Two-dimensional transport commutator bound by the Fourier l1 gradient sum

For each nonzero $k\in {\mathbb{Z}}^2$, choose $j$ with $\mathrm{\mid }{k}_j\mathrm{\mid }\ge \mathrm{\mid }k\mathrm{\mid }\mathrm{/}\sqrt{2}$. Integrating the coefficient formula by parts $N$ times in $x_j$ (each boundary term vanishes by periodicity) gives $\mathrm{\mid }\widehat{u}(k)\mathrm{\mid }\le C_N(1+\mathrm{\mid }k\mathrm{\mid }{)}^{-N}$ for every integer $N\ge 1$, with $C_N$ controlled by the suprema of the $N$th derivatives of $u$. The unit squares centered on lattice points in $n\le \mathrm{\mid }k\mathrm{\mid }<n+1$ have disjoint interiors and lie in the annulus of radii $\max (0,n-1)$ and $n+2$. For $n\ge 2$ its area is $\pi ((n+2{)}^2-(n-1{)}^2)=\pi (6n+3)\le 9\pi n$, and the $n=1$ disk of radius $3$ also has area $9\pi$; hence the shell has at most $9\pi n$ lattice points. Choose $N=m+5$. The shell sums of $(1+\mathrm{\mid }k\mathrm{\mid }{)}^{m+1}\mathrm{\mid }\widehat{u}(k)\mathrm{\mid }$ are then bounded by $C_m{n}^{-3}$, so the Fourier series of $u$, its first derivatives, ${\mathrm{\Lambda }}^{m}u$, and the first derivatives of ${\mathrm{\Lambda }}^{m}u$ converge absolutely. Thus the two products in the commutator have absolutely convergent coefficient convolutions, and the product formula and multiplier definition give $\widehat{[{\mathrm{\Lambda }}^m,u\cdot \mathrm{\nabla }]u}(k)=i{\sum }_{v+\mathrm{\ell }=k}(w_k^m-w_{\mathrm{\ell }}^m)(\widehat{u}(v)\cdot \mathrm{\ell })\widehat{u}(\mathrm{\ell })\mathrm{.}$

For $x\in {\mathbb{R}}^2$, $w_x=(1+\mathrm{\mid }x{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}$ satisfies $\mathrm{\mid }\mathrm{\nabla }{w}_x^m\mathrm{\mid }=m\mathrm{\mid }x\mathrm{\mid }{w}_x^{m-2}\le m{w}_x^{m-1}$. The mean-value theorem along the segment from $\mathrm{\ell }$ to $\mathrm{\ell }+v$, together with $w_{\mathrm{\ell }+\theta v}\le w_{\mathrm{\ell }}+w_v$ and $(a+b{)}^{m-1}\le C_m(a^{m-1}+b^{m-1})$, therefore gives $\mathrm{\mid }{w}_{\mathrm{\ell }+v}^m-w_{\mathrm{\ell }}^m\mathrm{\mid }\le C_m\mathrm{\mid }v\mathrm{\mid }(w_v^{m-1}+w_{\mathrm{\ell }}^{m-1})\mathrm{.}$

Apply the bound in step 2.1 to each summand in step 1.1 and use $\mathrm{\mid }\widehat{u}(v)\cdot \mathrm{\ell }\mathrm{\mid }\le \mathrm{\mid }\widehat{u}(v)\mathrm{\mid }\mathrm{\mid }\mathrm{\ell }\mathrm{\mid }$. The term containing $w_v^{m-1}$ is bounded pointwise by $C_m(w_v^m\mathrm{\mid }\widehat{u}(v)\mathrm{\mid })(\mathrm{\mid }\mathrm{\ell }\mathrm{\mid }\mathrm{\mid }\widehat{u}(\mathrm{\ell })\mathrm{\mid })$ because $\mathrm{\mid }v\mathrm{\mid }\le w_v$. The term containing $w_{\mathrm{\ell }}^{m-1}$ is bounded by $C_m(\mathrm{\mid }v\mathrm{\mid }\mathrm{\mid }\widehat{u}(v)\mathrm{\mid })(w_{\mathrm{\ell }}^m\mathrm{\mid }\widehat{u}(\mathrm{\ell })\mathrm{\mid })$ because $\mathrm{\mid }\mathrm{\ell }\mathrm{\mid }\le w_{\mathrm{\ell }}$. Summing over the two output components changes only the dimension-dependent constant.

For the first convolution, set $a_v=w_v^m\mathrm{\mid }\widehat{u}(v)\mathrm{\mid }$ and $b_{\mathrm{\ell }}=\mathrm{\mid }\mathrm{\ell }\mathrm{\mid }\mathrm{\mid }\widehat{u}(\mathrm{\ell })\mathrm{\mid }$. Then [F1] gives $\mathrm{\parallel }a{\mathrm{\parallel }}_{{\mathrm{\ell }}^2}=\mathrm{\parallel }u{\mathrm{\parallel }}_{H^m}$ and [F3] gives $\mathrm{\parallel }b{\mathrm{\parallel }}_{{\mathrm{\ell }}^1}=S(u)$, so $\mathrm{\parallel }a\ast b{\mathrm{\parallel }}_{{\mathrm{\ell }}^2}\le \mathrm{\parallel }a{\mathrm{\parallel }}_{{\mathrm{\ell }}^2}\mathrm{\parallel }b{\mathrm{\parallel }}_{{\mathrm{\ell }}^1}=\mathrm{\parallel }u{\mathrm{\parallel }}_{H^m}S(u)$. For the second convolution take $c_v=\mathrm{\mid }v\mathrm{\mid }\mathrm{\mid }\widehat{u}(v)\mathrm{\mid }$ and $d_{\mathrm{\ell }}=w_{\mathrm{\ell }}^m\mathrm{\mid }\widehat{u}(\mathrm{\ell })\mathrm{\mid }$; then $\mathrm{\parallel }c{\mathrm{\parallel }}_{{\mathrm{\ell }}^1}=S(u)$ and $\mathrm{\parallel }d{\mathrm{\parallel }}_{{\mathrm{\ell }}^2}=\mathrm{\parallel }u{\mathrm{\parallel }}_{H^m}$, giving the same bound by $\mathrm{\parallel }c\ast d{\mathrm{\parallel }}_{{\mathrm{\ell }}^2}\le \mathrm{\parallel }c{\mathrm{\parallel }}_{{\mathrm{\ell }}^1}\mathrm{\parallel }d{\mathrm{\parallel }}_{{\mathrm{\ell }}^2}$. Summing the finitely many vector components and applying Parseval [F2] proves the asserted $L^2$ estimate.