Notation 2.3. Given a set of frequencies $k_1,...,k_m \in \mathbb{Z}$, we denote by $k_{(1)},...,k_{(m)}$ a rearrangement of the $k_j's$ such that

\begin{align*} |k_{(1)}| \geq |k_{(2)}| \geq ... \geq |k_{(m)}|. \end{align*}

Proof. For $\psi : Y \to [0,\infty]$ a non-negative measurable function, we have by definition of a push-forward measure:

\begin{align*} \int_Y \psi(v) d\Upsilon_\# (f d\mu)(v) = \int_X \psi(\Upsilon(u)) f(u) d\mu(u) = \int_Y \psi(v) (f \circ \Upsilon^{-1} )(v) d\Upsilon_\# d\mu(v), \end{align*}

which indeed implies that $\Upsilon_\# (f d\mu) =(f \circ \Upsilon^{-1}) \Upsilon_\# \mu $.

Proof. Since $\Phi^N_t$ is bijective, with inverse $\Phi^N_{-t}$, Lemma 3.4 gives:

\begin{align*} (\Phi^N_t)_\# \mu_{s,R,N} = (\Phi^N_t)_\# \big( \chi_R(\mathcal E_N(u)) d\mu_s \big) = \chi_R(\mathcal E_N(\Phi^N_{-t}u))(\Phi^N_t)_\# \mu_s. \end{align*}

By (1.10) and the conservation $\mathcal E_N(\Phi^N_{-t}u) = \mathcal E_N(u)$, we deduce that:

\begin{align*} (\Phi^N_t)_\# \mu_{s,R,N} = \chi_R(\mathcal E_N(u))g_{s,N,t} d\mu_s = g_{s,N,t} d\mu_{s,R,N}. \end{align*}

as desired.

Proof. Let us consider the universal constants $C_0,c_1 \gt 0$ from Proposition 3.1. Introducing $\delta = c_1 (\frac{M}{C_0})^{1-p}$, we know that for every initial data $v_0 \in H^{\sigma}$,

\begin{align*} \| v_0 \|_{H^{\sigma}} \leq \frac{M}{C_0} \implies \sup_{-\delta \leq t \leq \delta} \| \Phi^N_t v_0 \|_{H^{\sigma}} \leq M. \end{align*}

Thus, using the additivity of the flow, and denoting $m:= \lfloor \frac{T}{\delta}\rfloor$, we have:

\begin{align*} \bigcap_{k=-m}^{m} \big\{u_0 : \| \Phi^N_{k \delta} u_0\|_{H^{\sigma}} \leq \frac{M}{C_0} \big\} \subset \big\{u_0 : \sup_{|t| \leq T} \| \Phi^N_t \|_{H^{\sigma}} \leq M \big\}. \end{align*}

As a consequence,

\begin{align*} &\mu_{s,R,N}\big(u_0 : \sup_{|t| \leq T} \| \Phi_t^Nu_0 \|_{H^{\sigma}} \gt M \big) \leq \sum_{k=-m}^m \mu_{s,R,N}\big( u_0 : \| \Phi^N_{k \delta} u_0\|_{H^{\sigma}} \gt \frac{M}{C_0} \big) \\ & = \sum_{k=-m}^m (\Phi^N_{k \delta})_\#\mu_{s,R,N}\big( \| u_0\|_{H^{\sigma}} \gt \frac{M}{C_0} \big) \leq \sum_{k=-m}^m \| g_{s,N,k\delta} \|_{L^2(d\mu_{s,R,N})} \mu_{s,R,N}\big(\| u_0\|_{H^{\sigma}} \gt \frac{M}{C_0} \big)^{\frac{1}{2}}. \end{align*}

Now, on the one hand, $\| g_{s,N,k\delta} \|_{L^2(d\mu_{s,R,N})} \leq e^{C_{s,R} (1+T)^A}$ by (3.3); and on the other hand, we have the standard Gaussian tail estimate (see for example Fernique’s theorem in [Reference Bogachev4] or [Reference Kuo24]):

\begin{align*} \mu_{s,R,N}\big(\| u_0\|_{H^{\sigma}} \gt \frac{M}{C_0} \big) \leq \mu_s\big(\| u_0\|_{H^{\sigma}} \gt \frac{M}{C_0} \big) \lesssim e^{-\alpha M^2}, \end{align*}

for some constant α > 0. Using this in the estimate above yields:

\begin{align*} \mu_{s,R,N}\big(u_0 : \sup_{|t| \leq T} \| \Phi_t^Nu_0 \|_{H^{\sigma}} \gt M \big) &\lesssim (\frac{T}{\delta} + 1) e^{C_{s,R} (1+T)^A} e^{-\alpha M^2} \\ &\lesssim (1+T)e^{C_{s,R} (1+T)^A} (1+M)^{p-1}e^{-\alpha M^2}. \end{align*}

Hence, taking $C_{s,R} \gt 0$ slightly larger and α > 0 slightly smaller leads to (3.4).

Proof. Without loss of generality, we work on the interval $[0,T]$. We introduce an increasing sequence $\{R_l\}_{l \in \mathbb{N}}$ of positive numbers tending to infinity. Our strategy consists in constructing first, for every R_l, and all integers M such that $K \subset B_M^{H^{\sigma}}$, Borel subsets $\Sigma_{K,T,M,R_l}$ of K, such that:

\begin{align*} \lim_{M \to \infty} \mu_{s,R_l}(K \setminus \Sigma_{K,T,M,R_l}) &= 0 & &\text{and,} & \forall u_0 \in \Sigma_{K,T,M,R_l}&: I_{max}(u_0) \supset [0,T]. \end{align*}

Indeed, if we do so, then the following Borel subset of K:

\begin{align*} \Sigma_{K,T} := \bigcup_{l,M} \Sigma_{K,T,M,R_l}, \end{align*}

would satisfy:

\begin{align*} &\mu_s(K \setminus \Sigma_{K,T}) = \mu_s\big( \bigcap_{l,M} K \setminus \Sigma_{K,T,M,R_l} \big) = \lim_{l' \to \infty} \mu_{s,R_{l'}} \big( \bigcap_{l,M} K \setminus \Sigma_{K,T,M,R_l} \big) \\ &\leq \liminf_{l' \to \infty} \mu_{s,R_{l'}} \big( \bigcap_{M} K \setminus \Sigma_{K,T,M,R_{l'}} \big) \leq \liminf_{l' \to \infty} \big[ \liminf_{M \to \infty } \mu_{s,R_{l'}} (K \setminus \Sigma_{K,T,M,R_{l'}}) \big] =0. \end{align*}

Therefore, this would complete the proof.

