Does $\partial\overline{\partial}f=0$ imply $f\equiv c$ for particular kind of $f$? Hi! 
Let $f\in C^{2,\alpha}\left( \mathbb{C}^{m}\setminus \overline{B_{R}},\mathbb{R} \right)$ with $m\geq 2$, $R>0$  and s.t. $f$ has an expansion of type
$$f=1+\mathcal{O}\left( \frac{1}{|z|} \right)$$
Suppose, moreover, that
$$\partial\overline{\partial}f=\sum_{i,j=1}^{m}\left(\partial_{i}\overline{\partial}_{j}f \right)dz^{i}\wedge \overline{dz^{j}}\equiv 0$$
Does it follow that 
$$f\equiv c \qquad c\in \mathbb{R}$$
I'm quite sure of this fact but i fear there are subtleties i don't see, so if it is true are there  references for this fact?
Thank you in advance. 
 A: Yes, this is true.  The point is that $\partial\bar\partial f = 0$ on the complement $C$ of a ball in $\mathbb{C}^n$ ($n\ge 2$) implies that $f = h_+ + \overline{h_-}$ for some functions $h_\pm$ that are holomorphic on $C$.  (They are unique up to adding a constant to one and subtracting it from the other.)  Now Hartogs' extension theorem implies that $h_\pm$ extend to be holomorphic on all of $\mathbb{C}^n$.  Your boundedness assumption now says that the real and imaginary parts of $f$, which are harmonic, are bounded, which implies, by Liouville's Theorem, that they are constant.
Requested explanation:  Since $d(\partial f) = (\partial + \bar\partial)(\partial f) = -\partial\bar\partial f =0$, it follows that $\partial f$ is a closed holomorphic $(1,0)$-form  on $C$ and hence is of the form $\partial f =\partial h_+ = dh_+$ for some holomorphic function $h_+$ on $C$, unique up to an additive constant. (NB: $C$ is simply connected, since $n\ge 2$, so closed $1$-forms on $C$ are exact.)  Since $\partial (f-h_+) = 0$, it follows that the function $f-h_+$ must be antiholomorphic, so $f = h_+ + \overline{h_-}$ for some holomorphic function $h_-$.
