Does a boundary for $\mathbb R^n_+$ exist?

Question

I'm trying to prove the following prompt:

Let $u(x)\in C^2(\mathbb R^n_+)\cap C^1(\overline{R^n_+})$ be a solution of $$\begin{cases}-\Delta u(x) = 0 & \text{ in }\mathbb R^n_+, \text{ and}\\\left.{\partial u\over\partial x_n}\right|_{x_n=0}=0.\end{cases} \tag{$\star$}$$ If $|u(x)|\le C < \infty$, then $u(x)$ is constant. ($\mathbb R^n_+$ means that $x_n > 0$.

This problem seems to scream the use the Strong Maximum Principle for the Laplace equation. But I need to know: does $\mathbb R^n_+$ have a boundary? Here was my attempt at the proof of the prompt:

PROOF: Since $u(x)\in C^2(\mathbb R^n_+)\cap C^1(\overline{\mathbb R^n_+})$ is a solution of $(\star)$, we have that $u(x)$ is harmonic within $\mathbb R^n_+$ and so $$\max_{\overline{R^n_+}}u = \max_{\partial R^n_+}u.\tag{!!!}$$ Also, since $|u(x)|\le C < \infty$, it is clear that there must exist some $x_0\in\mathbb R^n_+$ such that $\displaystyle u(x_0) = \max_{\overline{R^n_+}}u$. Because $\mathbb R^n_+$ is connected $(***)$, we can say furthermore by the Strong Maximum Principle that $u(x)$ is constant.

I feel like the answer is no, $R^n_+$ does not have a boundary, which renders everything after and including step $(!!!)$ invalid. Secondly, is it necessarily true that $R^n_+$ is connected? Intuition suggests that it is indeed connected, but the argument there cannot be applied since you can't connect every point through the origin. However, I imagine you can still find a path between any two points.

Answer 1

I am sorry to have a comment on your proof: it seems that your proof is not very suitable, because most time we always use the maximum principle in a bounded domain, and you require to consider the behavior of the harmonic function at infinity provided you use the maximum principle in a unbounded domain, sometimes it is rather difficult. Just as @Pedro Tamaroff's comments above, probably it is better to apply the reflection principle. I have a proof as below, please take a look if necessary and any comments are welcome.

We need a familiar estimate for the gradient of harmonic functions as following.

Lemma. Let $\Omega$ be a domain in $\mathbf R^n,$ and suppose that $u:\Omega\to\mathbf R$ is a harmonic function. Then for any $x\in\Omega$ and $r>0$ with $B_r(x)\subset\Omega,$ there exists a positive $C,$ which is only dependent on the dimension $n,$ such that $$|\nabla u(x)|\leq \frac{C}{r^{n+1}}\|u\|_{L^1(B_r(x))},$$ here $u$'s gradient is denoted by $\nabla u.$

For any $x=(x_1,\cdots,x_{n-1},x_n)\in \overline{\mathbf R^{n}_+},$ we set $$v(x_1,\cdots,x_{n-1},x_n)=\frac{\partial u}{\partial x_n}(x).$$ Note that $u\in C^2(\mathbf R^n_{+})\cap C^1(\overline{\mathbf R_{+}^n})$ and $u$ is harmonic in $\mathbf R^n_+,$ hence $v$ is well defined and $v\in C^2(\mathbf R^n_+)\cap C(\overline{\mathbf R_+^n})$ satisfies the equation$$\cases{\Delta v=0, &in $\mathbf R^n_+$,\cr v=0, &on $\{x_n=0\}$.\cr}$$ Now we note that $u$ is bounded on the half space $\overline{\mathbf R_+^n},$ thanks to the estimate above, we obtain that $v$ is also bounded on $\overline{\mathbf R_+^n}.$ Then applying the reflection principle, we can define a new harmonic function $v^\ast$ by $$v^\ast(x)=\cases{v(x), &$x_n\geq0$,\cr -v(x_1,\cdots,x_{n-1},-x_n), &$\{x_n<0\}$.\cr}$$Using Poisson's integral formula, it is easy to check that $v^\ast$ is indeed harmonic in the whole space $\mathbf R^n,$ and obviously $v^\ast$ is bounded in the entire space. Liouville's theorem implies that $v$ is constant and so $v$ vanishes in $\mathbf R^n.$ Thus $u$ is constant, this is our desired result.