2 Preliminaries

In this section, we present the notations and definitions which are used throughout this article.

2.1 Notations

In this subsection, we list the notations which we use throughout this paper. $C,c$ are the various time independent constants which depend on $s$ unless specified. $a+$ and $a-$ represent $a+\unicode[STIX]{x1D716}$ and $a-\unicode[STIX]{x1D716}$, respectively for arbitrary small $\unicode[STIX]{x1D716}>0$. $A\lesssim B$ denotes the estimate of the form $A\leqslant CB$. Similarly, $A\sim B$ denotes $A\lesssim B$ and $B\gtrsim A$.

Define $(dk)_{\unicode[STIX]{x1D706}}$ to be normalized counting measure on $\mathbb{Z}/\unicode[STIX]{x1D706}$:

$$\begin{eqnarray}\int \unicode[STIX]{x1D719}(k)(dk)_{\unicode[STIX]{x1D706}}=\frac{1}{\unicode[STIX]{x1D706}}\mathop{\sum }_{k\in \mathbb{Z}/\unicode[STIX]{x1D706}}\unicode[STIX]{x1D719}(k).\end{eqnarray}$$

Let $\hat{f}(k)$ and $\tilde{f}(k,\unicode[STIX]{x1D70F})$ denotes the Fourier transform of $f(x,t)$ in $x$ and in $x$ and $t$, respectively. We define the Sobolev space $H^{s}([0,\unicode[STIX]{x1D706}])$ with the norm

$$\begin{eqnarray}\Vert f\Vert _{H^{s}}=\Vert \hat{f}(k)\langle k\rangle ^{s}\Vert _{L^{2}((dk)_{\unicode[STIX]{x1D706}})},\end{eqnarray}$$

where $\langle \cdot \rangle =(1+|\cdot |)$. For details see [Reference Colliander, Keel, Staffilani, Takaoka and Tao4], [Reference Tsugawa23]. We define the space $X^{s,b}$ equipped with the norm

$$\begin{eqnarray}\Vert u\Vert _{X^{s,b}}=\Vert \langle k\rangle ^{s}\langle \unicode[STIX]{x1D70F}-4\unicode[STIX]{x1D70B}^{2}k^{3}\rangle \tilde{u} (k,\unicode[STIX]{x1D70F})\Vert _{L^{2}((dk)_{\unicode[STIX]{x1D706}}\,d\unicode[STIX]{x1D70F})}.\end{eqnarray}$$

We often study the KdV and mKdV equation in $X^{s,1/2}$ space but it hardly controls the norm $L_{t}^{\infty }H_{x}^{s}$, see [Reference Bourgain1], [Reference Colliander, Keel, Staffilani, Takaoka and Tao4], [Reference Tsugawa23]. To ensure the continuity of the solution, we define a slightly smaller space with the norm

$$\begin{eqnarray}\Vert u\Vert _{Y^{s}}=\Vert u\Vert _{X^{s,1/2}}+\Vert \langle k\rangle ^{s}\tilde{u} (k,\unicode[STIX]{x1D70F})\Vert _{L^{2}((dk)_{\unicode[STIX]{x1D706}})L^{1}(d\unicode[STIX]{x1D70F})}.\end{eqnarray}$$

$Z^{s}$ space is defined via the norm

$$\begin{eqnarray}\Vert u\Vert _{Z^{s}}=\Vert u\Vert _{X^{s,-1/2}}+\Vert \langle k\rangle ^{s}\langle \unicode[STIX]{x1D70F}-4\unicode[STIX]{x1D70B}^{2}k^{3}\rangle ^{-1}\tilde{u} (k,\unicode[STIX]{x1D70F})\Vert _{L^{2}((dk)_{\unicode[STIX]{x1D706}})L^{1}(d\unicode[STIX]{x1D70F})}.\end{eqnarray}$$

For the time interval $[t_{1},t_{2}]$, we define the restricted spaces $X_{([0,\unicode[STIX]{x1D706}]\times [t_{1},t_{2}])}^{s,b}$ and $Y_{([0,\unicode[STIX]{x1D706}]\times [t_{1},t_{2}])}^{s}$ equipped with the norms

$$\begin{eqnarray}\displaystyle \Vert u\Vert _{X_{([0,\unicode[STIX]{x1D706}]\times [t_{1},t_{2}])}^{s,b}} & = & \displaystyle \inf \{\Vert U\Vert _{X^{s,b}}:U|_{([0,\unicode[STIX]{x1D706}]\times [t_{1},t_{2}])}=u\},\nonumber\\ \displaystyle \Vert u\Vert _{Y_{([0,\unicode[STIX]{x1D706}]\times [t_{1},t_{2}])}^{s}} & = & \displaystyle \inf \{\Vert U\Vert _{Y^{s}}:U|_{([0,\unicode[STIX]{x1D706}]\times [t_{1},t_{2}])}=u\}.\nonumber\end{eqnarray}$$

We state the mean value theorem as follows:

Lemma 2.1. If $a$ is controlled by $b$ and $|k_{1}|\ll |k_{2}|$, then

$$\begin{eqnarray}a(k_{1}+k_{2})-a(k_{2})=O\left(|k_{1}|\frac{b(k_{2})}{|k_{2}|}\right).\end{eqnarray}$$

For details see [Reference Colliander, Keel, Staffilani, Takaoka and Tao4, Section 4].

2.2 Rescaling

In this subsection, we rescale the mKdV equation. We can rewrite equations (1.5)–(1.6) in $\unicode[STIX]{x1D706}$-rescaled form as follows:

(2.1)$$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x2202}_{t}v+\unicode[STIX]{x2202}_{xxx}v+6\left(v^{2}-\frac{1}{2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D706}}\Vert v\Vert _{L^{2}}^{2}\right)\unicode[STIX]{x2202}_{x}v+\unicode[STIX]{x1D706}^{-3}\unicode[STIX]{x1D6FE}v=\unicode[STIX]{x1D706}^{-3}g,\end{eqnarray}$$

(2.2)$$\begin{eqnarray}\displaystyle & & \displaystyle v(x,t_{0})=v_{t_{0}}(x),\end{eqnarray}$$

where

$$\begin{eqnarray}\displaystyle g(x,t) & = & \displaystyle \unicode[STIX]{x1D706}^{-1}F(\unicode[STIX]{x1D706}^{-1}x,\unicode[STIX]{x1D706}^{-3}t),\nonumber\\ \displaystyle v_{t_{0}}(x) & = & \displaystyle \unicode[STIX]{x1D706}^{-1}u(\unicode[STIX]{x1D706}^{-1}x,\unicode[STIX]{x1D706}^{-3}t_{0}),\nonumber\end{eqnarray}$$

for initial time $t_{0}$. If $u$ is the solution of the equations (1.5)–(1.6), then $v(x,t)=\unicode[STIX]{x1D706}^{-1}u(\unicode[STIX]{x1D706}^{-1}x,\unicode[STIX]{x1D706}^{-3}t)$ is the solution of the equations (2.1)–(2.2).

Rescaling is helpful in proving the local in time result as well as a priori estimate.

2.3 I-Operator

We define an operator $I$ which plays an important role for the $I$-method. Let $\unicode[STIX]{x1D719}:\mathbb{R}\rightarrow \mathbb{R}$ be a smooth monotone $\mathbb{R}$-valued function defined as:

$$\begin{eqnarray}\unicode[STIX]{x1D719}(k)=\left\{\begin{array}{@{}ll@{}}1\quad & |k|<1,\\ |k|^{s-1}\quad & |k|>2.\end{array}\right.\end{eqnarray}$$

Then, we define $m(k)=\unicode[STIX]{x1D719}(k/N)$, so that

$$\begin{eqnarray}m(k)=\left\{\begin{array}{@{}ll@{}}1\quad & |k|<N,\\ |k|^{s-1}N^{1-s}\quad & |k|>2N,\end{array}\right.\end{eqnarray}$$

where we fix $N$ to be a large cut-off. We define the operator $I$ as:

$$\begin{eqnarray}\widehat{Iu}(k)=m(k)\hat{u} (k).\end{eqnarray}$$

We can rescale the operator $I$ as follows:

$$\begin{eqnarray}\widehat{I^{\prime }u}(k)=m^{\prime }(k)\hat{u} (k),\end{eqnarray}$$

where $m^{\prime }(k/\unicode[STIX]{x1D706})=m(k)$. Let $N^{\prime }=N/\unicode[STIX]{x1D706}$. Then

$$\begin{eqnarray}m^{\prime }(k)=\left\{\begin{array}{@{}ll@{}}1\quad & |k|<N^{\prime },\\ |k|^{s-1}N^{\prime (1-s)}\quad & |k|>2N^{\prime }.\end{array}\right.\end{eqnarray}$$

We use the rescaled $I$-operator for proving the local results for mKdV equation in time. Moreover, for proving a priori estimate, we also use the same operator.

2.4 Strichartz estimate

Strichartz estimate plays an important role for the proof of the trilinear estimate. Bourgain, in [Reference Bourgain1], proved the $L^{4}$ Strichartz estimate for mKdV equation. In the present article, we use the same estimate. We list the following result:

Proposition 2.2. Let $b>1/3$. Then, we have

$$\begin{eqnarray}\Vert u\Vert _{L^{4}(\mathbb{R}\times \unicode[STIX]{x1D706}\mathbb{T})}\lesssim \Vert u\Vert _{X^{0,b}}.\end{eqnarray}$$

Corollary 2.3. We have

$$\begin{eqnarray}\Vert v\Vert _{L^{8}(\mathbb{R};L^{8/3}(\unicode[STIX]{x1D706}\mathbb{T}))}\lesssim \Vert v\Vert _{X^{0,5/12+}}.\end{eqnarray}$$

Proof. It is a consequence of Proposition 2.2, the embedding $X^{0,1/2+}{\hookrightarrow}L_{t}^{\infty }L_{x}^{2}$ and interpolation.◻

2.5 Local well-posedness

In this subsection, we state the local result in time which can be proved by using the contraction mapping. Let $\unicode[STIX]{x1D702}(t)\in C_{0}^{\infty }$ be a cut-off function such that:

$$\begin{eqnarray}\unicode[STIX]{x1D702}(t)=\left\{\begin{array}{@{}ll@{}}1\quad & \text{if}~|t|\leqslant 1,\\ 0\quad & \text{if}~|t|>2.\end{array}\right.\end{eqnarray}$$

Let

$$\begin{eqnarray}D_{\unicode[STIX]{x1D706}}(t)f(x)=\int e^{2i\unicode[STIX]{x1D70B}kx}e^{-(2i\unicode[STIX]{x1D70B}k)^{3}t}\hat{f}(k)(dk)_{\unicode[STIX]{x1D706}}.\end{eqnarray}$$

We assume the following well known lemmas:

Lemma 2.4. For any $s\in \mathbb{R}$

$$\begin{eqnarray}\Vert \unicode[STIX]{x1D702}(t)D_{\unicode[STIX]{x1D706}}(t)w\Vert _{Y^{s}}\lesssim \Vert w\Vert _{H^{s}}.\end{eqnarray}$$

Lemma 2.5. Let $F\in Z^{s}.$ Then

$$\begin{eqnarray}\left\Vert \unicode[STIX]{x1D702}(t)\int _{0}^{t}D_{\unicode[STIX]{x1D706}}(t-t^{\prime })F(t^{\prime })\,dt^{\prime }\right\Vert _{Y^{s}}\lesssim \Vert F\Vert _{Z^{s}}.\end{eqnarray}$$

For the proof of Lemmas 2.4 and 2.5 see [Reference Colliander, Keel, Staffilani, Takaoka and Tao4].

Proposition 2.6. Let $1/2\leqslant s<1$. Then the IVP (2.1)–(2.2) is locally well-posed for the initial data $v_{t_{0}}$ satisfying $I^{\prime }v_{t_{0}}\in {\dot{H}}^{1}(\mathbb{T})$ and $g\in {\dot{H}}^{1}(\mathbb{T})$. In particular, there exists a unique solution on the time interval $[t_{0},t_{0}+\unicode[STIX]{x1D6FF}]$ with the lifespan $\unicode[STIX]{x1D6FF}\sim (\Vert I^{\prime }v_{t_{0}}\Vert _{H^{1}}+\unicode[STIX]{x1D706}^{-3}\Vert g\Vert _{H^{1}}+\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3})^{-\unicode[STIX]{x1D6FC}}$ for some $\unicode[STIX]{x1D6FC}>0$ and the solution satisfies

$$\begin{eqnarray}\displaystyle \Vert I^{\prime }v\Vert _{Y^{1}([0,\unicode[STIX]{x1D706}]\times [t_{0},t_{0}+\unicode[STIX]{x1D6FF}])} & {\lesssim} & \displaystyle \Vert I^{\prime }v_{t_{0}}\Vert _{H^{1}}+\unicode[STIX]{x1D706}^{-3}\Vert I^{\prime }g\Vert _{H^{1}},\nonumber\\ \displaystyle \sup _{t_{0}\leqslant t\leqslant t_{0}+\unicode[STIX]{x1D6FF}}^{}\Vert I^{\prime }v(t)\Vert _{H^{1}} & {\lesssim} & \displaystyle \Vert I^{\prime }v_{t_{0}}\Vert _{H^{1}}+\unicode[STIX]{x1D706}^{-3}\Vert I^{\prime }g\Vert _{H^{1}}.\nonumber\end{eqnarray}$$

Remark 2.7. Note that

$$\begin{eqnarray}\displaystyle g(x,t) & = & \displaystyle \unicode[STIX]{x1D706}^{-1}F(\unicode[STIX]{x1D706}^{-1}x,\unicode[STIX]{x1D706}^{-3}t)\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D706}^{-1}f\left(\unicode[STIX]{x1D706}^{-1}x+\frac{1}{2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D706}}\int _{0}^{\unicode[STIX]{x1D706}^{-3}t}\Vert u(\unicode[STIX]{x1D70F})\Vert _{L^{2}}^{2}\,d\unicode[STIX]{x1D70F}\right).\nonumber\end{eqnarray}$$

