Let $B \subseteq \mathbb{R}^m$ be an open ball of radius $1$, centered at the origin of the coordinate system, and $u$ a smooth harmonic function on the complement of the ball $B$. Now, if one assumes that $u$ is rapidly decreasing at infinity in a sense that $\lim_{|x| \to \infty} |x|^n u(x) = 0$ for any $n \ge 0$, does it follow that $u$ is in fact trivial, $u(x) = 0$ for all $x$?
The only argument that falls to my mind goes via Poisson integral formula for the exterior Dirichlet problem (cf. Axler, Burdon, Ramey: Harmonic Function Theory, Chapter 4), $$u(x) = \oint_{\partial B} u(y) \, \frac{|x|^2 - 1}{|x - y|^m} \, \mathrm{d} \sigma(y) \, .$$ Taylor series of the integrand around the point $x$ leads to a power series representation of the function $u(x)$ with nonpositive powers of $|x|$ (I'm being sketchy here), and the fall-off assumption implies that in fact all the coefficients in the series are zero.
Does this make any sense? Given that the assertion is true, is there any more elegant way to prove it?
I had the same idea as @leoli1 in the comment, and here I want to write out the details.
Consider the inversion $I$ on ${\mathbb R}^m\backslash\{0\}$ defined by $$ I(x) = x^* = \frac{x}{|x|^2}. $$ Note that $|x^*|=\frac{1}{|x|}$. The Kelvin transform of your $u$ on $B^c$, the complement of the unit ball, is $$ u^*(x) = |x|^{2-m} u(x^*) $$ on $B^*$, the ball minus the origin. The usual transformation law is $$ \Delta (u^*(x)) = |x|^{-2-m} (\Delta u)(x^*), $$ so $u^*$ is harmonic on $B^*$.
Now for all $n\geq 0$, $$ \lim_{x\to 0} \frac{u^*(x)}{|x|^n} = \lim_{|x^*|\to \infty} |x^*|^{n+m-2} u(x^*) = 0. $$ With $n=0$, we see $u^*(x)$ extends continuously to 0 at the origin, and the extension would be a harmonic function on $B$. (A weaker conditions that $|x|^{m-2}u^*(x)\to 0$ as $x\to 0$ suffices. See Axler, Burdon, Ramey: Harmonic Function Theory, Thm 10.5.)
A harmonic function is analytic. Furthermore all the partial derivatives of $u^*$ at 0 must be 0, by the above limit again. So $u^*=0$, and $u=0$.