$u$ is harmonic in $\mathbb{R}^n_+$ and $u=0$ on the boundary. I wish to extend $u$ to a harmonic function on $\mathbb{R}^n$. Suppose I defined $u(x_1,...,x_{n-1},x_n)=u(x_1,...,x_{n-1},-x_n)$ for $x_n\leq0$. Clearly $u$ would be harmonic on the lower half space. But what about the boundary? If $u$ is harmonic at a point $\xi$ on the boundary it must satisfy the mean value property i.e. $$0=u(\xi)=\frac{1}{w_n}\int_{|x|=1}u(\xi+rx)dS_x$$ But for that to hold, shouldn't I define $u(x_1,...,x_{n-1},x_n)=-u(x_1,...,x_{n-1},-x_n)$?

  • $\begingroup$ Your reasoning seems sound to me. $\endgroup$ – DisintegratingByParts Nov 11 '19 at 17:51
  • $\begingroup$ @DisintegratingByParts I have seen people go with the first extension rather than the second one. Why is that? $\endgroup$ – Hrit Roy Nov 14 '19 at 13:02

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