Proof. The proof of Proposition 2.6 follows along the same lines as for KdV equation given in [Reference Tsugawa23] with the help of trilinear estimate given in Proposition 3.8. The only difference arises in the estimate of $g$ as it depends on unknown $u$. To deal with this issue, we define a new metric. Indeed, let

$$\begin{eqnarray}B=\{w\in X^{1,1/2}:\Vert w\Vert _{X^{1,1/2}}\lesssim (\Vert I^{\prime }v_{0}\Vert _{H^{1}}+\unicode[STIX]{x1D706}^{-3}\Vert I^{\prime }g\Vert _{H^{1}})\}\end{eqnarray}$$

and define the metric

$$\begin{eqnarray}d(w,w^{\prime })=\Vert w-w^{\prime }\Vert _{X^{0,1/2}}+\Vert v-v^{\prime }\Vert _{X^{0,1/2}},\end{eqnarray}$$

for $I^{\prime }v=w$. As $X^{0,1/2}$ is reflexive, the ball $B$ is complete with respect to the metric $d$: for details see [Reference Kato15, 9.14 and Lemma 7.3]. Therefore, it is enough to show

$$\begin{eqnarray}\displaystyle & & \displaystyle \Vert N(v,w)-N(v^{\prime },w^{\prime })\Vert _{Y^{0}}\lesssim \Vert \unicode[STIX]{x1D702}(t)(P(v,w)-P(v^{\prime },w^{\prime }))\Vert _{Z^{0}}\nonumber\\ \displaystyle & & \displaystyle \quad \lesssim (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}+(\Vert I^{\prime }v_{0}\Vert _{H^{1}}+\unicode[STIX]{x1D706}^{-3}\Vert I^{\prime }g\Vert _{H^{1}})^{2}+\unicode[STIX]{x1D706}^{-3}\Vert I^{\prime }g\Vert _{H^{1}})\nonumber\\ \displaystyle & & \displaystyle (\Vert w-w^{\prime }\Vert _{X^{0,1/2}}+\Vert v-v^{\prime }\Vert _{X^{0,1/2}}),\nonumber\end{eqnarray}$$

where

$$\begin{eqnarray}N(w)=\unicode[STIX]{x1D702}(t)D_{\unicode[STIX]{x1D706}}(t)I^{\prime }v_{0}-\unicode[STIX]{x1D702}(t)\int D_{\unicode[STIX]{x1D706}}(t-t^{\prime })\unicode[STIX]{x1D702}(t^{\prime })P(t^{\prime })\,dt^{\prime }\end{eqnarray}$$

with

$$\begin{eqnarray}P(v,w)=6I^{\prime }\left(v^{2}-\frac{1}{2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D706}}\Vert v\Vert _{L^{2}}^{2}\right)\unicode[STIX]{x2202}_{x}v+\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}w-\unicode[STIX]{x1D706}^{-3}I^{\prime }g.\end{eqnarray}$$

As the metric consists of both $w$ and $v$ terms, we consider the pair of equations as:

(2.3)$$\begin{eqnarray}\displaystyle & & \displaystyle \qquad \unicode[STIX]{x2202}_{t}v+\unicode[STIX]{x2202}_{xxx}v+6\left(v^{2}-\frac{1}{2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D706}}\Vert v\Vert _{L^{2}}^{2}\right)\unicode[STIX]{x2202}_{x}v+\unicode[STIX]{x1D706}^{-3}\unicode[STIX]{x1D6FE}v=\unicode[STIX]{x1D706}^{-3}g,\end{eqnarray}$$

(2.4)$$\begin{eqnarray}\displaystyle & & \displaystyle \qquad \unicode[STIX]{x2202}_{t}w+\unicode[STIX]{x2202}_{xxx}w+6I^{\prime }\left(v^{2}-\frac{1}{2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D706}}\Vert v\Vert _{L^{2}}^{2}\right)\unicode[STIX]{x2202}_{x}(I^{\prime })^{-1}w+\unicode[STIX]{x1D706}^{-3}\unicode[STIX]{x1D6FE}w=\unicode[STIX]{x1D706}^{-3}I^{\prime }g.\end{eqnarray}$$

The estimate of $v$ in $H^{s}$ follows from that of $w$ in $H^{1}$ because $\Vert v\Vert _{H^{s}}\lesssim \Vert w\Vert _{H^{1}}$. Therefore, we do not need to assume extra condition on ball for the variable “$v$”. Let

$$\begin{eqnarray}g^{\prime }(x,t)=\unicode[STIX]{x1D706}^{-1}f\left(\unicode[STIX]{x1D706}^{-1}x+\frac{1}{2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D706}}\int _{0}^{\unicode[STIX]{x1D706}^{-3}t}\Vert u^{\prime }(\unicode[STIX]{x1D70F})\Vert _{L^{2}}^{2}\,d\unicode[STIX]{x1D70F}\right).\end{eqnarray}$$

At first, we consider the external forcing term for equation (2.3) as:

$$\begin{eqnarray}\displaystyle \Vert I^{\prime }g-I^{\prime }g^{\prime }\Vert _{X^{0,-1/2}} & {\lesssim} & \displaystyle \Vert I^{\prime }g-I^{\prime }g^{\prime }\Vert _{L^{2}}\nonumber\\ \displaystyle & = & \displaystyle \left\Vert \unicode[STIX]{x1D706}^{-1}I^{\prime }f\left(\unicode[STIX]{x1D706}^{-1}x+\int _{0}^{\unicode[STIX]{x1D706}^{-3}t}\Vert \unicode[STIX]{x1D706}v(\unicode[STIX]{x1D706}\cdot ,\unicode[STIX]{x1D706}^{3}\unicode[STIX]{x1D70F})\Vert _{L^{2}}^{2}\,d\unicode[STIX]{x1D70F}\right)\right.\nonumber\\ \displaystyle & & \displaystyle -\left.\unicode[STIX]{x1D706}^{-1}I^{\prime }f\left(\unicode[STIX]{x1D706}^{-1}x+\int _{0}^{\unicode[STIX]{x1D706}^{-3}t}\Vert \unicode[STIX]{x1D706}v^{\prime }(\unicode[STIX]{x1D706}\cdot ,\unicode[STIX]{x1D706}^{3}\unicode[STIX]{x1D70F})\Vert _{L^{2}}^{2}\,d\unicode[STIX]{x1D70F}\right)\right\Vert _{L^{2}}\nonumber\\ \displaystyle & {\lesssim} & \displaystyle \left\Vert \unicode[STIX]{x1D706}^{-1}\int _{0}^{1}\frac{d}{d\unicode[STIX]{x1D703}}I^{\prime }f(\unicode[STIX]{x1D706}^{-1}x+\unicode[STIX]{x1D703}\unicode[STIX]{x1D6FC}(t)+(1-\unicode[STIX]{x1D703})\unicode[STIX]{x1D6FD}(t))\,d\unicode[STIX]{x1D703}\right\Vert _{L^{2}}\nonumber\end{eqnarray}$$

where

$$\begin{eqnarray}\unicode[STIX]{x1D6FC}(t)=\int _{0}^{\unicode[STIX]{x1D706}^{-3}t}\Vert \unicode[STIX]{x1D706}v(\unicode[STIX]{x1D706}\cdot ,\unicode[STIX]{x1D706}^{3}\unicode[STIX]{x1D70F})\Vert _{L^{2}}^{2}\,d\unicode[STIX]{x1D70F}\quad \text{and}\quad \unicode[STIX]{x1D6FD}(t)=\int _{0}^{\unicode[STIX]{x1D706}^{-3}t}\Vert \unicode[STIX]{x1D706}v^{\prime }(\unicode[STIX]{x1D706}\cdot ,\unicode[STIX]{x1D706}^{3}\unicode[STIX]{x1D70F})\Vert _{L^{2}}^{2}\,d\unicode[STIX]{x1D70F}.\end{eqnarray}$$

Now from mean value theorem, we get

$$\begin{eqnarray}\Vert I^{\prime }g-I^{\prime }g^{\prime }\Vert _{L^{2}}\lesssim \Vert I^{\prime }g\Vert _{H^{1}}\Vert v-v^{\prime }\Vert _{X^{0,1/2}}.\end{eqnarray}$$

Similarly for Equation (2.4), we get

$$\begin{eqnarray}\Vert g-g^{\prime }\Vert _{L^{2}}\lesssim \Vert g\Vert _{H^{1}}\Vert v-v^{\prime }\Vert _{X^{0,1/2}}.\end{eqnarray}$$

The nonlinear term can be estimated similarly to Proposition 3.8 below (see Remark 3.7 and the argument in Section 6). Hence, we can use the contraction principle. This shows that the solution $I^{\prime }v\in X^{1,1/2}.$ We need to show that the solution belongs to $Y^{1}$. But from Proposition 3.8, the nonlinear term of the integral equation belongs to $Y^{1}$. In the same way, we can verify other two terms of integral equation by using Schwarz inequality. Therefore, the solution $I^{\prime }v\in Y^{1}$.◻

3 Trilinear estimate

Define an operator $J$ such that

(3.1)$$\begin{eqnarray}\displaystyle & & \displaystyle \qquad {\hat{J}}[u,v,w]\nonumber\\ \displaystyle & & \displaystyle \qquad \quad =i\frac{k}{3}\mathop{\sum }_{\substack{ k_{1}+k_{2}+k_{3}=k \\ (k_{1}+k_{2})(k_{2}+k_{3})(k_{3}+k_{1})\neq 0}}\hat{u} (k_{1})\hat{v}(k_{2}){\hat{w}}(k_{3})-ik\hat{u} (k)\hat{v}(k){\hat{w}}(-k).\end{eqnarray}$$

We establish the following trilinear estimate for $J$:

Proposition 3.1. Let $s\geqslant 1/2$ and $u$, $v$, $w\in X^{s,1/2}$ are $\unicode[STIX]{x1D706}$-periodic in $x$ variable. Then, we have

(3.2)$$\begin{eqnarray}\Vert J[u,v,w]\Vert _{X^{s,-1/2}}\leqslant C\Vert u\Vert _{X^{s,1/2}}\Vert v\Vert _{X^{s,1/2}}\Vert w\Vert _{X^{s,1/2}}.\end{eqnarray}$$

Remark 3.2. We note that if $u$ is real valued, then

(3.3)$$\begin{eqnarray}J[u,u,u]=\left(u^{2}-\frac{1}{2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D706}}\Vert u\Vert _{L^{2}}^{2}\right)\unicode[STIX]{x2202}_{x}u\end{eqnarray}$$

yields the nonlinearity of mKdV. The first term and the second term of (3.1) can be estimated in $H^{s}$ for $s\geqslant 1/4$ and $s\geqslant 1/2$, respectively. So, the bound $s=1/2$ comes from the second term. Therefore, the nonlinearity of mKdV equation can be controlled if we prove Proposition 3.1. Simple computations yield

$$\begin{eqnarray}\displaystyle {\mathcal{F}}_{x}\left[\left(u^{2}-\frac{1}{2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D706}}\Vert u\Vert _{L^{2}}^{2}\right)\unicode[STIX]{x2202}_{x}u\right] & = & \displaystyle i\mathop{\sum }_{\substack{ k_{1}+k_{2}+k_{3}=k \\ (k_{1}+k_{2})\neq 0}}\hat{u} (k_{1})\hat{u} (k_{2})k_{3}\hat{u} (k_{3})\nonumber\\ \displaystyle & = & \displaystyle i\bigg\{\!\mathop{\sum }_{\substack{ k_{1}+k_{2}+k_{3}=k \\ (k_{1}+k_{2})(k_{2}+k_{3})(k_{3}+k_{1})\neq 0}}\hat{u} (k_{1})\hat{u} (k_{2})k_{3}\hat{u} (k_{3})\nonumber\\ \displaystyle & & \displaystyle +\,\mathop{\sum }_{\substack{ k_{1}+k_{2}+k_{3}=k \\ (k_{1}+k_{2})(k_{3}+k_{1})\neq 0 \\ (k_{2}+k_{3})=0}}\hat{u} (k_{1})\hat{u} (-k_{3})k_{3}\hat{u} (k_{3})\nonumber\\ \displaystyle & & \displaystyle +\,\mathop{\sum }_{\substack{ k_{1}+k_{2}+k_{3}=k \\ (k_{1}+k_{2})(k_{2}+k_{3})\neq 0 \\ (k_{3}+k_{1})=0}}\hat{u} (-k_{3})\hat{u} (k_{2})k_{3}\hat{u} (k_{3})\nonumber\\ \displaystyle & & \displaystyle +\,\mathop{\sum }_{\substack{ k_{1}+k_{2}+k_{3}=k \\ (k_{1}+k_{2})\neq 0 \\ (k_{2}+k_{3})=(k_{3}+k_{1})=0}}k_{3}\hat{u} (k_{1})\hat{u} (-k_{3})^{2}\bigg\}\nonumber\\ \displaystyle & = & \displaystyle i\frac{k}{3}\bigg\{\!\mathop{\sum }_{\substack{ k_{1}+k_{2}+k_{3}=k \\ (k_{1}+k_{2})\neq 0}}\hat{u} (k_{1})\hat{u} (k_{2})\hat{u} (k_{3})\bigg\}\nonumber\\ \displaystyle & & \displaystyle -\,ik|\hat{u} (k)|^{2}\hat{u} (k).\nonumber\end{eqnarray}$$

Remark 3.3. If $u$ is a complex-valued function, then we have only to consider

$$\begin{eqnarray}\left(|u|^{2}-\frac{1}{2\unicode[STIX]{x1D70B}}\Vert u\Vert _{L^{2}}^{2}\right)\unicode[STIX]{x2202}_{x}u-\frac{i}{2\unicode[STIX]{x1D70B}}\text{Im}\langle \unicode[STIX]{x2202}_{x}u,u\rangle _{L^{2}}u\end{eqnarray}$$

instead of the left hand side of the above equality. It yields the nonlinearity of the complex mKdV. This expression can be used to separate the resonant and nonresonant frequencies for complex mKdV equation which is helpful in the proof of trilinear estimate.

Proof of Proposition 3.1.

We first consider the trilinear estimate corresponding to nonresonant frequencies. We claim that

$$\begin{eqnarray}\left\Vert i\frac{k}{3}\int _{\substack{ k_{1}+k_{2}+k_{3}=k \\ (k_{1}+k_{2})(k_{2}+k_{3})(k_{3}+k_{1})\neq 0}}\hat{u} _{1}(k_{1})\hat{u} _{2}(k_{2})\hat{u} _{3}(k_{3})\right\Vert _{X^{s,-1/2}}\lesssim \mathop{\prod }_{i=1}^{3}\Vert u_{i}\Vert _{X^{s,1/2}}.\end{eqnarray}$$

From duality, it is enough to show

(3.4)$$\begin{eqnarray}\displaystyle & & \displaystyle \qquad \left|\int _{\substack{ k_{1}+k_{2}+k_{3}+k_{4}=0 \\ (k_{1}+k_{2})(k_{2}+k_{3})(k_{3}+k_{1})\neq 0}}\langle k_{4}\rangle \int _{\mathop{\sum }_{i=1}^{4}\unicode[STIX]{x1D70F}_{i}=0}\mathop{\prod }_{i=1}^{4}\tilde{u} _{i}(k_{i},\unicode[STIX]{x1D70F}_{i})(dk_{i})_{\unicode[STIX]{x1D706}}\,d\unicode[STIX]{x1D70F}_{i}\right|\nonumber\\ \displaystyle & & \displaystyle \qquad \quad \lesssim \mathop{\prod }_{i=1}^{3}\Vert u_{i}\Vert _{X^{s,1/2}}\Vert u_{4}\Vert _{X^{-s,1/2}}.\end{eqnarray}$$

Consider LHS of (3.4) and let the region of the first integration to be “$\ast$” and region of the second integration is denoted by “$\ast \ast$”. Define $\unicode[STIX]{x1D70E}_{i}=\unicode[STIX]{x1D70F}_{i}-4\unicode[STIX]{x1D70B}k_{i}^{3}$ for $1\leqslant i\leqslant 4$. We divide this estimate into the following four cases:

1. (1) $|\unicode[STIX]{x1D70E}_{4}|=\max \{|\unicode[STIX]{x1D70E}_{i}|~\text{for}~1\leqslant i\leqslant 4\}$.

2. (2) $|\unicode[STIX]{x1D70E}_{3}|=\max \{|\unicode[STIX]{x1D70E}_{i}|~\text{for}~1\leqslant i\leqslant 4\}$.

