Pome517F0BB04B80Proposition 15.20.8.517F0BB04B80 For the symmetric projective data of theorem 15.20.4, let $u_\lambda $ solve
On the open band $\{ -1<r<1\} $, $u_\lambda -u_*$ is constant. Both potentials are $C^{1,1}$ there and fail to be $C^2$ at every point of the two hypersurfaces $r=\pm t_*$. In particular, none of these positive-exponent solutions is a globally $C^2$ potential.
Proof. Uniqueness and invariance of the data imply that $u_\lambda $ is $K$-invariant. Write $u_\lambda +\rho =G_\lambda (r)$ on $U$. The positivity argument in lemma 15.20.5 uses only boundedness and invariance, so $G_\lambda $ is locally Lipschitz and $h_\lambda =G_\lambda '$ is locally of bounded variation, with
Its weak flux equation is the same one, with the prescribed coefficient replaced by
Indeed, the scalar convolution and signed-product argument of that lemma is unchanged, and $k_\lambda $ is exactly the radial coefficient of the right-hand measure. Since $G_\lambda $ is bounded on compact $r$-intervals, $k_\lambda $ is locally bounded. Hence the right side of the flux equation has no atoms. The same strictly positive jump integral eliminates every jump of $h_\lambda $. Thus $h_\lambda $ is continuous and $G_\lambda $ is $C^1$. The coefficient $k_\lambda $ is consequently continuous. Where $A(r,h_\lambda )>0$, inversion of the $C^1$ flux gives
On $-1<r<1$ the coefficient $k_\lambda $ is identically zero. Exchange invariance gives $h_\lambda (0)=0$ and oddness. On the positive half-band, the first-contact and no-departure arguments in the proof of theorem 15.20.4 apply verbatim: the first branch solves $h'=-z_0$, reaches $-p$ at $t_*$, and cannot leave that lower boundary to its right because $(h+p)'=p'-z_0<0$ on every proposed departure interval. Therefore
This is the derivative of $u_*+\rho $. Its primitive differs from $G_\lambda $ by one constant on the connected band. The claimed local $C^{1,1}$ regularity follows from the explicit slope.
At $t_*\in (0,1)$ the two one-sided derivatives of this slope are
These unequal derivatives prevent the scalar primitive from having a second derivative at $t_*$. Through every point of that level set, the submersion $r$ provides a smooth curve parametrized by $r$, transferring the failure to $u_\lambda $. Smooth $\rho $ cannot cancel the jump. Oddness gives the conclusion at $-t_*$, and the identical slope proves it for $u_*$. $\square$
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