In what follows, we construct such sets $\Sigma_{K,T,M,R_l}$. Let then $l\in \mathbb{N}$, and let M be an integer such that $K \subset B^{H^{\sigma}}_M$. We consider $\delta \sim (1+M)^{p-1}$ the local existence time from Proposition 3.1 associated to initial data of size smaller than M + 1. We denote $m := \lfloor \frac{T}{\delta} \rfloor$. Furthermore, thanks to Proposition 5.2, we introduce an integer N ₀ such that:

(3.5)\begin{align} N \geq N_0 \implies \forall A \in \mathcal B(H^{\sigma}), |\mu_{s,R_l}(A) - \mu_{s,R_l,N}(A)| \leq \frac{1}{mM} \end{align}

Thanks to the approximation by the truncated flow (see Proposition 3.2), we know that there exists $N_1 \geq N_0$ such that:

\begin{align*} N \geq N_1 \implies \sup_{u_0 \in K} \sup_{0 \leq t \leq \delta} \| \Phi_t u_0 - \Phi^N_t u_0 \|_{H^{\sigma}} \leq 1, \end{align*}

Hence, introducing the set:

\begin{align*} K_1 := K \cap \big\{ \sup_{0 \leq t \leq \delta} \| \Phi^{N_1}_t u_0 \|_{H^{\sigma}} \leq M \big\}, \end{align*}

we deduce that for every $u_0 \in K_1$, $\| \Phi_\delta u_0 \|_{H^{\sigma}} \leq M + 1$, which ensures by Proposition 3.1 that the solution generated by u ₀ lives on $[0,2\delta]$, that is $I_{max}(u_0) \supset [0,2\delta]$. Next, since K ₁ is compact (it is the intersection of the compact K with a closed set), we can apply Proposition 3.2 as above (but now with K ₁ and $[0,2 \delta]$), and find $N_2\geq N_0$ such that every initial data in:

\begin{align*} K_2 :=K \cap \big\{ \sup_{0 \leq t \leq \delta} \| \Phi^{N_1}_t u_0 \|_{H^{\sigma}} \leq M \big\} \cap \big\{ \sup_{0 \leq t \leq 2\delta} \| \Phi^{N_2}_t u_0 \|_{H^{\sigma}} \leq M \big\}, \end{align*}

generates a solution that lives on $[0,3\delta]$. Thus, proceeding by induction (recalling that $m = \lfloor \frac{T}{\delta} \rfloor$), we are able to construct $N_m,...,N_1$, all greater that N ₀, such that every initial data in the set:

\begin{align*} \Sigma_{K,T,M,R_l} := \bigcap_{k=1}^{m} K \cap \big\{\sup_{0 \leq t \leq k\delta} \| \Phi^{N_k}_t u_0 \|_{H^{\sigma}} \leq M \big\}, \end{align*}

yields a solution on $[0,T]$. Note that this set depends on R_l since we chose every N_k greater than N ₀, and that N ₀ itself depends on R_l (see (3.5)).

Next, we estimate the $\mu_{s,R_l}$-measure of this set. Since every N_k is greater than N ₀, we can write (thanks to (3.5)):

\begin{align*} \mu_{s,R_l}(K \setminus \Sigma_{K,T,M,R_l}) &\leq \sum_{k=1}^m \mu_{s,R_l}\big(\sup_{0 \leq t \leq k\delta} \| \Phi^{N_k}_t u_0 \|_{H^{\sigma}} \gt M\big) \\ & \leq \frac{1}{M} + \sum_{k=1}^m \mu_{s,R_l,N_k}\big(\sup_{0 \leq t \leq k\delta} \| \Phi^{N_k}_t u_0 \|_{H^{\sigma}} \gt M\big). \end{align*}

Applying now (3.4), leads to:

\begin{align*} \mu_{s,R_l}(K \setminus \Sigma_{K,T,M,R_l}) \leq \frac{1}{M} + C_{s,R_l,T} M^{p-1} e^{-\alpha M^2}, \end{align*}

which indeed tends to 0 as $M \to \infty$. This completes the proof.

Proof of Theorem 1.1

First, we recall that $\sigma \lt s - \frac{1}{2}$ is fixed and close to $s-\frac{1}{2}$. Then, we consider $\sigma' \in (\sigma ,s-\frac{1}{2})$. For every $n \in \mathbb{N}$, we introduce $B^{H^{\sigma'}}_n$, the closed centered ball of radius n in $H^{\sigma'}(\mathbb{T})$, which is compact in $H^{\sigma}(\mathbb{T})$. Moreover, we introduce an increasing sequence $\{T_k\}_{k \in \mathbb{N}}$ of positive time tending to infinity. Applying Lemma 3.8, we consider for every $n,k \in \mathbb{N}$:

\begin{align*} \Sigma_{n,k} &:= \Sigma_{B^{H^{\sigma'}}_n, T_k} & &\text{and then we introduce:} & \Sigma &:= \bigcap_{k \in \mathbb{N}} \bigcup_{n \in \mathbb{N}} \Sigma_{n,k}. \end{align*}

