 Solution

For $0<r<1$, $r\log r<0$. The printed upper bound is negative whenever $\Omega \ne 0$, so it cannot bound a nonnegative difference. The correct small-distance modulus is $r\log (1/r)$, and the radius in the hint must also be positive.

Write $h=|x_1-x_2|\le e^{-1}$, $x_3=(x_1+x_2)/2$, and $M=h\log (1/h)$. Then $h\le M\le e^{-1}$. Denote the planar vorticity velocity kernel by $K$. It satisfies $|K(z)|\le C/|z|$ and $|DK(z)|\le C/|z{|}^2$. Split the integral at radii $M$ and $1$ about $x_3$.

On $|y-x_3|\ge 1$, every point of the segment joining $x_1$ and $x_2$ remains at least $|y-x_3|-h/2\ge (1-h/2)|y-x_3|$ from $y$. The mean value theorem bounds the kernel difference by $Ch/|y-x_3{|}^2$, giving a contribution at most $Ch\Vert \Omega {\Vert }_1$.

On $M\le |y-x_3|\le 1$, the same estimate holds because $h/2\le M/2$. Polar integration givesHere $\log (1/M)\le \log (1/h)$ because $M\ge h$.

On $|y-x_3|<M$, use the sum of the two kernel magnitudes instead of their derivatives. This disk lies inside a disk of radius $M+h/2\le 3M/2$ about either $x_i$. Hence its contribution is bounded by $CM\Vert \Omega {\Vert }_{\infty }$. Combining the three regions yieldsAt coincident points the difference is zero. At larger distances use the boundedness from part (d). A convenient global log-Lipschitz modulus isgiving $|U[\Omega ](x)-U[\Omega ](y)|\le C(\Vert \Omega {\Vert }_1+\Vert \Omega {\Vert }_{\infty })\mu (|x-y|)$. This is positive, continuous and nondecreasing. It also proves continuity of $U[\Omega ]$.