5359pt(\textnormal{because
$\alpha(\sigma-1)<\alpha(s-3/2)$ })\end{split}$

Plugging this into (12.6), we obtain

$\mathfrak{Q}(M_{1},...,P_{5})\lesssim
N_{(1)}^{-2\sigma+\alpha(s-\frac{3}{2})}M_{(1)}^{2s-2+\varepsilon}M_{(3)}^{2}\left(P_{(2)}P_{(3)}P_{(4)}P_{(5)}\right)^{\frac{1}{2}}(N_{(3)}N_{(4)}...N_{(10)})^{-1}\left\lVert
u\right\rVert_{H^{1}}^{8+\alpha}\left\lVert
u\right\rVert^{2-\alpha}_{H^{\sigma}}$

Using the facts $M_{(1)}\lesssim N_{(1)}$ and $N_{(1)}\sim N_{(2)}$, and also
Remark 7.2, we deduce that

$\begin{split}\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}\mathfrak{Q}&(M_{1},...,P_{5})\\\
&\lesssim_{\varepsilon,R}N_{(1)}^{2(s-\sigma-1)+\varepsilon+\alpha(s-\frac{3}{2})}M_{(3)}^{2}\left(P_{(2)}P_{(3)}P_{(4)}P_{(5)}\right)^{\frac{1}{2}}N^{2}_{(1)}(\overbrace{N_{(1)}N_{(2)}N_{(3)}N_{(4)}...N_{(10)}}^{=M_{2}...M_{6}P_{1}...P_{5}})^{-1}\left\lVert
u\right\rVert^{2-\alpha}_{H^{\sigma}}\\\ &\lesssim
N_{(1)}^{2(s-\sigma)+\varepsilon+\alpha(s-\frac{3}{2})}M_{(3)}^{2}(M_{2}...M_{6})^{-1}P_{(1)}^{-1}\left(P_{(2)}P_{(3)}P_{(4)}P_{(5)}\right)^{-\frac{1}{2}}\left\lVert
u\right\rVert^{2-\alpha}_{H^{\sigma}}\\\
&\lesssim\begin{cases}N_{(1)}^{2(s-\sigma)-\frac{3}{2}+\varepsilon+\alpha(s-\frac{3}{2})}\left\lVert
u\right\rVert^{2-\alpha}_{H^{\sigma}},\hskip 54.06006pt\textnormal{if
$(N_{(1)},N_{(2)})\in\\{P_{1},...,P_{5}\\}^{2}$}\\\
N_{(1)}^{2(s-\sigma-1)+\varepsilon+\alpha(s-\frac{3}{2})}N_{(3)}N_{(4)}^{-\frac{1}{2}}\left\lVert
u\right\rVert^{2-\alpha}_{H^{\sigma}},\hskip 11.38092pt\textnormal{if
$(N_{(1)},N_{(2)})\notin\\{P_{1},...,P_{5}\\}^{2}$}\end{cases}\end{split}$
(12.8)

Here, we used estimates on
$M_{(3)}^{2}(M_{2}...M_{6})^{-1}P_{(1)}^{-1}\left(P_{(2)}P_{(3)}P_{(4)}P_{(5)}\right)^{-\frac{1}{2}}$
in terms of the $N_{j}$, which are gathered in the following lemma.

######  Lemma 12.4.

  * –

If $(N_{(1)},N_{(2)})\in\\{P_{1},...,P_{5}\\}^{2}$, then

$M_{(3)}^{2}(M_{2}...M_{6})^{-1}P_{(1)}^{-1}\left(P_{(2)}P_{(3)}P_{(4)}P_{(5)}\right)^{-\frac{1}{2}}\lesssim
N_{(1)}^{-\frac{3}{2}}$

  * –

If $(N_{(1)},N_{(2)})\notin\\{P_{1},...,P_{5}\\}^{2}$, then

$M_{(3)}^{2}(M_{2}...M_{6})^{-1}P_{(1)}^{-1}\left(P_{(2)}P_{(3)}P_{(4)}P_{(5)}\right)^{-\frac{1}{2}}\lesssim
N_{(1)}^{-2}N_{(3)}N_{(4)}^{-\frac{1}{2}}$

For clarity, we postpone the proof of Lemma 12.4 for the end of this section.
To conclude the first step of the proof, using (9.3) in (12.8) yields:

$\left\lVert\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}\mathfrak{Q}(M_{1},...,P_{5})\right\rVert_{\mathbb{L}^{p}(d\mu_{s})}\lesssim\begin{cases}N_{(1)}^{2(s-\sigma)-\frac{3}{2}+\varepsilon+\alpha(s-\frac{3}{2})}p^{\frac{2-\alpha}{2}},\hskip
54.06006pt\textnormal{if $(N_{(1)},N_{(2)})\in\\{P_{1},...,P_{5}\\}^{2}$}\\\
N_{(1)}^{2(s-\sigma-1)+\varepsilon+\alpha(s-\frac{3}{2})}N_{(3)}N_{(4)}^{-\frac{1}{2}}p^{\frac{2-\alpha}{2}},\hskip
11.38092pt\textnormal{if
$(N_{(1)},N_{(2)})\notin\\{P_{1},...,P_{5}\\}^{2}$}\end{cases}$ (12.9)

where the constant depends on $s,R,\sigma$ and $\varepsilon$. This is the
deterministic estimate we will work with later in the proof. We stress the
fact that (12.9) is true for $\mathfrak{Q}$ and also for $\mathfrak{Q}^{(D)}$
and $\mathfrak{Q}^{(W)}$ (with the exact same proof).

Step 2, Estimates for $\mathcal{Q}^{(D)}$ and $\mathcal{Q}^{(W)}$ :

•Let us begin with the estimate for $\mathcal{Q}^{(D)}$. Recall that in
$\mathcal{Q}^{(D)}$ we sum over the set $\mathcal{I}_{D}$ defined in (12.3).
And, the constraints in $\mathcal{I}_{D}$ imply that the dyadic integers (from
the dyadic decomposition) must satisfy one of the following conditions :

$\displaystyle(N_{(1)},N_{(2)})$ $\displaystyle\in\\{P_{1},...,P_{5}\\}^{2}$
or $\displaystyle N_{(3)}$ $\displaystyle\lesssim N_{(1)}^{1-\delta_{0}}$ or
$\displaystyle N_{(4)}$ $\displaystyle\gtrsim N_{(3)}^{\delta_{0}}$

In the first situation, when $(N_{(1)},N_{(2)})\in\\{P_{1},...,P_{5}\\}^{2}$,
(12.9) (with $\mathfrak{Q}^{(D)}$ instead of $\mathfrak{Q}$) yields:

$\left\lVert\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}\mathfrak{Q}^{(D)}(M_{1},...,P_{5})\right\rVert_{\mathbb{L}^{p}(d\mu_{s})}\lesssim
N_{(1)}^{2(s-\sigma)-\frac{3}{2}+\varepsilon+\alpha(s-\frac{3}{2})}p^{\frac{2-\alpha}{2}}$

And, the exponent of $N_{(1)}$ satisfy :

$\begin{cases}2(s-\sigma)-\frac{3}{2}+\varepsilon+\alpha(s-\frac{3}{2})>-\frac{1}{2}\\\
2(s-\sigma)-\frac{3}{2}+\varepsilon+\alpha(s-\frac{3}{2})\longrightarrow-\frac{1}{2}\hskip
5.69046pt\text{as}\begin{cases}\varepsilon\rightarrow 0\\\ \alpha\rightarrow
0\\\ \sigma\rightarrow s-\frac{1}{2}\end{cases}\end{cases}$

Thus, for $\varepsilon$ and $\alpha$ close enough to $0$ and $\sigma$ close
enough to $s-\frac{1}{2}$, we have

$\left\lVert\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}\mathfrak{Q}^{(D)}(M_{1},...,P_{5})\right\rVert_{\mathbb{L}^{p}(d\mu_{s})}\lesssim
N_{(1)}^{-}p^{\frac{2-\alpha}{2}}$

Now, in the second situation when
$(N_{(1},N_{(2)})\notin\\{P_{1},...,P_{5}\\}^{2}$ and : $N_{(3)}\lesssim
N_{(1)}^{1-\delta_{0}}$ or $N_{(4)}\gtrsim N_{(3)}^{\delta_{0}}$ , (12.9)
yields :

$\left\lVert\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}\mathfrak{Q}^{(D)}(M_{1},...,P_{5})\right\rVert_{\mathbb{L}^{p}(d\mu_{s})}\lesssim
N_{(1)}^{2(s-\sigma-1)+\varepsilon+\alpha(s-\frac{3}{2})}N_{(3)}N_{(4)}^{-\frac{1}{2}}p^{\frac{2-\alpha}{2}}$

And since the exponent of $N_{(1)}$ satisfy :

$\begin{cases}2(s-\sigma-1)+\varepsilon+\alpha(s-\frac{3}{2})>-1\\\
2(s-\sigma-1)+\varepsilon+\alpha(s-\frac{3}{2})\longrightarrow-1\hskip
5.69046pt\text{as}\begin{cases}\varepsilon\rightarrow 0\\\ \alpha\rightarrow
0\\\ \sigma\rightarrow s-\frac{1}{2}\end{cases}\end{cases}$

we deduce that for a fixed $\delta_{0}\in(0,1)$ and for $\varepsilon$ and
$\alpha$ close enough to $0$ and $\sigma$ close enough to $s-\frac{1}{2}$, we
have again

$\left\lVert\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}\mathfrak{Q}^{(D)}(M_{1},...,P_{5})\right\rVert_{\mathbb{L}^{p}(d\mu_{s})}\lesssim
N_{(1)}^{-}p^{\frac{2-\alpha}{2}}$

Finally, summing over the dyadic integers $M_{1},...,M_{6},P_{1},...,P_{5}$,
this estimate for both situations leads to

$\left\lVert\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}\mathcal{Q}^{(D)}\right\rVert_{\mathbb{L}^{p}(d\mu_{s})}\lesssim
p^{\frac{2-\alpha}{2}}$

which is the desired estimate.