Let us now observe that if $u_0 \in \Sigma$, then for every $k\in\mathbb{N}$, $I_{max}(u_0) \subset [-T_k, T_k]$. It means that u ₀ generates a global solution of (pNLS). In other words, Σ is a subset of the global well-posedness set $\mathcal G$ (defined in (1.5)). Next, we prove that Σ is of full µ_s-measure. It is sufficient to prove that for every k, $\cup_n \Sigma_{n,k}$ is of full µ_s-measure. Since we know that $H^{\sigma'}(\mathbb{T})$ is of full µ_s-measure, we can write:

\begin{align*} \mu_s \big( H^{\sigma} \setminus \bigcup_{n \in \mathbb{N}} \Sigma_{n,k}\big) &= \mu_s \big( H^{\sigma'} \setminus \bigcup_{n \in \mathbb{N}} \Sigma_{n,k}\big) = \lim_{n' \to \infty} \mu_s \big( B^{H^{\sigma'}}_{n'} \setminus \bigcup_{n \in \mathbb{N}} \Sigma_{n,k} \big) \\ & \leq \liminf_{n' \to \infty} \mu_s( B^{H^{\sigma'}}_{n'} \setminus \Sigma_{n',k} ) = 0. \end{align*}

As a consequence, $\mu_s(\Sigma^c) = 0$, and a fortiori $\mu_s(\mathcal G^c) = 0$. This concludes the proof.

Proof of Theorem 1.2

Let $q \in (1,\infty)$ and $q' \in (1,\infty)$ its conjugate exponent. We want to apply Lemma 4.1 with E the Banach space $L^{q'}(d\mu_s)$, whose dual is identified to $L^q(d\mu_s)$, and D the linear subspace of bounded continuous function on $H^{\sigma}(\mathbb{T})$ (which is indeed dense in $L^{q'}(d\mu_s)$).

Let R > 0. Let $\psi : H^{\sigma} \to \mathbb{R}$ a bounded continuous function. Then, for $N \in \mathbb{N}$, we have from the definition of the Radon-Nikodym derivative (1.10) and the invariance of $\mathcal E_N$ by $\Phi^N_t$,

(4.1)\begin{align} \int \psi(u) \chi_R(\mathcal E_N(u)) g_{s,N,t} d\mu_s = \int \psi(\Phi^N_t u) \chi_R(\mathcal E_N(u)) d\mu_s. \end{align}

Thanks to both (3.1) and (5.1), which in particular ensure that $\psi(\Phi^N_t u)$ and $\chi_R(\mathcal E_N(u))$ converge in $L^2(d\mu_s)$ to $\psi(\Phi_t(u))$ and $\chi_R(\mathcal E(u))$ respectively, passing to the limit $N \to \infty$ in the equality above leads to:

(4.2)\begin{align} \lim_{N \to \infty} \int \psi(u) \chi_R(\mathcal E_N(u)) g_{s,N,t} d\mu_s = \int \psi(\Phi_t u) \chi_R(\mathcal E(u)) d\mu_s = \int \psi(u) d(\Phi_t)_\# d\mu_{s,R}. \end{align}

Moreover, thanks to (3.3), we know that that the sequence $\chi_R(\mathcal E_N(u)) g_{s,N,t}$ is uniformly bounded in $L^q(d\mu_s)$. Hence, with (4.1) and (4.2), we can appeal to Lemma 4.1, and use the standard isomorphism $L^q(d\mu_s) \simeq (L^{q'}(d\mu_s))'$, to conclude that there exists $g_{s,t,R} \in L^q(d\mu_s)$ such that:

(4.3)\begin{align} (\Phi_t)_\#\mu_{s,R} &= g_{s,t,R} d\mu_s & &\text{with:} & \| g_{s,t,R}\|_{L^q(d\mu_s)} \leq \exp \big( C_{s,R,q}(1+|t|)^{A} \big), \end{align}

and,

(4.4)\begin{align} \chi_R(\mathcal E_N(u)) g_{s,N,t} \longrightarrow g_{s,t,R}, \text{weakly in}\ L^q(d\mu_s). \end{align}

Since R > 0 is arbitrary, we deduce that $(\Phi_t)_\# \mu_s$ is absolutely continuous with respect to µ_s; indeed, if a Borel set A is such that $\mu_s(A)=0$, then:

\begin{align*} (\Phi_t)_\# \mu_s (A) = \lim_{R \to \infty}(\Phi_t)_\# \mu_{s,R} (A) \leq \liminf_{R \to \infty} \| g_{s,t,R}\|_{L^q(d\mu_s)} \mu_s(A)^{\frac{1}{q'}} = 0. \end{align*}

Invoking now the Radon-Nikodym theorem, there exists a non-negative measurable function $g_{s,t}$ such that:

(4.5)\begin{align} (\Phi_t)_\# \mu_s = g_{s,t} \mu_s. \end{align}

Let us now prove that $\mathcal E(u) = \mathcal E(\Phi_tu)$ µ_s-almost everywhere. Again, we consider $\psi : H^{\sigma} \to \mathbb{R}$ a bounded continuous function, and we write:

\begin{align*} &\int \psi(u) \mathcal E(\Phi_t(u)) d\mu_{s,R} = \int \psi(\Phi_{-t}(u)) \mathcal E(u) d (\Phi_t)_\# \mu_{s,R} \\ & = \int \psi(\Phi_{-t}(u)) \mathcal E(u) g_{t,s,R} d\mu_s = \lim_{N \to \infty} \int \psi(\Phi^N_{-t}(u)) \mathcal E_N(u) \chi_R(\mathcal E_N(u)) g_{s,N,t} d\mu_s. \end{align*}

This limit is justified thanks to the weak convergence in $L^q(d\mu_s)$ in (4.4) and the convergence in $L^q(d\mu_s)$ of $\psi(\Phi^N_{-t}(u)) \mathcal E_N(u) $ to $\psi(\Phi_{-t}(u)) \mathcal E(u)$ (see the propositions 3.2 and 5.1). Using now the invariance of $\mathcal E_N$ by the truncated flow, and again Proposition 5.1, we can continue the equality above as:

\begin{align*} \int \psi&(u) \mathcal E(\Phi_t(u)) d\mu_{s,R} = \lim_{N \to \infty} \int \psi(u) \mathcal E_N(u) \chi_R(\mathcal E_N(u)) d\mu_s \\ & = \int \psi(u) \mathcal E(u) \chi_R(\mathcal E(u)) d\mu_s = \int \psi(u) \mathcal E(u)d\mu_{s,R}. \end{align*}

Since ψ is arbitrary, we can deduce that $\mathcal E(\Phi_t(u)) = \mathcal E(u)$ $\mu_{s,R}$-almost everywhere; and then, since R is also arbitrary, we can conclude that this equality holds µ_s-almost everywhere.