3. (3) $|\unicode[STIX]{x1D70E}_{2}|=\max \{|\unicode[STIX]{x1D70E}_{i}|~\text{for}~1\leqslant i\leqslant 4\}$.

4. (4) $|\unicode[STIX]{x1D70E}_{1}|=\max \{|\unicode[STIX]{x1D70E}_{i}|~\text{for}~1\leqslant i\leqslant 4\}$.

From symmetry, we can assume that $|k_{1}|\geqslant |k_{2}|\geqslant |k_{3}|$. We only consider Case 1 and Case 2, because the other cases can be treated in a similar way.

For Case 1, we multiply and divide LHS of (3.4) by $\langle k_{4}\rangle ^{s}\langle \unicode[STIX]{x1D70E}_{4}\rangle ^{-1/2}$ to get

(3.5)$$\begin{eqnarray}\left|\int _{\ast }\int _{\ast \ast }\langle k_{4}\rangle ^{s+1}\langle \unicode[STIX]{x1D70E}_{4}\rangle ^{-1/2}\tilde{u} _{1}\tilde{u} _{2}\tilde{u} _{3}(\langle k_{4}\rangle ^{-s}\langle \unicode[STIX]{x1D70E}_{4}\rangle ^{1/2}\tilde{u} _{4})\right|.\end{eqnarray}$$

As we know, $k_{1}+k_{2}+k_{3}+k_{4}=0$ and $\unicode[STIX]{x1D70F}_{1}+\unicode[STIX]{x1D70F}_{2}+\unicode[STIX]{x1D70F}_{3}+\unicode[STIX]{x1D70F}_{4}=0$, from simple calculations, we have

(3.6)$$\begin{eqnarray}\langle \unicode[STIX]{x1D70E}_{4}\rangle \gtrsim 3(|k_{1}+k_{2}\Vert k_{2}+k_{3}\Vert k_{3}+k_{1}|)\sim 3(|k_{2}+k_{3}\Vert k_{3}+k_{4}\Vert k_{4}+k_{2}|).\end{eqnarray}$$

We can again subdivide Case 1 into three cases:

1. (1a) $|k_{1}|\sim |k_{2}|\sim |k_{3}|$.

2. (1b) $|k_{1}|\gg |k_{2}|\gtrsim |k_{3}|$.

3. (1c) $|k_{1}|\sim |k_{2}|\gg |k_{3}|$.

Case 1a. We do not use duality and show the estimate (3.2) directly by the following lemma:

Lemma 3.4. For $s\geqslant 1/2,$ we have

(3.7)$$\begin{eqnarray}\Vert \unicode[STIX]{x2202}_{x}M(u,u,u)\Vert _{X^{s,-1/2}}\lesssim \Vert u\Vert _{X^{s,1/2}}^{3},\end{eqnarray}$$

where

$$\begin{eqnarray}{\mathcal{F}}_{x}[M(u,v,w)]=\mathop{\sum }_{\substack{ k_{1}+k_{2}+k_{3}=k \\ |k_{1}|\sim |k_{2}|\sim |k_{3}|}}\hat{u} (k_{1})\hat{v}(k_{2}){\hat{w}}(k_{3}),\end{eqnarray}$$

and ${\mathcal{F}}_{x}$ denotes the Fourier transform in $x$ variable.

Proof. We only show (3.7) for $s=1/2$. We have

$$\begin{eqnarray}\displaystyle & & \displaystyle \Vert \unicode[STIX]{x2202}_{x}M(u,u,u)\Vert _{X^{1/2,-1/2}}\nonumber\\ \displaystyle & & \displaystyle \quad \sim \left(\int _{k}\langle k\rangle ^{3}\left(\int _{-\infty }^{\infty }\langle \unicode[STIX]{x1D70E}\rangle ^{-1}|{\mathcal{F}}_{x,t}[M(u,u,u)]|^{2}\,d\unicode[STIX]{x1D70F}\right)(dk)_{\unicode[STIX]{x1D706}}\right)^{1/2}\nonumber\\ \displaystyle & & \displaystyle \quad \sim \Vert (\langle k\rangle ^{1/2}|\tilde{u} |)^{3}\langle \unicode[STIX]{x1D70E}\rangle ^{-1/2}\Vert _{L^{2}(\unicode[STIX]{x1D706}\mathbb{T}\times \mathbb{R})},\nonumber\end{eqnarray}$$

where ${\mathcal{F}}_{x,t}$ is the Fourier transform in both $x$ and $t$ variables. Let $\tilde{v}(k,\unicode[STIX]{x1D70F})=\langle k\rangle ^{1/2}|\tilde{u} (k,\unicode[STIX]{x1D70F})|$. From Strichartz’s estimate (Proposition 2.2) and its dual, we get

$$\begin{eqnarray}\displaystyle \hspace{57.0pt}\Vert (\langle k\rangle ^{1/2}|\tilde{u} |)^{3}\langle \unicode[STIX]{x1D70E}\rangle ^{-1/2}\Vert _{L^{2}(\unicode[STIX]{x1D706}\mathbb{T}\times \mathbb{R})} & {\lesssim} & \displaystyle \Vert v^{3}\Vert _{X^{0,-1/2}}\nonumber\\ \displaystyle & {\lesssim} & \displaystyle \Vert v^{3}\Vert _{L^{4/3}(\unicode[STIX]{x1D706}\mathbb{T}\times \mathbb{R})}\nonumber\\ \displaystyle & {\lesssim} & \displaystyle \Vert v\Vert _{L^{4}(\unicode[STIX]{x1D706}\mathbb{T}\times \mathbb{R})}^{3}\nonumber\\ \displaystyle & {\lesssim} & \displaystyle \Vert u\Vert _{_{X^{1/2,1/2}}}^{3}.\hspace{66.0pt}\square\nonumber\end{eqnarray}$$

Case 1b. We may assume that the size of the Fourier support of $u_{j}$ satisfies

$$\begin{eqnarray}\displaystyle |k_{1}| & {\sim} & \displaystyle |k_{4}|\gg |k_{2}|,|k_{3}|,\nonumber\\ \displaystyle |\unicode[STIX]{x1D70E}_{4}| & {\gtrsim} & \displaystyle |k_{2}+k_{3}|\,|k_{3}+k_{4}|\,|k_{4}+k_{2}|,\nonumber\\ \displaystyle \frac{1}{\unicode[STIX]{x1D706}} & {\leqslant} & \displaystyle |k_{2}+k_{3}|.\nonumber\end{eqnarray}$$

We prove the following estimate of the quadrilinear functional on $\mathbb{R}\times \unicode[STIX]{x1D706}\mathbb{T}$ with parameter $\unicode[STIX]{x1D706}\geqslant 1$.

Lemma 3.5. Under the above conditions, we have

(3.8)$$\begin{eqnarray}\displaystyle & & \displaystyle \left|\int _{\ast }\int _{\ast \ast }\langle k_{4}\rangle ^{1+s}\langle \unicode[STIX]{x1D70E}_{4}\rangle ^{-1/2}\tilde{u} _{1}\tilde{u} _{2}\tilde{u} _{3}(\langle k_{4}\rangle ^{-s}\langle \unicode[STIX]{x1D70E}_{4}\rangle ^{1/2}\tilde{u} _{4})\right|\nonumber\\ \displaystyle & & \displaystyle \quad \lesssim \Vert u_{1}\Vert _{X^{s,1/2}}\Vert u_{2}\Vert _{X^{0,1/2}}\Vert u_{3}\Vert _{X^{0,1/2}}\Vert u_{4}\Vert _{X^{-s,1/2}}.\end{eqnarray}$$

Proof. We first note that

(3.9)$$\begin{eqnarray}|\unicode[STIX]{x1D70E}_{4}|\gtrsim |k_{2}+k_{3}|\,|k_{1}|^{2}.\end{eqnarray}$$

From the Plancherel theorem, inequality (3.9) and the Sobolev embedding, the left side of (3.8) can be bounded as follows:

(3.10)$$\begin{eqnarray}\displaystyle & & \displaystyle \left|\int _{\ast }\int _{\ast \ast }\langle k_{4}\rangle ^{1+s}\langle \unicode[STIX]{x1D70E}_{4}\rangle ^{-1/2}\tilde{u} _{1}\tilde{u} _{2}\tilde{u} _{3}(\langle k_{4}\rangle ^{-s}\langle \unicode[STIX]{x1D70E}_{4}\rangle ^{1/2}\tilde{u} _{4})\right|\nonumber\\ \displaystyle & & \displaystyle \quad \lesssim \int _{\ast }\int _{\ast \ast }\langle k_{1}\rangle ^{s}|\tilde{u} _{1}(k_{1})|(|k_{2}+k_{3}|^{-1/2}|\tilde{u} _{2}(k_{2})|\,|\tilde{u} _{3}(k_{3})|)\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,|\unicode[STIX]{x1D70E}_{4}|^{1/2}|k_{4}|^{-s}|\tilde{u} _{4}(k_{4})|\,d\unicode[STIX]{x1D70F}\nonumber\\ \displaystyle & & \displaystyle \quad \lesssim \Vert D_{x}^{s}v_{1}\Vert _{L^{4}(\mathbb{R}\times \unicode[STIX]{x1D706}\mathbb{T})}\Vert D_{x}^{-1/2}(v_{2}v_{3})\Vert _{L^{4}(\mathbb{R}\times \unicode[STIX]{x1D706}\mathbb{T})}\Vert v_{4}\Vert _{X^{-s,1/2}}\nonumber\\ \displaystyle & & \displaystyle \quad \lesssim \Vert v_{1}\Vert _{X^{s,1/3+}}\Vert (v_{2}v_{3})\Vert _{L^{4}(\mathbb{R};L^{4/3}(\unicode[STIX]{x1D706}\mathbb{T}))}\Vert v_{4}\Vert _{X^{-s,1/2}},\end{eqnarray}$$

where $\tilde{v_{j}}=|\tilde{u_{j}}|$. Furthermore, by Hölder’s inequality and Corollary 2.3, we have

$$\begin{eqnarray}\displaystyle \hspace{48.0pt} & & \displaystyle \left|\int _{\ast }\int _{\ast \ast }\langle k_{4}\rangle ^{1+s}\langle \unicode[STIX]{x1D70E}_{4}\rangle ^{-1/2}\tilde{u} _{1}\tilde{u} _{2}\tilde{u} _{3}(\langle k_{4}\rangle ^{-s}\langle \unicode[STIX]{x1D70E}_{4}\rangle ^{1/2}\tilde{u} _{4})\right|\nonumber\\ \displaystyle & & \displaystyle \quad \lesssim \Vert v_{1}\Vert _{X^{s,1/3+}}\Vert v_{2}\Vert _{X^{0,5/12+}}\Vert v_{3}\Vert _{X^{0,5/12+}}\Vert v_{4}\Vert _{X^{-s,1/2}}.\hspace{33.60004pt}\square\nonumber\end{eqnarray}$$

Case 1c. Inequality (3.9) becomes

$$\begin{eqnarray}|\unicode[STIX]{x1D70E}_{4}|\gtrsim |k_{1}+k_{2}|\,|k_{1}|^{2}.\end{eqnarray}$$

Therefore, we can estimate Case 1c similarly to Case 1b.

Case 2. For this case, we divide the same subcases as in Case 1.

Case 2a. By the duality argument, Case 2a is in fact similar to Case 1a.

Case 2b. For this case, we wish to estimate

$$\begin{eqnarray}\left|\int _{\ast }\int _{\ast \ast }\langle k_{4}\rangle ^{1+s}\langle \unicode[STIX]{x1D70E}_{3}\rangle ^{-1/2}\tilde{u} _{1}\tilde{u} _{2}(\langle \unicode[STIX]{x1D70E}_{3}\rangle ^{1/2}\tilde{u} _{3})(\langle k_{4}\rangle ^{-s}\tilde{u} _{4})\right|.\end{eqnarray}$$

We also have

$$\begin{eqnarray}|\unicode[STIX]{x1D70E}_{3}|\gtrsim |k_{2}+k_{3}|\,|k_{1}|^{2}.\end{eqnarray}$$

Therefore,

$$\begin{eqnarray}\displaystyle & & \displaystyle \left|\int _{\ast }\int _{\ast \ast }\langle k_{4}\rangle ^{1+s}\langle \unicode[STIX]{x1D70E}_{3}\rangle ^{-1/2}\tilde{u} _{1}\tilde{u} _{2}(\langle \unicode[STIX]{x1D70E}_{3}\rangle ^{1/2}\tilde{u} _{3})(\langle k_{4}\rangle ^{-s}\tilde{u} _{4})\right|\nonumber\\ \displaystyle & & \displaystyle \quad \lesssim \iint _{\mathbb{R}\times \unicode[STIX]{x1D706}\mathbb{T}}(D_{x}^{s}v_{1})\cdot D_{x}^{-1/2}[v_{2}(\unicode[STIX]{x1D6EC}^{1/2}v_{3})]\cdot (D_{x}^{-s}v_{4})\,dt\,\text{d}x,\nonumber\end{eqnarray}$$

where ${\mathcal{F}}_{t,x}(\unicode[STIX]{x1D6EC}^{1/2}v_{3}):=\langle \unicode[STIX]{x1D70E}_{3}\rangle ^{1/2}|\tilde{u} _{3}|$. From Hölder’s inequality, Sobolev embedding in $x$, Proposition 2.2 and Corollary 2.3, we have

$$\begin{eqnarray}\displaystyle & & \displaystyle \iint _{\mathbb{R}\times \unicode[STIX]{x1D706}\mathbb{T}}(D_{x}^{s}v_{1})\cdot D_{x}^{-1/2}[v_{2}(\unicode[STIX]{x1D6EC}^{1/2}v_{3})]\cdot (D_{x}^{-s}v_{4})\,dt\,dx\nonumber\\ \displaystyle & & \displaystyle \quad \lesssim \Vert v_{1}\Vert _{X^{s,1/3+}}\Vert D_{x}^{-1/2}[v_{2}(\unicode[STIX]{x1D6EC}^{1/2}v_{3})]\Vert _{L_{t}^{8/5}L_{x}^{8/3}}\Vert (D_{x}^{-s}v_{4})\Vert _{L_{t}^{8}L_{x}^{8/3}}\nonumber\\ \displaystyle & & \displaystyle \quad \lesssim \Vert v_{1}\Vert _{X^{s,1/3+}}\Vert v_{2}\Vert _{L_{t}^{8}L_{x}^{8/3}}\Vert (\unicode[STIX]{x1D6EC}^{1/2}v_{3})\Vert _{L_{t,x}^{2}}\Vert v_{4}\Vert _{X^{-s,1/2}}\nonumber\\ \displaystyle & & \displaystyle \quad \lesssim \Vert v_{1}\Vert _{X^{s,1/3+}}\Vert v_{2}\Vert _{X^{0,5/12+}}\Vert v_{3}\Vert _{X^{0,1/2}}\Vert v_{4}\Vert _{X^{-s,1/2}}.\nonumber\end{eqnarray}$$

Case 2c. This case can be handled similarly to Case 1b or Case 1c.

