L1 consistency of the mollified Biot-Savart kernel

On $\mathrm{\mid }x\mathrm{\mid }\le 2\epsilon$, [F2] and $\mathrm{\mid }K(x)\mathrm{\mid }\le (2\pi \mathrm{\mid }x\mathrm{\mid }{)}^{-1}$ give ${\int }_{\mathrm{\mid }x\mathrm{\mid }\le 2\epsilon }\mathrm{\mid }K(x)\mathrm{\mid }dx\le C\epsilon$. Also $\mathrm{\mid }{K}_{\epsilon }(x)\mathrm{\mid }\le \mathrm{\parallel }{\rho }_{\epsilon }{\mathrm{\parallel }}_{\mathrm{\infty }}{\int }_{\mathrm{\mid }w\mathrm{\mid }\le 3\epsilon }\mathrm{\mid }K(w)\mathrm{\mid }dw\le C_{\rho }\mathrm{/}\epsilon ,$ so its integral over this disk is at most $C_{\rho }\epsilon$.

First, each component $z_j{\rho }_{\epsilon }(z)$ is integrable because it is bounded and compactly supported. Radiality makes it odd under the reflection $z\mapsto -z$; applying [F4] to that reflection, whose Jacobian determinant has absolute value one, gives $\int z_j{\rho }_{\epsilon }(z)dz=-\int z_j{\rho }_{\epsilon }(z)dz$, hence every first-moment component is zero. Now fix $\mathrm{\mid }x\mathrm{\mid }>2\epsilon$ and $\mathrm{\mid }z\mathrm{\mid }\le \epsilon$. The segment from $x$ to $x-z$ is contained in the open ball $B_{\mathrm{\mid }x\mathrm{\mid }\mathrm{/}2}(x)$, which is convex and avoids the origin, and each second coordinate derivative of $K(w)=w^{\perp }\mathrm{/}(2\pi \mathrm{\mid }w{\mathrm{\mid }}^2)$ is bounded there by $C\mathrm{\mid }x{\mathrm{\mid }}^{-3}$: direct differentiation gives terms of the forms $c{w}_i\mathrm{\mid }w{\mathrm{\mid }}^{-4}$ and $c{w}_i{w}_j{w}_k\mathrm{\mid }w{\mathrm{\mid }}^{-6}$. Apply the scalar Taylor theorem [F3] to each component of $K$ along this segment. The coordinate remainders are each bounded by $C\mathrm{\mid }z{\mathrm{\mid }}^2\mathrm{\mid }x{\mathrm{\mid }}^{-3}$, so [F5] gives the vector bound $K(x-z)=K(x)-DK(x)z+R(x,z)$ with $\mathrm{\mid }R(x,z)\mathrm{\mid }\le C\mathrm{\mid }z{\mathrm{\mid }}^2\mathrm{\mid }x{\mathrm{\mid }}^{-3}$. Integrating in $z$ against ${\rho }_{\epsilon }$, the linear term vanishes by the just-proved first-moment identity, and $\int \mathrm{\mid }z{\mathrm{\mid }}^2{\rho }_{\epsilon }(z)dz\le {\epsilon }^2$. Hence $\mathrm{\mid }{K}_{\epsilon }(x)-K(x)\mathrm{\mid }\le C{\epsilon }^2\mathrm{\mid }x{\mathrm{\mid }}^{-3}$.

Integrating the bound in step 1.2 by [F2] over $\mathrm{\mid }x\mathrm{\mid }>2\epsilon$ gives at most $C{\epsilon }^2{\int }_{2\epsilon }^{\mathrm{\infty }}{r}^{-3}rdr\sigma (S^1)\le C\epsilon \mathrm{.}$ Together with step 1.1 this proves the stated $L^1$ estimate.

For each $x$, the convolution difference is absolutely bounded by $\int \mathrm{\mid }{K}_{\epsilon }(x-y)-K(x-y)\mathrm{\mid }\mathrm{\mid }\omega (y)\mathrm{\mid }dy\le \mathrm{\parallel }{K}_{\epsilon }-K{\mathrm{\parallel }}_1\mathrm{\parallel }\omega {\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{.}$ Taking the essential supremum in $x$ and using step 2.1 proves the force estimate.