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

Question

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.

Answer 1

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 $h_-$.