For the resonant part (the second term of operator $J$ (3.1)), the proof is similar to Lemma 3.4 with $M$ defined in the formula (3.7) changes to the following:

$$\begin{eqnarray}\hspace{114.0pt}{\mathcal{F}}_{x}[M(u,u,u)]=|\hat{u} (k)|^{3}.\hspace{98.39996pt}\square\end{eqnarray}$$

Remark 3.6. The argument in Cases 1b, 1c, 2b, 2c in above proof actually shows the following estimate. For $s\geqslant 0$, it holds that

$$\begin{eqnarray}\mathop{\left\Vert {\mathcal{F}}^{-1}\left[k\int _{\unicode[STIX]{x1D6FA}_{k}}\mathop{\prod }_{j=1}^{3}\hat{u} _{j}(k_{j},t)(dk_{j})_{\unicode[STIX]{x1D706}}\right]\right\Vert }\nolimits_{X^{s,-1/2}}\lesssim \Vert u_{1}\Vert _{X^{s,1/2}}\Vert u_{2}\Vert _{X^{0,1/2}}\Vert u_{3}\Vert _{X^{0,1/2}},\end{eqnarray}$$

where

$$\begin{eqnarray}\unicode[STIX]{x1D6FA}_{k}:=\left\{\begin{array}{@{}lr@{}} & k_{1}+k_{2}+k_{3}=k\\ (k_{1},k_{2},k_{3})\in (\mathbb{Z}/\unicode[STIX]{x1D706})^{3}: & (k_{1}+k_{2})(k_{2}+k_{3})(k_{3}+k_{1})\neq 0\\ & |k_{1}|\geqslant |k_{2}|\geqslant |k_{3}|,|k_{1}|\gg |k_{3}|\end{array}\right\}.\end{eqnarray}$$

Remark 3.7. We can verify the estimate:

(3.11)$$\begin{eqnarray}\displaystyle \Vert J[u,v,w]\Vert _{X^{0,-1/2}} & {\lesssim} & \displaystyle \min (\Vert u\Vert _{X^{s,1/2}}\Vert v\Vert _{X^{s,1/2}}\Vert w\Vert _{X^{0,1/2}},\nonumber\\ \displaystyle & & \displaystyle \Vert u\Vert _{X^{s,1/2}}\Vert v\Vert _{X^{0,1/2}}\Vert w\Vert _{X^{s,1/2}},\nonumber\\ \displaystyle & & \displaystyle \Vert u\Vert _{X^{0,1/2}}\Vert v\Vert _{X^{s,1/2}}\Vert w\Vert _{X^{s,1/2}} ),\end{eqnarray}$$

for $s\geqslant 1/2$, or equivalently

$$\begin{eqnarray}\displaystyle & & \displaystyle \max (\Vert D_{x}^{-s}J[D_{x}^{s}u,v,w]\Vert _{X^{s,-1/2}},\nonumber\\ \displaystyle & & \displaystyle \quad \Vert D_{x}^{-s}J[u,D_{x}^{s}v,w]\Vert _{X^{s,-1/2}},\nonumber\\ \displaystyle & & \displaystyle \quad \Vert D_{x}^{-s}J[u,v,D_{x}^{s}w]\Vert _{X^{s,-1/2}} )\lesssim \Vert u\Vert _{X^{s,1/2}}\Vert v\Vert _{X^{s,1/2}}\Vert w\Vert _{X^{s,1/2}},\nonumber\end{eqnarray}$$

holds from Proposition 3.1.

Now, we prove the trilinear estimate corresponding to the function space $Z^{s}$:

Proposition 3.8. For $s\geqslant 1/2$ and $u$, $v$, $w\in X^{s,1/2}$, we have

(3.12)$$\begin{eqnarray}\Vert J[u,v,w]\Vert _{Z^{s}}\leqslant C\Vert u\Vert _{X^{s,1/2}}\Vert v\Vert _{X^{s,1/2}}\Vert w\Vert _{X^{s,1/2}}.\end{eqnarray}$$

Proof. From Proposition 3.1, it is enough to show

$$\begin{eqnarray}\Vert \langle k\rangle ^{s}\langle \unicode[STIX]{x1D70E}\rangle ^{-1}J[u,v,w]\Vert _{L_{(dk)_{\unicode[STIX]{x1D706}}}^{2}L_{d\unicode[STIX]{x1D70F}}^{1}}\leqslant C\Vert u\Vert _{X^{s,1/2}}\Vert v\Vert _{X^{s,1/2}}\Vert w\Vert _{X^{s,1/2}}.\end{eqnarray}$$

Similar to Proposition 3.1, we also divide this problem into the following four cases.

1. (1) $|\unicode[STIX]{x1D70E}|=\max \{|\unicode[STIX]{x1D70E}|,|\unicode[STIX]{x1D70E}_{i}|~\text{for}~1\leqslant i\leqslant 3\}$.

2. (2) $|\unicode[STIX]{x1D70E}_{1}|=\max \{|\unicode[STIX]{x1D70E}|,|\unicode[STIX]{x1D70E}_{i}|~\text{for}~1\leqslant i\leqslant 3\}$.

3. (3) $|\unicode[STIX]{x1D70E}_{2}|=\max \{|\unicode[STIX]{x1D70E}|,|\unicode[STIX]{x1D70E}_{i}|~\text{for}~1\leqslant i\leqslant 3\}$.

4. (4) $|\unicode[STIX]{x1D70E}_{3}|=\max \{|\unicode[STIX]{x1D70E}|,|\unicode[STIX]{x1D70E}_{i}|~\text{for}~1\leqslant i\leqslant 3\}$.

Case 1 is the worst one. Indeed, otherwise we have by Schwarz’s inequality,

$$\begin{eqnarray}\displaystyle & & \displaystyle \!\!\!\left\Vert \langle k\rangle ^{s}\langle k\rangle \langle \unicode[STIX]{x1D70E}\rangle ^{-1}\mathop{\sum }_{k}\hat{u} _{1}\hat{u} _{2}\hat{u} _{3}\right\Vert _{L_{(dk)_{\unicode[STIX]{x1D706}}}^{2}L_{\unicode[STIX]{x1D70F}}^{1}}\nonumber\\ \displaystyle & & \displaystyle \!\!\!\!\quad \lesssim \left\Vert \left(\int _{-\infty }^{\infty }\frac{1}{\langle \unicode[STIX]{x1D70E}\rangle ^{2(1/2+\unicode[STIX]{x1D716})}}\,d\unicode[STIX]{x1D70F}\!\right)^{\!1/2}\!\left(\int _{-\infty }^{\infty }\frac{\langle k\rangle ^{2s}\langle k\rangle ^{2}}{\langle \unicode[STIX]{x1D70E}\rangle ^{2(1/2-\unicode[STIX]{x1D716})}}\left|\mathop{\sum }_{k}\hat{u} _{1}\hat{u} _{2}\hat{u} _{3}\right|^{2}d\unicode[STIX]{x1D70F}\!\right)^{\!1/2}\right\Vert _{L_{(dk)_{\unicode[STIX]{x1D706}}}^{2}}\!\!.\nonumber\\ \displaystyle & & \displaystyle \!\!\!\!\quad \lesssim C\left\Vert \frac{\langle k\rangle \langle k\rangle ^{s}}{\langle \unicode[STIX]{x1D70E}\rangle ^{(1/2-\unicode[STIX]{x1D716})}}\mathop{\sum }_{k}\hat{u} _{1}\hat{u} _{2}\hat{u} _{3}\right\Vert _{L_{(dk)_{\unicode[STIX]{x1D706}}}^{2}L_{\unicode[STIX]{x1D70F}}^{2}}\nonumber\end{eqnarray}$$

and hence it reduces to the same proof as in Proposition 3.1. Therefore, we only have to prove Case 1. From symmetry, assume that $|k_{1}|\geqslant |k_{2}|\geqslant |k_{3}|$. We divide Case 1 into further three cases as follows:

1. (1a) $|k_{1}|\sim |k_{2}|\sim |k_{3}|$.

2. (1b) $|k_{1}|\gg |k_{2}|\gtrsim |k_{3}|$.

3. (1c) $|k_{1}|\sim |k_{2}|\gg |k_{3}|$.

Case 1a. By the Schwarz’s inequality, we have

$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{-\infty }^{\infty }\langle \unicode[STIX]{x1D70E}\rangle ^{-1}|{\mathcal{F}}_{t,x}[M(u,u,u)]|\,d\unicode[STIX]{x1D70F}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant \left(\int _{-\infty }^{\infty }\langle \unicode[STIX]{x1D70E}\rangle ^{-1-\unicode[STIX]{x1D716}}\,d\unicode[STIX]{x1D70F}\right)^{1/2}\left(\int _{-\infty }^{\infty }\langle \unicode[STIX]{x1D70E}\rangle ^{-1+\unicode[STIX]{x1D716}}|{\mathcal{F}}_{t,x}[M(u,u,u)]|^{2}\,d\unicode[STIX]{x1D70F}\right)^{1/2},\nonumber\end{eqnarray}$$

where $M$ is defined in (3.7). This case is reduced to Lemma 3.4.

Case 1b. In this case, we can clearly see that $|\unicode[STIX]{x1D70E}|\gtrsim |k_{2}+k_{3}|\,|k|^{2}$. This implies that

$$\begin{eqnarray}\langle k\rangle ^{1+s}\langle \unicode[STIX]{x1D70E}\rangle ^{-1}\lesssim \langle k_{1}\rangle ^{s}\langle k_{2}+k_{3}\rangle ^{-2/5}|k_{2}+k_{3}|^{-7/10}\langle \unicode[STIX]{x1D70E}_{1}\rangle ^{-1/10}\langle \unicode[STIX]{x1D70E}_{2}\rangle ^{-1/10}\langle \unicode[STIX]{x1D70E}_{3}\rangle ^{-1/10}.\end{eqnarray}$$

Remark 3.9. Note that above argument works only when $|k_{1}|\gtrsim 1$. However, the low frequency case ($|k_{j}|\ll 1$ for all $j$) is very easy for this problem and we can omit it.

Thus, it suffices to estimate

$$\begin{eqnarray}\displaystyle & & \displaystyle \Vert (\langle k_{1}\rangle ^{s}\langle \unicode[STIX]{x1D70E}_{1}\rangle ^{-1/10}\tilde{u} _{1})\nonumber\\ \displaystyle & & \displaystyle \quad \ast \,[\langle k^{\prime }\rangle ^{-2/5}|k^{\prime }|^{-7/10}1_{k^{\prime }\neq 0}\{(\langle \unicode[STIX]{x1D70E}_{2}\rangle ^{-1/10}\tilde{u} _{2})\ast (\langle \unicode[STIX]{x1D70E}_{3}\rangle ^{-1/10}\tilde{u} _{3})\}]\Vert _{L_{(dk)_{\unicode[STIX]{x1D706}}}^{2}L_{\unicode[STIX]{x1D70F}}^{1}},\nonumber\end{eqnarray}$$

where $\ast$ means convolution in $\unicode[STIX]{x1D70F},k$. Since, $\langle k^{\prime }\rangle ^{-2/5}|k^{\prime }|^{-7/10}1_{k^{\prime }\neq 0}$ is integrable with respect to $(dk^{\prime })_{\unicode[STIX]{x1D706}}$ and bound is uniform in $\unicode[STIX]{x1D706}$, we apply Young’s inequality in $\unicode[STIX]{x1D70F},k$ and Hölder’s inequality in $k$ to estimate it by

$$\begin{eqnarray}\Vert \langle k\rangle ^{s}\langle \unicode[STIX]{x1D70E}\rangle ^{-1/10}\tilde{u} _{1}\Vert _{L_{(dk)_{\unicode[STIX]{x1D706}}}^{2}L_{\unicode[STIX]{x1D70F}}^{1}}\Vert \langle \unicode[STIX]{x1D70E}\rangle ^{-1/10}\tilde{u} _{2}\Vert _{L_{(dk)_{\unicode[STIX]{x1D706}}}^{2}L_{\unicode[STIX]{x1D70F}}^{1}}\Vert \langle \unicode[STIX]{x1D70E}\rangle ^{-1/10}\tilde{u} _{3}\Vert _{L_{(dk)_{\unicode[STIX]{x1D706}}}^{2}L_{\unicode[STIX]{x1D70F}}^{1}},\end{eqnarray}$$

which is dominated by $\Vert u_{1}\Vert _{X^{s,1/2}}\Vert u_{2}\Vert _{X^{0,1/2}}\Vert u_{3}\Vert _{X^{0,1/2}}$.

Case 1c. This case is analogous. The only difference is that in this case we have $|\unicode[STIX]{x1D70E}|\gtrsim |k_{1}+k_{2}|\,|k_{1}|^{2}$, and thus $u_{1},u_{2}$ play the role of $u_{2},u_{3}$ in Case 1b.

The estimate for the resonant term follows in the same way as Case 1a. ◻

Remark 3.10. The following estimate has been shown by Remark 3.6 and the argument in Cases 1b, 1c of the above proof: for any $s\geqslant 0$, we have

$$\begin{eqnarray}\mathop{\left\Vert {\mathcal{F}}^{-1}\left[k\int _{\unicode[STIX]{x1D6FA}_{k}}\mathop{\prod }_{j=1}^{3}\hat{u} _{j}(k_{j},t)(dk_{j})_{\unicode[STIX]{x1D706}}\right]\right\Vert }\nolimits_{Z^{s}}\lesssim \Vert u_{1}\Vert _{X^{s,1/2}}\Vert u_{2}\Vert _{X^{0,1/2}}\Vert u_{3}\Vert _{X^{0,1/2}},\end{eqnarray}$$

where $\unicode[STIX]{x1D6FA}_{k}$ is given in Remark 3.6.

