Existence of harmonic conjugate in several variables

Question

Let $\Omega$ be a domain in $\mathbb{C}^m$. A function $f:\Omega\mapsto\mathbb{R}$ is pluriharmonic if $\frac{{\partial}^2f}{\partial z_{i}\partial\bar{z_{j}}}=0$ $\forall~i,j\in\{1,...,m\}$ on $\Omega$ . When $m=1$, $f$ locally possesses a harmonic conjugate i.e. $\exists$ a harmonic function $g$ s.t. $f+\imath g$ is holomorphic (Note that for $m=1$, pluriharmonic functions are harmonic). Does the same holds for $m>1?$

Answer 1

You probably want pluriharmonic, which is not your condition $\frac{\partial^2 f}{\partial z_j \partial \bar{z}_j} = 0$ for all $j=1,...,m$. For example, $z_1 \bar{z}_2+\bar{z}_1 z_2$ is a counterexample. It is not the real part of a holmorphic function, in particular it is not harmonic on the $z_1 = z_2$ complex line. You really need harmonic along every complex line (your definition had better be invariant under complex linear transformations if it is to make sense).

You want $\frac{\partial^2 f}{\partial z_j \partial \bar{z}_k} = 0$ for all $j,k=1,...,m$. Then $f$ is called pluriharmonic, and then it does indeed posses, locally, a conjugate function. I find it easiest to write the power series for $f$ in terms of $z$ and $\bar{z}$. For example at the origin:

$\displaystyle f(z,\bar{z}) = \sum_{\alpha,\beta} c_{\alpha\beta} z^\alpha \bar{z}^\beta$,

where $\alpha$ and $\beta$ are multiindexes. It is common notation to write $f$ then as a function of $z$ and $\bar{z}$, as it formally is just that. The condition of pluriharmonicity guarantees no mixed terms in the expansion. Therefore

$f(z,\bar{z}) = h(z) + \overline{g(z)}$,

where $h$ and $g$ are holomorphic functions. If we assume that $f$ vanishes at the origin, then plugging in $0$ for $\bar{z}$ in the expansion, obtains $h$:

$f(z,0) = h(z)$,

(be careful what that means, we're no longer evaluating $f$ but it's expansion which is really a function of twice as many variables) And if $f$ was real valued originally, then $f = 2 \operatorname{Re} h$.