•Next, in order to finish the proof of Lemma 12.1, we need the same estimate
for the contribution $\mathcal{Q}^{(W)}$. For this term, while the estimate
(12.9) is not conclusive, we are in a situation where we can make use of
Wiener chaos. Recall that in $\mathcal{Q}^{(W)}$, we sum over the set
$\mathcal{I}_{W}$ defined in (12.4). Even though the following method is very
similar to the one used for estimating $\mathcal{R}^{(W)}$ in Section 11 (see
the proof of Lemma 11.1), we re-perform the analysis in detail.

As usual, we start by decomposing $\mathcal{Q}^{(W)}$ as161616For the analysis
below, we need to keep the complex conjugation bars, so here we don’t use
$\mathfrak{Q}^{(W)}$
$\mathcal{Q}^{(W)}=\sum_{M_{1},...,M_{6},P1,...,P_{5}}\mathcal{Q}^{(W)}_{M_{1},...,M_{6},P1,...,P_{5}}$,
where :

$\mathcal{Q}^{(W)}_{M_{1},...,M_{6},P1,...,P_{5}}:=\sum_{\mathcal{I}_{W}}\frac{\psi_{2s}(\vec{k})}{\Omega(\vec{k})}w_{p_{1}}\overline{w_{p_{2}}}...w_{p_{5}}\overline{w_{k_{2}}}...\overline{w_{k_{6}}}\left(\prod_{j=1}^{6}\mathds{1}_{|k_{j}|\sim
M_{j}}\right)\left(\prod_{j=1}^{5}\mathds{1}_{p_{j}\sim P_{j}}\right)$

Henceforth, we denote $w_{k_{j}}^{M_{j}}:=\mathds{1}_{|k_{j}|\sim
N_{j}}w_{k_{j}}$ and $w_{p_{j}}^{P_{j}}:=\mathds{1}_{|p_{j}|\sim
P_{j}}w_{p_{j}}$ for better readability.

Without loss of generality, we only prove the estimate for the contribution in
$\mathcal{Q}^{(W)}$ where the three highest frequencies are $k_{2},k_{3}$ and
$k_{4}$, because the other cases are identical.
Again we denote $\mathcal{B}_{\ll N_{(3)}}$ the $\sigma$-algebra generated by
Gaussians $\left(g_{k}\right)_{|k|\leq N_{(3)}/100}$. Since we only need to
consider the contribution when $N_{(1)}$ is large, we can assume that
$N_{(1)}$ is sufficiently large to ensure $N_{(4)}\leq N_{(3)}/100$. This
follows from the constraints in $\mathcal{I}_{W}$ which imply $N_{(3)}\gtrsim
N_{(1)}^{1-\delta_{0}}$ and $N_{(4)}\lesssim N_{(3)}^{\delta_{0}}$. Thus, for
$N_{(1)}$ large enough, we have $N_{(3)}^{\delta_{0}}\ll N_{(3)}$, ensuring
that $N_{(4)}\leq N_{(3)}/100$.
As a consequence, we have that the random variables :

$\begin{split}&w_{k_{2}}^{M_{2}},w_{k_{3}}^{M_{3}},w_{k_{4}}^{M_{4}}\hskip
5.69046pt\textnormal{are independent of $\mathcal{B}_{\ll N_{(3)}}$},\\\
&w_{k_{5}}^{M_{5}},w_{k_{6}}^{M_{6}},w_{p_{1}}^{P_{1}},w_{p_{2}}^{P_{2}},w_{p_{3}}^{P_{3}},w_{p_{4}}^{P_{4}},w_{p_{5}}^{P_{5}}\hskip
5.69046pt\textnormal{are $\mathcal{B}_{\ll N_{(3)}}$ mesurable}.\end{split}$

Now, identically to (11.7), we have

$\left\lVert\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}\mathcal{Q}_{M_{1},...,M_{6},P_{1},...,P_{5}}^{(W)}\right\rVert_{\mathbb{L}^{p}(d\mu_{s})}\leq\Big{\|}\big{\|}\mathcal{Q}_{M_{1},...,M_{6},P_{1},...,P_{5}}^{(W)}\big{\|}_{\mathbb{L}^{p}(d\mu_{s}|\mathcal{B}_{\ll
N_{(3)}})}\cdot\mathds{1}_{B_{R}^{H^{1}}}(P_{N_{(3)/100}}u)\Big{\|}_{\mathbb{L}^{\infty}(d\mu_{s})}$
(12.10)

And, the conditional Wiener-chaos estimate from Lemma 9.9 (with $m=3$),
followed by Lemma 9.10 combined with the absence of paring (see Paragraph
12.1.1), allows us to obtain :

$\begin{split}&\left\lVert\mathcal{Q}_{M_{1},...,M_{6},P_{1},...,P_{5}}^{(W)}\right\rVert_{\mathbb{L}^{p}(d\mu_{s}|\mathcal{B}_{\ll
N_{(3)}})}\lesssim
p^{\frac{3}{2}}\left\lVert\mathcal{Q}_{M_{1},...,M_{6},P_{1},...,P_{5}}^{(W)}\right\rVert_{\mathbb{L}^{2}(d\mu_{s}|\mathcal{B}_{\ll
N_{(3)}})}\\\ &\lesssim
p^{\frac{3}{2}}(N_{(1)}N_{(2)}N_{(3)})^{-s}\bigg{(}\sum_{k_{2},k_{3},k_{4}}\Big{|}\sum_{\begin{subarray}{c}k_{1},k_{5},k_{6}\\\
p_{1},p_{2},p_{3},p_{4},p_{5}\end{subarray}}C(\vec{k},\vec{p})\cdot\frac{\psi_{2s}(\vec{k})}{\Omega(\vec{k})}w_{p_{1}}^{P_{1}}\overline{w_{p_{2}}^{P_{2}}}w_{p_{3}}^{P_{3}}\overline{w_{p_{4}}^{P_{4}}}w_{p_{5}}^{P_{5}}w_{k_{5}}^{M_{5}}\overline{w_{k_{6}}^{M_{6}}}\Big{|}^{2}\bigg{)}^{\frac{1}{2}}\end{split}$
(12.11)

where we gathered all the constraints into the term :

$C(\vec{k},\vec{p}):=\mathds{1}_{k_{1}-k_{2}+...-k_{6}=0}\cdot\mathds{1}_{p_{1}-p_{2}+...+p_{5}=k_{1}}\cdot\mathds{1}_{\Omega(\vec{k})\neq
0}\cdot\prod_{j=1}^{6}\mathds{1}_{|k_{j}|\sim
M_{j}}\prod_{j=1}^{5}\mathds{1}_{|p_{j}|\sim P_{j}}$

Next, by Cauchy-Schwarz,

$\begin{split}&\sum_{k_{2},k_{3},k_{4}}\bigg{|}\sum_{\begin{subarray}{c}k_{1},k_{5},k_{6}\\\
p_{1},p_{2},p_{3},p_{4},p_{5}\end{subarray}}C(\vec{k},\vec{p})\cdot\frac{\psi_{2s}(\vec{k})}{\Omega(\vec{k})}w_{p_{1}}^{P_{1}}...w_{p_{5}}^{P_{5}}w_{k_{5}}^{M_{5}}\overline{w_{k_{6}}^{M_{6}}}\bigg{|}^{2}\\\
&\lesssim\sum_{k_{2},k_{3},k_{4}}\bigg{(}\sum_{\begin{subarray}{c}k_{1},k_{5},k_{6}\\\
p_{1},p_{2},p_{2},p_{4},p_{5}\end{subarray}}C(\vec{k},\vec{p})\cdot\Big{|}\frac{\psi_{2s}(\vec{k})}{\Omega(\vec{k})}\Big{|}^{2}|w_{p_{1}}^{P_{1}}...w_{p_{5}}^{P_{5}}w_{k_{5}}^{M_{5}}|^{2}\bigg{)}\bigg{(}\sum_{\begin{subarray}{c}k_{1},k_{5},k_{6}\\\
p_{1},p_{2},p_{2},p_{4},p_{5}\end{subarray}}C(\vec{k},\vec{p})|w_{k_{6}}^{M_{6}}|^{2}\bigg{)}\end{split}$

From Lemma 9.4 we get
$\left|\frac{\psi_{2s}(\vec{k})}{\Omega(\vec{k})}\right|\lesssim|k_{(1)}|^{2s-2}(1+\frac{|k_{(3)}|^{2}}{|\Omega(\vec{k})|})\lesssim|k_{(1)}|^{2s-2}|k_{(3)}|^{2}\lesssim
N_{(1)}^{2s-2}N_{(3)}^{2}$. Then, on the one hand:

$\begin{split}\sum_{\begin{subarray}{c}k_{1},k_{5},k_{6}\\\
p_{1},p_{2},p_{2},p_{4},p_{5}\end{subarray}}C(\vec{k},\vec{p})\cdot&\Big{|}\frac{\psi_{2s}(\vec{k})}{\Omega(\vec{k})}\Big{|}^{2}|w_{p_{1}}^{P_{1}}...w_{p_{5}}^{P_{5}}w_{k_{5}}^{M_{5}}|^{2}\lesssim(N_{(1)}^{2s-2}N_{(3)}^{2})^{2}\sum_{\begin{subarray}{c}k_{1},k_{5},k_{6}\\\
p_{1},p_{2},p_{2},p_{4},p_{5}\end{subarray}}C(\vec{k},\vec{p})\cdot|w_{p_{1}}^{P_{1}}...w_{p_{5}}^{P_{5}}w_{k_{5}}^{M_{5}}|^{2}\\\
&\lesssim(N_{(1)}^{2s-2}N_{(3)}^{2})^{2}\sum_{\begin{subarray}{c}k_{5}\\\
p_{1},p_{2},p_{2},p_{4},p_{5}\end{subarray}}|w_{p_{1}}^{P_{1}}...w_{p_{5}}^{P_{5}}w_{k_{5}}^{M_{5}}|^{2}\cdot\underbrace{\sum_{k_{1},k_{6}}\mathds{1}_{k_{6}=k_{1}-k_{2}...+k_{5}}\cdot\mathds{1}_{p_{1}-p_{2}+...+p_{5}=k_{1}}}_{\leq
1}\\\ &\lesssim(N_{(1)}^{2s-2}N_{(3)}^{2}\left\lVert
w^{P_{1}}\right\rVert_{\mathbb{L}^{2}}...\left\lVert
w^{P_{5}}\right\rVert_{\mathbb{L}^{2}}\left\lVert
w^{M_{5}}\right\rVert_{\mathbb{L}^{2}})^{2}\\\
&\lesssim\left(N_{(1)}^{2s-2}N_{(3)}^{2}(P_{1}...P_{5}M_{5})^{-1}\left\lVert
w^{P_{1}}\right\rVert_{H^{1}}...\left\lVert
w^{P_{5}}\right\rVert_{H^{1}}\left\lVert
w^{M_{5}}\right\rVert_{H^{1}}\right)^{2}\end{split}$