4 A priori estimate

In this section, we show a priori estimate of the solution to the mKdV equation which are needed for the proof of Theorem 1.1. The energy for the mKdV equation is given as:

$$\begin{eqnarray}E(u)=\int (\unicode[STIX]{x2202}_{x}u)^{2}-(u)^{4}\,dx.\end{eqnarray}$$

For the operator $I^{\prime }$, we have

$$\begin{eqnarray}E(I^{\prime }v)=\int (\unicode[STIX]{x2202}_{x}I^{\prime }v)^{2}-(I^{\prime }v)^{4}\,dx.\end{eqnarray}$$

From equations (2.1)–(2.2), we obtain

(4.1)$$\begin{eqnarray}\displaystyle \quad \frac{d}{dt}\Vert v(t)\Vert _{L^{2}}^{2} & = & \displaystyle 2\int v(\unicode[STIX]{x1D706}^{-3}g-\unicode[STIX]{x1D706}^{-3}\unicode[STIX]{x1D6FE}v),\end{eqnarray}$$

(4.2)$$\begin{eqnarray}\displaystyle \frac{d(E(I^{\prime }v))}{dt} & = & \displaystyle \left[\int (-2\unicode[STIX]{x2202}_{x}^{2}I^{\prime }v-4(I^{\prime }v)^{3})(-\unicode[STIX]{x2202}_{x}^{3}I^{\prime }v-2\unicode[STIX]{x2202}_{x}I^{\prime }v^{3})\right]\nonumber\\ \displaystyle & & \displaystyle +\left[\int (-2\unicode[STIX]{x2202}_{x}^{2}I^{\prime }v-4(I^{\prime }v)^{3})(\unicode[STIX]{x1D706}^{-3}I^{\prime }g-\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}I^{\prime }v)\right].\end{eqnarray}$$

We establish the following lemma:

Lemma 4.1. Let $v$ be the solution of IVP (2.1)–(2.2) for $t\in [0,T^{\prime }]$. Then, we have

(4.3)$$\begin{eqnarray}\Vert I^{\prime }v(T^{\prime })\Vert _{L^{2}}^{2}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}T^{\prime })\leqslant C_{1}\left(\Vert v(0)\Vert _{L^{2}}^{2}+\frac{1}{\unicode[STIX]{x1D6FE}^{2}}\Vert g\Vert _{L^{2}}^{2}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}T^{\prime })\right)\end{eqnarray}$$

and

(4.4)$$\begin{eqnarray}\displaystyle & & \displaystyle \Vert I^{\prime }v(T^{\prime })\Vert _{{\dot{H}}^{1}}^{2}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}T^{\prime })\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant C_{1}\biggl(\Vert I^{\prime }v(0)\Vert _{{\dot{H}}^{1}}^{2}+\frac{1}{\unicode[STIX]{x1D6FE}^{2}}\Vert I^{\prime }g\Vert _{L_{T^{\prime }}^{\infty }{\dot{H}}^{1}}^{2}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}T^{\prime })\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\Vert v(0)\Vert _{L^{2}}^{6}+\frac{1}{\unicode[STIX]{x1D6FE}^{6}}\Vert g\Vert _{L^{2}}^{6}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}T^{\prime })\biggr)+\left|\int _{0}^{T^{\prime }}M(t)\,dt\right|,\end{eqnarray}$$

where

$$\begin{eqnarray}M(t)=\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}t)\int _{\unicode[STIX]{x1D706}\mathbb{ T}}\{-2\unicode[STIX]{x2202}_{x}^{2}I^{\prime }v-4(I^{\prime }v)^{3}\}\{-2\unicode[STIX]{x2202}_{x}I^{\prime }v^{3}-\unicode[STIX]{x2202}_{x}^{3}I^{\prime }v\}.\end{eqnarray}$$

Proof. From (4.1), we have

$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{d}{dt}(\Vert v(t)\Vert _{L^{2}}^{2}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}t))\nonumber\\ \displaystyle & & \displaystyle \quad =\left(-\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}\Vert v(t)\Vert _{L^{2}}^{2}+2\unicode[STIX]{x1D706}^{-3}\int _{\unicode[STIX]{x1D706}\mathbb{T}}v(t)g(t)\,dx\right)\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}t)\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant \frac{\unicode[STIX]{x1D706}^{-3}}{\unicode[STIX]{x1D6FE}}\Vert g\Vert _{L^{2}}^{2}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}t).\nonumber\end{eqnarray}$$

Integrating over $[0,T^{\prime }]$ and from the definition of operator $I$, we get (4.3).

From equation (4.2), we get

$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{d}{dt}(E(I^{\prime }v(t))\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}t))\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{d}{dt}E(I^{\prime }v(t))\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}t)+\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}E(I^{\prime }v(t))\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}t)\nonumber\\ \displaystyle & & \displaystyle \quad =\left[\int (-2\unicode[STIX]{x2202}_{x}^{2}I^{\prime }v-4(I^{\prime }v)^{3})(-\unicode[STIX]{x2202}_{x}^{3}I^{\prime }v-\unicode[STIX]{x2202}_{x}I^{\prime }v^{3})\right]\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}t)\nonumber\\ \displaystyle & & \displaystyle \qquad +\left[\int (-2\unicode[STIX]{x2202}_{x}^{2}I^{\prime }v-4(I^{\prime }v)^{3})(\unicode[STIX]{x1D706}^{-3}I^{\prime }g-\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}I^{\prime }v)\right]\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}t)\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}t)\int (\unicode[STIX]{x2202}_{x}I^{\prime }v)^{2}-(I^{\prime }v)^{4}\nonumber\\ \displaystyle & & \displaystyle \quad =M(t)+\biggl[\int -2\unicode[STIX]{x1D706}^{-3}\unicode[STIX]{x2202}_{x}^{2}I^{\prime }vI^{\prime }g-4\unicode[STIX]{x1D706}^{-3}(I^{\prime }v)^{3}I^{\prime }g\nonumber\\ \displaystyle & & \displaystyle \qquad -\,\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}(\unicode[STIX]{x2202}_{x}I^{\prime }v)^{2}+3(I^{\prime }v)^{4}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}\biggr]\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}t).\nonumber\end{eqnarray}$$

Put the value of $E$, integrate over $[0,T^{\prime }]$, take absolute value on both side and from Gagliardo–Nirenberg inequality, we get

$$\begin{eqnarray}\displaystyle & & \displaystyle \left(\Vert I^{\prime }v(T^{\prime })\Vert _{{\dot{H}}^{1}}^{2}-\Vert I^{\prime }v(T^{\prime })\Vert _{L^{4}}^{4}\right)\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}T^{\prime })\nonumber\\ \displaystyle & & \displaystyle \quad =\Vert I^{\prime }v(0)\Vert _{{\dot{H}}^{1}}^{2}-\Vert I^{\prime }v(0)\Vert _{L^{4}}^{4}+\int _{0}^{T^{\prime }}M(t^{\prime })dt^{\prime }+\int _{0}^{T^{\prime }}\biggl[\int -2\unicode[STIX]{x1D706}^{-3}\unicode[STIX]{x2202}_{x}^{2}I^{\prime }vI^{\prime }g\nonumber\\ \displaystyle & & \displaystyle \qquad -\,4\unicode[STIX]{x1D706}^{-3}(I^{\prime }v)^{3}I^{\prime }g-\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}(\unicode[STIX]{x2202}_{x}I^{\prime }v)^{2}+3(I^{\prime }v)^{4}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}\biggr]\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}t^{\prime })\,dt^{\prime }\nonumber\\ \displaystyle & & \displaystyle \quad \lesssim \Vert I^{\prime }v(0)\Vert _{{\dot{H}}^{1}}^{2}+\left|\int _{0}^{T^{\prime }}M(t^{\prime })\,dt^{\prime }\right|+\unicode[STIX]{x1D706}^{-3}\int _{0}^{T^{\prime }}[\Vert I^{\prime }g\Vert _{{\dot{H}}^{1}}\Vert I^{\prime }v(t^{\prime })\Vert _{{\dot{H}}^{1}}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\Vert I^{\prime }v(t^{\prime })\Vert _{{\dot{H}}^{1}}\Vert I^{\prime }v(t^{\prime })\Vert _{L^{2}}^{2}\Vert I^{\prime }g\Vert _{L^{2}}-\unicode[STIX]{x1D6FE}\Vert I^{\prime }v(t^{\prime })\Vert _{{\dot{H}}^{1}}^{2}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\unicode[STIX]{x1D6FE}\Vert I^{\prime }v(t^{\prime })\Vert _{{\dot{H}}^{1}}\Vert I^{\prime }v(t^{\prime })\Vert _{L^{2}}^{3}]\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}t^{\prime })\,dt^{\prime }.\nonumber\end{eqnarray}$$

From Young’s inequality, we have

$$\begin{eqnarray}\displaystyle & & \displaystyle \Vert I^{\prime }v(T^{\prime })\Vert _{{\dot{H}}^{1}}^{2}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}T^{\prime })\lesssim \Vert I^{\prime }v(0)\Vert _{{\dot{H}}^{1}}^{2}+\frac{1}{\unicode[STIX]{x1D6FE}^{2}}\Vert I^{\prime }g\Vert _{{\dot{H}}^{1}}^{2}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}T^{\prime })\nonumber\\ \displaystyle & & \displaystyle \quad +\left|\int _{0}^{T^{\prime }}M(t^{\prime })\,dt^{\prime }\right|+\Vert I^{\prime }v(T^{\prime })\Vert _{L^{2}}^{6}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}T^{\prime })\nonumber\\ \displaystyle & & \displaystyle \quad +\int _{0}^{T^{\prime }}\left(\Vert I^{\prime }v(t^{\prime })\Vert _{L^{2}}^{6}+\frac{1}{\unicode[STIX]{x1D6FE}^{2}}\Vert I^{\prime }v(t^{\prime })\Vert _{L^{2}}^{4}\Vert I^{\prime }g\Vert _{L^{2}}^{2}\right)\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}t^{\prime })\,dt^{\prime }.\nonumber\end{eqnarray}$$

From inequality (4.3) we get

$$\begin{eqnarray}\displaystyle & & \displaystyle \left(\Vert I^{\prime }v(t^{\prime })\Vert _{L^{2}}^{6}+\frac{1}{\unicode[STIX]{x1D6FE}^{2}}\Vert I^{\prime }v(t^{\prime })\Vert _{L^{2}}^{4}\Vert I^{\prime }g\Vert _{L^{2}}^{2}\right)\nonumber\\ \displaystyle & & \displaystyle \quad \lesssim \Vert v(0)\Vert _{L^{2}}^{6}\exp (-3\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}t^{\prime })+\frac{1}{\unicode[STIX]{x1D6FE}^{6}}\Vert I^{\prime }g\Vert _{L^{2}}^{6},\nonumber\end{eqnarray}$$

and hence we obtain inequality (4.4).◻

We need to state the following Leibniz rule type lemma:

Lemma 4.3. For $b\geqslant 0$, we have

$$\begin{eqnarray}\displaystyle & & \displaystyle \Vert f(t)g(x,t)\Vert _{X^{s,b}}\lesssim \Vert \hat{f}\Vert _{L^{1}}\Vert g\Vert _{X^{s,b}}+\Vert f\Vert _{H_{t}^{b}}\Vert \langle k\rangle ^{s}\tilde{g}\Vert _{L_{(dk)_{\unicode[STIX]{x1D706}}}^{2}L_{d\unicode[STIX]{x1D70F}}^{1}}.\nonumber\\ \displaystyle & & \displaystyle \Vert \langle k\rangle ^{s}{\mathcal{F}}_{t,x}[f(t)g(x,t)]\Vert _{L_{(dk)_{\unicode[STIX]{x1D706}}}^{2}L_{\unicode[STIX]{x1D70F}}^{1}}\lesssim \Vert \hat{f}\Vert _{L^{1}}\Vert \langle k\rangle ^{s}\tilde{g}\Vert _{L_{(dk)_{\unicode[STIX]{x1D706}}}^{2}L_{\unicode[STIX]{x1D70F}}^{1}}.\nonumber\end{eqnarray}$$

Proof. We prove first statement and the proof for second statement is almost the same as that for $X^{s,0}$. Assume that $\unicode[STIX]{x1D70F}=\unicode[STIX]{x1D70F}_{1}+\unicode[STIX]{x1D70F}_{2}$. Then

$$\begin{eqnarray}\langle \unicode[STIX]{x1D70F}-k^{3}\rangle ^{b}\lesssim \langle \unicode[STIX]{x1D70F}_{1}\rangle ^{b}+\langle \unicode[STIX]{x1D70F}-\unicode[STIX]{x1D70F}_{1}-k^{3}\rangle ^{b}.\end{eqnarray}$$

Hence

$$\begin{eqnarray}\displaystyle & & \displaystyle \langle \unicode[STIX]{x1D70F}-k^{3}\rangle ^{b}\langle k\rangle ^{s}{\mathcal{F}}[f(t)g(x,t)]=\langle \unicode[STIX]{x1D70F}-k^{3}\rangle ^{b}\langle k\rangle ^{s}\int _{\unicode[STIX]{x1D70F}_{1}}\hat{f}(\unicode[STIX]{x1D70F}_{1})\tilde{g}(k,\unicode[STIX]{x1D70F}-\unicode[STIX]{x1D70F}_{1})\,d\unicode[STIX]{x1D70F}_{1}\nonumber\\ \displaystyle & & \displaystyle \quad \lesssim \langle k\rangle ^{s}\int _{\unicode[STIX]{x1D70F}_{1}}\langle \unicode[STIX]{x1D70F}_{1}\rangle ^{b}|\hat{f}(\unicode[STIX]{x1D70F}_{1})\tilde{g}(k,\unicode[STIX]{x1D70F}-\unicode[STIX]{x1D70F}_{1})|+\langle \unicode[STIX]{x1D70F}-\unicode[STIX]{x1D70F}_{1}-k^{3}\rangle ^{b}|\hat{f}(\unicode[STIX]{x1D70F}_{1})\tilde{g}(k,\unicode[STIX]{x1D70F}-\unicode[STIX]{x1D70F}_{1})|\,d\unicode[STIX]{x1D70F}_{1}.\nonumber\end{eqnarray}$$

After taking $L^{2}$ norm, we get

$$\begin{eqnarray}\displaystyle & & \displaystyle \Vert \langle \unicode[STIX]{x1D70F}-k^{3}\rangle ^{b}\langle k\rangle ^{s}{\mathcal{F}}[f(t)g(x,t)]\Vert _{L_{k,\unicode[STIX]{x1D70F}}^{2}}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant \Vert \langle k\rangle ^{s}(\langle \unicode[STIX]{x1D70F}_{1}\rangle ^{b}\hat{f})\ast \tilde{g}\Vert _{L_{t,k}^{2}}+\Vert \langle k\rangle ^{s}\hat{f}\ast (\langle \unicode[STIX]{x1D70F}-k^{3}\rangle ^{b}\tilde{g})\Vert _{L_{t,k}^{2}}.\nonumber\end{eqnarray}$$

From Young’s inequality in $\unicode[STIX]{x1D70F}$, we obtain

$$\begin{eqnarray}\displaystyle \hspace{66.0pt} & & \displaystyle \Vert \langle k\rangle ^{s}(\langle \unicode[STIX]{x1D70F}_{1}\rangle ^{b}\hat{f})\ast \tilde{g}\Vert _{L_{\unicode[STIX]{x1D70F}}^{2}}+\Vert \langle k\rangle ^{s}\hat{f}\ast (\langle \unicode[STIX]{x1D70F}-k^{3}\rangle ^{b}\tilde{g})\Vert _{L_{\unicode[STIX]{x1D70F}}^{2}}\nonumber\\ \displaystyle & & \displaystyle \quad \lesssim \Vert \hat{f}\Vert _{L^{1}}\Vert g\Vert _{X^{s,b}}+\Vert f\Vert _{H_{t}^{b}}\Vert \langle k\rangle ^{s}\tilde{g}\Vert _{L_{(dk)_{\unicode[STIX]{x1D706}}}^{2}L_{d\unicode[STIX]{x1D70F}}^{1}}.\hspace{51.60004pt}\square\nonumber\end{eqnarray}$$

Similarly to [Reference Tsugawa23, Proposition 3.1], we finally have the following proposition:

Proposition 4.4. Let $1/2\leqslant s<1$. Let $T>0$ is given, $\unicode[STIX]{x1D716}>0$ be sufficiently small and $u$ be a solution of IVP (1.5)–(1.6) on $[0,T]$. Assume that $N^{(1/2)(1-\unicode[STIX]{x1D716})}\geqslant \unicode[STIX]{x1D6FE},N^{\unicode[STIX]{x1D716}}\geqslant C_{2}T$ and

$$\begin{eqnarray}\displaystyle \left(\Vert u(0)\Vert _{L^{2}}^{2}+\frac{1}{\unicode[STIX]{x1D6FE}^{2}}\Vert f\Vert _{L^{2}}^{2}\exp (\unicode[STIX]{x1D6FE}T)\right) & {\leqslant} & \displaystyle N^{(1/6)(1-\unicode[STIX]{x1D716})}\nonumber\\ \displaystyle \left(\Vert Iu(0)\Vert _{{\dot{H}}^{1}}^{2}+\frac{1}{\unicode[STIX]{x1D6FE}^{2}}\Vert If\Vert _{{\dot{H}}^{1}}^{2}\exp (\unicode[STIX]{x1D6FE}T)\right) & {\leqslant} & \displaystyle N^{(1/6)(1-\unicode[STIX]{x1D716})}.\nonumber\end{eqnarray}$$

Then, we have

$$\begin{eqnarray}\displaystyle \Vert Iu(T)\Vert _{L^{2}}^{2}\exp (\unicode[STIX]{x1D6FE}T) & {\leqslant} & \displaystyle C_{1}\biggl(\Vert u(0)\Vert _{L^{2}}^{2}+\frac{1}{\unicode[STIX]{x1D6FE}^{2}}\Vert f\Vert _{L^{2}}^{2}\exp (\unicode[STIX]{x1D6FE}T)\biggr),\nonumber\\ \displaystyle \Vert Iu(T)\Vert _{{\dot{H}}^{1}}^{2}\exp (\unicode[STIX]{x1D6FE}T) & {\leqslant} & \displaystyle C_{1}\biggl(\Vert Iu(0)\Vert _{{\dot{H}}^{1}}^{2}+\Vert u(0)\Vert _{L^{2}}^{6}+\frac{1}{\unicode[STIX]{x1D6FE}^{6}}\Vert f\Vert _{L^{2}}^{6}\exp (\unicode[STIX]{x1D6FE}T)\nonumber\\ \displaystyle & & \displaystyle +\,\frac{1}{\unicode[STIX]{x1D6FE}^{2}}\Vert If\Vert _{{\dot{H}}^{1}}^{2}\exp (\unicode[STIX]{x1D6FE}T)\biggr)\nonumber\\ \displaystyle & & \displaystyle +\,\biggl(\Vert Iu(0)\Vert _{H^{1}}^{2}+\frac{1}{\unicode[STIX]{x1D6FE}^{2}}\Vert If\Vert _{H^{1}}^{2}\exp (\unicode[STIX]{x1D6FE}T)\biggr),\nonumber\end{eqnarray}$$

where $C_{1}$ is the constant given in Lemma 4.1 and is independent of $N$ and $T$.

