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Let $\{\Omega_{j}\}_{j\in\mathbb{N}}$ be a sequence of smooth bounded domains in $\mathbb{C}^{n}$ such that $\Omega_{j}$ converges to a smooth bounded domain $\Omega$ in the sense that the defining function of $\Omega_{j}$, say $\rho_{j}$ converges to the defining function of $\Omega~~$,say $\rho$ uniformly on each compact subset of $\mathbb{C}^{n}$, where a defining function of a smooth bounded domain $D$ is defined as a smooth real-valued function say $\psi$ on $\mathbb{C}^{n}$ such that $\psi$ satisfies the following: $$ \begin{align} D =\{z\in\mathbb{C}^{n}:\psi(z)<0\}&,\\ \partial D =\{z\in\mathbb{C}^{n}:\psi(z)=0\}&,\text{ and}\\ \operatorname{grad}(\rho)\neq0\text{ on }\partial D.& \end{align}$$

This notion of convergence implies the Hausdorff convergence of open domains. Precisely if I say: For any compact set $K\subset\Omega$, then $K\subset\Omega_{j}$ for sufficiently large $j$. Also, for any compact set $L\subset\mathbb{C}^{n}\setminus\overline{\Omega}$, then $L\subset\mathbb{C}^{n}\setminus\overline{\Omega_{j}}$. My question is: for fixed $x_{0}\in \Omega$ can we imply that there exists a positive constant $c$ such that $\min\{P_{j}(x_{0},y):y\in\partial\Omega_{j}\}\geq c$ for large $j$, where $P_{j}$ denotes the Poisson kernel of $\Omega_{j}$. It would be a great help if any suggestions I get.

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    $\begingroup$ Upon reflection, I don't see how your mode of convergence implies Hausdorff convergence either. Perhaps you meant "uniformly on each compact subset of $\mathbb{C}^n$" rather than of $D$? $\endgroup$
    – Kostya_I
    Commented Apr 13, 2023 at 8:33
  • $\begingroup$ Yeah, you are right. I have made a typo. I will modify it. Thanks for pointing it out. $\endgroup$
    – Naruto
    Commented Apr 13, 2023 at 9:13
  • $\begingroup$ I believe this does not change the answer: your convergence is just Hausdorff convergence, by a variant of the argument given below (you can choose $\rho_j$ to coincide with $\rho$ outside a small nbh of $\partial D$, and to be small in this nbh.) If you additionally require uniform convergence of enough derivatives of $\rho_j$, then you have uniform convergence of Poisson kernels, so the result is true. $\endgroup$
    – Kostya_I
    Commented Apr 13, 2023 at 9:55
  • $\begingroup$ Could you please explain it? I am not able to follow you. $\endgroup$
    – Naruto
    Commented Apr 13, 2023 at 10:38
  • $\begingroup$ I will explain the convergence in a question. $\endgroup$
    – Naruto
    Commented Apr 13, 2023 at 10:39

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No. Take $\Omega$ to be, say, the unit disc $B_1$ in $\mathbb{C}$ with the defining function $\psi(z)=|z|^2-1$. In fact, any sequence of smooth domains $\Omega_j$ such $B_{1-\frac{1}{j}}\subset\Omega_j\subset B_1$ converges to $\Omega$ in your sense: if $\psi_j$ are defining functions for $\Omega_j$, then, using a partition of unity, you can modify them so that they coincide with $\psi$ on $B_{1-\frac{2}{j}}$.

Now, if $\Omega_j$ have long fjords, then the Poisson kernel inside those fjords can be arbitrarily small. Indeed, imagine the fjord is a strip divided into equal squares by segments $l_1,l_2,\dots $, with $l_1$ closest to the dead-end. Then, $$\max_{y\in l_n} G(x_0,y)\leq \max_{y\in l_n} \mathrm{hm}(y,l_{n+1})\cdot\max_{y\in l_{n+1}} G(x_0,y)\leq c\cdot \max_{y\in l_{n+1}} G(x_0,y),$$ where the harmonic measure is taken in the part of the fjord cut out by $l_{n+1}$, and $c<1$ is the corresponding maximum when one replaces that part by a half-infinite strip. Consequently, the Green's function inside the fjord is exponentially small in the aspect ratio of the fjord, and by Harnack estimate so is its normal derivative.

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  • $\begingroup$ How? I did not understand what you were saying. If it has long fjords, why can the Poisson kernel inside those fjords be made arbitrarily small? Could you please explain it? $\endgroup$
    – Naruto
    Commented Apr 13, 2023 at 9:19
  • $\begingroup$ What do you mean by fjords I am not able to follow your language and how then the inequality has come. I don't know much about Poisson's theory. I know at a very basic level. Could you please please explain according to that? $\endgroup$
    – Naruto
    Commented Apr 13, 2023 at 10:50
  • $\begingroup$ @Naruto, I think if you want someone to explain you the basics, Mathoverflow is not a right forum. $\endgroup$
    – Kostya_I
    Commented Apr 13, 2023 at 12:03

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