Exterior-sphere domains need not have global C2alpha Dirichlet solutions

Let $U=\{(r\cos \theta ,r\sin \theta ):0<r<1,0<\theta <2\pi \mathrm{/}3\}$. It is bounded and Lipschitz: its boundary consists of two line segments and a circular arc, meeting at finitely many corners of angles $2\pi \mathrm{/}3$ and $\pi \mathrm{/}2$. At a smooth boundary point, a sufficiently small ball tangent on the exterior side is disjoint from $U$; on the circular arc one can use a ball outside the unit disk. At a corner of interior angle $\beta <\pi$, choose a ball tangent at the corner with its center on the exterior angle bisector. Directions from the vertex to points in this ball lie within angle $\pi \mathrm{/}2$ of the center direction, while the sector directions are at angular distance at least $\pi -\beta \mathrm{/}2>\pi \mathrm{/}2$; hence the ball is exterior. Therefore $U$ satisfies the exterior sphere condition everywhere.

Let $\chi$ be the profile supplied by [F3]. Set $u(r,\theta )=\chi (r)r^{3\mathrm{/}2}\sin (3\theta \mathrm{/}2),L=\mathrm{\Delta },f=\mathrm{\Delta }u,g=0.$ Since $r>0$ on $U$, $u\in C^{\mathrm{\infty }}(U)$. It extends continuously by zero at the vertex, vanishes on both straight sides because the sine factor vanishes, and vanishes on the circular arc because $\chi (1)=0$. To compute $\mathrm{\Delta }w$ for any $C^2$ function in local polar coordinates, write $c=\cos \theta$ and $s=\sin \theta$. The chain rule [F5] gives $w_{xx}=c^2{w}_{rr}+\frac{s^2}{r}{w}_r+\frac{s^2}{r^2}{w}_{\theta \theta }-\frac{2sc}{r}{w}_{r\theta }+\frac{2sc}{r^2}{w}_{\theta },w_{yy}=s^2{w}_{rr}+\frac{c^2}{r}{w}_r+\frac{c^2}{r^2}{w}_{\theta \theta }+\frac{2sc}{r}{w}_{r\theta }-\frac{2sc}{r^2}{w}_{\theta }\mathrm{.}$ By [F4], adding yields $\mathrm{\Delta }w=w_{rr}+r^{-1}{w}_r+r^{-2}{w}_{\theta \theta }$. For $w=r^{3\mathrm{/}2}\sin (3\theta \mathrm{/}2)$, these derivatives are $w_r=\frac{3}{2}{r}^{1\mathrm{/}2}\sin (3\theta \mathrm{/}2)$, $w_{rr}=\frac{3}{4}{r}^{-1\mathrm{/}2}\sin (3\theta \mathrm{/}2)$, and $w_{\theta \theta }=-\frac{9}{4}{r}^{3\mathrm{/}2}\sin (3\theta \mathrm{/}2)$, so the polar Laplacian is zero on $r<1\mathrm{/}4$. On the annulus where $\chi$ varies, $r$ is bounded away from zero and all factors are smooth; outside $r<1\mathrm{/}2$, $u=0$. Thus $f$ is smooth and supported in $1\mathrm{/}4\le r\le 1\mathrm{/}2$. The domain $U$ is convex, so boundedness of $\mathrm{\nabla }f$ gives a Lipschitz bound on $f$ along each segment in $U$; since $U$ is bounded and $0<\alpha <1$, this implies $f\in C^{0,\alpha }(U)$.

Along the ray $\theta =\pi \mathrm{/}3$ and for $0<r<1\mathrm{/}4$, ${\mathrm{\partial }}_{rr}u=(3\mathrm{/}4)r^{-1\mathrm{/}2}$, which is unbounded as $r\downarrow 0$. A bounded global $C^{2,\alpha }$ norm bounds every Cartesian second derivative and hence every unit-direction second derivative, so [F1] implies $u\mathrm{\notin }{C}^{2,\alpha }(U)$. The coefficients of $L=\mathrm{\Delta }$ are smooth, the ellipticity constant is $1$, and the expanded zero-order coefficient is $0\le 0$.

The displayed $u$ is a classical solution in $C^2(U)\cap C^0(\bar{U})$ of $\mathrm{\Delta }u=f$ with zero Dirichlet data. Any other solution in that class differs from $u$ by a zero-boundary harmonic function, which vanishes by [F2]. Hence it is the unique classical solution, and no solution in the asserted global $C^{2,\alpha }(U)\cap C^0(\bar{U})$ class exists. This refutes the assertion.