Truncated Fourier formula for derivatives of the Bogovskii operator

For $0<\delta <ϵ<1\mathrm{/}2$, pair $T_{\varphi ,j}^{\delta }f-T_{\varphi ,j}^{ϵ}f$ with a Schwartz test function $\zeta$. This difference integrates only over $\delta \le t\le ϵ$, where the change of variables $x=y+t(z-y)$ and Fubini are absolutely justified. The inner pairing contains $t^{-1}{\mathrm{\partial }}_j\varphi (z)\zeta (y+t(z-y))$. Integration by parts in $z$ changes its integral to $-\int \varphi (z){\mathrm{\partial }}_j\zeta (y+t(z-y))dz$; the factor $t^{-1}$ has canceled. Its absolute value is at most $\mathrm{\parallel }\varphi {\mathrm{\parallel }}_1\mathrm{\parallel }\mathrm{\nabla }\zeta {\mathrm{\parallel }}_{\mathrm{\infty }}$, uniformly in $y,t$. Thus the paired difference is bounded by $(ϵ-\delta )\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }\varphi {\mathrm{\parallel }}_1\mathrm{\parallel }\mathrm{\nabla }\zeta {\mathrm{\parallel }}_{\mathrm{\infty }}$, so $T_{\varphi ,j}^{ϵ}f$ has a distributional limit as $ϵ\downarrow 0$.

For $ϵ<t<1\mathrm{/}2$, Fourier transform in $x$ and use $x=y+t(z-y)$, whose Jacobian is $t^d$. The factor $t^{-d}$ cancels the Jacobian and ${\mathrm{\partial }}_{x_j}\varphi (z)=t^{-1}{\mathrm{\partial }}_{z_j}\varphi (z)$. Integration by parts in $z$ gives $\widehat{{\mathrm{\partial }}_j\varphi }(t\xi )=2\pi it{\xi }_j\widehat{\varphi }(t\xi )$ by [F1]; the remaining $y$ integral is $\widehat{f}((1-t)\xi )$. Hence the truncated transform is $2\pi i{\xi }_j{\int }_{ϵ}^{1\mathrm{/}2}\widehat{\varphi }(t\xi )\widehat{f}((1-t)\xi )dt$. The integrals converge pointwise in $\xi$, and their absolute values are bounded uniformly in $ϵ$ by $\pi \mathrm{\mid }{\xi }_j\mathrm{\mid }\mathrm{\parallel }\varphi {\mathrm{\parallel }}_1\mathrm{\parallel }f{\mathrm{\parallel }}_1$. Against any Schwartz test function this is an integrable majorant, so [F5] gives convergence in ${\mathcal{S}}^{\mathrm{\prime }}$. Step 1.1 gives convergence in ${\mathcal{S}}^{\mathrm{\prime }}$ as well, since its bound uses the Schwartz seminorm $\mathrm{\parallel }\mathrm{\nabla }\zeta {\mathrm{\parallel }}_{\mathrm{\infty }}$. Fourier transformation therefore identifies the displayed integral with the same limit.

For $ϵ>0$, let ${\mathcal{B}}_U^{ϵ}f$ be the defining integral truncated to $t\ge ϵ$. For fixed $x\mathrm{\ne }y$, the kernel integrand is zero for all sufficiently small $t$, because nonzero values require $\mathrm{\mid }x-y\mathrm{\mid }\mathrm{/}t=\mathrm{\mid }z-y\mathrm{\mid }\le R:=\mathrm{diam}(U)$. By [F4], the absolute $t$ integral is bounded by $C\mathrm{\mid }x-y{\mathrm{\mid }}^{1-d}{\mathbf{1}}_{0<\mathrm{\mid }x-y\mathrm{\mid }\le R}$. Its product with $\mathrm{\mid }f(y)\mathrm{\mid }$ is integrable in $(x,y)$ by the annular bound and Fubini. Applying [F5] to this absolute majorant proves convergence of the truncated fields to ${\mathcal{B}}_{U}f$ in $L^1({\mathbb{R}}^d)$ and hence in distributions; no sign condition on $\omega$ is required. Rewrite the $k$-th component using $x_k-y_k=t(z_k-y_k)$ from [F2] as the integral of $[z_k\omega (z)-y_k\omega (z)]f(y)$ against $dt\mathrm{/}{t}^d$. Differentiating the truncated fields in $x_j$, then using step 1.1 for the distributional limit, gives $T_{x_k\omega ,j}f-T_{\omega ,j}(y_{k}f)$; the minus sign comes from the second term.