\documentclass[reqno]{amsart} \usepackage{amssymb} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2008(2008), No. 71, pp. 1--10.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2008 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2008/71\hfil Power series solution] {Power series solution for the modified KdV equation} \author[T. Nguyen\hfil EJDE-2008/71\hfilneg] {Tu Nguyen} \address{Tu Nguyen \newline Department of Mathematics, University of Chicago, 5734 S. University Ave., Chicago, IL 60637, USA} \email{tu@math.uchicago.edu} \thanks{Submitted April 10, 2008. Published May 13, 2008.} \subjclass[2000]{35Q53} \keywords{Local well-posedness; KdV equation} \begin{abstract} We use the method developed by Christ \cite{MR2333210} to prove local well-posedness of a modified Korteweg de Vries equation in $\mathcal{F}L^{s,p}$ spaces. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{corollary}[theorem]{Corollary} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{definition}[theorem]{Definition} \newtheorem{lemma}[theorem]{Lemma} \section{Introduction} The modified Korteweg de Vries (mKdV) equation on a torus $\mathbb{T}$ has the form $$\begin{gathered} \partial_{t}u+\partial_{x}^{3}u+u^{2}\partial_{x}u=0 \\ u(\cdot,0)=u_{0} \end{gathered}\label{eq:mKdV}$$ where $(x,t)\in\mathbb{T}\times\mathbb{R}$, $u$ is a real-valued function. If $u$ is a smooth solution of \eqref{eq:mKdV}, then $\| u(\cdot,t)\|_{L^{2}(\mathbb{T})}=\| u_{0}\|_{L^{2}(\mathbb{T})}$ for all $t$; therefore, $\widetilde{u}(x,t)=u(x+\frac{1}{2\pi}\| u_{0}\|_{L^{2}(\mathbb{T})}^{2}t,t)$ is a solution of $$\begin{gathered} \partial_{t}u+\partial_{x}^{3}u+\Big(u^{2}-\frac{1}{2\pi} \int_{\mathbb{T}}u^{2}(x,t)dx\Big)\partial_{x}u=0\\ u(\cdot,0)=u_{0} \end{gathered}\label{eq:modified mKdV}$$ Thus, \eqref{eq:modified mKdV} and \eqref{eq:mKdV} are essentially equivalent. Using Fourier restriction norm method, Bourgain \cite{MR1215780} proved that \eqref{eq:modified mKdV} is locally well-posed for initial data $u_{0}\in H^{s}(\mathbb{T})$ when $s\geq1/2$, and the solution map is uniformly continuous. In \cite{MR1466164}, he also showed that the solution map is not $C^{3}$ in $H^{s}(\mathbb{T})$ when $s<1/2$. Takaoka and Tsutsumi \cite{MR2097834} proved local-wellposedness of \eqref{eq:modified mKdV} when $1/2>s>3/8$, and they showed that solution map is not uniformly continuous for this range of $s$. For \eqref{eq:mKdV}, Kappeler and Topalov \cite{MR2131061} used inverse scattering method to show wellposedness when $s\geq0$ and Christ, Colliander and Tao \cite{MR2018661} showed that uniformly continuous dependence on the initial data does not hold when $s<1/2$. Thus, there is a gap between known local well-posedness results and the space $H^{-1/2}(\mathbb{T})$ suggested by the standard scaling