Circle-averaged negative Bourgain estimate for rough vector fields

Direct expansion gives ${\zeta }_{\pm }\cdot {\zeta }_{\pm }=0$, $\mathrm{\mid }{\zeta }_{\pm }{\mathrm{\mid }}^2=2s^2$, and ${\zeta }_{+}+{\zeta }_{-}=ik$. For either sign, writing $a=\xi \cdot \eta$, $b=\xi \cdot {\eta }^{\perp }$, the symbol satisfies $\mathrm{\mid }{p}_{\zeta }(\xi )\mathrm{\mid }\asymp \mid \mathrm{\mid }\xi {\mathrm{\mid }}^2+k\cdot \xi \pm 2r(s)b\mid +2s\mathrm{\mid }a\mathrm{\mid }$, with absolute constants. Put $R=\mathrm{\mid }\xi \mathrm{\mid }$, $A=\mathrm{\mid }{\mathrm{\Pi }}_P\xi \mathrm{\mid }$, and $W(s,\eta ,\xi )=R^2\mathrm{/}(s+\mathrm{\mid }{p}_{\zeta }(\xi )\mathrm{\mid })$. We prove the uniform kernel bound ${\lambda }^{-1}{\int }_{S^1}{\int }_{\lambda }^{2\lambda }Wdsd\eta \le C(1+(1+\mathrm{\mid }k{\mathrm{\mid }}^2)\mathrm{/}\lambda )$. If $R\le 100\sqrt{1+\mathrm{\mid }k{\mathrm{\mid }}^2}$, the pointwise bound $W\le R^2\mathrm{/}\lambda$ proves it. Otherwise $\mathrm{\mid }k\cdot \xi \mathrm{\mid }\le R^2\mathrm{/}100$. If $A\le R^2\mathrm{/}(100\lambda )$, then $2r(s)\mathrm{\mid }b\mathrm{\mid }\le 4\lambda A\le R^2\mathrm{/}25$ and the real part has absolute value at least $R^2\mathrm{/}2$, giving $W\le 2$. Thus only the case $A>R^2\mathrm{/}(100\lambda )$ remains; here $R^2\mathrm{/}A<100\lambda$.

In this last case, on $\mathrm{\mid }a\mathrm{\mid }\ge A\mathrm{/}2$ the imaginary part gives $W\le R^2\mathrm{/}(\lambda A)\le 100$. On the remaining angular set, partition into shells $E_j=\{2^{-j-2}A<\mathrm{\mid }a\mathrm{\mid }\le 2^{-j-1}A\}$, $j\ge 0$, ignoring the measure-zero set $a=0$. Each shell has arclength at most $C{2}^{-j}$, since $a=A\cos \theta$ and $\mathrm{\mid }\sin \theta \mathrm{\mid }\ge \sqrt{3}\mathrm{/}2$ there. On each shell, $\mathrm{\mid }b\mathrm{\mid }\ge \sqrt{3}A\mathrm{/}2$ and the denominator is bounded below, up to an absolute constant, by $\mathrm{\mid }B\pm 2r(s)b\mathrm{\mid }+\lambda (1+2^{-j-1}A)$, where $B=R^2+k\cdot \xi$ is independent of $s,\eta$. Since $dr\mathrm{/}ds=s\mathrm{/}r\ge 1$ and the $r$-interval is contained in $[\lambda \mathrm{/}2,2\lambda ]$, integration in $s$ is bounded by integration in $r$ over this latter interval. For any interval $I$ of length at most $2\lambda$ and $c>0$, splitting at the zero of the affine function gives ${\int }_I(\mathrm{\mid }B+2br\mathrm{\mid }+c{)}^{-1}dr\le C\mathrm{\mid }b{\mathrm{\mid }}^{-1}\log (1+\lambda \mathrm{\mid }b\mathrm{\mid }\mathrm{/}c)$. Therefore ${\int }_{\lambda }^{2\lambda }Wds\le C(R^2\mathrm{/}A)\log (1+CA\mathrm{/}(1+2^{-j-1}A))\le C\lambda (j+1)$. Multiplying by the shell measure and summing ${\sum }_{j\ge 0}(j+1)2^{-j}<\mathrm{\infty }$ proves the kernel bound of step 1.1. This proof treats all $A$, including $A=0$ via its earlier case, and its constants are independent of $k,P$.

Initially let $f$ be Schwartz. By [F1], $\mathrm{\parallel }\varphi \mathrm{div}f{\mathrm{\parallel }}_{{\dot{X}}_{\zeta }^{-1\mathrm{/}2}}^2\le C\int \mathrm{\mid }\xi \cdot \widehat{f}(\xi ){\mathrm{\mid }}^2\mathrm{/}(s+\mathrm{\mid }{p}_{\zeta }(\xi )\mathrm{\mid })d\xi \le C\int W(s,\eta ,\xi )\mathrm{\mid }\widehat{f}(\xi ){\mathrm{\mid }}^{2}d\xi$. Interchange the nonnegative integrals using [F2] and apply step 2.1 and Plancherel. This proves the first estimate. For $f\in H^1$, the simpler denominator bound $s+\mathrm{\mid }{p}_{\zeta }\mathrm{\mid }\ge s\ge \lambda$ gives $\mathrm{\parallel }\varphi \mathrm{div}f{\mathrm{\parallel }}_{{\dot{X}}_{\zeta }^{-1\mathrm{/}2}}^2\le C{\lambda }^{-1}\mathrm{\parallel }\mathrm{\nabla }f{\mathrm{\parallel }}_2^2$; averaging over the parameter interval of length $\lambda$ and circle length $2\pi$ proves the second estimate. The same estimates hold for general $L^2$ and $H^1$ fields by Fourier truncation and smooth approximation. The limits match the distribution $\varphi \mathrm{div}f$: convergence in the weighted negative space implies distributional convergence when tested against Schwartz functions, by reciprocal-weight Cauchy–Schwarz, and $L^2$ convergence of $f$ already implies distributional convergence of its derivative.

Write $f=f_h+(f-f_h)$ with $f_h={\rho }_h\ast f$. Differentiation under the convolution, first for smooth fields and then by $L^2$ approximation, gives $\mathrm{\parallel }\mathrm{\nabla }{f}_h{\mathrm{\parallel }}_2\le \mathrm{\parallel }\mathrm{\nabla }{\rho }_h{\mathrm{\parallel }}_1\mathrm{\parallel }f{\mathrm{\parallel }}_2=C_{\rho }{h}^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_2$ by [F3]. Also $\mathrm{\parallel }f-f_h{\mathrm{\parallel }}_2\le {\omega }_2(f,h)$: at each $x$ apply the integral Cauchy–Schwarz inequality to the probability density ${\rho }_h(y)dy$, then integrate in $x$ and interchange nonnegative integrals to bound the squared norm by $\int {\rho }_h(y)\mathrm{\parallel }f-f(\cdot -y){\mathrm{\parallel }}_2^{2}dy$. The support of ${\rho }_h$ is inside $\mathrm{\mid }y\mathrm{\mid }\le h$. Since $\mathrm{\parallel }v+w{\mathrm{\parallel }}^2\le 2\mathrm{\parallel }v{\mathrm{\parallel }}^2+2\mathrm{\parallel }w{\mathrm{\parallel }}^2$, apply the $H^1$ estimate of step 3.1 to $f_h$ and its $L^2$ estimate to $f-f_h$. This yields the claimed mollified estimate after enlarging the constant.