PomeProposition 11.46.20.7D9147EC36A1 For every $k\ge 2$ and $0<\alpha <1$, there is a nonnegative radial probability density $q_0$, smooth off $o$ and along the hyperplane, which equals $q_\alpha $ near $o$ and vanishes on a nonempty open annulus. Its solution has ordinary type $u_\alpha $ and a globally smooth residual, constant near $o$.
11.46.19 Smooth residuals with an open density-zero annulus309EDE002FAE
Proof. Choose a smooth radial $h$ with $0\le h\le 1$, equal to zero on a nonempty finite annulus and equal to one near $o$ and outside a larger finite annulus. Choose a nonzero nonnegative smooth radial cutoff $\chi $ supported in a disjoint finite annulus where $h=1$. Set
\[ C=\frac{\int (1-h)q_\alpha \kappa ^ k}{\int \chi q_\alpha \kappa ^ k}>0, \qquad q_0=q_\alpha (h+C\chi ). \]
This is nonnegative and has integral one. It is smooth at every point off $o$, including the hyperplane, and equals $q_\alpha $ near $o$. It vanishes on the chosen open annulus and is still in no $L^ p$, $p>1$. Theorem 11.46.8 applies, with zero relative density error on a neighborhood of $o$. There the exact central fluxes agree, so $Q=Q_\alpha $ and $R'=0$. Hence $R$ is constant near $o$ and extends smoothly across it. It is smooth everywhere else by the positive-root inverse, including the entire zero annulus. The solution retains its genuine pole; this smoothness assertion concerns the residual. $\square$
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