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. 



Does $\partial\overline{\partial}f=0$ imply $f\equiv c$ for particular kind of $f$?
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 $h\_-$.


