Paraproduct and Fourier series

Question

I know that it's possible to define the paraproduct $T_a u$ when $a=a(x)\in L^{\infty}$ and $u\in H^s$ and in this case $T_a u\in H^s$.

Remark: $T_a u$ is the pseudo differential operator with symbol $\sigma^{\psi}_a(x,\xi)=(\mathcal{F}^{-1}_{\eta}\psi(\eta,\xi)\star a)(x)$, where $\psi$ is an admissible cut off function. Considering the Bony's choice for the admissible cut off function, we can write $T_a u$ as a part of the product $a u$ with $a$ and $u$ written with the Paley-Littlewood decomposition.

I want to prove the following estimate $$\lVert T_au\rVert_{H^s}\le C \lVert a \rVert_{\infty}(1+\lVert u \rVert_{H^s})$$ using only Fourier series, but it's not clear to me how i can define $T_a u$. I know only that there is a way separating a Fourier series in a "high frequency" part and a "low frequency" part.

Answer 1

Let us assume then that $a,u$ are measurable $\mathbb Z^n$ periodic functions, so that $$ \hat u(k)=\int_{[0,1]^n}e^{-2iπ x\cdot k} u(x) dx. $$ We have, $$ u=\sum_{\nu\in \mathbb N}u_\nu,\quad u_0=\sum_{\vert k\vert\le 1}\hat u(k) e^{2iπ x\cdot k},\quad\text{for $\nu\ge 1$, }\quad u_\nu=\sum_{2^\nu\le \vert k\vert< 2^{\nu+1}}\hat u(k) e^{2iπ x\cdot k}, $$ and we define for a fixed $N_0$ $$ T_a u=\sum_{\nu\ge N_0}\underbrace{\bigl(\sum_{\mu\le \nu-N_0} a_µ\bigr)}_{S_{\nu-N_0}(a)}u_\nu. $$ The Fourier transform of $S_{\nu-N_0}(a)$ is supported in a ball with center 0, radius $2^{\nu-N_0+1}$ so that the Fourier transform of the product $S_{\nu-N_0}(a) u_\nu$ is located in the ring where $\vert k\vert\sim 2^\nu$. As a result $$ \Vert T_a u\Vert_{H^s}^2\sim\sum_\nu 2^{2s\nu}\Vert S_{\nu-N_0}(a) u_\nu\Vert_{L^2}^2. $$ Moreover $S_{\nu-N_0}(a)$ is the convolution of a function in $L^1$ (with norm bounded independently of $\nu$) with $a$ so that $ \Vert S_\nu(a)\Vert_{L^\infty}\lesssim \Vert a\Vert_{L^\infty}, $ implying $$\Vert T_a u\Vert_{H^s}^2\lesssim \sum_\nu\Vert a\Vert_{L^\infty}^2 2^{2s\nu}\Vert u_\nu\Vert^2_{L^2}\sim \Vert a\Vert_{L^\infty}^2 \Vert u\Vert^2_{H^s}, $$ which is the sought result.