argument. Recently, Gr\"unrock and Vega \cite{MR0003} showed local well-posedness of the mKdV equation on $\mathbb{R}$ with initial data in $\widehat{H_{s}^{r}}(\mathbb{R}):=\{f\in\mathcal{D}'(\mathbb{R}) :\| f\|_{\widehat{H_{s}^{r}}}:=\| \langle \cdot\rangle ^{s} \hat{f}(\cdot)\|_{L^{r'}}<\infty\},$ when $2\geq r>1$ and $s\geq\frac{1}{2}-\frac{1}{2r}$. (for $r>\frac{4}{3}$, this was obtained by Gr\"unrock \cite{MR2096258}). This is an extension of the result of Kenig, Ponce and Vega \cite{MR1211741} that local-wellposedness holds in $H^{s}(\mathbb{R})$ when $s\geq1/4$. Furthermore, as $\widehat{H_{s}^{r}}$ scales like $H^{\sigma}$ with $\sigma=s+\frac{1}{2}-\frac{1}{r}$, this result covers spaces that have scaling exponent $-\frac{1}{2}+$. There is also a related recent work of Gr\"unrock and Herr \cite{MR0002} on the derivative nonlinear Schr\"odinger equation on $\mathbb{T}$. Both \cite{MR0003} and \cite{MR0002} used a version of Bourgain's method. In this paper, we apply the new method of solution developed by Christ \cite{MR2333210} to investigate the local well-posedness of \eqref{eq:modified mKdV} with initial data in $\mathcal{F}L^{s,p}(\mathbb{T}) :=\{f\in\mathcal{D}'(\mathbb{T}):\| f\|_{\mathcal{F}L^{s,p}} :=\| \langle \cdot\rangle ^{s}\hat{f}(\cdot)\|_{l^{p}}<\infty\}.$ Let $B(0,R)$ be the ball of radius $R$ centered at $0$ in $\mathcal{F}L^{s,p}(\mathbb{T})$. Our main result is the following. \begin{theorem} \label{thm1.1} Suppose $s\geq1/2$, $1\leq p < \infty$ and $p'(s+1/4)>1$. Let $W$ be the solution map for smooth initial data of \eqref{eq:modified mKdV}. Then for any $R>0$ there is $T>0$ such that the solution map $W$ extends to a uniformly continuous map from $B(0,R)$ to $C([0,T],\mathcal{F}L^{s,p}(\mathbb{T}))$. \end{theorem} We note that the $\mathcal{F}L^{s,p}(\mathbb{T})$ spaces that are covered by Theorem 1.1 have scaling index $\frac{1}{4}+$. The restriction $s\geq1/2$ is due to the presence of the derivative in the nonlinear term, and is only used to bound the operator $S_{2}$ in section 3. The same restriction on $s$ is also required in the work on the derivative nonlinear Schr\"odinger equation on $\mathbb{T}$ by Gr\"unrock and Herr \cite{MR0002}. We believe that the range of $p$ in Theorem 1.1 is not sharp. Concerning \eqref{eq:mKdV}, we have the following result. \begin{corollary} \label{coro1.2} Suppose $s\geq1/2$, $1\leq p < \infty$ and $p'(s+1/4)>1$. Let $\widetilde{W}$ be the solution map for smooth initial data of \eqref{eq:mKdV}. Then for any $R>0$ there is $T>0$ such that for any $c>0$, the solution map $\widetilde{W}$ extends to a uniformly continuous map from $B(0,R)\cap\left\{ \varphi:\| \varphi\|_{L^{2}}=c\right\} \subset\mathcal{F}L^{s,p}(\mathbb{T})$ to $C([0,T],\mathcal{F}L^{s,p}(\mathbb{T}))$. \end{corollary} As in \cite{MR2333210}, the solution map $W$ obtained in Theorem 1.1 gives a weak solution of \eqref{eq:modified mKdV} in the following sense. Let $T_{N}$ be defined by $T_{N}u=(\chi_{[-N,N]}\widehat{u})^{\vee}$. Let $\mathcal{N}u:=\left(u^{2}-\frac{1}{2\pi}\int_{\mathbb{T}}u^{2}(x,t)dx\right)\partial_{x}u$ be the limit in $C([0,T],\mathcal{D}'(\mathbb{T}))$ of $\mathcal{N}(T_{N}u)$ as $N\rightarrow\infty$, provided it exists. \begin{proposition} \label{prop1.3} Let $s$ and $p$ be as in Theorem 1.1. Let $\varphi\in\mathcal{F}L^{s,p}$ and $u:=W\varphi\in C([0,T],\mathcal{F}L^{s,p})$. Then $\mathcal{N}u$ exists and $u$ satisfies \eqref{eq:modified mKdV} in the sense of distribution in $(0,T)\times\mathbb{T}$. \end{proposition} To prove these results, we formally expand the solution map into a sum of multilinear operators. These multilinear operators are described in the section 2. Then we will show that if $u(\cdot,0)\in\mathcal{F}L^{s,p}$ then the sum of these operators converges in $\mathcal{F}L^{s,p}$ for small time $t$, when $s$ and $p$ satisfy the conditions of Theorem 1.1. Furthermore, this gives a weak solution of \eqref{eq:modified mKdV}, justifying our formal derivation. \section{Multilinear operators} We rewrite \eqref{eq:modified mKdV} as a system of ordinary differential equations of the spatial Fourier series of $u$ (see \cite[formula (1.9)]{MR2097834}, and \cite[Lemma 8.16]{MR1215780}). \label{eq:first} \begin{aligned} &\frac{d\hat{u}(n,t)}{dt}-in^{3}\hat{u}(n,t)\\ & = -i\sum_{n_{1}+n_{2}+n_{3}=n}\hat{u}(n_{1},t)\hat{u}(n_{2},t)n_{3} \hat{u}(n_{3},t) +i\sum_{n_{1}}\hat{u}(n_{1},t)\hat{u}(-n_{1},t)n\hat{u}(n,t) \\ & = \frac{-in}{3}\sum_{n_{1}+n_{2}+n_{3}=n}^{*}\hat{u}(n_{1},t) \hat{u}(n_{2},t)\hat{u}(n_{3},t) +in\hat{u}(n,t)\hat{u}(-n,t)\hat{u}(n,t), \end{aligned} where the star means the sum is taken over the triples satisfying $n_{j}\ne n$, $j=1,2,3$. We note that these are precisely the triples with $\sigma(n_{1},n_{2},n_{3})\ne0$. Let $a(n,t)=e^{-in^{3}t}\hat{u}(n,t)$, then $a_{n}(t)$ satisfy \begin{align*} \frac{da(n,t)}{dt} &=-\frac{in}{3}\sum_{n_{1}+n_{2}+n_{3}=n}^{*}e^{i\sigma(n_{1}, n_{2},n_{3})t}a(n_{1},t)a(n_{2},t)a(n_{3},t)\\ &\quad +ina(n,t)a(-n,t)a(n,t), \end{align*} where $\sigma(n_{1},n_{2},n_{3})=n_{1}^{3}+n_{2}^{3}+n_{3}^{3}- (n_{1}+n_{2}+n_{3})^{3}=-3(n_{1}+n_{2})(n_{2}+n_{3})(n_{3}+n_{1}).$ Or, in integral form, \label{eq:a integral sum} \begin{aligned} a(n,t) & = a(n,0)-\frac{in}{3}\int_{0}^{t}\sum_{n_{1}+n_{2}+n_{3}=n}^{*} e^{i\sigma(n_{1},n_{2},n_{3})s}a(n_{1},s)a(n_{2},s)a(n_{3},s)ds\\ &\quad +in\int_{0}^{t} |a(n,s) |^{2}a(n,s)ds. \end{aligned} If, $a$ is sufficiently nice, say $a\in C([0,T],l^{1})$ (which is the case if $u\in C([0,T],H^{s}(\mathbb{T}))$ for large $s$) then we can exchange the order of the integration and summation