Quantitative concentration of the ball Poisson kernel

Work under [F4], put $\kappa :=R^2-\mathrm{\mid }x-a{\mathrm{\mid }}^2$ (positive because $x$ is interior) and $C:=\{y\in \mathrm{\partial }{B}_R(a):\mathrm{\mid }y-p\mathrm{\mid }<\delta \}$, $D:=\mathrm{\partial }{B}_R(a)\setminus C$. By [F2] the two masses $M_C:={\int }_C{P}_{R,a}(x,\cdot )dS$ and $M_D:={\int }_D{P}_{R,a}(x,\cdot )dS$ are finite and add to ${\int }_{\mathrm{\partial }{B}_R(a)}{P}_{R,a}(x,\cdot )dS=1$ by [F1].

For $y\in D$ one has $\mathrm{\mid }y-p\mathrm{\mid }\ge \delta$, hence $\mathrm{\mid }x-y\mathrm{\mid }\ge \mathrm{\mid }y-p\mathrm{\mid }-\mathrm{\mid }x-p\mathrm{\mid }>\delta \mathrm{/}2$ and $1\mathrm{/}\mathrm{\mid }x-y{\mathrm{\mid }}^n\le 2^n{\delta }^{-n}$; therefore $P_{R,a}(x,y)\le \kappa 2^n\mathrm{/}(R{\omega }_{n-1}{\delta }^n)$ by [F1].

Integrating the bound of step 2.1 over $D$ and using [F2] with $\mathrm{\mid }\mathrm{\partial }{B}_R(a)\mathrm{\mid }={\omega }_{n-1}{R}^{n-1}$ gives $M_D\le (\kappa 2^n\mathrm{/}(R{\omega }_{n-1}{\delta }^n)){\omega }_{n-1}{R}^{n-1}=2^n{R}^{n-2}{\delta }^{-n}\kappa =C_{n,R}{\delta }^{-n}\kappa$.

Hence $M_C=1-M_D\ge 1-C_{n,R}{\delta }^{-n}\kappa$ by step 1.1, and $M_C\le 1$ by the unit-mass identity of [F1]; as $\kappa =R^2-\mathrm{\mid }x-a{\mathrm{\mid }}^2\to 0$ when $x\to p$ (because $\mathrm{\mid }p-a\mathrm{\mid }=R$), the cap mass tends to one and the mass $M_D$ outside the cap tends to zero.

This computation is the mass-split content of the cap/complement estimate [F3] read on constant data: for $g\equiv 1$ one has ${\omega }_{g,p}=0$, $\parallel g{\parallel }_{\mathrm{\infty }}=1$ and $U_g(x)=1$ by [F1], so [F3] reduces to the trivial inequality $0\le 2^{n+1}{R}^{n-2}{\delta }^{-n}\kappa$; the genuine concentration information for kernel mass alone is exactly the bound of step 3.1 and the limit of step 4.1. Both assertions of the statement are therefore proved.