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So new to complex geometry/dolbeault cohomology, and I have basic question.

If $\partial \overline{\partial} f = 0$, then do we know that $f$ is constant?

If $\partial (\overline{\partial} f) = 0$, I know this means $\overline{\partial} f $ is anti-holomorphic. From $\overline{\partial} f = \frac{\partial f}{\partial \overline{z}_i} d\overline{z}_i$, that means $\frac{\partial f}{\partial \overline{z}_i}$ are anti-holomorphic.

This implies (I think) that $f$ is anti-holomorphic.

However, using fact that $\partial \overline{\partial} f = - \overline{\partial}\partial f$, we get that $f$ is also holomorphic, so $f$ must be constant.

Michael Albanese
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user308973
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A function $f$ satisfying $\partial\bar{\partial}f = 0$ is not necessarily constant. For example, any holomorphic function on $\mathbb{C}$ satisfies $\bar{\partial}f = 0$, so it also satisfies $\partial\bar{\partial}f = 0$.

If $M$ is a complex manifold, a function $f : M \to \mathbb{C}$ satisfying $\partial\bar{\partial}f = 0$ is called pluriharmonic. If $M$ is compact, then every pluriharmonic function is constant; see here.

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Michael Albanese
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