and on the other hand,

$\sum_{\begin{subarray}{c}k_{1},k_{2},k_{3},k_{4},k_{5},k_{6}\\\
p_{1},p_{2},p_{2},p_{4},p_{5}\end{subarray}}C(\vec{k},\vec{p})|w_{k_{6}}^{N_{6}}|^{2}=\sum_{k_{6}}|w_{k_{6}}^{N_{6}}|^{2}\cdot\sum_{\begin{subarray}{c}k_{1},k_{2},k_{3},k_{4},k_{5}\\\
p_{1},p_{2},p_{2},p_{4},p_{5}\end{subarray}}C(\vec{k},\vec{p})\lesssim
N_{(2)}N_{(3)}P_{1}...P_{5}M_{5}\cdot(M_{6}^{-1}\left\lVert
w^{M_{6}}\right\rVert_{H^{1}})^{2}$

where we used the counting bound from Lemma 9.3. We deduce from the two
inequalities above that

$\begin{split}\biggl{(}\sum_{k_{2},k_{3},k_{4}}\big{|}\sum_{\begin{subarray}{c}k_{1},k_{5},k_{6}\\\
p_{1},p_{2},p_{3},p_{4},p_{5}\end{subarray}}&C(\vec{k},\vec{p})\cdot\frac{\psi_{2s}(\vec{k})}{\Omega(\vec{k})}w_{p_{1}}^{P_{1}}\overline{w_{p_{2}}^{P_{2}}}w_{p_{3}}^{P_{3}}\overline{w_{p_{4}}^{P_{4}}}w_{p_{5}}^{P_{5}}w_{k_{5}}^{M_{5}}\overline{w_{k_{6}}^{M_{6}}}\big{|}^{2}\biggr{)}^{\frac{1}{2}}\\\
&\lesssim
N_{(1)}^{2s-2}N_{(3)}^{2}N_{(1)}^{\frac{1}{2}}N_{(3)}^{\frac{1}{2}}\left\lVert
w^{P_{1}}\right\rVert_{H^{1}}...\left\lVert
w^{P_{5}}\right\rVert_{H^{1}}\left\lVert
w^{M_{5}}\right\rVert_{H^{1}}\left\lVert w^{M_{6}}\right\rVert_{H^{1}}\\\
&\lesssim N_{(1)}^{2s-\frac{3}{2}}N_{(3)}^{\frac{5}{2}}\left\lVert
w^{P_{1}}\right\rVert_{H^{1}}...\left\lVert
w^{P_{5}}\right\rVert_{H^{1}}\left\lVert
w^{M_{5}}\right\rVert_{H^{1}}\left\lVert
w^{M_{6}}\right\rVert_{H^{1}}\end{split}$

Coming back to (12.11), we deduce that

$\left\lVert\mathcal{Q}_{M_{1},...,M_{6},P_{1},...,P_{5}}^{(W)}\right\rVert_{\mathbb{L}^{p}(d\mu_{s}|\mathcal{B}_{\ll
N_{(3)}})}\lesssim
p^{\frac{3}{2}}N_{(1)}^{-\frac{3}{2}}N_{(3)}^{\frac{5}{2}-s}\left\lVert
w^{P_{1}}\right\rVert_{H^{1}}...\left\lVert
w^{P_{5}}\right\rVert_{H^{1}}\left\lVert
w^{M_{5}}\right\rVert_{H^{1}}\left\lVert w^{M_{6}}\right\rVert_{H^{1}}$

And plugging this into (12.10), we obtain :

$\left\lVert\mathds{1}_{\\{\mathcal{C}(u)\leq
R\\}}\mathcal{Q}_{M_{1},...,M_{6},P_{1},...,P_{5}}^{(W)}\right\rVert_{\mathbb{L}^{p}(d\mu_{s})}\lesssim_{R}p^{\frac{3}{2}}N_{(1)}^{-\frac{3}{2}}N_{(3)}^{\frac{5}{2}-s}$

Interpolating the above inequality with (12.9) (more precisely (12.9) with
$\mathfrak{Q}^{(W)}$ instead of $\mathfrak{Q}$), we can conclude in the same
way as what we did with $\mathcal{R}^{(W)}$ in the proof of Lemma 11.1 (in
particular using again Lemma 11.5). ∎

In order to complete the proof of Lemma 12.1, we provide in the next paragraph
a proof of the technical Lemma 12.4.

##### Proof of Lemma 12.4

###### Proof of Lemma 12.4.

Recall that we use Notations 12.2.
•Assume first that $(N_{(1)},N_{(2)})\in\\{P_{1},...,P_{5}\\}^{2}$. Then,
$P_{(1)}^{-1}P_{(2)}^{-\frac{1}{2}}\lesssim N_{(1)}^{-\frac{3}{2}}$. Moreover,
we have

$M_{(3)}^{2}(M_{2}...M_{6})^{-1}\leq
M_{(3)}^{2}(M_{(2)}M_{(3)}...M_{(6)})^{-1}\leq(M_{(4)}M_{(5)}M_{(6)})^{-1}\lesssim
1$

Combining these two inequalities, we obtain

$M_{(3)}^{2}(M_{2}...M_{6})^{-1}P_{(1)}^{-1}\left(P_{(2)}P_{(3)}P_{(4)}P_{(5)}\right)^{-\frac{1}{2}}\lesssim
N_{(1)}^{-\frac{3}{2}}(P_{(3)}P_{(4)}P_{(5)})^{-\frac{1}{2}}\lesssim
N_{(1)}^{-\frac{3}{2}}$

which is the desired estimate.

•Now, assume that $(N_{(1)},N_{(2)})\notin\\{P_{1},...,P_{5}\\}^{2}$. It means
that :

$\begin{split}(N_{(1)},N_{(2)})\in\\{M_{2},...,M_{6}\\}^{2}\hskip
5.69046pt&\textbf{or}\left(\hskip
5.69046ptN_{(1)}\in\\{M_{2},...,M_{6}\\}\hskip 5.69046pt\textbf{and}\hskip
5.69046ptN_{(2)}\in\\{P_{1},...,P_{5}\\}\hskip 5.69046pt\right)\hskip
5.69046pt\\\ &\textbf{or}\left(\hskip
5.69046ptN_{(2)}\in\\{M_{2},...,M_{6}\\}\hskip 5.69046pt\textbf{and}\hskip
5.69046ptN_{(1)}\in\\{P_{1},...,P_{5}\\}\hskip 5.69046pt\right)\end{split}$

– Fristly, we suppose that $(N_{(1)},N_{(2)})\in\\{M_{2},...,M_{6}\\}^{2}$.
Then, we note that $N_{(3)}\in\\{M_{2},...,M_{6}\\}$ or $P_{(1)}=N_{(3)}$. And
also,

$N_{(4)}\in\\{M_{2},...,M_{6}\\}\hskip 11.38092pt\textbf{or}\hskip
11.38092ptN_{(4)}\in\\{P_{1},...,P_{5}\\}$

Considering these facts (and still the fact that $N_{(1)}\sim N_{(2)}$), we
deduce that

$M_{(3)}^{2}(M_{2}...M_{6})^{-1}P_{(1)}^{-1}\left(P_{(2)}P_{(3)}P_{(4)}P_{(5)}\right)^{-\frac{1}{2}}\lesssim
M_{(3)}^{2}N_{(1)}^{-2}N_{(3)}^{-1}N_{(4)}^{-\frac{1}{2}}$

Using then the fact that $M_{(3)}\lesssim N_{(3)}$, we have

$M_{(3)}^{2}(M_{2}...M_{6})^{-1}P_{(1)}^{-1}\left(P_{(2)}P_{(3)}P_{(4)}P_{(5)}\right)^{-\frac{1}{2}}\lesssim
N_{(1)}^{-2}N_{(3)}N_{(4)}^{-\frac{1}{2}}$

which is the desired inequality.
–Secondly, we suppose that $N_{(1)}\in\\{M_{2},...,M_{6}\\}\hskip
5.69046pt\textbf{and}\hskip 5.69046ptN_{(2)}\in\\{P_{1},...,P_{5}\\}$. We
consider two cases.
$a)$ The first one is when $M_{1}\in\\{M_{(1)},M_{(2)},M_{(3)}\\}$. In that
case, $M_{(4)},M_{(5)},M_{(6)},P_{(2)},...,P_{(5)}$ are seven indices out of
the ten indices $N_{j}$ (that would not have been the case if
$M_{(1)}\in\\{M_{(4)},M_{(5)},M_{(6)}\\}$, because according to Notations
12.2, $M_{1}$ is not one of the $N_{j}$). As a consequence,

$\max\\{M_{(4)},M_{(5)},M_{(6)},P_{(2)},...,P_{(5)}\\}\geq N_{(4)}\hskip
8.5359pt\textnormal{so,}\hskip
8.5359pt(M_{(4)}M_{(5)}M_{(6)})^{-1}(P_{(2)}...P_{(5)})^{-\frac{1}{2}}\lesssim
N_{(4)}^{-\frac{1}{2}}$

On the other hand, using the fact that $M_{(1)}\sim M_{(2)}$, we have

$M_{(3)}^{2}(M_{2}M_{3}...M_{6})^{-1}\lesssim
M_{(3)}^{2}(M_{(2)}M_{(3)}...M_{(6)})^{-1}\lesssim
M_{(1)}^{-1}M_{(3)}(M_{(4)}M_{(5)}M_{(6)})^{-1}$