Finally, thanks to (4.5), the invariance of $\mathcal E$ by the flow, and Lemma 3.4, we have for every R > 0Footnote ¹:

\begin{align*} (\Phi_t)_\# \mu_{s,R} = (\Phi_t)_\# \big( \chi_R(\mathcal E(u)) d\mu_s\big) = \chi_R(\mathcal E(\Phi_{-t}u)) g_{s,t} d\mu_s = \chi_R(\mathcal E(u)) g_{s,t} d\mu_s = g_{s,t} d\mu_{s,R}. \end{align*}

Using (4.3) and (4.4), we also obtain:

\begin{align*} \| \chi_R(\mathcal E(u)) g_{s,t}\|_{L^q(d\mu_s)} \leq \exp \big( C_{s,R,q}(1+|t|)^{M_0} \big), \end{align*}

and,

\begin{align*} \chi_R(\mathcal E_N(u)) g_{s,N,t} \longrightarrow \chi_R(\mathcal E(u)) g_{s,t} \text{weakly in} L^q(d\mu_s). \end{align*}

This completes the proof.

Proof of Corollary 1.4

For T > 0, proceeding as in the proof of (3.4), we have the flow tail estimate:

\begin{align*} \mu_{s,R}\big(u_0 : \sup_{|t| \leq T} \| \Phi_tu_0 \|_{H^{\sigma}} \gt M \big) \leq e^{C_{s,R} (1+T)^A} e^{-\alpha M^2}, \end{align*}

for any M > 0. As a consequence, for $B \gt \frac{A}{2}$:

\begin{align*} \sum_{T \in \mathbb{N}} \mu_{s,R}\big(u_0 : \sup_{|t| \leq T} \| \Phi_t u_0 \|_{H^{\sigma}} \gt T^{B} \big) \lt +\infty. \end{align*}

By the Borel-Cantelli lemma, it implies that the set formed by the elements u ₀ such that $ \sup_{|t| \leq T} \| \Phi_t u_0 \|_{H^{\sigma}} \gt T^{B}$ for infinitely many $T \in \mathbb{N}$ is of zero $\mu_{s,R}$-measure. It means that for $\mu_{s,R}$-almost every u ₀, there exists $T_0= T_0(u_0)\in \mathbb{N}$ such that:

\begin{align*} T \geq T_0 \implies \sup_{|t| \leq T} \| \Phi_t u_0 \|_{H^{\sigma}} \leq T^{B}. \end{align*}

This implies (1.16) for $\mu_{s,R}$-almost every u ₀, and since R > 0 is arbitrary, it also holds for µ_s-almost every u ₀.

Notation 5.3. For a real number γ, the symbol $\gamma+$ refers to a real number greater than γ that can be chosen arbitrarily close to it.

Proof. For commodity, we denote $V_N := \pi_N V$ and $Y_N := \pi_N Y$. First, by assumption, we have:

\begin{align*} \mathcal E_N(U) = \| U \|_{L^2(\mathbb{T})}^2 + \Big| \frac{1}{p+1} \| U \|_{L^{p+1}(\mathbb{T})}^{p+1} + \big( \frac{1}{2} \| \partial_x U \|^2_{L^2(\mathbb{T})} - \sigma_N \big) \Big| \leq R. \end{align*}

So, as an immediate consequence,

(5.6)\begin{align} \| U \|_{L^2}^2 \leq R \end{align}

Next, we expand the term $\frac{1}{2} \| \partial_x U \|^2_{L^2} - \sigma_N$ as:

\begin{align*} \frac{1}{2} \| \partial_x U \|^2_{L^2} - \sigma_N = \frac{1}{2} \| \partial_x V_N \|^2_{L^2} + \text{Re} \int_{\mathbb{T}} \partial_x V_N \overline{\partial_x Y_N}dx + \frac{1}{2} \| \partial_x Y_N \|^2_{L^2} - \sigma_N, \end{align*}

and we deduce that:

\begin{multline*} \frac{1}{2} \| \partial_x V_N\|^2_{L^2} \leq \mathcal E_N(U) + \big| \int_{\mathbb{T}} \partial_x V_N \overline{\partial_x Y_N}dx \big| + \big| \frac{1}{2} \| \partial_x Y_N \|^2_{L^2} - \sigma_N \big| \\ \leq R + \| Y_N\|_{H^{s-\frac{1}{2}-}} \|V_N \|_{H^{\frac{5}{2}-s +}} + \big| \frac{1}{2} \| \partial_x Y_N \|^2_{L^2} - \sigma_N \big|. \end{multline*}

Using (5.6), it follows that

(5.7)\begin{align} \frac{1}{2} \| V_N\|_{H^1}^2 \leq 3 R + \| Y_N\|_{H^{s-\frac{1}{2}-}} \|V_N \|_{H^{\frac{5}{2}-s +}} + 2 \|Y_N\|_{L^2}^2 + \big| \frac{1}{2} \| \partial_x Y_N \|^2_{L^2} - \sigma_N \big|. \end{align}