to obtain \label{eq:a sum integral} \begin{aligned} a(n,t) & = a(n,0)-\frac{in}{3}\sum_{n_{1}+n_{2} +n_{3}=n}^{*}\int_{0}^{t}e^{i\sigma(n_{1},n_{2},n_{3})s}a(n_{1},s) a(n_{2},s)a(n_{3},s)ds\\ &\quad +in\int_{0}^{t}|a(n,s)|^{2}a(n,s)ds. \end{aligned} Replacing the $a(n_{j},s)$ in the right hand side by their equations obtained using \eqref{eq:a sum integral}, we get \label{eq:a sum integral first step} \begin{aligned} a(n,t) & = a(n,0)-\frac{in}{3}\sum_{n_{1}+n_{2}+n_{3}=n}^{*} a(n_{1},0)a(n_{2},0)a(n_{3},0)\int_{0}^{t}e^{i\sigma(n_{1},n_{2},n_{3})s}ds\\ & \quad +in|a(n,0)|^{2}a(n,0)\int_{0}^{t}ds +\textrm{ additional terms }\\ & = a(n,0)-\frac{n}{3}\sum_{n_{1}+n_{2}+n_{3}=n}^{*} \frac{a(n_{1},0)a(n_{2},0)a(n_{3},0)}{\sigma(n_{1},n_{2},n_{3})} (e^{i\sigma(n_{1},n_{2},n_{3})t}-1)\\ & \quad +int|a(n,0)|^{2}a(n,0)+\textrm{ additional terms } \end{aligned} The additional terms are those which depend not only on $a(\cdot,0)$. An example of the additional terms is \begin{align*} &-\frac{nn_{3}}{9}\!\sum_{n_{1}+n_{2}+n_{3}=n}^{*}a(n_{1},0)a(n_{2},0) \!\sum_{m_{1}+m_{2}+m_{3}=n_{3}}^{*}\int_{0}^{t} e^{i\sigma(n_{1},n_{2},n_{3})s}\int_{0}^{s} e^{i\sigma(m_{1},m_{2},m_{3})s'}\\ &\times a(m_{1},s')a(m_{2},s')a(m_{3},s')ds'ds \end{align*} Then we can again use \eqref{eq:a sum integral} for each appearance of $a(m,\cdot)$ in the additional terms, and obtain more and more complicated additional terms. We refer to section 2 of \cite{MR2333210} for more detailed description of these additional terms. Continuing this process indefinitely, we get a formal expansion of $a(n,t)$ as a sum of multilinear operators of $a(\cdot,0)$. We will now describe these operators and then show that their sum converges. Again, we refer to section 3 of \cite{MR2333210} for more detailed explanations. Each of our multilinear operators will be associated to a tree, which has the property that each of its node has either zero or three children. We will only consider trees with this property. If a node $v$ of $T$ has three children, they will be denoted by $v_{1},v_{2},v_{3}$. We denote by $T^{0}$ the set of non-terminal nodes of $T$, and $T^{\infty}$ the set of terminal nodes of $T$. Clearly, if $|T|=3k+1$ then $|T^{0}|=k$ and $|T^{\infty}|=2k+1$. \begin{definition} \label{def2.1} \rm Let $T$ be a tree. Then $\mathcal{J}(T)$ is the set of $j\in\mathbb{Z}^{T}$ such that if $v\in T^{0}$ then $j_{v}=j_{v_{1}}+j_{v_{2}}+j_{v_{3}},$ and either $j_{v_{i}}\ne j_{v}$ for all $i$, or $j_{v_{1}}=-j_{v_{2}}=j_{v_{3}}=j_{v}$. We will denote by $v(T)$ be the root of $T$ and $j(T)=j(v(T))$. For $j\in\mathcal{J}(T)$ and $v\in T^{0}$, $\sigma(j,v):=\sigma(j(v_{1}),j(v_{2}),j(v_{3})).