Combining these two inequalities and recalling that $M_{(1)},P_{(1)}\sim
N_{(1)}$, we obtain

$M_{(3)}^{2}(M_{2}...M_{6})^{-1}P_{(1)}^{-1}\left(P_{(2)}P_{(3)}P_{(4)}P_{(5)}\right)^{-\frac{1}{2}}\lesssim
N_{(1)}^{-2}M_{(3)}N_{(4)}^{-\frac{1}{2}}\lesssim
N_{(1)}^{-2}N_{(3)}N_{(4)}^{-\frac{1}{2}}$

$b)$ The second case we consider is when
$M_{1}\in\\{M_{(4)},M_{(5)},M_{(6)}\\}$. In that case,
$\\{M_{(1)},M_{(2)},M_{(3)}\\}\subset\\{M_{2},M_{3},...,M_{6}\\}$. Thus,

$(M_{2}...M_{6})^{-1}\lesssim(M_{(1)}M_{(2)}M_{(3)}M_{(5)}M_{(6)})^{-1}$

so,

$M_{(3)}^{2}(M_{2}...M_{6})^{-1}\lesssim N_{(1)}^{-2}M_{(3)}$

and,

$M_{(3)}^{2}(M_{2}...M_{6})^{-1}P_{(1)}^{-1}\left(P_{(2)}P_{(3)}P_{(4)}P_{(5)}\right)^{-\frac{1}{2}}\lesssim
N_{(1)}^{-3}M_{(3)}\lesssim N_{(1)}^{-3}N_{(3)}$

which is an even better estimate than the one we desired.
–Finally, the case $N_{(2)}\in\\{M_{2},...,M_{6}\\}\hskip
5.69046pt\textbf{and}\hskip 5.69046ptN_{(1)}\in\\{P_{1},...,P_{5}\\}$ is
identical to the previous one.
Hence, the proof of Lemma 12.4 is finished.

∎

##  Appendix A Construction and properties of the flow and the truncated flow

In this appendix, we study the local and global wellposedness in
$H^{\sigma}(\mathbb{T})$, $\sigma\geq 1$, for both equation (1.1) and
truncated equation (2.1). We also study approximation properties of the
truncated flow along with its structure.

###  A.1. Local and global wellposedness

We begin this paragraph by recalling some facts about the space :

$W(\mathbb{T}):=\\{u\in\mathcal{D}^{\prime}(\mathbb{T}):\hskip
2.84544pt\underset{k\in\mathbb{Z}}{\sum}\hskip
2.84544pt|\widehat{u}(k)|<+\infty\\}$

of absolutely convergent Fourier series, called the Wiener algebra, and
equipped with the norm

$\left\lVert
u\right\rVert_{W(\mathbb{T})}:=\underset{k\in\mathbb{Z}}{\sum}\hskip
2.84544pt|\widehat{u}(k)|$

  1. (1)

$W(\mathbb{T})$ is a Banach algebra with the estimate

$\left\lVert uv\right\rVert_{W(\mathbb{T})}\leq\left\lVert
u\right\rVert_{W(\mathbb{T})}\left\lVert v\right\rVert_{W(\mathbb{T})}$

  2. (2)

For every $\sigma\geq 0$, $W(\mathbb{T})\cap H^{\sigma}(\mathbb{T})$ is an
algebra. Furthermore, there exists $C_{\sigma}>0$ such that

$\left\lVert uv\right\rVert_{H^{\sigma}}\leq C_{\sigma}(\left\lVert
u\right\rVert_{H^{\sigma}}\left\lVert
v\right\rVert_{W(\mathbb{T})}+\left\lVert
v\right\rVert_{H^{\sigma}}\left\lVert u\right\rVert_{W(\mathbb{T})})$

In particular, if $\sigma>\frac{1}{2}$, $H^{\sigma}(\mathbb{T})$ is a Banach
algebra and there exists $C^{\prime}_{\sigma}>0$ such that

$\left\lVert uv\right\rVert_{H^{\sigma}}\leq C^{\prime}_{\sigma}\left\lVert
u\right\rVert_{H^{\sigma}}\left\lVert v\right\rVert_{H^{\sigma}}$

We are now ready to prove the following local existence theorem :

######  Theorem A.1.

Let $\sigma\geq 1$. Let $t_{0}\in\mathbb{R}$. Both equation (1.1) and
truncated equation (2.1) are locally well-posed in $H^{\sigma}(\mathbb{T})$,
in the sense that for any $R_{0}>0$, there exists $T_{0}>0$ such that for
every $\left\lVert u\right\rVert_{H^{\sigma}}\leq R_{0}$ there exists a unique
$u\in\mathcal{C}([t_{0}-T_{0},t_{0}+T_{0}],H^{\sigma}(\mathbb{T}))$ that
satisfies the Duhamel formula :

$u(t)=e^{i(t-t_{0})\partial_{x}^{2}}u_{0}-i\int_{t_{0}}^{t}e^{i(t-\tau)\partial_{x}^{2}}(|u(\tau)|^{4}u(\tau))d\tau$
(A.1)

for all $|t-t_{0}|\leq T_{0}$. And, for every $N\in\mathbb{N}$, there exists a
unique $v_{N}\in\mathcal{C}([t_{0}-T_{0},t_{0}+T_{0}],H^{\sigma}(\mathbb{T}))$
that satisfies the Duhamel formula :

$v_{N}(t)=e^{i(t-t_{0})\partial_{x}^{2}}u_{0}-i\int_{t_{0}}^{t}e^{i(t-\tau)\partial_{x}^{2}}\Pi_{N}(|\Pi_{N}v_{N}(\tau)|^{4}\Pi_{N}v_{N}(\tau))d\tau$
(A.2)

for all $|t-t_{0}|\leq T_{0}$.

######  Remark A.2.

It is important to notice that the existence time $T_{0}>0$ is the same for
both NLS and truncated equation and does not depend on the integer
$N\in\mathbb{N}$.

###### Proof.

To show Theorem A.1, we apply a fixed point argument. Fix $R_{0}>0$ and
$||u_{0}||_{H^{\sigma}}\leq R_{0}$. For $T>0$, we denote

$X_{T}:=\mathcal{C}([t_{0}-T,t_{0},+T],H^{\sigma}(\mathbb{T}))$

which is a Banach space when endowed with the sup norm

$\left\lVert
u\right\rVert_{X_{T}}:=\underset{\tau\in[t_{0}-T,t_{0},+T]}{\textnormal{sup}}\left\lVert
u(\tau)\right\rVert_{H^{\sigma}}$

For every $N\in\mathbb{N}\cup\\{\infty\\}$, let us consider the well-defined
map

$\begin{split}\Gamma_{N}:\hskip 2.84544pt&X_{T}\longrightarrow X_{T}\\\
&u\longmapsto
e^{i(t-t_{0})\partial_{x}^{2}}u_{0}-i\int_{t_{0}}^{t}e^{i(t-\tau)\partial_{x}^{2}}\Pi_{N}(|\Pi_{N}u(\tau)|^{4}\Pi_{N}u(\tau))d\tau\end{split}$

where by convention $\Pi_{\infty}=id$. We recall the crucial point that
$||\Pi_{N}u||_{H^{\sigma}}\leq\left\lVert u\right\rVert_{H^{\sigma}}$. Thanks
to that point, every estimate below are uniform in $N$ so that every constant
below does not depend on N.

Using now the facts that $H^{\sigma}(\mathbb{T})$ is an algebra (since
$\sigma\geq 1>\frac{1}{2}$) and $e^{i(t-t_{0})\partial_{x}^{2}}$ is a linear
isometry on $H^{\sigma}(\mathbb{T})$, we have for some constant $C_{0}>0$,

$\left\lVert\Gamma_{N}u\right\rVert_{X_{T}}\leq\left\lVert
u_{0}\right\rVert_{H^{\sigma}}+C_{0}T\left\lVert u\right\rVert_{X_{T}}^{5}$
(A.3)

In addition, using the multilinearity of
$(u_{1},u_{2},u_{3},u_{4},u_{5})\longmapsto
u_{1}\overline{u_{2}}u_{3}\overline{u_{4}}u_{5}$ we have for some constant
$C_{1}>0$,

$\begin{split}\left\lVert\Gamma_{N}u-\Gamma_{N}v\right\rVert_{X_{T}}&=\left\lVert\int_{t_{0}}^{t}e^{i(t-\tau)\partial_{x}^{2}}\Pi_{N}(|\Pi_{N}u(\tau)|^{4}\Pi_{N}u(\tau)-|\Pi_{N}v(\tau)|^{4}\Pi_{N}v(\tau))d\tau\right\rVert_{X_{T}}\\\
&\leq C_{1}T(\left\lVert u\right\rVert^{4}_{X_{T}}+\left\lVert
v\right\rVert^{4}_{X_{T}})\left\lVert u-v\right\rVert_{X_{T}}\end{split}$
(A.4)

Let $R>2R_{0}$ (for example $R=1+2R_{0}$) and
$T_{0}:=\frac{1}{3\text{max}(C_{0},C_{1})R^{4}}$. Fix $0<T\leq T_{0}$ and
denote by $\overline{B}_{R}(T)$ the closed centered ball in $X_{T}$ of radius
$R$. We get from (A.3) and (A.4) and our choice of the parameters $R$ and
$T_{0}$ that for any $u,v\in\overline{B}_{R}(T)$,

$||\Gamma_{N}u||_{\overline{B}_{R}(T)}\leq\frac{R}{2}+\frac{R}{2}\leq R$ (A.5)

and,

$||\Gamma_{N}u-\Gamma_{N}v||_{\overline{B}_{R}(T)}\leq 2C_{1}TR^{4}\left\lVert
u-v\right\rVert_{\overline{B}_{R}(T)}\leq\frac{2}{3}\left\lVert
u-v\right\rVert_{\overline{B}_{R}(T)}$ (A.6)

These two estimates imply that the map $\Gamma_{N}$ is a contraction from (the
complete space) $\overline{B}_{R}(T)$ to itself. Hence, applying the Banach’s
fixed point theorem to the map

$\Gamma_{N}:\overline{B}_{R}(T)\longrightarrow\overline{B}_{R}(T)$

leads to the existence part of Theorem A.1. However, through classical
considerations, we can prove that the uniqueness holds in the entire space
$\mathcal{C}([t_{0}-T,t_{0}+T],H^{\sigma}(\mathbb{T}))$. So the proof is
completed. ∎

Putting together the local solutions from Theorem A.1, we obtain the two
following corollaries.