We interpolate now $H^{\frac{5}{2}-s+}$ between H ¹ and H^s (note that $ 1 \leq \frac{5}{2}-s \lt \frac{5}{4} \lt s$) and then apply Young’s inequality:

(5.8)\begin{align} \| Y_N\|_{H^{s-\frac{1}{2}-}} \|V_N \|_{H^{\frac{5}{2}-s +}} \leq & \| Y_N\|_{H^{s-\frac{1}{2}-}} \|V_N \|^{2 {s-\frac{5}{4}}{s-1} - }_{H^1} \|V_N \|_{H^s}^{\frac{\frac{3}{2}-s}{s-1}+}\nonumber\\ & \quad \leq C \| Y_N\|^M_{H^{s-\frac{1}{2}-}} + \frac{1}{4} \|V_N \|^2_{H^1} + \|V_N \|^{2(3-2s)+}_{H^s}, \end{align}

where $C,M \gt 0$ are large constants. Here we used the following identities:

\begin{align*} \begin{cases} \frac{5}{2}-s+ = \big(1- (\frac{\frac{3}{2}-s}{s-1} +)\big) \cdot 1 + \big( \frac{\frac{3}{2}-s}{s-1} + \big) \cdot s \text{(for the interpolation)}\\ 1 = ( \frac{s-\frac{5}{4}}{s-1}+) + (1 - \frac{s-\frac{5}{4}}{s-1} -) + \frac{1}{M} \text{(for Young's inequality)}. \end{cases} \end{align*}

We emphasize that the second identity is valid since M can be chosen arbitrary large.

Finally, plugging (5.8) into (5.7) yields:

\begin{align*} \frac{1}{4} \| V_N\|_{H^1}^2 \leq 3 R + \|V_N \|^{2(3-2s)+}_{H^s} + C \| Y_N\|^M_{H^{s-\frac{1}{2}-}} + 2 \|Y_N\|_{L^2}^2 + \big| \frac{1}{2} \| \partial_x Y_N \|^2_{L^2} - \sigma_N \big|. \end{align*}

Hence,

(5.9)\begin{align} \| V_N\|_{H^1} \leq W_{s,R,N}(1 + \| V_N \|_{H^s})^{3-2s+}, \end{align}

where,

(5.10)\begin{align} W_{s,R,N} \lesssim \big( R + \| Y_N\|^M_{H^{s-\frac{1}{2}-}} + \|Y_N\|_{L^2}^2 + \big| \frac{1}{2} \| \partial_x Y_N \|^2_{L^2} - \sigma_N \big| \big)^{\frac{1}{2}}. \end{align}

Let us now conclude the proof. Fix $\alpha \lt s- \frac{1}{2}$. Then by interpolation, by (5.6), and by (5.9), we can write:

\begin{multline*} \| U \|_{H^\alpha} \leq \| Y_N \|_{H^\alpha} + \| V_N \|_{H^\alpha} \leq \| Y_N \|_{H^\alpha} + \| V_N \|_{L^2}^{1-\alpha} \| V_N \|_{H^1}^\alpha \\ \leq \| Y_N \|_{H^\alpha} + (\| U \|_{L^2} + \| Y_N \|_{L^2})^{1-\alpha} W_{s,R,N}^\alpha (1 + \| V_N \|_{H^s})^{\alpha(3-2s)+} \\ \leq W_{s,R,N} (1 + \| V_N \|_{H^s})^{\alpha(3-2s)+}, \end{multline*}

where,

\begin{align*} W_{s,R,N} = \| Y_N \|_{H^\alpha} + ( \sqrt{R} + \| Y_N \|_{L^2})^{1-\alpha} W_{s,R,N}^\alpha. \end{align*}

Thanks to Lemma 2.7, (5.2), and the expression of $W_{s,R,N}$ in (5.10), we indeed obtain that the $W_{s,R,N}$ satisfy the condition (5.4).

To obtain (5.5), we use the standard estimate:

\begin{align*} \|\pi_N(|\pi_N U|^{p-1}\pi_N U)\|_{H^{\frac{1}{2} + \varepsilon }(\mathbb{T})} \leq C_{\varepsilon} \| U \|_{H^{\frac{1}{2} + \varepsilon }(\mathbb{T})} \| U \|^{p-1}_{H^{\frac{1}{2}}(\mathbb{T})}, \end{align*}

and then apply (5.3). This completes the proof.

Proof. Let $p \geq 7$ (odd). For commodity we assume that $N_1 \geq ... \geq N_{p+1}$, that is $N_j = N_{(j)}$.

• First, we prove (6.5). We rearrange the sum as:

\begin{align*} \sum_{k_1-k_2+...-k_{p+1}=0 } | \Psi_{2s}(\vec{k}) | \prod_{j=1}^{p+1} | f_{k_j}^{(j)} | = \sum_{k_7,...,k_{p+1}} \prod_{j=7}^{p+1} | f_{k_j}^{(j)} | \sum_{\substack{k_1-k_2+...-k_6 \\ = -k_7 +...+k_{p+1}}} | \Psi_{2s}(\vec{k}) | \prod_{j=1}^{6} | f_{k_j}^{(j)} | \end{align*}

and we know that (Lemma 7.5 in [Reference Knezevitch22]):

\begin{align*} \sum_{\substack{k_1-k_2+...-k_6 \\ = -k_7 +...+k_{p+1}}} | \Psi_{2s}(\vec{k}) | \prod_{j=1}^{6} | f_{k_j}^{(j)} | \leq C_s N_{(1)}^{2(s-1)+} N_{(3)}^{2} \prod_{j=1}^{6} \| f^{(j)}_{k_j}\|_{l^2}. \end{align*}

Thus,

\begin{align*} \sum_{k_1-k_2+...-k_{p+1}=0 } | \Psi_{2s}(\vec{k}) | \prod_{j=1}^{p+1} | f_{k_j}^{(j)} | \leq C_s N_{(1)}^{2(s-1)+} N_{(3)}^{2} \prod_{j=1}^{6} \| f^{(j)}_{k_j}\|_{l^2} \sum_{k_7,...,k_{p+1}} \prod_{j=7}^{p+1} | f_{k_j}^{(j)} |. \end{align*}

Using now the Cauchy-Schwarz inequality, and the localization of the $f^{(j)}_{k_j}$, leads to (6.5).