$ Also define $\mathcal{R}(T,t)=\{s\in\mathbb{R}_{+}^{T^{0}}:\text{ if }v1. \end{proposition} \begin{proof} In the proof, all the sums are taken over the triples (n_{1},n_{2},n_{3}) that satisfy the additional property that n_{i}\ne n, for all 1\leq i\leq3. Clearly, we can assume n>0. Note that if say |n_{1}|\geq5n then as |n_{2}+n_{3}|=|n-n_{1}|\geq4n, at least one of n_{2} and n_{3} has absolute value bigger than 2n. Also, we cannot have |n_{i}|\leq n/4 for all i. Thus, up to permutation, there are four cases. \begin{enumerate} \item |n_{1}|,|n_{2}|,|n_{3}|\in[n/4,5n] \item |n_{1}|,|n_{2}|\in[n/4,5n], |n_{3}|\leq n/4 \item |n_{1}|\in[n/4,5n], |n_{2}|,|n_{3}|\leq n/4 \item |n_{1}|,|n_{2}|\geq2n \end{enumerate} By Lemma \ref{lem3.3}, it suffices to show that in each of these four regions, for some i\ne j the l_{i,j}^{p'}-norm of m is bounded. \noindent\textbf{Case 1.} As 3n=\sum(n-n_{i}) for some index i, say i=3, we must have |n-n_{3}|\sim n. Since we also have |n_{1}|,|n_{2}|\gtrsim n, \[ |m(n_{1},n_{2},n_{3})|\lesssim\frac{\langle n\rangle ^{1/2-s}}{\langle n_{3} \rangle ^{s}|(n-n_{1})(n-n_{2})|^{1/2}}.$ We will use the inequality $|\frac{1}{n_{3}(n-n_{2})}|=|\frac{1}{n_{1}} \Big(\frac{1}{n_{3}}-\frac{1}{n-n_{2}}\Big)|\leq\frac{1}{|n_{1}|} \Big(\frac{1}{|n_{3}|}+\frac{1}{|n-n_{2}|}\Big).$ \noindent (1) If $1/4\leq s\leq1/2$: then $\langle n_{3}\rangle ^{p'(1/2-s)} \lesssim\langle n\rangle ^{p'(1/2-s)}$, so \begin{align*} \| m\|_{l_{1,2}^{p'}}^{p'} & \lesssim \sum_{|n_{1}|\leq5n}\frac{\langle n\rangle ^{p'(1/2-s)}} {|n-n_{1}|^{p'/2}}\sum_{|n_{2}|\leq5n} \frac{\langle n_{3}\rangle ^{p'(1/2-s)}} {\left(\langle n_{3}\rangle |n-n_{2}|\right)^{p'/2}}\\ & \lesssim \sum_{|n_{1}|\leq5n}\frac{\langle n\rangle ^{p'(1/2-s)}} {|n-n_{1}|^{p'/2}}\sum_{|n_{2}|\leq 5n}\! \frac{\langle n\rangle ^{p'(1/2-s)}}{|n_{1}|^{p'/2}} \Big(\frac{1}{|n-n_{2}|^{p'/2}}+\frac{1}{|n-n_{1}-n_{2}|^{p'/2}}\Big)\\ & \lesssim \sum_{|n_{1}|\leq5n}\frac{\langle n\rangle ^{p'(1-2s)}A_{n}} {|(n-n_{1})n_{1}|^{p'/2}}\\ & \lesssim \langle n\rangle ^{p'(1-2s)}A_{n}\sum_{|n_{1}|\leq5n} \Big(\frac{1}{n}(\frac{1}{|n-n_{1}|}+\frac{1}{|n_{1}|})\Big)^{p'/2}\\ & \lesssim \langle n\rangle ^{p'(1/2-2s)}A_{n}^{2}. \end{align*} where $\sum_{02 \end{cases} \] we easily check that$\langle n\rangle ^{(1/2-2s)p'}A_{n}^{2}$is bounded by a constant, under our hypothesis on$s$and$p'$. \noindent(2) If$s>1/2$: then$\langle n-n_{2}\rangle ^{p'(s-1/2)}\lesssim\langle n\rangle ^{p'(s-1/2)}, so \begin{align*} \| m\|_{l_{1,2}^{p'}}^{p'} & \lesssim \sum_{|n_{1}|\leq5n}\frac{\langle n\rangle ^{p'(1/2-s)}} {|n-n_{1}|^{p'/2}}\sum_{|n_{2}|\leq5n}\frac{\langle n-n_{2} \rangle ^{p'(s-1/2)}}{\left(\langle n_{3}\rangle |n-n_{2}|\right)^{p's}}\\ & \lesssim \sum_{|n_{1}|\leq5n}\frac{\langle n\rangle ^{p'(1/2-s)}} {|n-n_{1}|^{p'/2}}\sum_{|n_{2}|\leq5n}\frac{\langle n \rangle ^{p'(s-1/2)}}{|n_{1}|^{p's}}\Big(\frac{1}{|n-n_{2}|^{p's}} +\frac{1}{|n-n_{1}-n_{2}|^{p's}}\Big)\\ & \lesssim \sum_{|n_{1}|\leq5n}\frac{B_{n}}{|n-n_{1}|^{p'/2}|n_{1}|^{p's}}\\ & \lesssim B_{n}\sum_{|n_{1}|\leq5n}|n-n_{1}|^{p'(s-1/2)} \Big(\frac{1}{n}(\frac{1}{|n-n_{1}|}+\frac{1}{|n_{1}|})\Big)^{p's}\\ & \lesssim \langle n\rangle ^{-p'/2}B_{n}^{2}. \end{align*} whereB_{n}=\sum_{01 \end{cases} \] we easily check that $\langle n\rangle ^{-p'/2}B_{n}^{2}$ is bounded by a constant, under our hypothesis on $s$ and $p'$. \noindent\textbf{Case 2} This case can be treated in exactly the same way as the first case, except when $n_{3}=0$. In the region $n_{3}=0$, \begin{align*} \| m\|_{l_{1,3}^{p'}}^{p'} & \lesssim \sum_{n_{1}}\frac{\langle n \rangle ^{p'(1/2-s)}}{|n_{1}(n-n_{1})|^{p'/2}} \leq\sum_{n_{1}}\langle n\rangle ^{-p's} \Big(\frac{1}{|n_{1}|^{p'/2}}+\frac{1}{|n-n_{1}|^{p'/2}}\Big)\\ & \lesssim \langle n\rangle ^{-p's}A_{n}\lesssim1 \end{align*} \noindent\textbf{Case 3} As $|n_{1}|,|n-n_{2}|,|n-n_{3}|\sim n$, $|m(n_{1},n_{2},n_{3})|\lesssim\frac{1}{\langle n_{2}\rangle ^{s} \langle n_{3}\rangle ^{s}|n_{2}+n_{3}|^{1/2}}.$ Without loss of generality, we assume that $|n_{3}|\geq|n_{2}|$. \noindent (1) If $|n_{2}|<|n_{3}|/2$: \begin{align*} \| m\|_{l_{2,3}^{p'}}^{p'} & \lesssim \sum_{0\leq|n_{2}|\leq n/4}\frac{1}{\langle n_{2} \rangle ^{p's}}\sum_{n/4\geq|n_{3}|>2n_{2}} \frac{1}{\langle n_{3}\rangle ^{p'(s+1/2)}}\\ & \lesssim \sum_{0\leq|n_{2}|\leq n/4}\frac{1}{\langle n_{2} \rangle ^{p'(2s+1/2)-1}} \lesssim 1 \end{align*} if $(s+1/4)p'>1$. \noindent(2) If $|n_{2}|\geq|n_{3}|/2$: \begin{align*} \| m\|_{l_{2,3}^{p'}}^{p'} & \lesssim \sum_{|n_{3}|\leq n/4}\frac{1}{\langle n_{3} \rangle ^{2p's}}\sum_{|n_{3}|\geq n_{2}\geq|n_{3}|/2} \frac{1}{\langle n_{3}+n_{2}\rangle ^{p'/2}}\\ & \lesssim \sum_{|n_{3}|\leq n/4}\frac{1}{\langle n_{3} \rangle ^{2p's}}\max\{\log\langle n_{3}\rangle ,\langle n_{3} \rangle ^{-p'/2+1}\}\\ & \lesssim \sum_{|n_{3}|\leq n/4}\frac{\log\langle n_{3} \rangle }{\langle n_{3}\rangle ^{2p's}}+\sum_{|n_{3}| \leq n/4}\frac{1}{\langle n_{3}\rangle ^{p'(2s+1/2)-1}}\lesssim1 \end{align*} as $2p's\geq p'(s+1/4)>1$. \noindent\textbf{Case 4} $|n_{1}|,|n_{2}|>2n$: Note that in this case, $|n_{1}|\sim|n-n_{1}|$ and $|n_{2}|\sim|n-n_{3}|$. \noindent(1) If $|n_{3}|,|n-n_{3}|\geq n/2:$ we have $|m(n_{1},n_{2},n_{3})|\lesssim\frac{\langle n\rangle ^{1/2}} {\langle n_{1}\rangle ^{s+1/2}\langle n_{2}\rangle ^{s+1/2}},$ hence \begin{align*} \| m\|_{l_{1,2}^{p'}}^{p'} & \lesssim \langle n\rangle ^{p'/2}\sum_{|n_{1}|,|n_{2}|>2n} \frac{1}{\langle n_{1}\rangle ^{p'(s+1/2)}\langle n_{2} \rangle ^{p'(s+1/2)}}\\ & \lesssim \frac{\langle n\rangle ^{p'/2}}{\langle 2n \rangle ^{p'(2s+1)-2}}\lesssim1. \end{align*} \noindent (2) If $|n_{3}|2n} \frac{n^{p'(s+1/2)}}{\langle n_{1}\rangle ^{p'(2s+1)}} \lesssim\frac{B_{n}}{n^{p'(s+1/2)-1}}\lesssim1 \] \noindent(3) If$|n-n_{3}|