######  Corollary A.3.

Let $\sigma\geq 1$. For every $u_{0}\in H^{\sigma}(\mathbb{T})$, there exists
a unique maximal solution
$u\in\mathcal{C}(I_{max}(u_{0}),H^{\sigma}(\mathbb{T}))$ of (1.1) where
$I_{max}(u_{0})$ is an open interval containing 0. Moreover, if
$I_{max}(u_{0})$ is strictly included in $\mathbb{R}$, then

$\left\lVert u(t)\right\rVert_{H^{\sigma}}\longrightarrow+\infty$

as $t\longrightarrow\partial I_{max}(u_{0})$.

######  Corollary A.4.

Let $\sigma\geq 1$. For every $u_{0}\in H^{\sigma}(\mathbb{T})$ and every
$N\in\mathbb{N}$, there exists a unique maximal solution
$u_{N}\in\mathcal{C}(I_{max,N}(u_{0}),H^{\sigma}(\mathbb{T}))$ of (2.1) where
$I_{max,N}(u_{0})$ is an open interval containing 0. Moreover, if
$I_{max,N}(u_{0})$ is strictly included in $\mathbb{R}$, then

$||u_{N}(t)||_{H^{\sigma}}\longrightarrow+\infty$

as $t\longrightarrow\partial I_{max,N}(u_{0})$.

Thus, to obtain that the solutions of equations (1.1) and (2.1) are global, it
suffices to show that the $H^{\sigma}(\mathbb{T})$-norm of the solutions do
not blow up in finite time. Fortunately, we have :

######  Proposition A.5.

Let $u_{0}\in H^{\sigma}(\mathbb{T})$. There exists a constant $C_{0}>0$ only
depending on $\left\lVert u\right\rVert_{H^{1}(\mathbb{T})}$ and $\sigma\geq
1$, such that

$\left\lVert u(t)\right\rVert_{H^{\sigma}}\leq\left\lVert
u\right\rVert_{H^{\sigma}}e^{C_{0}|t|},\hskip 2.84544pt\hskip
2.84544pt\textnormal{for all}\hskip 2.84544ptt\in I_{max}(u_{0})$ (A.7)

and,

$||u_{N}(t)||_{H^{\sigma}}\leq\left\lVert
u\right\rVert_{H^{\sigma}}e^{C_{0}|t|},\hskip 2.84544pt\hskip
2.84544pt\textnormal{for all}\hskip 2.84544ptt\in I_{max,N}(u_{0})$ (A.8)

where $u$ is the maximal solution of (1.1) and $u_{N}$ is the maximal solution
of (2.1). Moreover, we can explicitly choose $C_{0}\leq
C_{\sigma}(1+\left\lVert u\right\rVert_{H^{1}(\mathbb{T})})^{12}$ where
$C_{\sigma}>0$ depends only on $\sigma$.

###### Proof.

We use the convention $\Pi_{\infty}=id$, $u_{\infty}=u$ and
$I_{max,\infty}(u_{0})=I_{max}(u_{0})$. Let $N\in\mathbb{N}\cup\\{\infty\\}$
and $t\in I_{max,N}(u_{0})$. Passing to the $H^{\sigma}$-norm in the Duhamel
formula

$u_{N}(t)=e^{i(t)\partial_{x}^{2}}u_{0}-i\int_{0}^{t}e^{i(t-\tau)\partial_{x}^{2}}\Pi_{N}(|\Pi_{N}u_{N}(\tau)|^{4}\Pi_{N}u_{N}(\tau))d\tau$

we get that, for some constant $C>0$ (only depending on $\sigma$),

$\begin{split}||u_{N}(t)||_{H^{\sigma}}&\leq\left\lVert
u\right\rVert_{H^{\sigma}}+\mid\int_{0}^{t}\left\lVert|\Pi_{N}u_{N}(\tau)|^{4}\Pi_{N}u_{N}(\tau)\right\rVert_{H^{\sigma}}d\tau\mid\\\
&\leq\left\lVert u\right\rVert_{H^{\sigma}}+C\mid\int_{0}^{t}\left\lVert
u_{N}(\tau)\right\rVert_{H^{\sigma}}\left\lVert
u_{N}(\tau)\right\rVert^{4}_{W(\mathbb{T})}d\tau\mid\end{split}$ (A.9)

The crucial point of the proof is that $\left\lVert
u_{N}(\tau)\right\rVert_{W(\mathbb{T})}$ is bounded by the
$H^{1}(\mathbb{T})$-norm of the initial data. Indeed, using the mass and
Hamiltonian conservation

$\begin{split}\left\lVert
u_{N}(\tau)\right\rVert_{W(\mathbb{T})}&\leq\left\lVert
u_{N}(\tau)\right\rVert_{H^{1}(\mathbb{T})}\\\ &\leq(\left\lVert
u_{N}(\tau)\right\rVert^{2}_{\mathbb{L}^{2}}+\left\lVert\partial_{x}u_{N}(\tau)\right\rVert^{2}_{\mathbb{L}^{2}})^{\frac{1}{2}}\\\
&\lesssim\left(\left\lVert
u_{N}(\tau)\right\rVert^{2}_{\mathbb{L}^{2}}+\left(\frac{1}{2}\left\lVert\partial_{x}u_{N}(\tau)\right\rVert^{2}_{\mathbb{L}^{2}}+\frac{1}{6}\left\lVert
u_{N}(\tau)\right\rVert^{6}_{\mathbb{L}^{6}}\right)\right)^{\frac{1}{2}}\\\
&\lesssim(\left\lVert
u\right\rVert^{2}_{\mathbb{L}^{2}}+\frac{1}{2}\left\lVert\partial_{x}u_{0}\right\rVert^{2}_{\mathbb{L}^{2}}+\frac{1}{6}\left\lVert
u\right\rVert^{6}_{\mathbb{L}^{6}})^{\frac{1}{2}}\\\ &\leq
C^{\prime}_{0}\end{split}$

where we can choose $C^{\prime}_{0}\sim(1+\left\lVert
u\right\rVert_{H^{1}(\mathbb{T})})^{3}$, because $H^{1}(\mathbb{T})$ embeds
continuously in $\mathbb{L}^{6}(\mathbb{T})$. Now, plugging this bound into
(A.9), we get

$||u_{N}(t)||_{H^{\sigma}}\leq\left\lVert
u\right\rVert_{H^{\sigma}}+C_{0}\mid\int_{0}^{t}\left\lVert
u_{N}(\tau)\right\rVert_{H^{\sigma}}d\tau\mid$

where we can choose $C_{0}\sim(1+\left\lVert
u\right\rVert_{H^{1}(\mathbb{T})})^{12}$. Then, applying Gronwall’s
inequality, we obtain

$||u_{N}(t)||_{H^{\sigma}}\leq\left\lVert
u\right\rVert_{H^{\sigma}}e^{C_{0}|t|}$

which is the desired bound. ∎

######  Theorem A.6.

For every $u_{0}\in H^{\sigma}(\mathbb{T})$, each maximal solution of (1.1)
and (2.1) is global. In other words,

$I_{max}(u_{0}),I_{max,N}(u_{0})=+\infty$

for all $N\in\mathbb{N}$.

###### Proof.

The estimates from Proposition A.5 imply that the $H^{\sigma}$-norm of both
solutions does not blow up in finite time. Therefore, Corollaries A.3 and A.4
lead to the result. ∎

###  A.2. Regularity of the flows and approximation properties

We recall that we denote by $\Phi(t)$ the flow of (1.1) and by $\Phi_{N}(t)$
the flow of the truncated equation (2.1). We also use the notation
$\Phi_{\infty}(t)=\Phi(t)$.

######  Proposition A.7.

Let $\sigma\geq 1$. Let $R>0$ and $T>0$. There exists a constant
$\Lambda(R,T)>0$ only depending on $T,R$ and $\sigma$, such that for any
$u_{0}\in B_{R}^{H^{\sigma}}$,

$\sup_{|t|\leq
T}\left\lVert\Phi(t)u_{0}\right\rVert_{H^{\sigma}}+\sup_{|t|\leq
T}\left\lVert\Phi_{N}(t)u_{0}\right\rVert_{H^{\sigma}}\leq\Lambda(R,T),\hskip
2.84544pt\hskip 2.84544pt\forall N\in\mathbb{N}$

###### Proof.

This is a consequence of Proposition A.5. For example we can take

$\Lambda(R,T)=Re^{C_{\sigma}(1+R)^{12}T}$

for a certain constant $C_{\sigma}>0$ only depending on $\sigma$. ∎

We continue this paragraph with two significant regularity results for the
flow of (1.1) and (2.1).

######  Proposition A.8.

Let $t\in\mathbb{R}$. For every $N\in\mathbb{N}\cup\\{\infty\\}$, the map

$\begin{split}\Phi_{N}(t):\hskip
2.84544pt&H^{\sigma}(\mathbb{T})\longrightarrow H^{\sigma}(\mathbb{T})\\\
&u\longmapsto\Phi_{N}(t)u\end{split}$

is continuous.