• Second, we prove (6.6). Here, we are in the regime $N_{(4)} \ll N_{(3)}$, so we are in a position to use the lower bound on Ω from Lemma 2.5. Besides, thanks to Remark 2.6, we see that the case $\Psi_{2s}(\vec{k})=\Psi^{(0)}_{2s}(\vec{k})$ leads to a zero-contribution, so we only need to treat the case $\Psi_{2s}(\vec{k})=\Psi^{(1)}_{2s}(\vec{k})$. Now, combining (2.4) and (2.6), we can write:

\begin{align*} \sum_{k_1-k_2+...-k_{p+1}=0 } | \Psi_{2s}(\vec{k}) | \prod_{j=1}^{p+1} | f_{k_j}^{(j)} | \leq C_s N_{(1)}^{2(s-1)} \sum_{k_3,...,k_{p+1}} \prod_{j=3}^{p+1} | f_{k_j}^{(j)} | \sum_{k_2} |f^{(2)}_{k_2}f^{(1)}_{k_2 - k_3 + ... +k_{p+1}} |. \end{align*}

Applying the Cauchy-Schwarz inequality first in the k ₂ summation and then in the $k_3,....,k_{p+1}$ summation leads to (6.6).

Proof. Lemma 3.4 combined with (3.2) gives:

\begin{align*} (\Phi^N_t)_\# \rho_{s,R,N} &= (\Phi^N_t)_\# \big( e^{-R_{s,N}(u)} d\mu_{s,R,N}\big) = e^{-R_{s,N}(\Phi^N_{-t}u)} g_{s,N,t} d\mu_{s,R,N} \\ &= e^{-(R_{s,N}(\Phi^N_{-t}u) - R_{s,N}(u))} g_{s,N,t} d\rho_{s,R,N}, \end{align*}

which is the desired equality recalling the definitions (1.10) and (6.2).

Proof. First, using the definition of the Radon-Nikodym derivative, we have for $\tau \in [0,t]$:

\begin{align*} \begin{split} \| f_{s,N,\tau} & \|^q_{L^q(\rho_{s,R,N})} = \int f_{s,N,\tau}^{q-1} f_{s,N,\tau} d\rho_{s,R,N} \\ & = \int \exp\big(-(q-1)(E_{s,N}(\Phi^N_{-\tau}u) - E_{s,N}(u)) \big) (\Phi^N_\tau)_\# d\rho_{s,R,N} \\ & =\int \exp\big((q-1)(E_{s,N}(\Phi^N_{\tau}u) - E_{s,N}(u)) \big) d\rho_{s,R,N}. \end{split} \end{align*}

Then, we introduce a parameter $n \in \mathbb{N}$, which is intended to tend to infinity. By subdividing $[0,\tau]$ into intervals of length $\frac{\tau}{n}$, applying Jensen’s inequality, using again the definition of the Radon-Nikodym derivative, and then Hölder’s inequality (considering qʹ the conjugate exponent of q), we obtain:

\begin{align*} \begin{split} \| &f_{s,N,\tau} \|^q_{L^q(\rho_{s,R,N})} = \int \exp\big( (q-1)n \sum_{k=0}^{n-1} \frac{1}{n}( E_{s,N}(\Phi^N_{(k+1)\frac{\tau}{n}}u) - E_{s,N}(\Phi^N_{k\frac{\tau}{n}}u))\big) d\rho_{s,R,N} \\ & \leq \sum_{k=0}^{n-1} \frac{1}{n} \int \exp \big( (q-1)n ( E_{s,N}(\Phi^N_{\frac{\tau}{n}} \Phi^N_{k\frac{\tau}{n}}u) - E_{s,N}(\Phi^N_{k\frac{\tau}{n}}u)) \big) d\rho_{s,R,N} \\ & = \sum_{k=0}^{n-1} \frac{1}{n} \int \exp \big( (q-1)n ( E_{s,N}( \Phi^N_{\frac{\tau}{n}}u) - E_{s,N}(u)) \big) f_{s,N,\frac{k \tau}{n}} d\rho_{s,R,N} \\ & \leq \sum_{k=0}^{n-1} \frac{1}{n} \big[ \int \exp \big( q n ( E_{s,N}( \Phi^N_{\frac{\tau}{n}}u) - E_{s,N}(u)) \big) d\rho_{s,R,N} \big]^{\frac{1}{q'}} \| f_{s,N,\frac{k \tau}{n}} \|_{L^q(d\rho_{s,R,N})} \\ & \leq \sup_{t' \in [0,t]}\| f_{s,N,t'} \|_{L^q(d\rho_{s,R,N})} \big[ \int \exp \big( q n ( E_{s,N}( \Phi^N_{\frac{\tau}{n}}u) - E_{s,N}(u)) \big) d\rho_{s,R,N} \big]^{\frac{1}{q'}}. \end{split} \end{align*}

Thanks to (6.4), we obtain by passing to the limit $n \to \infty$:

(7.5)\begin{align} \| f_{s,N,\tau} \|^q_{L^q(\rho_{s,R,N})} \leq \sup_{t' \in [0,t]}\| f_{s,N,t'} \|_{L^q(d\rho_{s,R,N})} \big[ \int \exp \big( q \tau Q_{s,N}(u) \big) d\rho_{s,R,N} \big]^{\frac{1}{q'}}, \end{align}

and then, passing to the sup over $\tau \in [0,t]$ yields:

\begin{align*} \sup_{\tau \in [0,t]} \| f_{s,N,\tau} \|^q_{L^q(\rho_{s,R,N})} \leq \sup_{\tau \in [0,t]} \int \exp \big( q \tau Q_{s,N}(u) \big) d\rho_{s,R,N} \leq \int \exp \big( q |t Q_{s,N}(u)| \big) d\rho_{s,R,N}, \end{align*}

which is the desired inequality. To conclude, let us justify the passage to the limit in (7.5). Here, we can apply the dominated convergence theorem; indeed:

\begin{align*} n | E_{s,N}( \Phi^N_{\frac{\tau}{n}}u) - E_{s,N}(u)| \leq | \tau | \sup_{t' \in [0,\frac{\tau}{n}]} |\frac{d}{dt'} E_{s,N}(\Phi^N_{t'}u)| = | \tau | \sup_{t' \in [0,\frac{\tau}{n}]} |Q_{s,N}(\Phi^N_{t'}u)|, \end{align*}

and thanks to the $L^2(\mathbb{T})$-norm conservation, the fact that the $L^2(\mathbb{T})$-norm is bounded in the integral due to the cutoff $\chi_R \circ \mathcal E_N$, and the expression of $Q_{s,N}$ in (6.3), we can then write:

\begin{align*} n | E_{s,N}( \Phi^N_{\frac{\tau}{n}}u) - E_{s,N}(u)| \leq |\tau |C_{R,N} \lt + \infty, \end{align*}

where $C_{R,N} \gt 0$ is independent of n.

Notations 7.6. (Preparation for the proof of Lemma 7.5)

Let $N\in \mathbb{N}$. Firstly, for $V \in H^s(\mathbb{T})$, we denote:

\begin{align*} U := \pi_N (V + Y),\ \text{where}\ Y \sim \mu_s\ \text{is the Gaussian random variable defined in (1.4)}. \end{align*}

Secondly, for $\mathcal F_{s,N} \in \{\mathcal M_{s,N}, \mathcal T_{s,N}, \mathcal N_{s,N} \}$, and considering $F_N : U \mapsto \pi_N(|\pi_N U|^{p-1}\pi_N U)$, we observe that there exists a unique

\begin{align*} (\Psi_{2s},G) \in \{(\Psi^{(0)}_{2s},id) \} \cup \{\Psi^{(1)}_{2s} \} \times \{id, F_N\} \end{align*}

such that:

\begin{align*} \mathcal F_{s,N}(U)= \sum_{k_1-k_2+...-k_{p+1}=0} \Psi_{2s}(\vec{k}) G_{k_1}\overline{U_{k_2}}...\overline{U_{k_{p+1}}}, \end{align*}

with the abuse of notation $G = G(U)$, that we use for commodity (and where $G_k, U_k$ correspond to the k-th Fourier coefficient). This concise formulation for $\mathcal F_{s,N}$ will allow us to avoid treating the terms $\mathcal M_{s,N}, \mathcal T_{s,N}$, and $\mathcal N_{s,N}$ separately.

Thirdly, for a set of dyadic integers $(N_1,...,N_{p+1}) \in (2^{\mathbb{N}})^{p+1}$, we denote $\vec{N} = (N_1,...,N_{p+1})$, and we consider the dyadic block for $\mathcal F_{s,N}$:

(7.8)\begin{align} \mathcal F_{s,\vec{N}}(U) := \sum_{k_1-k_2+...-k_{p+1}=0} \Psi_{2s}(\vec{k}) G^{N_1}_{k_1}\overline{U^{N_2}_{k_2}}...\overline{U^{N_{p+1}}_{k_{p+1}}}, \end{align}

where, for $W: \mathbb{T} \to \mathbb{C}$ and $L \in 2^{\mathbb{N}}$, we adopt the notation $W^L = P_{L} \pi_N W$, with P_L the projector onto frequencies of size ∼ L.

Finally, we denote $N_{(1)} \geq ... \geq N_{(p+1)}$ a non-increasing rearrangement of the dyadic integers $N_1,...,N_{p+1}$.

Proof of Lemma 7.5

Let $N \in \mathbb{N}$ and $\mathcal F_{s,N} \in \{\mathcal M_{s,N}, \mathcal T_{s,N}, \mathcal N_{s,N} \}$. We adopt the Notations 7.6. We start by applying the simplified Boué-Dupuis formula (2.7) to the function $\mathcal F = \lambda \mathbb{1}_{\mathcal E_N \leq R}|\mathcal F_{s,N}|$:

\begin{align*} \log \int e^{\lambda | \mathcal F_{s,N}(u)|}d\mu_{s,R,N} &\leq \log \int e^{\mathbb{1}_{\mathcal E_N(u) \leq R} \lambda | \mathcal F_{s,N}(u)|} d\mu_s \\ & \leq \mathbb{E} \big[\sup_{V \in H^s} \{\lambda \mathbb{1}_{\mathcal E_N(U) \leq R}|\mathcal F_{s,N}(U)| - \frac{1}{2}\left\lVert V \right\rVert^2_{H^s} \} \big]. \end{align*}

Then, decomposing $\mathcal F_{s,N}$ into dyadic blocks, we obtain:

(7.9)\begin{align} \log \int e^{\lambda | \mathcal F_{s,N}(u)|}d\mu_{s,R,N} \leq \mathbb{E} \big[\sup_{V \in H^s} \{\lambda \mathbb{1}_{\mathcal E_N(U) \leq R}\sum_{\vec{N}} | \mathcal F_{s,\vec{N}}(U)| - \frac{1}{2}\left\lVert V \right\rVert^2_{H^s} \} \big]. \end{align}

Now, our goal is to prove that:

(7.10)\begin{align} \mathbb{1}_{\mathcal E_N(U) \leq R}|\mathcal F_{s,\vec{N}}(U)| \leq N_{(1)}^{0-} X_{s,R,N} (1 + \| V \|_{H^s})^{2-}, \end{align}

with $X_{s,R,N}$ a non-negative random variable satisfying:

(7.11)\begin{align} \forall m \gt 0, \mathbb{E} [ X_{s,R,N}^m] \leq C_{s,R,m} \lt +\infty. \end{align}

Indeed, if we do so, then we will be able to continue (7.9) as follows:

\begin{align*} \log \int e^{\lambda | \mathcal F_{s,N}(u)|}d\mu_{s,R,N} \lesssim \mathbb{E}\big[ \sup_{V \in H^s} \lambda X_{s,R,N}(1 + \| V \|_{H^s})^{2-} - \frac{1}{2}\left\lVert V \right\rVert^2_{H^s} \big] \lesssim \lambda^A \mathbb{E}[X_{s,R,N}^A], \end{align*}

for A > 0 large; and finally using the property (7.11) in the estimate above will ensure (7.6).

