Cap and complement estimate for the ball Poisson integral

Work under [F6] and set $\kappa :=R^2-\mathrm{\mid }x-a{\mathrm{\mid }}^2$, $C:=\{y\in \mathrm{\partial }{B}_R(a):\mathrm{\mid }y-p\mathrm{\mid }<\delta \}$ and $D:=\mathrm{\partial }{B}_R(a)\setminus C$. Since $x$ is interior, $\mathrm{\mid }x-a\mathrm{\mid }<R$ and $\kappa >0$. By [F3] the numbers $\parallel g{\parallel }_{\mathrm{\infty }}$ and ${\omega }_{g,p}(\delta )$ are finite, and the integrand $y\mapsto P_{R,a}(x,y)g(y)$ is Borel with $\mathrm{\mid }{P}_{R,a}(x,y)g(y)\mathrm{\mid }\le \parallel g{\parallel }_{\mathrm{\infty }}{\sup }_{z\in \mathrm{\partial }{B}_R(a)}{P}_{R,a}(x,z)$, a finite bound by [F1] and [F3]; the sphere has finite surface measure by [F4], so $U_g(x)$ is absolutely convergent.

For $y\in D$ one has $\mathrm{\mid }y-p\mathrm{\mid }\ge \delta$, hence $\mathrm{\mid }x-y\mathrm{\mid }\ge \mathrm{\mid }y-p\mathrm{\mid }-\mathrm{\mid }x-p\mathrm{\mid }>\delta -\delta \mathrm{/}2=\delta \mathrm{/}2$, so $1\mathrm{/}\mathrm{\mid }x-y{\mathrm{\mid }}^n\le 2^n{\delta }^{-n}$.

The cap carries mass at most one: $C$ is open in $\mathrm{\partial }{B}_R(a)$, hence Borel, $0\le P_{R,a}(x,\cdot ){\mathbf{1}}_C\le P_{R,a}(x,\cdot )$ pointwise by [F1], and the surface integral is monotone by [F2]; therefore ${\int }_C{P}_{R,a}(x,\cdot )dS\le {\int }_{\mathrm{\partial }{B}_R(a)}{P}_{R,a}(x,\cdot )dS=1$ by the unit-mass clause of [F1].

The modulus vanishes at small scales: $g$ is continuous at $p$ on the sphere, so for every $\eta >0$ there is ${\delta }_0>0$ with $\mathrm{\mid }g(y)-g(p)\mathrm{\mid }<\eta$ whenever $y\in \mathrm{\partial }{B}_R(a)$ and $\mathrm{\mid }y-p\mathrm{\mid }<{\delta }_0$; the set over which the supremum in ${\omega }_{g,p}({\delta }_0)$ is taken is nonempty by [F3], so $0\le {\omega }_{g,p}({\delta }_0)\le \eta$.

Consequently, for every $y\in D$, [F1] and step 2.1 give $P_{R,a}(x,y)=\kappa \mathrm{/}(R{\omega }_{n-1}\mathrm{\mid }x-y{\mathrm{\mid }}^n)\le \kappa 2^n\mathrm{/}(R{\omega }_{n-1}{\delta }^n)$, and also $\mathrm{\mid }g(y)-g(p)\mathrm{\mid }\le \mathrm{\mid }g(y)\mathrm{\mid }+\mathrm{\mid }g(p)\mathrm{\mid }\le 2\parallel g{\parallel }_{\mathrm{\infty }}$ by [F3].

The complement carries little mass: by [F2], [F4] and step 3.1, ${\int }_D{P}_{R,a}(x,\cdot )dS\le \frac{\kappa 2^n}{R{\omega }_{n-1}{\delta }^n}\mathrm{\mid }\mathrm{\partial }{B}_R(a)\mathrm{\mid }=\frac{\kappa 2^n}{R{\omega }_{n-1}{\delta }^n}\cdot {\omega }_{n-1}{R}^{n-1}=2^n{R}^{n-2}{\delta }^{-n}\kappa$.

Splitting by [F2] and [F5] and bounding each piece, $\mathrm{\mid }{U}_g(x)-g(p)\mathrm{\mid }\le {\int }_C{P}_{R,a}(x,\cdot )\mathrm{\mid }g-g(p)\mathrm{\mid }dS+{\int }_D{P}_{R,a}(x,\cdot )\mathrm{\mid }g-g(p)\mathrm{\mid }dS\le {\omega }_{g,p}(\delta ){\int }_C{P}_{R,a}(x,\cdot )dS+2\parallel g{\parallel }_{\mathrm{\infty }}{\int }_D{P}_{R,a}(x,\cdot )dS\le {\omega }_{g,p}(\delta )+2^{n+1}{R}^{n-2}{\delta }^{-n}\parallel g{\parallel }_{\mathrm{\infty }}\kappa$, which is the displayed estimate.

Therefore $U_g(x)\to g(p)$ as $x\to p$ from inside: given $\eta >0$, choose ${\delta }_0$ as in step 2.3, keep it fixed and let $x\to p$ with $\mathrm{\mid }x-p\mathrm{\mid }<{\delta }_0\mathrm{/}2$; step 5.1 gives $\mathrm{\mid }{U}_g(x)-g(p)\mathrm{\mid }\le \eta +2^{n+1}{R}^{n-2}{\delta }_0^{-n}\parallel g{\parallel }_{\mathrm{\infty }}(R^2-\mathrm{\mid }x-a{\mathrm{\mid }}^2)$, and $R^2-\mathrm{\mid }x-a{\mathrm{\mid }}^2\to R^2-\mathrm{\mid }p-a{\mathrm{\mid }}^2=0$ because $\mathrm{\mid }p-a\mathrm{\mid }=R$, so ${\mathrm{lim\hspace{0.17em}sup}}_{x\to p}\mathrm{\mid }{U}_g(x)-g(p)\mathrm{\mid }\le \eta$.

Since $\eta >0$ was arbitrary, the limsup in step 6.1 is zero; thus the displayed estimate holds for all admissible $x,\delta$ and the integral tends to $g(p)$ as $x\to p$ from inside the ball, which proves both assertions of the statement. The argument uses the kernel formula, its positivity and its unit mass, but never the ball Dirichlet solution theorem, so no circularity arises with the later boundary-trace theorems.