We can rescale Proposition 4.4 by taking $\unicode[STIX]{x1D706}=N^{(1/6)(1-\unicode[STIX]{x1D716})}$, $N^{\prime }=N/\unicode[STIX]{x1D706}$, $T^{\prime }=\unicode[STIX]{x1D706}^{3}T$. Also, we note that $\Vert I^{\prime }v\Vert _{{\dot{H}}^{1}}^{2}=\unicode[STIX]{x1D706}^{-3}\Vert Iu\Vert _{{\dot{H}}^{1}}^{2}$, $\Vert I^{\prime }g\Vert _{{\dot{H}}^{1}}^{2}=\unicode[STIX]{x1D706}^{-3}\Vert If\Vert _{{\dot{H}}^{1}}^{2}$. We rewrite Proposition 4.4 as follows:

Proposition 4.6. Let $1/2\leqslant s<1$, $T^{\prime }>0$ is given and let $v$ be a solution of IVP (2.1)–(2.2) on $[0,T^{\prime }]$. Assume that $\unicode[STIX]{x1D706}^{3}\geqslant \unicode[STIX]{x1D6FE}$ and that for suitable $C_{2}>0$, $N^{\prime }\geqslant C_{2}T^{\prime }\unicode[STIX]{x1D706}^{2}$ and

$$\begin{eqnarray}\displaystyle \left(\Vert v(0)\Vert _{L^{2}}^{2}+\frac{1}{\unicode[STIX]{x1D6FE}^{2}}\Vert g\Vert _{L^{2}}^{2}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}T^{\prime })\right) & {\leqslant} & \displaystyle 1\nonumber\\ \displaystyle \left(\Vert I^{\prime }v(0)\Vert _{{\dot{H}}^{1}}^{2}+\frac{1}{\unicode[STIX]{x1D6FE}^{2}}\Vert I^{\prime }g\Vert _{{\dot{H}}^{1}}^{2}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}T^{\prime })\right) & {\leqslant} & \displaystyle 1.\nonumber\end{eqnarray}$$

Then, we have

$$\begin{eqnarray}\displaystyle \Vert I^{\prime }v(T^{\prime })\Vert _{L^{2}}^{2}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}T^{\prime }) & {\leqslant} & \displaystyle C_{1}\left(\!\Vert v(0)\Vert _{L^{2}}^{2}+\frac{1}{\unicode[STIX]{x1D6FE}^{2}}\Vert g\Vert _{L^{2}}^{2}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}T^{\prime })\!\right)\nonumber\\ \displaystyle \Vert I^{\prime }v(T^{\prime })\Vert _{{\dot{H}}^{1}}^{2}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}T^{\prime }) & {\leqslant} & \displaystyle C_{1}\left(\!\Vert I^{\prime }v(0)\Vert _{{\dot{H}}^{1}}^{2}+\frac{1}{\unicode[STIX]{x1D6FE}^{2}}\Vert I^{\prime }g\Vert _{{\dot{H}}^{1}}^{2}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}T^{\prime })\right.\nonumber\\ \displaystyle & & \displaystyle +\left.\Vert v(0)\Vert _{L^{2}}^{6}+\frac{1}{\unicode[STIX]{x1D6FE}^{6}}\Vert g\Vert _{L_{T^{\prime }}^{\infty }L^{2}}^{6}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}T^{\prime })\!\right)\nonumber\\ \displaystyle & & \displaystyle +\,\unicode[STIX]{x1D706}^{-2}\left(\!\Vert I^{\prime }v(0)\Vert _{H^{1}}^{2}+\frac{1}{\unicode[STIX]{x1D6FE}^{2}}\Vert I^{\prime }g\Vert _{H^{1}}^{2}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}T^{\prime })\!\right)\!.\nonumber\end{eqnarray}$$

Proof of Proposition 4.6.

Take $\unicode[STIX]{x1D6FF}>0$ and $j\in \mathbb{N}$ such that $\unicode[STIX]{x1D6FF}j=T^{\prime }$ where $\unicode[STIX]{x1D6FF}\sim (\Vert I^{\prime }v(0)\Vert _{H^{1}}+\Vert g\Vert _{H^{1}}+\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3})^{-\unicode[STIX]{x1D6FC}}$, $\unicode[STIX]{x1D6FC}>0$. For $0\leqslant m\leqslant j$, $m\in \mathbb{Z}$, we prove

(4.5)$$\begin{eqnarray}\displaystyle & & \displaystyle \Vert I^{\prime }v(m\unicode[STIX]{x1D6FF})\Vert _{{\dot{H}}^{1}}^{2}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}m\unicode[STIX]{x1D6FF})\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant C_{1}\left(\Vert I^{\prime }v(0)\Vert _{{\dot{H}}^{1}}^{2}+\Vert v(0)\Vert _{L^{2}}^{6}+\frac{1}{\unicode[STIX]{x1D6FE}^{2}}\Vert I^{\prime }g\Vert _{{\dot{H}}^{1}}^{2}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}m\unicode[STIX]{x1D6FF})\right.\nonumber\\ \displaystyle & & \displaystyle \qquad +\left.\frac{1}{\unicode[STIX]{x1D6FE}^{6}}\Vert g\Vert _{L^{2}}^{6}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}k\unicode[STIX]{x1D6FF})\right)\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\unicode[STIX]{x1D706}^{-2}\left(\Vert I^{\prime }v(0)\Vert _{H^{1}}^{2}+\frac{1}{\unicode[STIX]{x1D6FE}^{2}}\Vert I^{\prime }g\Vert _{H^{1}}^{2}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}T^{\prime })\right)\end{eqnarray}$$

by induction. Note that, by the assumption, (4.5) would imply that

$$\begin{eqnarray}\displaystyle & & \displaystyle \Vert I^{\prime }v(m\unicode[STIX]{x1D6FF})\Vert _{{\dot{H}}^{1}}^{2}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}m\unicode[STIX]{x1D6FF})\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant (2C_{1}+1)\left(\Vert I^{\prime }v(0)\Vert _{{\dot{H}}^{1}}^{2}+\frac{1}{\unicode[STIX]{x1D6FE}^{2}}\Vert I^{\prime }g\Vert _{{\dot{H}}^{1}}^{2}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}m\unicode[STIX]{x1D6FF})\right)\leqslant 2C_{1}+1\nonumber\end{eqnarray}$$

and, combined with (4.3), that

$$\begin{eqnarray}\displaystyle & & \displaystyle (\Vert I^{\prime }v(m\unicode[STIX]{x1D6FF})\Vert _{H^{1}}^{2}+\unicode[STIX]{x1D706}^{-6}\Vert I^{\prime }g\Vert _{H^{1}}^{2})\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}m\unicode[STIX]{x1D6FF})\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant (3C_{1}+2)\left(\Vert I^{\prime }v(0)\Vert _{{\dot{H}}^{1}}^{2}+\frac{1}{\unicode[STIX]{x1D6FE}^{2}}\Vert I^{\prime }g\Vert _{{\dot{H}}^{1}}^{2}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}m\unicode[STIX]{x1D6FF})\right)\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant 3C_{1}+2.\nonumber\end{eqnarray}$$

For $m=0$, (4.5) holds trivially. We assume (4.5) holds true for $0\leqslant m\leqslant l$ where $0\leqslant l\leqslant j-1$. From Lemma 4.1, we have

(4.6)$$\begin{eqnarray}\displaystyle & & \displaystyle \Vert I^{\prime }v((l+1)\unicode[STIX]{x1D6FF})\Vert _{{\dot{H}}^{1}}^{2}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}(l+1)\unicode[STIX]{x1D6FF})\leqslant C_{1}\biggl(\Vert I^{\prime }v(0)\Vert _{{\dot{H}}^{1}}^{2}+\Vert v(0)\Vert _{L^{2}}^{6}\nonumber\\ \displaystyle & & \displaystyle \quad +\,\frac{1}{\unicode[STIX]{x1D6FE}^{2}}\Vert I^{\prime }g\Vert _{{\dot{H}}^{1}}^{2}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}(l+1)\unicode[STIX]{x1D6FF})+\frac{1}{\unicode[STIX]{x1D6FE}^{6}}\Vert g\Vert _{L^{2}}^{6}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}(l+1)\unicode[STIX]{x1D6FF})\nonumber\\ \displaystyle & & \displaystyle \quad +\left|\int _{0}^{(l+1)\unicode[STIX]{x1D6FF}}M(t)\,dt\right|.\end{eqnarray}$$

Therefore, it suffices to prove

$$\begin{eqnarray}\left|\int _{0}^{(l+1)\unicode[STIX]{x1D6FF}}M(t)\,dt\right|\leqslant \unicode[STIX]{x1D706}^{-2}\left(\Vert I^{\prime }v(0)\Vert _{H^{1}}+\frac{1}{\unicode[STIX]{x1D6FE}^{2}}\Vert I^{\prime }g\Vert _{H^{1}}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}T^{\prime })\right).\end{eqnarray}$$

We have the following estimate:

Lemma 4.8.

$$\begin{eqnarray}\displaystyle & & \displaystyle \left|\int _{0}^{T^{\prime }}\int _{\unicode[STIX]{x1D706}\mathbb{T}}\{-2\unicode[STIX]{x2202}_{x}^{2}I^{\prime }v-4(I^{\prime }v)^{3}\}\{-2\unicode[STIX]{x2202}_{x}I^{\prime }v^{3}-\unicode[STIX]{x2202}_{x}^{3}I^{\prime }v\}\,dt\right|\nonumber\\ \displaystyle & & \displaystyle \quad \lesssim (N^{\prime })^{-1}\Vert I^{\prime }v\Vert _{Y_{T^{\prime }}^{1}}\Vert I^{\prime }v\Vert _{X_{T^{\prime }}^{1,1/2}}^{3}+(N^{\prime })^{-2}\Vert I^{\prime }v\Vert _{X_{T^{\prime }}^{1,1/2}}^{6}.\nonumber\end{eqnarray}$$

We prove Lemma 4.8 in last section.

Lemma 4.8 implies that

$$\begin{eqnarray}\displaystyle \left|\int _{0}^{(l+1)\unicode[STIX]{x1D6FF}}\!M(t)\,dt\right| & {\leqslant} & \displaystyle \mathop{\sum }_{k=0}^{l}\left|\int _{k\unicode[STIX]{x1D6FF}}^{(k+1)\unicode[STIX]{x1D6FF}}M(t)\,dt\right|,\nonumber\\ \displaystyle & {\lesssim} & \displaystyle (N^{\prime })^{-1}\mathop{\sum }_{k=0}^{l}\left\Vert \exp \left(\frac{1}{4}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}t\right)I^{\prime }v\right\Vert _{Y_{([0,\unicode[STIX]{x1D706}]\times [k\unicode[STIX]{x1D6FF},(k+1)\unicode[STIX]{x1D6FF}])}^{1}}^{4}\nonumber\\ \displaystyle & & \displaystyle +\,(N^{\prime })^{-2}\mathop{\sum }_{k=0}^{l}\left\Vert \exp \left(\frac{1}{6}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}t\right)I^{\prime }v\right\Vert _{X_{([0,\unicode[STIX]{x1D706}]\times [k\unicode[STIX]{x1D6FF},(k+1)\unicode[STIX]{x1D6FF}])}^{1,1/2}}^{6}.\nonumber\end{eqnarray}$$

From Lemma 4.3, we obtain

$$\begin{eqnarray}\displaystyle \!\left|\int _{0}^{(l+1)\unicode[STIX]{x1D6FF}}\!M(t)\,dt\right| & {\lesssim} & \displaystyle (N^{\prime })^{-1}\mathop{\sum }_{k=0}^{l}\left\Vert \exp \left(\frac{1}{4}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}t\right)\right\Vert _{(k)}^{4}\Vert I^{\prime }v\Vert _{Y_{([0,\unicode[STIX]{x1D706}]\times [k\unicode[STIX]{x1D6FF},(k+1)\unicode[STIX]{x1D6FF}])}^{1}}^{4}\nonumber\\ \displaystyle \! & & \displaystyle +\,(N^{\prime })^{-2}\mathop{\sum }_{k=0}^{l}\left\Vert \exp \left(\frac{1}{6}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}t\right)\right\Vert _{(k)}^{6}\Vert I^{\prime }v\Vert _{X_{([0,\unicode[STIX]{x1D706}]\times [k\unicode[STIX]{x1D6FF},(k+1)\unicode[STIX]{x1D6FF}])}^{1,1/2}}^{6},\nonumber\end{eqnarray}$$

where

$$\begin{eqnarray}\Vert f(t)\Vert _{(k)}\lesssim \Vert {\mathcal{F}}_{t}[\unicode[STIX]{x1D702}(t-k\unicode[STIX]{x1D6FF})f(t)]\Vert _{L^{1}}+\Vert \unicode[STIX]{x1D702}(t-k\unicode[STIX]{x1D6FF})f(t)\Vert _{H^{1/2}}.\end{eqnarray}$$

From simple computations, we can verify that $\Vert e^{at}\Vert _{(k)}\lesssim e^{ak\unicode[STIX]{x1D6FF}}$ uniformly in $0<a\leqslant 1$, $\unicode[STIX]{x1D6FF}>0$ and $k\in \mathbb{Z}$, and we have