###### Proof.

For better readability, we only provide the proof for $\Phi(t)$. Let $u,v\in
H^{\sigma}(\mathbb{T})$. Without loss of generality, we can assume that
$\left\lVert u\right\rVert_{H^{\sigma}}\leq R$ and $\left\lVert
v\right\rVert_{H^{\sigma}}\leq R$ for some $R>0$. We invoke $\Lambda(R,t)>0$
from the Proposition A.7. Firstly, using the Duhamel formula and passing to
the $H^{\sigma}(\mathbb{T})$-norm we get that for all $s\leq t$

$\left\lVert\Phi(s)v-\Phi(s)u\right\rVert_{H^{\sigma}}\leq\left\lVert
v-u\right\rVert_{H^{\sigma}}+C\Lambda(R,t)^{4}\mid\int_{0}^{s}\left\lVert\Phi(\tau)v-\Phi(\tau)u\right\rVert_{H^{\sigma}}d\tau\mid$

Secondly, the Gronwall’s inequality yields

$\left\lVert\Phi(t)v-\Phi(t)u\right\rVert_{H^{\sigma}}\leq\left\lVert
v-u\right\rVert_{H^{\sigma}}e^{C\Lambda(R,t)^{4}|t|}$

In particular,

$\left\lVert\Phi(t)v-\Phi(t)u\right\rVert_{H^{\sigma}}\underset{v\rightarrow
u}{\longrightarrow}0$

∎

We can go even further with the following proposition:

######  Proposition A.9.

For every $N\in\mathbb{N}\cup\\{\infty\\}$, the map

$\begin{split}\Phi_{N}:\hskip 2.84544pt\mathbb{R}&\times
H^{\sigma}(\mathbb{T})\longrightarrow H^{\sigma}(\mathbb{T})\\\
&(t,u)\longmapsto\Phi_{N}(t)u\end{split}$

is continuous.

###### Proof.

For better readability, we only provide the proof for $\Phi$. We fix
$t\in\mathbb{R}$ and $u\in H^{\sigma}(\mathbb{T})$ and we show that

$\left\lVert\Phi(t+h)(u+\delta
u)-\Phi(t)u\right\rVert_{H^{\sigma}}\longrightarrow 0,\hskip
14.22636pt\text{as $h\rightarrow 0$ and $\delta u\rightarrow 0$}$

We denote $F(\tau,v):=|\Phi(\tau)v|^{4}\Phi(\tau)v$. Let us compute the
difference $\Phi(t+h)(u+\delta u)-\Phi(t)u$ using the Duhamel formula.

$\begin{split}&\Phi(t+h)(u+\delta u)-\Phi(t)u=\Bigl{(}\Phi(t+h)(u+\delta
u)-\Phi(t+h)u\Bigr{)}+\Bigl{(}\Phi(t+h)u-\Phi(t)u\Bigr{)}\\\
&=e^{i(t+h)\partial_{x}^{2}}\delta
u-i\int_{0}^{t+h}e^{i(t+h-\tau)\partial_{x}^{2}}(F(\tau,u+\delta
u)-F(\tau,u))d\tau\\\
&+e^{it\partial_{x}^{2}}(e^{ih\partial_{x}^{2}}-1)u-i\int_{0}^{t+h}e^{i(t+h-\tau)\partial_{x}^{2}}F(\tau,u)d\tau+i\int_{0}^{t}e^{i(t-\tau)\partial_{x}^{2}}F(\tau,u)d\tau\\\
&=e^{i(t+h)\partial_{x}^{2}}\delta
u+e^{it\partial_{x}^{2}}(e^{ih\partial_{x}^{2}}-1)u-i\int_{0}^{t+h}e^{i(t+h-\tau)\partial_{x}^{2}}(F(\tau,u+\delta
u)-F(\tau,u))d\tau\\\
&-i\int_{t}^{t+h}e^{i(t+h-\tau)\partial_{x}^{2}}F(\tau,u)d\tau-i\int_{0}^{t}e^{i(t-\tau)\partial_{x}^{2}}(e^{ih\partial_{x}^{2}}-1)F(\tau,u)d\tau\end{split}$

Furthermore, without loss of generality we can assume that $t+h\leq t+1=:T$
and $\left\lVert u+\delta u\right\rVert_{H^{\sigma}}\leq\left\lVert
u\right\rVert_{H^{\sigma}}+1=:R$. Now, we invoke $\Lambda(R,T)>0$ from
Proposition A.7. Passing to the $H^{\sigma}(\mathbb{T})$-norm in the above
formula we have

$\begin{split}\left\lVert\Phi(t+h)(u+\delta
u)-\Phi(t)u\right\rVert_{H^{\sigma}}&\leq\left\lVert\delta
u\right\rVert_{H^{\sigma}}+\left\lVert(e^{ih\partial_{x}^{2}}-1)u\right\rVert_{H^{\sigma}}\\\
&+C\Lambda(R,T)^{4}\mid\int_{0}^{T}\left\lVert\Phi(\tau)(u+\delta
u)-\Phi(\tau)u\right\rVert_{H^{\sigma}}d\tau\mid\\\
&+|h|\Lambda(R,T)^{5}+\mid\int_{0}^{t}\left\lVert(e^{ih\partial_{x}^{2}}-1)F(\tau,u)\right\rVert_{H^{\sigma}}d\tau\mid\end{split}$

Using the continuity of $v\longmapsto\Phi(\tau)v$ (Proposition A.8) and the
dominated convergence theorem, we deduce that the right hand side tends to $0$
as $h\rightarrow 0$ and $\delta u\rightarrow 0$. So the proof of Proposition
A.9 is complete. ∎

######  Proposition A.10 (Approximation property).

Let $\sigma\geq 1$. Let $K$ be a compact subset of $H^{\sigma}(\mathbb{T})$
and $T>0$. Then, uniformly in $|t|\leq T$ and $u_{0}\in K$,

$\lim_{N\rightarrow+\infty}\left\lVert\Phi(t)u_{0}-\Phi_{N}(t)u_{0}\right\rVert_{H^{\sigma}}=0,$

In other words,

$\sup_{|t|\leq T}\hskip 2.84544pt\sup_{u_{0}\in K}\hskip
2.84544pt\left\lVert\Phi(t)u_{0}-\Phi_{N}(t)u_{0}\right\rVert_{H^{\sigma}}\underset{N\rightarrow+\infty}{\longrightarrow}0$

###### Proof.

First, from the compactness of $K$, we invoke $R>0$ such that $K\subset
B_{R}^{H^{\sigma}}$, where $B_{R}^{H^{\sigma}}$ is the closed centered ball of
radius $R$ in $H^{\sigma}(\mathbb{T})$. We then invoke $\Lambda(R,T)>0$ from
Proposition A.7. Now, we set $R_{1}:=1+2\Lambda(R,T)$ and invoke
$\delta=\delta(R_{1})>0$ the local existence time associated to $R_{1}$ from
the local theory (Theorem A.1). These parameters ensure that for every
$N\in\mathbb{N}\cup\\{\infty\\}$ and $u_{0}\in B_{R}^{H^{\sigma}}$, the
Duhamel map,

$\begin{split}\Gamma_{N}:\hskip
2.84544pt&\overline{B}_{R_{1}}(\delta)\longrightarrow\overline{B}_{R_{1}}(\delta)\\\
&u\longmapsto
e^{it\partial_{x}^{2}}u_{0}-i\int_{0}^{t}e^{i(t-\tau)\partial_{x}^{2}}\Pi_{N}(|\Pi_{N}u(\tau)|^{4}\Pi_{N}u(\tau))d\tau\end{split}$

is a contraction, with a universal contraction coefficient $0<\gamma<1$ (for
example $\gamma=\frac{2}{3}$ as in the proof of Theorem A.1), where
$\overline{B}_{R_{1}}(\delta)$ is the closed centered ball in
$\mathcal{C}([-\delta,\delta],H^{\sigma}(\mathbb{T}))$ of radius $R_{1}$.

Step 1: Local-time convergence

Firstly, we prove the local property

$\sup_{|t|\leq\delta}\hskip 2.84544pt\sup_{u_{0}\in K}\hskip
2.84544pt\left\lVert\Phi(t)u_{0}-\Phi_{N}(t)u_{0}\right\rVert_{H^{\sigma}}\underset{N\rightarrow+\infty}{\longrightarrow}0$
(A.10)

Let $u_{0}\in B_{R}^{H^{\sigma}}$. We denote $u(t)=\Phi(t)u_{0}$ and
$u_{N}(t)=\Phi_{N}(t)u_{0}$. Since $u$ and $u_{N}$ are respectively the fixed
point of $\Gamma_{\infty}$ and $\Gamma_{N}$, we have

$\begin{split}\left\lVert
u-u_{N}\right\rVert_{\overline{B}_{R_{1}}(\delta)}&=\left\lVert\Gamma_{\infty}u-\Gamma_{N}u_{N}\right\rVert_{\overline{B}_{R_{1}}(\delta)}\\\
&\leq\left\lVert\Gamma_{\infty}u-\Gamma_{N}u\right\rVert_{\overline{B}_{R_{1}}(\delta)}+\left\lVert\Gamma_{N}u-\Gamma_{N}u_{N}\right\rVert_{\overline{B}_{R_{1}}(\delta)}\\\
&\leq\left\lVert\Gamma_{\infty}u-\Gamma_{N}u\right\rVert_{\overline{B}_{R_{1}}(\delta)}+\gamma\left\lVert
u-u_{N}\right\rVert_{\overline{B}_{R_{1}}(\delta)}\end{split}$

Therefore,

$\left\lVert
u-u_{N}\right\rVert_{\overline{B}_{R_{1}}(\delta)}\leq\frac{1}{1-\gamma}\left\lVert\Gamma_{\infty}u-\Gamma_{N}u\right\rVert_{\overline{B}_{R_{1}}(\delta)}$

Hence, to prove (A.10), it suffices to show that

$\sup_{u_{0}\in
K}\left\lVert\Gamma_{\infty}u-\Gamma_{N}u\right\rVert_{\overline{B}_{R_{1}}(\delta)}\underset{N\rightarrow+\infty}{\longrightarrow}0$

Next, for every $|t|\leq\delta$ we have,

$\begin{split}&\Gamma_{\infty}u(t)-\Gamma_{N}u(t)=-i\int_{0}^{t}e^{i(t-\tau)\partial_{x}^{2}}(|u(\tau)|^{4}u(\tau)-\Pi_{N}(|\Pi_{N}u(\tau)|^{4}\Pi_{N}u(\tau)))d\tau\\\
&=-i\int_{0}^{t}e^{i(t-\tau)\partial_{x}^{2}}\Pi_{N}^{\perp}(|u(\tau)|^{4}u(\tau))d\tau-i\int_{0}^{t}e^{i(t-\tau)\partial_{x}^{2}}\Pi_{N}(|u(\tau)|^{4}u(\tau)-|\Pi_{N}u(\tau)|^{4}\Pi_{N}u(\tau))d\tau\end{split}$

