Suppose $n\geq 2$, set $B=B_r(0)\subset \mathbb{R}^n$, $B^+=\{x\in B|x_n>0\}$, $H=\{x\in B|x_n=0\}$, Let $u\in C^\infty(B^+)\cap C^1(B^+\cup H)$ be a solution of the following oblique derivative problem:
$$\begin{cases}
\Delta u=0 & \text{in} & B^+,\\
au+\beta\cdot\nabla u=0 & \text{on} & H,
\end{cases}$$
where the constant $a\leq 0$, $\beta$ is a constant vector with the $n$-th component $\beta_n>0$. Then $u$ can be uniquely extended to a harmoic function on $B'=B_{\lambda r}(0)$, where $\lambda=\beta_n/|\beta|$. In fact, let $v=au+\beta\cdot\nabla u$ on $B^+\cup H$, and $v(x',-x_n)=-v(x',x_n)$, then $v$ is harmonic on $B$; for any $x_0\in H$, let
$$u(x_0+t\beta)=e^{-at}(u(x_0)+\int_0^t v(x_0+s\beta)e^{as}ds)$$
whenever $x_0+t\beta\in B$, then it's easy to check $u$ is smooth and harmonic in $B'$, and coincides with the previous $u$ in $B'\cap \{x_n\geq 0\}$.
The question is: Does the following estimate necessarily hold, where $C$ is a positive constant independent of $u$?
$$\sup_{B'}|u|\leq C\sup_{B^+}|u|$$
I guessed it's true but could not find the answer. If it's indeed true, please give a sketch of proof, if not, I also want to see the reason.
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$\begingroup$ Erm... The points $x_0+t\beta: x_0\in H, t<0$ do not necessarily cover $B_-$, so your extension is not defined in the whole $B$, as written. $\endgroup$– fedjaCommented Jul 31, 2020 at 15:11
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$\begingroup$ No chance. Consider $n=2$, $\beta=(1,1)$. Then $H=[-1,1]$, right? Now just draw the picture. What do you see? I see a slanted strip that covers some part of the unit disk, but far from the whole unit disk. $\endgroup$– fedjaCommented Jul 31, 2020 at 16:14
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$\begingroup$ @fedja Well, I modified the question a little, which is enough for my purposes. $\endgroup$– lzcslCommented Jul 31, 2020 at 16:53
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