(4.7)$$\begin{eqnarray}\displaystyle \left|\int _{0}^{(l+1)\unicode[STIX]{x1D6FF}}M(t)\,dt\right| & {\lesssim} & \displaystyle (N^{\prime })^{-1}\mathop{\sum }_{k=0}^{l}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}k\unicode[STIX]{x1D6FF})\Vert I^{\prime }v\Vert _{Y_{[0,\unicode[STIX]{x1D706}]\times [k\unicode[STIX]{x1D6FF},(k+1)\unicode[STIX]{x1D6FF}]}^{1}}^{2}\nonumber\\ \displaystyle & & \displaystyle \times \left\{\Vert I^{\prime }v\Vert _{Y_{[0,\unicode[STIX]{x1D706}]\times [k\unicode[STIX]{x1D6FF},(k+1)\unicode[STIX]{x1D6FF}]}^{1}}^{2}+\Vert I^{\prime }v\Vert _{Y_{[0,\unicode[STIX]{x1D706}]\times [k\unicode[STIX]{x1D6FF},(k+1)\unicode[STIX]{x1D6FF}]}^{1}}^{4}\right\}.\end{eqnarray}$$

From the first inequality of Proposition 2.6, we have

(4.8)$$\begin{eqnarray}\displaystyle & & \displaystyle \exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}k\unicode[STIX]{x1D6FF})\Vert I^{\prime }v\Vert _{Y_{[0,\unicode[STIX]{x1D706}]\times [k\unicode[STIX]{x1D6FF},(k+1)\unicode[STIX]{x1D6FF}]}^{1}}^{2}\nonumber\\ \displaystyle & & \displaystyle \quad \lesssim \exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}k\unicode[STIX]{x1D6FF})(\Vert I^{\prime }v(k\unicode[STIX]{x1D6FF})\Vert _{H^{1}}^{2}+\unicode[STIX]{x1D706}^{-6}\Vert I^{\prime }g\Vert _{H^{1}}^{2})\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant (3C_{1}+2)\left(\Vert I^{\prime }v(0)\Vert _{H^{1}}^{2}+\frac{1}{\unicode[STIX]{x1D6FE}^{2}}\Vert I^{\prime }g\Vert _{H^{1}}^{2}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}T^{\prime })\right)\nonumber\\ \displaystyle & {\leqslant} & \displaystyle 3C_{1}+2.\end{eqnarray}$$

From inequalities (4.7), (4.8), we get

$$\begin{eqnarray}\displaystyle \left|\int _{0}^{(l+1)\unicode[STIX]{x1D6FF}}M(t)\,dt\right| & {\lesssim} & \displaystyle (C_{2}\unicode[STIX]{x1D706}^{2}T^{\prime })^{-1}(C_{1}^{2}+C_{1}^{3})(l+1)\biggl(\Vert I^{\prime }v(0)\Vert _{H^{1}}\nonumber\\ \displaystyle & & \displaystyle +\,\frac{1}{\unicode[STIX]{x1D6FE}^{2}}\Vert I^{\prime }g\Vert _{H^{1}}^{2}\exp (\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3}T^{\prime })\biggr).\nonumber\end{eqnarray}$$

Note that

$$\begin{eqnarray}\unicode[STIX]{x1D6FF}^{-1}\lesssim (\Vert I^{\prime }v(k\unicode[STIX]{x1D6FF})\Vert _{H^{1}}+\unicode[STIX]{x1D706}^{-3}\Vert g\Vert _{H^{1}}+\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D706}^{-3})^{\unicode[STIX]{x1D6FC}}\lesssim (C_{1}^{1/2}+1+\Vert f\Vert _{H^{1}})^{\unicode[STIX]{x1D6FC}}.\end{eqnarray}$$

We choose $C_{6}$ sufficiently large such that $(C_{2}T^{\prime })^{-1}(C_{1}^{2}+C_{1}^{3})(l+1)\leqslant (C_{2}\unicode[STIX]{x1D6FF})^{-1}(C_{1}^{2}+C_{1}^{3})\ll 1$, which leads to Proposition 4.6.◻

5 Proof of Theorem 1.1

In this section, we describe the proof of Theorem 1.1.

Proof of Theorem 1.1.

Let $0<\unicode[STIX]{x1D716}\ll 12s-11$ be fixed. We choose $T_{1}>0$ so that

(5.1)$$\begin{eqnarray}\displaystyle \exp (\unicode[STIX]{x1D6FE}T_{1}) & {>} & \displaystyle (\Vert u_{0}\Vert _{H^{s}}^{2}+\Vert u_{0}\Vert _{L^{2}}^{6})\left(\frac{1}{\unicode[STIX]{x1D6FE}^{2}}\Vert f\Vert _{H^{1}}^{2}+\frac{1}{\unicode[STIX]{x1D6FE}^{6}}\Vert f\Vert _{L^{2}}^{6}\right)^{-1}\nonumber\\ \displaystyle & & \displaystyle \times \max \bigg\{\unicode[STIX]{x1D6FE}^{4(1-s)/(1-\unicode[STIX]{x1D716})},(C_{2}T_{1})^{2(1-s)/\unicode[STIX]{x1D716}},\nonumber\\ \displaystyle & & \displaystyle (2\Vert u_{0}\Vert _{H^{s}}^{2})^{12(1-s)/(12s-11-\unicode[STIX]{x1D716})},\nonumber\\ \displaystyle & & \displaystyle (2\unicode[STIX]{x1D6FE}^{-2}\Vert f\Vert _{H^{1}}^{2}\exp (\unicode[STIX]{x1D6FE}T_{1}))^{6(-2s+2)/(1-\unicode[STIX]{x1D716})}\bigg\},\end{eqnarray}$$

which is possible as $6(-2s+2)/(1-\unicode[STIX]{x1D716})<1$. $T_{1}$ depends only on $\Vert u_{0}\Vert _{H^{s}}$, $\Vert f\Vert _{H^{1}}$ and $\unicode[STIX]{x1D6FE}$. Set

(5.2)$$\begin{eqnarray}\displaystyle N & = & \displaystyle \max \bigg\{\unicode[STIX]{x1D6FE}^{2/(1-\unicode[STIX]{x1D716})},(C_{2}T_{1})^{1/\unicode[STIX]{x1D716}},\nonumber\\ \displaystyle & & \displaystyle (2\Vert u_{0}\Vert _{H^{s}}^{2})^{6/(12s-11-\unicode[STIX]{x1D716})},(2\unicode[STIX]{x1D6FE}^{-2}\Vert f\Vert _{H^{1}}^{2}e^{\unicode[STIX]{x1D6FE}T_{1}})^{6/(1-\unicode[STIX]{x1D716})}\bigg\}.\end{eqnarray}$$

From the choice of $T_{1}$ and $N$, we know

$$\begin{eqnarray}N^{(1-\unicode[STIX]{x1D716})/2}\geqslant \unicode[STIX]{x1D6FE},\qquad N^{\unicode[STIX]{x1D716}}\geqslant C_{2}T_{1},\end{eqnarray}$$

and

$$\begin{eqnarray}\displaystyle & \Vert Iu_{0}\Vert _{H^{1}}^{2}\leqslant N^{2-2s}\Vert u_{0}\Vert _{H^{s}}^{2}\leqslant {\textstyle \frac{1}{2}}N^{(1-\unicode[STIX]{x1D716})/6}, & \displaystyle \nonumber\\ \displaystyle & \unicode[STIX]{x1D6FE}^{-2}\Vert If\Vert _{H^{1}}^{2}e^{\unicode[STIX]{x1D6FE}T_{1}}\leqslant {\textstyle \frac{1}{2}}N^{(1-\unicode[STIX]{x1D716})/6}. & \displaystyle \nonumber\end{eqnarray}$$

Hence, from Proposition 4.4, we get

$$\begin{eqnarray}\displaystyle & & \displaystyle \!\!\Vert u(T_{1})\Vert _{H^{s}}^{2}\leqslant \Vert Iu(T_{1})\Vert _{H^{1}}^{2}\nonumber\\ \displaystyle & & \displaystyle \!\!\!\quad \leqslant C_{3}\biggl(\Vert Iu_{0}\Vert _{H^{1}}^{2}\exp (-\unicode[STIX]{x1D6FE}T_{1})+\Vert u_{0}\Vert _{L^{2}}^{6}\exp (-\unicode[STIX]{x1D6FE}T_{1})+\frac{1}{\unicode[STIX]{x1D6FE}^{2}}\Vert If\Vert _{H^{1}}^{2}+\frac{1}{\unicode[STIX]{x1D6FE}^{6}}\Vert f\Vert _{L^{2}}^{6}\biggr)\nonumber\\ \displaystyle & & \displaystyle \!\!\!\quad \leqslant C_{3}\biggl(N^{2(1-s)}(\Vert u_{0}\Vert _{H^{s}}^{2}\exp (-\unicode[STIX]{x1D6FE}T_{1})+\Vert u_{0}\Vert _{L^{2}}^{6}\exp (-\unicode[STIX]{x1D6FE}T_{1}))\nonumber\\ \displaystyle & & \displaystyle \!\!\!\qquad +\,\frac{1}{\unicode[STIX]{x1D6FE}^{2}}\Vert f\Vert _{H^{1}}^{2}+\frac{1}{\unicode[STIX]{x1D6FE}^{6}}\Vert f\Vert _{L^{2}}^{6}\biggr).\nonumber\end{eqnarray}$$

From (5.1) and (5.2), we get

$$\begin{eqnarray}N^{2(1-s)}\exp (-\unicode[STIX]{x1D6FE}T_{1})(\Vert u_{0}\Vert _{H^{s}}^{2}+\Vert u_{0}\Vert _{L^{2}}^{6})<\frac{1}{\unicode[STIX]{x1D6FE}^{2}}\Vert f\Vert _{H^{1}}^{2}+\frac{1}{\unicode[STIX]{x1D6FE}^{6}}\Vert f\Vert _{L^{2}}^{6}\end{eqnarray}$$

which helps us give the bound

$$\begin{eqnarray}\Vert u(T_{1})\Vert _{H^{s}}^{2}\leqslant 2C_{3}\left(\frac{1}{\unicode[STIX]{x1D6FE}^{2}}\Vert f\Vert _{H^{1}}^{2}+\frac{1}{\unicode[STIX]{x1D6FE}^{6}}\Vert f\Vert _{L^{2}}^{6}\right)<K_{1},\end{eqnarray}$$

where $K_{1}$ depends only on $\Vert f\Vert _{H^{1}}$ and $\unicode[STIX]{x1D6FE}$. In the next place, one can fix any $T_{2}>0$ and solve mKdV equation on time interval $[T_{1},T_{1}+T_{2}]$ with initial data replaced by $u(T_{1})$. Let $K_{2}>0$ be sufficiently large such that

(5.3)$$\begin{eqnarray}\displaystyle K_{2}\exp (\unicode[STIX]{x1D6FE}t) & {>} & \displaystyle \max \bigg\{\unicode[STIX]{x1D6FE}^{4(1-s)/(1-\unicode[STIX]{x1D716})},(C_{2}t)^{2(1-s)/\unicode[STIX]{x1D716}},\nonumber\\ \displaystyle & & \displaystyle (2K_{1})^{12(1-s)/(12s-11-\unicode[STIX]{x1D716})},(2\unicode[STIX]{x1D6FE}^{-2}\Vert f\Vert _{H^{1}}^{2}\exp (\unicode[STIX]{x1D6FE}t))^{6(-2s+2)/(1-\unicode[STIX]{x1D716})}\bigg\},\end{eqnarray}$$

for any $t>0$. Set $N^{2(1-s)}=K_{2}\exp (\unicode[STIX]{x1D6FE}T_{2})$, then inequality (5.3) verifies the assumptions in Proposition 4.4 and hence we obtain

$$\begin{eqnarray}\displaystyle \Vert Iu(T_{1}+T_{2})\Vert _{H^{1}}^{2} & {\leqslant} & \displaystyle C\biggl(N^{2(1-s)}\Vert u(T_{1})\Vert _{H^{s}}^{2}\exp (-\unicode[STIX]{x1D6FE}T_{2})\nonumber\\ \displaystyle & & \displaystyle +\,\Vert u(T_{1})\Vert _{L^{2}}^{6}\exp (-\unicode[STIX]{x1D6FE}T_{2})+\frac{1}{\unicode[STIX]{x1D6FE}^{2}}\Vert f\Vert _{H^{1}}^{2}+\frac{1}{\unicode[STIX]{x1D6FE}^{6}}\Vert f\Vert _{L^{2}}^{6}\biggr)\nonumber\\ \displaystyle & {\leqslant} & \displaystyle C_{4}\biggl(K_{1}K_{2}+K_{1}^{2}+\frac{1}{\unicode[STIX]{x1D6FE}^{2}}\Vert f\Vert _{H^{1}}^{2}+\frac{1}{\unicode[STIX]{x1D6FE}^{4}}\Vert f\Vert _{L^{2}}^{6}\biggr)<K_{3}.\nonumber\end{eqnarray}$$

For $t>T_{1}$, we define the maps $L_{1}(t)$ and $L_{2}(t)$ as

$$\begin{eqnarray}\widehat{L_{1}(t)u_{0}}=\widehat{S(t)u_{0}}|_{|k|<N_{t}},\qquad \widehat{L_{2}(t)u_{0}}=\widehat{S(t)u_{0}}|_{|k|\geqslant N_{t}},\end{eqnarray}$$

where $S(t)u_{0}=u(t)$ and $N_{t}=\!K_{2}\exp (\unicode[STIX]{x1D6FE}(t-T_{1}))^{1/2(1-s)}$. It is easy to see that for $t>T_{1}$,

$$\begin{eqnarray}\displaystyle \Vert L_{1}(t)u_{0}\Vert _{H^{1}}^{2}\leqslant \Vert Iu(t)\Vert _{H^{1}}^{2} & {<} & \displaystyle K_{3},\nonumber\\ \displaystyle \Vert L_{2}(t)u_{0}\Vert _{H^{s}}^{2}\leqslant N^{2s-2}\Vert Iu(t)\Vert _{H^{1}}^{2} & {<} & \displaystyle K_{2}^{-1}K_{3}\exp (-\unicode[STIX]{x1D6FE}(t-T_{1})).\nonumber\end{eqnarray}$$

Hence we obtain Theorem 1.1 by taking $K=\max \{K_{3}^{1/2},K_{2}^{-1/2}K_{3}^{1/2}\}$.◻

6 Multilinear estimates

In this section, we prove the 4-linear and 6-linear estimates given in Lemma 4.8.

Proof of Lemma 4.8.