So passing to the sup norm we get,

$\begin{split}\left\lVert\Gamma_{\infty}u-\Gamma_{N}u\right\rVert_{\overline{B}_{R_{1}}(\delta)}&\leq\delta\hskip
2.84544pt\sup_{|\tau|\leq\delta}\left\lVert\Pi_{N}^{\perp}(|u(\tau)|^{4}u(\tau))\right\rVert_{H^{\sigma}}\\\
&+C\delta\hskip 2.84544pt\sup_{|\tau|\leq\delta}\left\lVert
u(\tau)-\Pi_{N}u(\tau)\right\rVert_{H^{\sigma}}(\left\lVert
u(\tau)\right\rVert_{H^{\sigma}}^{4}+\left\lVert\Pi_{N}u(\tau)\right\rVert_{H^{\sigma}}^{4})\\\
&\leq
C\delta(\sup_{|\tau|\leq\delta}\left\lVert\Pi_{N}^{\perp}(|u(\tau)|^{4}u(\tau))\right\rVert_{H^{\sigma}}+\sup_{|\tau|\leq\delta}\left\lVert\Pi_{N}^{\perp}u(\tau)\right\rVert_{H^{\sigma}}\Lambda(R,T)^{4})\end{split}$

Taking the supremum over $u_{0}\in K$ we then obtain,

$\begin{split}\sup_{u_{0}\in
K}&\left\lVert\Gamma_{\infty}u-\Gamma_{N}u\right\rVert_{\overline{B}_{R_{1}}(\delta)}\\\
&\leq C\delta(\sup_{u_{0}\in
K}\sup_{|\tau|\leq\delta}\left\lVert\Pi_{N}^{\perp}(|u(\tau)|^{4}u(\tau))\right\rVert_{H^{\sigma}}+\Lambda(R,T)^{4}\sup_{u_{0}\in
K}\sup_{|\tau|\leq\delta}\left\lVert\Pi_{N}^{\perp}u(\tau)\right\rVert_{H^{\sigma}})\end{split}$
(A.11)

Besides, based on a classical result in functional analysis (see Lemma 3.4),
$\Pi_{N}^{\perp}$ satisfies the key property of converging uniformly to $0$ on
compact sets as $N\rightarrow+\infty$. At the same time, the two following
sets

$\displaystyle K_{1}:$ $\displaystyle=\\{\Phi(\tau)u_{0}:\hskip
2.84544ptu_{0}\in K,\hskip 2.84544pt|\tau|\leq\delta\\},$ $\displaystyle
K_{2}:$ $\displaystyle=\\{|\Phi(\tau)u_{0}|^{4}\Phi(\tau)u_{0}:\hskip
2.84544ptu_{0}\in K,\hskip 2.84544pt|\tau|\leq\delta\\}$

are compacts in $H^{\sigma}(\mathbb{T})$. Indeed, it results from the facts
that the map

$\begin{split}&\mathbb{R}\times H^{\sigma}(\mathbb{T})\longrightarrow
H^{\sigma}(\mathbb{T})\\\ &(t,u_{0})\longmapsto\Phi(t)u_{0}\end{split}$

and the map

$\begin{split}&H^{\sigma}(\mathbb{T})^{5}\longrightarrow
H^{\sigma}(\mathbb{T})\\\ (u_{1},u_{2}&,u_{3},u_{4},u_{5})\longmapsto
u_{1}\overline{u_{2}}u_{3}\overline{u_{4}}u_{5}\end{split}$

are continuous and the fact that the image of a compact set under a continuous
map is compact. Now, rewriting (A.11) we get

$\begin{split}\sup_{u_{0}\in
K}\left\lVert\Gamma_{\infty}u-\Gamma_{N}u\right\rVert_{\overline{B}_{R_{1}}(\delta)}&\leq
C\delta(\sup_{w_{2}\in
K_{2}}\Pi_{N}^{\perp}w_{2}+\Lambda(R,T)^{4}\sup_{w_{1}\in
K_{1}}\Pi_{N}^{\perp}w_{1})\\\
&\underset{N\rightarrow\infty}{\longrightarrow}0\end{split}$

And this implies the desired local property (A.10).

Step 2: Long-time convergence

Secondly, we complete the proof of Proposition A.10 by iterating this local
argument. Let $m:=\lfloor\frac{T}{\delta}\rfloor+1$, and for any integer
$|k|\leq m$, let $I_{k}:=[k\delta-\delta,k\delta+\delta]$. We show that for
any $|k|\leq$ m,

$\sup_{t\in I_{k}}\hskip 2.84544pt\sup_{u_{0}\in K}\hskip
2.84544pt\left\lVert\Phi(t)u_{0}-\Phi_{N}(t)u_{0}\right\rVert_{H^{\sigma}}\underset{N\rightarrow+\infty}{\longrightarrow}0$
(A.12)

If we do so, the proof of Proposition A.10 will be completed. At this stage,
we know that (A.12) is true when $k=0$. Now, we assume that (A.12) is true for
some integer $|k|\leq m-1$ and we show that (A.12) is still true for every
integer $|k^{\prime}|=|k|+1$. Also, without loss of generality, we assume that
$k\geq 0$ and $k^{\prime}=k+1$. The crucial point here is that since we chose
$R_{1}=1+2\Lambda(R,T)$, we have that for every
$N\in\mathbb{N}\cup\\{\infty\\}$ and $u_{0}\in B_{R}^{H^{\sigma}}$, the
Duhamel map

$\begin{split}\Gamma_{N}:\hskip
2.84544pt&\overline{B}_{R_{1}}^{(k)}(\delta)\longrightarrow\overline{B}^{k}_{R_{1}}(\delta)\\\
&u\longmapsto
e^{i(t-t_{k})\partial_{x}^{2}}u_{N}(t_{k})-i\int_{t_{k}}^{t}e^{i(t-\tau)\partial_{x}^{2}}\Pi_{N}(|\Pi_{N}u(\tau)|^{4}\Pi_{N}u(\tau))d\tau\end{split}$

is a contraction with a universal contraction coefficient $0<\gamma<1$, where
$t_{k}:=k\delta$, $u_{\infty}(t_{k}):=u(t_{k})$, and
$\overline{B}_{R_{1}}^{(k)}(\delta)$ is the closed centered ball in
$\mathcal{C}([t_{k}-\delta,t_{k}+\delta],H^{\sigma}(\mathbb{T}))$ of radius
$R_{1}$.

Now, doing the same calculations as in the first step of the proof we obtain,

$\left\lVert
u-u_{N}\right\rVert_{\overline{B}^{(k)}_{R_{1}}(\delta)}\leq\frac{1}{1-\gamma}\left\lVert\Gamma_{\infty}u-\Gamma_{N}u\right\rVert_{\overline{B}^{(k)}_{R_{1}}(\delta)}$

and,

$\begin{split}\sup_{u_{0}\in
K}&\left\lVert\Gamma_{\infty}u-\Gamma_{N}u\right\rVert_{\overline{B}^{(k)}_{R_{1}}(\delta)}\leq\sup_{u_{0}\in
K}\sup_{t\in I_{k}}\left\lVert
e^{i(t-t_{k})\partial_{x}^{2}}(u(t_{k})-u_{N}(t_{k}))\right\rVert_{H^{\sigma}}\\\
&+C\delta(\sup_{u_{0}\in K}\sup_{\tau\in
I_{k}}\left\lVert\Pi_{N}^{\perp}(|u(\tau)|^{4}u(\tau))\right\rVert_{H^{\sigma}}+\Lambda(R,T)^{4}\sup_{u_{0}\in
K}\sup_{\tau\in
I_{k}}\left\lVert\Pi_{N}^{\perp}u(\tau)\right\rVert_{H^{\sigma}})\end{split}$

On the right hand side, the first term :

$\sup_{u_{0}\in K}\sup_{t\in I_{k}}\left\lVert
e^{i(t-t_{k})\partial_{x}^{2}}(u(t_{k})-u_{N}(t_{k}))\right\rVert_{H^{\sigma}}=\sup_{u_{0}\in
K}\left\lVert u(t_{k})-u_{N}(t_{k})\right\rVert_{H^{\sigma}}$

tends to $0$ as $N\longrightarrow+\infty$ by our assumption. And, we handle
the remaining term in the same way as in the first part of the proof. Finally
(A.12) is true for all $|k|\leq m$, so the proof of Proposition A.10 is
completed. ∎

######  Corollary A.11.

Let $\sigma\geq 1$, $T>0$ and let $K$ be a compact subset of
$H^{\sigma}(\mathbb{T})$. Then,

  1. (1)

for any $\varepsilon>0$, there exists $N_{0}\in\mathbb{N}$ such that for all
$N\geq N_{0}$

$\Phi_{N}(t)(K)\subset\Phi(t)(K)+B_{\varepsilon}^{H^{\sigma}}$

for all $|t|\leq T$.

  2. (2)

for any $\varepsilon>0$, there exists $N_{1}\in\mathbb{N}$ such that for all
$N\geq N_{1}$

$\Phi(t)(K)\subset\Phi_{N}(t)(K+B_{\varepsilon}^{H^{\sigma}})$

for all $|t|\leq T$.

###### Proof.

Let $\varepsilon>0$.

  * •

For the first point, we take $u_{0}\in K$ and we write

$\Phi_{N}(t)u_{0}=\Phi(t)u_{0}+\left(\Phi_{N}(t)u_{0}-\Phi(t)u_{0}\right)$
(A.13)

From Proposition A.10, there exists $N_{0}\in\mathbb{N}$ such that for all
$N\geq N_{0}$

$\sup_{|t|\leq T}\sup_{v_{0}\in
K}\left\lVert\Phi_{N}(t)v_{0}-\Phi(t)v_{0}\right\rVert_{H^{\sigma}}\leq\varepsilon$

Thus, for all $N\geq N_{0}$, $\Phi_{N}(t)u_{0}-\Phi(t)u_{0}\in
B_{\varepsilon}^{H^{\sigma}}$ for all $|t|\leq T$ and all $u_{0}\in K$. Coming
back to (A.13), it implies that for all $N\geq N_{0}$

$\Phi_{N}(t)u_{0}\in\Phi(t)(K)+B_{\varepsilon}^{H^{\sigma}}$

for all $u_{0}\in K$ and all $|t|\leq T$.

