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Feb 21, 2021 at 20:01 comment added Mateusz Kwaśnicki Wait a minute, you wrote that $w = 0$ on the inner boundary, did you not? Otherwise of course no such gradient estimate holds.
Feb 21, 2021 at 17:16 comment added Harish I am still not convinced, because we can have arbitrary boundary data on inner and outer bounary. Now say we do not have zero boundary data in inner boundary, then we will have non zero integration $\int_{\partial B_1}P_D(x,y) \varphi (y)d\sigma(y) \neq 0$. but as per your claim $P_D$ is zero, which is a contradiction.
Feb 15, 2021 at 21:14 comment added Mateusz Kwaśnicki It is perfectly true, $P_D(x, y)$ does vanish on the boundary — or, if one prefers this formulation, $P_D(x,y) = \delta_x(dy)$ when $x$ is on the boundary. (OK: this is not strictly true for irregular boundary points, where definitions vary. But here all boundary points are regular for the Dirichlet problem.)
Feb 15, 2021 at 13:11 comment added Harish my objection is the following: assuming that the poissin's kernel formulae(first equation) holds for the points on the inner boundaey. (Since it is taken to be true for gradient, hence must be true for values of $w$). We have $w=0$ for every $x\in \partial B_{kr}$ then $\int_{\partial B_1} P_D(x,y) \varphi(y) d\sigma(y)=0$ for every $\varphi$. Thus $P_D=0$ on $\partial B_1$. Which I suspect is not true.
Feb 13, 2021 at 22:49 history answered Mateusz Kwaśnicki CC BY-SA 4.0