Relations among harmonic and plurisubharmonic functions

Statement

On a domain $U\subseteq\mathbb C^d$ , let $H$ , $SH$ , $PSH$ , and $PH$ denote the harmonic, subharmonic, plurisubharmonic, and pluriharmonic functions, respectively. Then

PH=H\cap PSH,\qquad H\subseteq SH,\qquad PSH\subseteq SH.

Equivalently, inside $SH$ , the classes $H$ and $PSH$ overlap exactly in $PH$ . When $d=1$ , $PSH=SH$ and $PH=H$ .

Diagram

The set-theoretic picture for $d>1$ is

\begin{array}{c} SH\\[2pt] \supset\; H\ \cup\ PSH,\\[2pt] H\cap PSH=PH. \end{array}

Neither $H\subseteq PSH$ nor $PSH\subseteq H$ holds in general.

Why the intersection is pluriharmonic

For a smooth function, $PSH$ means that the Levi matrix is positive semidefinite, while harmonicity says that its trace is zero because

\Delta u=4\sum_{j=1}^d \frac{\partial^2u}{\partial z_j\partial\bar z_j}.

A positive-semidefinite Hermitian matrix with zero trace is zero. Hence a harmonic PSH function has vanishing Levi form and is pluriharmonic. The same conclusion holds without smoothness by distributional regularity.

Separating examples

On $\mathbb C^d$ , the function $u(z)=|z_1|^2$ is PSH but not harmonic. For $d\ge2$ , the function $v(z)=|z_1|^2-|z_2|^2$ is harmonic but not PSH. These examples expose the distinction: subharmonicity controls the trace of the complex Hessian, whereas plurisubharmonicity controls the whole Hermitian matrix.

References

Lars Hörmander, Notions of Convexity, Birkhäuser, 2007. DOI record. Relevant: Chapter 2.
Marek Klimek, Pluripotential Theory, Oxford University Press, 1991. Publisher record.