Implication of $\partial \overline{\partial} f = 0$ on compact complex manifold

Question

Let $X$ be a compact complex manifold and $f : X \rightarrow \mathbb R.$ Why does $\partial\overline{\partial}f=0$ imply that $f$ is constant?

I can see that $\partial\overline{\partial}f=0$ means that the $(0,1)$-form $\overline{\partial}f$ is anti-holomorphic, i.e., $$ \overline{\partial}f = \frac{\partial f}{\partial \overline{z}^j } d\overline{z}^j $$ where the $\frac{\partial f}{\partial \overline{z}^j }$ are anti-holomorphic. But I'm not sure how to proceed from here. Can I say that this implies $f$ is antiholomorphic and then use something similar to the result that the only holomorphic functions on a compact complex manifold are the constants?

Answer 1

The short answer is that $\partial \bar \partial$ is the Laplacian, whose solutions obey a maximum principle. That is, if $\partial \bar \partial f = \Delta f = 0$ on some region $\Omega$, then $f$ can achieve a local maximum only on the boundary of $\Omega$. But since $\Omega$ is a compact manifold, it has no boundary. And if $f$ has no local maxima on a compact manifold, then $f$ must be constant.

Edited to add: The above is slightly incorrect for manifolds of more than 1 complex dimension, because of course the Laplacian depends on choosing some Hermitian metric tensor. What is more correct is to say that a function $f$ which satisfies $\partial \bar \partial f = 0$ can always be written $f(\vec z, \vec z^*) = f^+(\vec z) + f^-(\vec z^*)$. Then you can apply what you know about (anti)holomorphic functions.