  * •

The second point is a consequence of the first one. Let $|t|\leq T$. From the
continuity of the map $\Phi(t)$ (see Proposition A.8) we have that
$\Phi(t)(K)$ is a compact subset of $H^{\sigma}(\mathbb{T})$. So from the
first point there exists $N_{1}\in\mathbb{N}$ such that for all $N\geq N_{1}$

$\Phi_{N}(-\tau)\left(\Phi(t)(K)\right)\subset\Phi(-\tau)\Phi(t)(K)+B_{\varepsilon}^{H^{\sigma}}$

for all $|\tau|\leq T$. In particular, for $\tau=t$ we have :

$\Phi_{N}(-t)\left(\Phi(t)(K)\right)\subset\Phi(-t)\Phi(t)(K)+B_{\varepsilon}^{H^{\sigma}}=K+B_{\varepsilon}^{H^{\sigma}}$

Applying $\Phi_{N}(t)$ to this we obtain

$\Phi(t)(K)\subset\Phi_{N}(t)\left(K+B_{\varepsilon}^{H^{\sigma}}\right)$

This concludes the proof of Corollary A.11. ∎

###  A.3. Structure of the truncated flow

We set

$\displaystyle E_{N}$ $\displaystyle:=\Pi_{N}\mathbb{L}^{2}(\mathbb{T})$
$\displaystyle E_{N}^{\perp}$
$\displaystyle:=\Pi_{N}^{\perp}\mathbb{L}^{2}(\mathbb{T})=(Id-\Pi_{N})\mathbb{L}^{2}(\mathbb{T})$

######  Proposition A.12.

  1. We have the following properties ;

  2. (1)

The truncated flow $\Phi_{N}(t)$ commute with the frequency projector
$\Pi_{N}$, that is

$\Phi_{N}(t)\circ\Pi_{N}=\Pi_{N}\circ\Phi_{N}(t)$

As a consequence, the truncated flow $\Phi_{N}(t)$ maps the finite-dimensional
space $E_{N}$ to itself. Moreover, the induced map

$\begin{split}\widetilde{\Phi}_{N}(t):\hskip 2.84544pt&E_{N}\longrightarrow
E_{N}\\\ &u_{0}\longmapsto\Phi_{N}(t)(u_{0})\end{split}$

is the flow of the ODE

$\begin{cases}i\partial_{t}u+\partial_{x}^{2}u=\Pi_{N}\left(|u|^{4}u\right)\\\
u|_{t=0}=u_{0}\in E_{N}\end{cases}$ (FNLS)

which can be seen as the finite-dimensional Hamiltonian equation on $E_{N}$ :

$\begin{cases}i\partial_{t}u=\frac{\partial H_{N}}{\partial\overline{u}}(u)\\\
u|_{t=0}=u_{0}\in E_{N}\end{cases}$

with Hamiltonian
$H_{N}(u):=\frac{1}{2}\left\lVert\partial_{x}^{2}u\right\rVert^{2}_{\mathbb{L}^{2}(\mathbb{T})}+\frac{1}{6}\left\lVert
u\right\rVert_{\mathbb{L}^{6}(\mathbb{T})}^{6}$, for $u\in E_{N}$.

  3. (2)

The truncated flow $\Phi_{N}(t)$ commute with $\Pi_{N}^{\perp}$, that is

$\Phi_{N}(t)\circ\Pi_{N}^{\perp}=\Pi_{N}^{\perp}\circ\Phi_{N}(t)$

As a consequence, the truncated flow $\Phi_{N}(t)$ maps the space
$E_{N}^{\perp}$ to itself. Moreover, the induced map

$\begin{split}\Phi^{\perp}_{N}(t):\hskip
2.84544pt&E_{N}^{\perp}\longrightarrow E_{N}^{\perp}\\\
&u_{0}\longmapsto\Phi_{N}(t)(u_{0})\end{split}$

is the solution of the linear Schrödinger equation

$\begin{cases}i\partial_{t}u+\partial_{x}^{2}u=0\\\ u|_{t=0}=u_{0}\in
E_{N}^{\perp}\end{cases}$

Hence, $\Phi^{\perp}_{N}(t)$ coincide with the linear operator
$e^{it\partial_{x}^{2}}$ on $E_{N}^{\perp}$.

  4. (3)

The truncated flow $\Phi_{N}(t)$ can be factorized as
$(\widetilde{\Phi}_{N}(t),e^{it\partial_{x}^{2}})$ on $E_{N}\times
E_{N}^{\perp}$. In other words, for every $u_{0}\in H^{\sigma}(\mathbb{T})$,

$\Phi_{N}(t)(u_{0})=\widetilde{\Phi}_{N}(t)\Pi_{N}u_{0}+e^{it\partial_{x}^{2}}\Pi_{N}^{\perp}u_{0}$

Let us now prove these three points.

###### Proof.

Let $u_{0}\in L^{2}(\mathbb{T})$.

  1. (1)

• $\Phi_{N}(t)$ is the flow of (2.1). And, when we apply $\Pi_{N}$ to equation
(2.1), we see that $\Pi_{N}\Phi_{N}(t)u_{0}$ is the solution of the equation

$\begin{cases}i\partial_{t}u+\partial_{x}^{2}u=\Pi_{N}\left(|\Pi_{N}u|^{4}\Pi_{N}u\right)=\Pi_{N}\left(|u|^{4}u\right)\\\
u|_{t=0}=\Pi_{N}u_{0}\end{cases}$

On the other hand, from the definition of the flow of (2.1), the solution of
the equation above is none other than $\Phi_{N}(t)\Pi_{N}u_{0}$ itself. This
means that

$\Pi_{N}\Phi_{N}(t)u_{0}=\Phi_{N}(t)\Pi_{N}u_{0}$

• Next, we show that (FNLS) can be seen as a finite-dimensional Hamiltonian
equation on $E_{N}$. Every element $u\in E_{N}$ can be decompose as

$u=v+iw$

where

$\begin{split}v&=\sum_{|k|\leq N}\textnormal{Re}(u_{k})e^{ikx}=:\sum_{|k|\leq
N}v_{k}e^{ikx}\in E_{N}\\\ w&=\sum_{|k|\leq
N}\textnormal{Im}(u_{k})e^{ikx}=:\sum_{|k|\leq N}w_{k}e^{ikx}\in
E_{N}\end{split}$

Furthermore, we invoke the operators

$\begin{split}&\frac{\partial}{\partial\overline{u}_{k}}:=\frac{\partial}{\partial
v_{k}}+i\frac{\partial}{\partial w_{k}}\\\
&\frac{\partial}{\partial\overline{u}}:=\frac{\partial}{\partial
v}+i\frac{\partial}{\partial w}=:\sum_{|k|\leq
N}e^{ikx}\frac{\partial}{\partial v_{k}}+i\sum_{|k|\leq
N}e^{ikx}\frac{\partial}{\partial w_{k}}\end{split}$

and also the function

$\begin{split}H_{N}:\hskip 2.84544pt&E_{N}\longrightarrow\mathbb{R}\\\
&u\longmapsto\frac{1}{2}\left\lVert\partial_{x}^{2}u\right\rVert^{2}_{\mathbb{L}^{2}(\mathbb{T})}+\frac{1}{6}\left\lVert
u\right\rVert_{\mathbb{L}^{6}(\mathbb{T})}^{6}\end{split}$

By performing elementary computations, we see that we can rewrite (FNLS) as

$\begin{cases}i\partial_{t}u=\frac{\partial H_{N}}{\partial\overline{u}}(u)\\\
u|_{t=0}=u_{0}\end{cases}$ (FNLS)

This means that (FNLS) is an Hamiltonian equation on $E_{N}$ with associated
Hamiltonian $H_{N}$.

  2. (2)

Once again, applying $\Pi_{N}^{\perp}$ to equation (2.1), we see that
$\Pi_{N}^{\perp}\Phi_{N}(t)u_{0}$ is the solution of the equation

$\begin{cases}i\partial_{t}u+\partial_{x}^{2}u=0\\\
u|_{t=0}=\Pi_{N}^{\perp}u_{0}\end{cases}$

This means that

$\Pi_{N}^{\perp}\Phi_{N}(t)u_{0}=e^{it\partial_{x}^{2}}\Pi_{N}^{\perp}u_{0}$
(A.14)

Furthermore, $\Pi_{N}\Phi_{N}(t)\Pi_{N}^{\perp}u_{0}$ is the solution of the
equation

$\begin{cases}i\partial_{t}u+\partial_{x}^{2}u=\Pi_{N}\left(|\Pi_{N}u|^{4}\Pi_{N}u\right)\\\
u|_{t=0}=0\end{cases}$

Thus, $\Pi_{N}\Phi_{N}(t)\Pi_{N}^{\perp}u_{0}$ is none other than $0$, and we
obtain from (A.14) that

$\Phi_{N}(t)\Pi_{N}^{\perp}u_{0}=\Pi_{N}^{\perp}\Phi_{N}(t)\Pi_{N}^{\perp}u_{0}=e^{it\partial_{x}^{2}}\Pi_{N}^{\perp}u_{0}=\Pi_{N}^{\perp}\Phi_{N}(t)u_{0}$

  3. (3)

As a consequence of the two previous points, we can write

$\begin{split}\Phi_{N}(t)u_{0}&=\Pi_{N}\Phi_{N}(t)u_{0}+\Pi_{N}^{\perp}\Phi_{N}(t)u_{0}\\\
&=\Phi_{N}(t)\Pi_{N}u_{0}+\Phi_{N}(t)\Pi_{N}^{\perp}u_{0}\\\
&=\underbrace{\widetilde{\Phi}_{N}(t)\Pi_{N}u_{0}}_{\in
E_{N}}+\underbrace{e^{it\partial_{x}^{2}}\Pi_{N}^{\perp}u_{0}}_{\in
E_{N}^{\perp}}\end{split}$

∎

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