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The Schwarz reflection principle says (Serge Lang, Complex Analysis, 1993):

Let $U^+$ be a connected open set in the upper half plane, and suppose that the boundary of $U^+$ contains an open interval $I$ of real numbers. Let $U^-$ be the reflection of $U^+$ across the real axis. If $f$ is a function on $U^+\cup I$, analytic on $U^+$ and continuous on I, and f is real valued on $I$, then $f$ has a unique analytic continuation on $U^+\cup I\cup U^-$.

I wonder if the assumption "$f$ is continuous on $I$" could be replaced by the weaker assumption that "$\text{Im}(f)$ is continuous on $I$", since the analogue reflection theorem for harmonic functions only needs such a weaker assumption. Or is there a counter-example?

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Yes, the continuity of $\operatorname{Im}f$ is enough. Let's write $f=u+iv$. As you noted, $v$ extends to a harmonic function $\hat v$ on $ U^+\cup I\cup U^-$ by odd reflection: $$\hat v(x+iy)=\begin{cases} v(x+iy)\quad & y>0 \\ 0\quad & y=0 \\ -v(x-iy)\quad & y<0 \end{cases} $$ Also extend $$\hat u(x+iy)=\begin{cases} u(x+iy)\quad & y>0 \\ u(x-iy)\quad & y<0 \end{cases} $$ leaving $\hat u$ undefined on $I$ for now. I claim that $\hat u$ extends continuously to $I$.

To this end, fix $z_0\in I$. Let $D$ be a disk centered at $z_0$ and contained in $U^+\cup I\cup U^-$. Since $\hat v$ is harmonic in $D$, there is a harmonic function $w$ in $D$ for which $\hat v$ is a conjugate, that is $$w_x=\hat v_y,\quad w_y=-\hat v_x $$ Observe that $\hat u$ also satisfies $$\hat u_x=\hat v_y,\quad \hat u_y=-\hat v_x $$ in $D\setminus I $. Therefore, the difference $\hat u-w $ is locally constant on $D\setminus I$. Since $D\setminus I$ consists of two connected components, there are constants $c_1,c_2$ such that $\hat u = w+c_1 $ in $D\cap U^+$ and $\hat u=w+c_2$ in $D\cap U^-$. Using the symmetry of $\hat u$ and the continuity of $w$, we obtain
$$c_1-c_2= \lim_{y\to 0 }\, \big\{ (\hat u(x+iy)-\hat u(x-iy)) - (w(x+iy)-w(x-iy))\big\}=0$$ Thus $\hat u=w+c$ in $D\setminus I$, which allows us to extend $\hat u$ continuously to $D$.

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