Ball Poisson integrals converge uniformly along radial boundary approaches

Work under [F3]. By [F2] the quantity $\parallel g{\parallel }_{\mathrm{\infty }}={\sup }_{\mathrm{\partial }{B}_R(a)}\mathrm{\mid }g\mathrm{\mid }$ is finite and $g$ is uniformly continuous on the sphere: for every $\epsilon >0$ there is $\delta >0$ with $\mathrm{\mid }g(y)-g(z)\mathrm{\mid }\le \epsilon$ whenever $y,z\in \mathrm{\partial }{B}_R(a)$ and $\mathrm{\mid }y-z\mathrm{\mid }<\delta$. In particular, for every $p\in \mathrm{\partial }{B}_R(a)$ and every cap radius $\delta$ with this property, ${\omega }_{g,p}(\delta )=\sup \{\mathrm{\mid }g(y)-g(p)\mathrm{\mid }:\mathrm{\mid }y-p\mathrm{\mid }<\delta \}\le \epsilon$ by [F1].

Fix such an $\epsilon$ and an associated $\delta >0$, and let $r<R$ with $R-r<\delta \mathrm{/}2$. For $p=a+R\theta$ and $x=a+r\theta$ with $\theta \in S^{n-1}$ we have $\mathrm{\mid }x-p\mathrm{\mid }=(R-r)\mathrm{\mid }\theta \mathrm{\mid }=R-r<\delta \mathrm{/}2$, so [F1] applies and gives $\mathrm{\mid }{U}_g(x)-g(p)\mathrm{\mid }\le \epsilon +2^{n+1}{R}^{n-2}{\delta }^{-n}\parallel g{\parallel }_{\mathrm{\infty }}(R^2-r^2)$.

Choose $r$ additionally so close to $R$ that $2^{n+1}{R}^{n-2}{\delta }^{-n}\parallel g{\parallel }_{\mathrm{\infty }}(R^2-r^2)\le \epsilon$; this is possible because $R^2-r^2\to 0$ as $r\uparrow R$. Then step 2.1 gives $\mathrm{\mid }{U}_g(a+r\theta )-g(a+R\theta )\mathrm{\mid }\le 2\epsilon$ for every $\theta \in S^{n-1}$ simultaneously, since neither the bound $\epsilon$ from [F2] nor the factor $R^2-r^2$ depends on $\theta$.

Taking the supremum over $\theta$ and letting $\epsilon \downarrow 0$ shows ${\sup }_{\theta }\mathrm{\mid }{U}_g(a+r\theta )-g(a+R\theta )\mathrm{\mid }\to 0$ as $r\uparrow R$, which is the assertion. The estimate used is the pointwise cap/complement bound; the ball Dirichlet solution theorem is not needed for this uniformity statement, and no structure of $U_g$ beyond the integral formula is used.