We have

$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{0}^{T^{\prime }}\int _{\unicode[STIX]{x1D706}\mathbb{T}}(-2\unicode[STIX]{x2202}_{x}^{2}I^{\prime }v-4(I^{\prime }v)^{3})(-\unicode[STIX]{x2202}_{x}^{3}I^{\prime }v-2\unicode[STIX]{x2202}_{x}I^{\prime }(v^{3}))\,dx\,dt,\nonumber\\ \displaystyle & & \displaystyle \quad =4\int _{0}^{T^{\prime }}\int _{\unicode[STIX]{x1D706}\mathbb{T}}\unicode[STIX]{x2202}_{x}^{3}I^{\prime }v[(I^{\prime }v)^{3}-I^{\prime }v^{3}]\,dx\,dt\nonumber\\ \displaystyle & & \displaystyle \qquad +\,8\int _{0}^{T^{\prime }}\int _{\unicode[STIX]{x1D706}\mathbb{T}}\unicode[STIX]{x2202}_{x}(I^{\prime }v)^{3}[(I^{\prime }v)^{3}-I^{\prime }v^{3}]\,dx\,dt,\nonumber\\ \displaystyle & & \displaystyle \quad =4I_{1}+8I_{2},\nonumber\end{eqnarray}$$

for any arbitrary $T^{\prime }>0$. From the Plancherel’s theorem

$$\begin{eqnarray}\displaystyle I_{1} & = & \displaystyle c\int _{0}^{T^{\prime }}dt\left[\int _{\unicode[STIX]{x1D6FA}_{1}}+\int _{\unicode[STIX]{x1D6FA}_{2}}\right]\!\left(1-\frac{m^{\prime }(k_{1}+k_{2}+k_{3})}{m^{\prime }(k_{1})m^{\prime }(k_{2})m^{\prime }(k_{3})}\right)k_{4}^{3}\mathop{\prod }_{i=1}^{4}\widehat{I^{\prime }v}(t,k_{i})(dk_{i})_{\unicode[STIX]{x1D706}}\nonumber\\ \displaystyle & = & \displaystyle I_{11}+I_{12},\nonumber\end{eqnarray}$$

where

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6FA}_{1} & = & \displaystyle \{k_{1}+k_{2}+k_{3}+k_{4}=0:(k_{1}+k_{2})(k_{2}+k_{3})(k_{3}+k_{1})\neq 0\}\nonumber\\ \displaystyle \unicode[STIX]{x1D6FA}_{2} & = & \displaystyle \{k_{1}+k_{2}+k_{3}+k_{4}=0:(k_{1}+k_{2})(k_{2}+k_{3})(k_{3}+k_{1})=0\}.\nonumber\end{eqnarray}$$

We use the following lemma given in [Reference Colliander, Keel, Staffilani, Takaoka and Tao5, Lemma 3.2]:

Lemma 6.1. We have

$$\begin{eqnarray}\left|\int _{0}^{T^{\prime }}\int _{\unicode[STIX]{x1D706}\mathbb{T}}f(x,t)g(x,t)\,dx\,dt\right|\lesssim \Vert f\Vert _{Y^{s}}\Vert g\Vert _{Z^{-s}}\end{eqnarray}$$

for any $T^{\prime }>0$, $\unicode[STIX]{x1D706}\geqslant 1$, $s\in \mathbb{R}$ and any $f,g$ on $\mathbb{R}\times \unicode[STIX]{x1D706}\mathbb{T}$.

Lemma 6.1 with $f=\unicode[STIX]{x2202}_{x}I^{\prime }v$ and $s=0$ reduces the estimate of $I_{11}$ to proving

(6.1)$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\left\Vert {\mathcal{F}}^{-1}\left[k^{2}\int _{\unicode[STIX]{x1D6FA}_{k}^{\prime }}\left(1-\frac{m^{\prime }(k)}{m^{\prime }(k_{1})m^{\prime }(k_{2})m^{\prime }(k_{3})}\right)\mathop{\prod }_{i=1}^{3}\widehat{w}(t,k_{i})(dk_{i})_{\unicode[STIX]{x1D706}}\right]\right\Vert }\nolimits_{Z^{0}}\nonumber\\ \displaystyle & & \displaystyle \quad \lesssim (N^{\prime })^{-1}\Vert w\Vert _{X^{1,1/2}}^{3},\end{eqnarray}$$

where

$$\begin{eqnarray}\unicode[STIX]{x1D6FA}_{k}^{\prime }:=\left\{\begin{array}{@{}lr@{}} & k_{1}+k_{2}+k_{3}=k\\ (k_{1},k_{2},k_{3})\in (\mathbb{Z}/\unicode[STIX]{x1D706})^{3}: & (k_{1}+k_{2})(k_{2}+k_{3})(k_{3}+k_{1})\neq 0\\ & |k_{1}|\geqslant |k_{2}|\geqslant |k_{3}|,|k_{1}|\geqslant N^{\prime }/3\end{array}\right\},\end{eqnarray}$$

and $w$ is an arbitrary extension of $I^{\prime }v\in X_{T^{\prime }}^{1,1/2}$ to $X^{1,1/2}$. Note that we may assume $|k_{1}|\geqslant |k_{2}|\geqslant |k_{3}|$ by symmetry and $|k_{1}|\geqslant N^{\prime }/3$ for the integral to be nonzero.

• ∙ Let $|k_{3}|\gtrsim N^{\prime }$. From Proposition 3.8 with $s=1/2$, inequality (6.1) is reduced to the following multiplier bound:

  $$\begin{eqnarray}|k|^{1/2}\left|1-\frac{m^{\prime }(k)}{m^{\prime }(k_{1})m^{\prime }(k_{2})m^{\prime }(k_{3})}\right|\lesssim (N^{\prime })^{-1}\langle k_{1}\rangle ^{1/2}\langle k_{2}\rangle ^{1/2}\langle k_{3}\rangle ^{1/2}\end{eqnarray}$$
  for any $(k_{1},k_{2},k_{3})\in \unicode[STIX]{x1D6FA}_{k}^{\prime }$ with $|k_{3}|\gtrsim N^{\prime }$. We have
  $$\begin{eqnarray}|k|^{1/2}\left|1-\frac{m^{\prime }(k)}{m^{\prime }(k_{1})m^{\prime }(k_{2})m^{\prime }(k_{3})}\right|\lesssim \frac{|k|^{1/2}m^{\prime }(k)}{m^{\prime }(k_{1})m^{\prime }(k_{2})m^{\prime }(k_{3})}\end{eqnarray}$$
  which implies (6.1) as
  $$\begin{eqnarray}|k|^{1/2}m^{\prime }(k)\lesssim |k_{1}|^{1/2}m^{\prime }(k_{1}),\qquad |k_{j}|^{1/2}m^{\prime }(k_{j})\gtrsim (N^{\prime })^{1/2}\quad (j=2,3).\end{eqnarray}$$

• ∙ Let $|k_{3}|\ll N^{\prime }$. Since $|k_{1}|\geqslant N^{\prime }/3$, we have $|k_{1}|\gg |k_{3}|$. By Remark 3.10 with $s=0$, the estimate (6.1) follows if we verify the following multiplier bound:

  (6.2)$$\begin{eqnarray}|k|\left|1-\frac{m^{\prime }(k)}{m^{\prime }(k_{1})m^{\prime }(k_{2})m^{\prime }(k_{3})}\right|\lesssim (N^{\prime })^{-1}\langle k_{1}\rangle \langle k_{2}\rangle \langle k_{3}\rangle\end{eqnarray}$$
  for any $(k_{1},k_{2},k_{3})\in \unicode[STIX]{x1D6FA}_{k}^{\prime }$ with $|k_{3}|\ll N^{\prime }$. To show (6.2), we divide it into two cases:

  1. – If $|k_{2}|\ll N^{\prime }$, then from mean value theorem LHS of (6.2) is bounded by $|k_{2}|$ and estimate (6.2) follows since $|k_{1}|\gtrsim N^{\prime }$.

  2. – If $|k_{2}|\gtrsim N^{\prime }$, then

     $$\begin{eqnarray}|k|\left|1-\frac{m^{\prime }(k)}{m^{\prime }(k_{1})m^{\prime }(k_{2})m^{\prime }(k_{3})}\right|\lesssim \frac{|k|m^{\prime }(k)}{m^{\prime }(k_{1})m^{\prime }(k_{2})m^{\prime }(k_{3})}.\end{eqnarray}$$
     Also, $|k|m^{\prime }(k)\lesssim |k_{1}|m^{\prime }(k_{1})$, $|k_{2}|m^{\prime }(k_{2})\gtrsim N^{\prime }$ and $|k_{3}|m^{\prime }(k_{3})\gtrsim 1$ which implies (6.2).

For $I_{12}$ note that in $\unicode[STIX]{x1D6FA}_{2}$

$$\begin{eqnarray}\left[\left(1-\frac{m^{\prime }(k_{1}+k_{2}+k_{3})}{m^{\prime }(k_{1})m^{\prime }(k_{2})m^{\prime }(k_{3})}\right)k_{4}^{3}\right]_{\text{sym}}=\frac{1}{4}\mathop{\sum }_{j=1}^{4}k_{j}^{3}-\frac{1}{m_{1}^{\prime }m_{2}^{\prime }m_{3}^{\prime }m_{4}^{\prime }}\mathop{\sum }_{j=1}^{4}k_{j}^{3}(m_{j}^{\prime })^{2}\end{eqnarray}$$

(see [Reference Colliander, Keel, Staffilani, Takaoka and Tao4, Section 4]) for details. Therefore, we have $I_{12}=0$.

For $I_{2}$ term, we use the following $L_{t,x}^{12}$ estimate:

Lemma 6.3.

$$\begin{eqnarray}\Vert v\Vert _{L^{12}(\mathbb{R}\times \unicode[STIX]{x1D706}\mathbb{T})}\lesssim \Vert v\Vert _{X^{1/2-,1/2-}}.\end{eqnarray}$$

Lemma 6.3 follows from Sobolev embedding. Now, we consider $I_{2}$. From the Fourier transformation, we get

$$\begin{eqnarray}\displaystyle I_{2} & = & \displaystyle \int _{0}^{T^{\prime }}\int _{0}^{\unicode[STIX]{x1D706}}\unicode[STIX]{x2202}_{x}(I^{\prime }v)^{3}[(I^{\prime }v)^{3}-I^{\prime }v^{3}]\,dx\,dt,\nonumber\\ \displaystyle & {\lesssim} & \displaystyle \int _{\substack{ \mathop{\sum }_{i=1}^{6}k_{i}=0}}\int _{\mathop{\sum }_{i=1}^{6}\unicode[STIX]{x1D70F}_{i}=0}\biggl|\langle k_{1}+k_{2}+k_{3}\rangle (\widetilde{w_{1}})(\widetilde{w_{2}})(\widetilde{w_{3}})\nonumber\\ \displaystyle & & \displaystyle \times \left(1-\frac{m(k_{4}+k_{5}+k_{6})}{m(k_{4})m(k_{5})m(k_{6})}\right)(\widetilde{w_{4}})(\widetilde{w_{5}})(\widetilde{w_{6}})\biggr|(dk_{i})_{\unicode[STIX]{x1D706}}\,d\unicode[STIX]{x1D70F}_{i},\nonumber\end{eqnarray}$$

where $\widetilde{w_{j}}=|(1_{[0,T^{\prime }]}I^{\prime }v)(k_{j},\unicode[STIX]{x1D70F}_{j})|$. We may suppose $\langle k_{1}\rangle =\max \{\langle k_{i}\rangle ,1\leqslant i\leqslant 3\}$ and $|k_{4}|\leqslant |k_{5}|\leqslant |k_{6}|$. We claim that

(6.3)$$\begin{eqnarray}\left|\left(1-\frac{m(k_{4}+k_{5}+k_{6})}{m(k_{4})m(k_{5})m(k_{6})}\right)\right|\lesssim (N^{\prime })^{-2}\langle k_{4}\rangle ^{1/2}\langle k_{5}\rangle \langle k_{6}\rangle\end{eqnarray}$$

for any $k_{4},k_{5},k_{6}$ with $|k_{4}|\leqslant |k_{5}|\leqslant |k_{6}|$. We prove (6.3) as follows:

• ∙ If $|k_{6}|\ll N^{\prime }$, then LHS of (6.3) is equal to zero.

• ∙ If $|k_{6}|\gtrsim N^{\prime }\gg |k_{5}|$. From mean value theorem, we have

  $$\begin{eqnarray}\displaystyle \left|\left(1-\frac{m(k_{4}+k_{5}+k_{6})}{m(k_{4})m(k_{5})m(k_{6})}\right)\right| & {\lesssim} & \displaystyle \left|\frac{m(k_{6})-m(k_{4}+k_{5}+k_{6})}{m(k_{6})}\right|\nonumber\\ \displaystyle & {\lesssim} & \displaystyle \frac{|k_{4}+k_{5}|}{|k_{6}|}\nonumber\end{eqnarray}$$
  which implies (6.3) from the fact that $|k_{6}|\gtrsim N^{\prime }$.

• ∙ Otherwise, we have

  $$\begin{eqnarray}\displaystyle \left|\left(1-\frac{m(k_{4}+k_{5}+k_{6})}{m(k_{4})m(k_{5})m(k_{6})}\right)\right| & {\lesssim} & \displaystyle \frac{1}{m(k_{4})m(k_{5})m(k_{6})}\nonumber\\ \displaystyle & {\lesssim} & \displaystyle \frac{|k_{4}|^{1/2}|k_{5}|\,|k_{6}|}{|k_{4}|^{1/2}|k_{5}|\,|k_{6}|m(k_{4})m(k_{5})m(k_{5})}.\nonumber\end{eqnarray}$$
  The map $|k|\rightarrow |k|^{a}m^{\prime }(|k|)$ is nondecreasing if $0\leqslant a\leqslant 1$, $1-a\leqslant s\leqslant 1$. Since $|k_{5}|,|k_{6}|\gtrsim N^{\prime }$, we have $|k_{5}|m^{\prime }(k_{5})\gtrsim N^{\prime },|k_{6}|m^{\prime }(k_{6})\gtrsim N^{\prime }$ and $|k_{4}|^{1/2}m^{\prime }(k_{4})\gtrsim 1$. This implies (6.3).

From Hölder’s inequality, Proposition 2.2 and Lemma 6.3, we have

$$\begin{eqnarray}\displaystyle |I_{2}| & {\lesssim} & \displaystyle (N^{\prime })^{-2}\Vert {\mathcal{F}}^{-1}(\langle k_{1}\rangle |\widehat{w}|)\Vert _{L_{x,t}^{4}}\Vert {\mathcal{F}}^{-1}(|\widehat{w}|)\Vert _{L_{x,t}^{12}}\Vert {\mathcal{F}}^{-1}(|\widehat{w}|)\Vert _{L_{x,t}^{12}}\nonumber\\ \displaystyle & & \displaystyle \times \,\Vert {\mathcal{F}}^{-1}(\langle k_{4}\rangle ^{1/2}|\widehat{w}|)\Vert _{L_{x,t}^{12}}\Vert {\mathcal{F}}^{-1}(\langle k_{5}\rangle |\widehat{w}|)\Vert _{L_{x,t}^{4}}\Vert {\mathcal{F}}^{-1}(\langle k_{6}\rangle |\widehat{w}|)\Vert _{L_{x,t}^{4}}\nonumber\\ \displaystyle & {\lesssim} & \displaystyle (N^{\prime })^{-2}\Vert I^{\prime }v\Vert _{X^{1,1/2}}^{6}.\nonumber\end{eqnarray}$$

At the last inequality, we have used $\Vert 1_{[0,T^{\prime }]}f\Vert _{X^{s,1/2-}}\lesssim \Vert f\Vert _{X^{s,1/2}}$, which can be shown by modifying the proof of Lemma 6.1.◻