Hence, let us turn to the proof of (7.10). In $\mathcal F_{s,\vec{N}}(U)$ (see  (7.8)), only N ₁ has a special position, due to the presence of G. In our analysis, the size of N ₁ with respect to the other frequencies does not play a crucial role, and in all cases we estimate G ^{N ₁} applying Proposition 5.4. Hence, for convenience, we assume in what follows that $N_j = N_{(j)}$ (for $j=1,...,p+1$), as the other cases are similar. Finally, throughout the proof we use the following fact:

\begin{align*} \big( k_1-k_2+...-k_{p+1}=0 \text{ and } \forall j, |k_j| \sim N_j \big) \implies N_{(1)} \sim N_{(2)},\ \text{that is } N_1 \sim N_2, \end{align*}

without explicitly referring to it.

Now, we consider the two following frequency regime:

\begin{align*} \mathcal F_{s,\vec{N}}(U) = \mathbb{1}_{N_4 \ll N_3} \mathcal F_{s,\vec{N}}(U) + \mathbb{1}_{N_4 \sim N_3} \mathcal F_{s,\vec{N}}(U), \end{align*}

and we treat separately each of these terms.

The first regime: When $N_4 \ll N_3$. We are here in a position to use the refined estimate (6.6). Doing so leads to:

\begin{align*} |\mathcal F_{s,\vec{N}}(U)| \lesssim N_{(1)}^{2(s-1)} \| G^{N_1} \|_{L^2} \| U^{N_2} \|_{L^2} \prod_{j=3}^{p+1} \| U^{N_j} \|_{H^{\frac{1}{2}}}. \end{align*}

Applying now (5.5) for G ^{N ₁} (true for $G=F_N$, and also clearly for G = id), (5.3) with $\alpha = 2(s-\frac{5}{4})$ to U ^{N ₂}, and (5.3) with $\alpha = \frac{1}{2}$ to $U^{N_j}$ ( $j \geq 3$), yields:

\begin{align*} \begin{split} \mathbb{1}_{\mathcal E_N(U) \leq R} |\mathcal F_{s,\vec{N}}(U)| &\lesssim N_1^{-\varepsilon} X_{s,R,N} (1+ \| V \|_{H^s})^{(3-2s) [\frac{p}{2} +\varepsilon + 2(s-\frac{5}{4}) + \frac{p-1}{2}] + } \\ & = N_1^{-\varepsilon} X_{s,R,N} (1+ \| V \|_{H^s})^{(3-2s) [2s +p -3 +\varepsilon] + }, \end{split} \end{align*}

where we gathered in $X_{s,R,N}$ all the positive random variables appearing when using Proposition 5.4. Thanks to this proposition, we indeed see that $X_{s,R,N}$ satisfies (7.11); and then, thanks to Lemma 2.1, the inequality above implies the desired estimate (7.10) for ɛ > 0 small enough.

The second regime: When $N_4 \sim N_3$. Here, we apply (6.5) and obtain:

\begin{align*} \mathbb{1}_{\mathcal E_N(U) \leq R} |\mathcal F_{s,\vec{N}}(U)| \lesssim \mathbb{1}_{\mathcal E_N(U) \leq R} N_1^{2(s-1) +} N_3^2 \| G^{N_1} \|_{L^2} \prod_{j=2}^{6} \| U^{N_j} \|_{L^2} \prod_{j=7}^{p+1} \| U^{N_j} \|_{H^{\frac{1}{2}}}. \end{align*}

In this inequality, we simply bound $\| U^{N_5} \|_{L^2}$ and $\| U^{N_6} \|_{L^2}$ by R (see (5.6)). Then, we use again (5.5) for G ^{N ₁}; and, we use (5.3) with $\alpha= \sigma = s -\frac{1}{2}-$ for U ^{N ₂} and U ^{N ₃}, and with $\alpha = \frac{1}{2}$ for $U^{N_j}$ (j = 4 and $j \geq 7$). In doing so we obtain:

\begin{align*} \begin{split} \mathbb{1}_{\mathcal E_N(U) \leq R} |\mathcal F_{s,\vec{N}}(U)| & \lesssim N_1^{s - 2 -\varepsilon + } N_3^{2 - s +} X_{s,R,N} (1+ \| V \|_{H^s})^{(3-2s) [ \frac{p}{2} + \varepsilon + 2(s - \frac{1}{2}) + \frac{p-4}{2}] + } \\ & \lesssim N_1^{-\varepsilon +} X_{s,R,N} (1+ \| V \|_{H^s})^{(3-2s) [2s + p - 3 +\varepsilon] + }, \end{split} \end{align*}

with again $X_{s,R,N}$ satisfying (7.11) (thanks to (5.4) in Proposition 5.4). Since we can replace in the inequality above $-\varepsilon +$ by $-\varepsilon + \delta_0$, where ɛ and δ ₀ can be chosen independently, we indeed obtain (thanks to Lemma 2.1) the desired estimate (7.10) (taking $\delta_0 \leq \frac{\varepsilon}{2}$ and ɛ > 0 small enough). The proof is now complete.