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I recently learned about the Schwarz reflection principle and I was wondering whether any of the assumptions can be weakened. Explicitly, let $\Omega\subseteq\mathbb{C}$ be open and connected such that $\Omega$ is symmetric about the real axis, i.e. $\forall~z\in\Omega$, $\bar z\in\Omega$. Define $\Omega^+=\{z\in\Omega\colon~\text{Im}(z)>0\}$ and $\Omega_0^+=\{z\in\Omega\colon~\text{Im}(z)\ge0\}$. Suppose that $u$ is a continuous function on $\Omega_0^+$ and harmonic on $\Omega^+$. The Schwarz reflection principle says we can extend $u$ to a harmonic function on $\Omega$ if the harmonic conjugate $v$ of $u$ is zero on $\Omega_0^+\cap\mathbb{R}$. Is the condition of $v$ being zero on $\Omega_0^+\cap\mathbb{R}$ necessary? I suspect it is, but I have not been able to construct any counterexamples. Any hints and/or suggestions are welcome.

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It's not necessary to assume $v=0$. You could assume $v=1$. Or that $u=0$ (or another constant) as here or here.

But you have to assume something like this, because otherwise the restriction of $u$ to $\Omega_0^+\cap \mathbb{R}$ need not be even differentiable, which precludes having a harmonic extension across the boundary (harmonic functions are infinitely differentiable, and moreover real-analytic). E.g., put the Weierstrass nowhere-differentiable function on the interval $[-1,1]$, extend it continuously to the upper half of unit circle, then harmonically to the upper half disk (any continuous boundary function has harmonic extension inside of the half-disk).

At a minimum, assume $u$ is real-analytic on the boundary; then there is a harmonic extension a little bit over the boundary (not to the entire mirror image of $\Omega^+$).

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