Let $f: \mathbb{R}^{n}\to \mathbb{R}$ be a harmonic function (i.e. $\Delta f = 0$) and suppose $\lim_{|x|\to \infty}f(x) = 0$. Is it true that $f$ must be identically zero?
It seems the answer is yes, by an application of the maximum principle. Because $\lim_{|x|\to \infty}f(x) = 0$, given $\varepsilon > 0$ there exists some $R > 0$ such that $|f(x)| \le \varepsilon$ whenever $|x| \ge R$. Take $\bar{\Omega} = \overline{B_{R}(0)}$ to be the closed ball of radius $R$ centered at the origin. Moreover, let $\Omega = \text{int} \bar{\Omega}$, which is open and nonemtpy. Then $f$ is harmonic on $\Omega$ and, by the maximum principle, it must have a maximum at $\partial \Omega$. This maximum is less or equal than $\varepsilon$ and, since the latter is arbitrary, it follows that $f = 0$.
Am I missing something?