# How is the maximum-minimum principle used here?

In the sequel, let $D\subset{\mathbb R}^m(m=2,3)$ be a bounded domain of class $C^2$. And assume that the boundary $\partial D$ is connected.

Exterior Dirichlet Problem. Find a function $u$ that is harmonic in ${\mathbb R}^m\setminus \bar D$, is continuous in ${\mathbb R}^m\setminus D$, and satisfies the boundary condition $$u = f\qquad\text{on}~\partial D,$$ where $f$ is a given continuous function. For $|x|\to\infty$ it is required that $$u(x)=O(1),\quad m=2,\qquad\text{and}\quad u(x)=o(1),\quad m=3,$$ uniformly for all directions.

I don't fully understand how the uniqueness of the solution (if exists) to this problem is proved.

For the $m=3$ case, the book only said that from the maximum-minimum principle, observing that $u(x)=o(1),~|x|\to\infty$, we obtain $u=0$ in ${\mathbb R}^3\setminus D$ when $f=0$.

Question: How is the maximum-minimum principle used here?

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Consider the sphere $\Omega_r:=\{x\in{\mathbb R}^3:|x|=r\}$. Let $$M_r:=\max_{x\in\Omega_r}~u(x),\qquad m_r:=\min_{x\in\Omega_r}u(x).$$ Since $u(x)=o(1)$ as $|x|\to\infty$, we have $\lim_{r\to\infty}M_r=0$ and $\lim_{r\to\infty}m_r=0$.
Consider $r\geq R$ such that $\Omega_{R}\subset{\mathbb R}^3\setminus D$. Denote $B_r:=\{x\in{\mathbb R}^3:|x|\leq r\}$. Now use the maximum-minimum principle on $({\mathbb R}^3\setminus D)\cap B_r$. Let $$\bar M_r:=\max_{x\in({\mathbb R}^3\setminus D)\cap B_r}~u(x),\qquad \bar m_r:=\min_{x\in({\mathbb R}^3\setminus D)\cap B_r}u(x).$$ When $f=0$, i.e., $u(x)=0$ on $\partial D$, we have $\lim_{r\to\infty}\bar M_r=0$ and $\lim_{r\to\infty}\bar m_r=0$ since $\bar M_r$ and $\bar m_r$ are obtained either on $\partial D$ or on $\Omega_r$.
For any $x\in {\mathbb R}^3\setminus D$, there exists $R>0$ such that $x\in({\mathbb R}^3\setminus D)\cap B_r$ for all $r\geq R$. And thus $\bar m_r\leq u(x)\leq \bar M_r$ for all $r\geq R$. Hence $u(x)=0$.