$U^+ \colon= \left\{x\in \mathbb{R}^n\mid |x| < 1, x_n>0\right\}$ is an open half-ball. Assume $u \in C^2 (\overline{U^+}$) is harmonic in $U^+$ with $u=0$ on $\partial U^+ \cap \{x_n=0\}$.
Set $v(x)=u(x)$ if $x_n \geq 0$ and $v(x)=-u(x_1,x_2,\dots, -x_n)$ if $x_n < 0$ for $x \in U=B^0(0,1)$ . Prove $v \in C^2$ ($U$), and thus $v$ is harmonic within $U$. (Evan's PDE Ch2.10)
Now assume only that $u\in C^2(U^{+}) \cap C(\overline{U^{+}})$. Show that $v$ is harmonic within $U$. ( Hint: Use Poisson's formula for the ball)
Best Answer
We still need the assumption that $u=0$ on $\partial U^+ \cap \{x_n=0\}$; otherwise $v$ is not continuous.
One option is to use the mean value characterization of harmonic functions: a continuous function $v$ is harmonic iff for every point $x$ there is $r>0$ such that $u(x)$ is equal to the average of $u$ on the sphere $\{x':|x'-x|=\rho\}$, for every $0<\rho<r$.
If you have the above characterization, the problem is easy: for $x$ with $x_n=0$ the spherical average is zero by symmetry.
But you can also use the Poisson integral formula as suggested. Apply it on the ball $B_r\{x: |x|<r\}$ for some $r\in (0,1)$, using $v$ as the boundary values. The Poisson integral defines a harmonic function $w$ in $B_r$ which agrees with $v$ on $\partial B_r$. By symmetry, $w=0$ when $x_n=0$. Thus, on the domain $B_r\cap\{x_n>0\}$ the functions $w$ and $u$ have the same boundary values. Since they are both harmonic, the maximum principle implies $w\equiv u $ in $B_r\cap\{x_n>0\}$. Since both $w$ and $v$ are antisymmetric (odd) in $x_n$, it follows that $w\equiv v$ in $B_r$.