Harmonic function reflection I'm  learning some harmonic function theory by reviewing some problems. I came across two:
1) Prove that a real harmonic function $u$ from $\mathbb{R}^n$ to $\mathbb{R}$ such that $u(x, 0) = 0$ for all
x ∈ $\mathbb{R}^{n-1}$ satisfies that $u(x,−y) = −u(x,y)$ for all $(x,y) \in \mathbb{R}^n$.
What I tried to use is Liouville's theorem applied to the function $f(x,y)= u(x, y)+ u(x, -y)$ which eventually would be zero but I realized that the problems doesn't tell me anything about boundedness of the function.
2) A similar problem to 1) but in $\mathbb{R}^2$ only difference is that instead of having harmonicity on the whole space I'm constrained to work on the open disk of radius $a$ centered at the origin. Same condition $u(x, 0) = 0$ for all
x ∈ $\mathbb{R}$. This should result i also having $u(x,−y) = −u(x,y)$.
In this case I also tried to use $f(x,y)= u(x, y)+ u(x, -y)$ but instead I think uniqueness of the solution of Laplace's equation could be used... any help?
 A: There are many ways to get the result. You could for example see that your $f$ is harmonic, and satisfies $D^\alpha f(x,0) = 0$ for all multi-indices $\alpha$, which yields the result since harmonic functions are analytic. That works the same in both situations, by the way. My preferred way is to use
$$v(x,y) = \begin{cases}\;\; u(x,y) &, y \geqslant 0\\ -u(x,-y) &, y < 0. \end{cases}$$
$v$ is a continuous function that has the mean value property, hence harmonic. Since $u$ and $v$ coincide on a non-empty open set, and their domain is connected, we deduce $v \equiv u$. This also works for both situations.
For the second situation only: Identify $\mathbb{R}^2$ with $\mathbb{C}$. On a disk, every real-valued harmonic function is the imaginary part of a holomorphic function, say $u(z) = \operatorname{Im} f(z)$. Since $f$ is real-valued on $\mathbb{R} \cap D_a(0)$, we have
$$f(z) = \overline{f(\overline{z})}$$
for all $z \in D_a(0)$ by the identity theorem for holomorphic functions, and that implies
$$u(x,-y) = \operatorname{Im} f(x-iy) = - \operatorname{Im} f(x+iy) = -u(x,y).$$
Since the second situation was singled out, I suspect it was intended to write $u$ as the imaginary (or perhaps real) part of a holomorphic function, but the reflection principle works fine in both situations, and more generally, on every domain that is symmetric with respect to the reflection $(x,y) \mapsto (x,-y)$.
