Subharmonic in any holomorphic coordinates = Plurisubharmonic?

An upper semi-continuous function $$u : \Omega \to \mathbb{R}$$, $$\Omega \subseteq \mathbb{C}^n$$ is said to be subharmonic if it satisfies the submean inequality $$u(a) \leq \mu_S(u;a,r)$$, where $$\mu_S(u;a,r)$$ is the mean value of $$u$$ on the sphere $$S(a,r)$$ with center $$a$$ and radius $$r$$ (for any $$a$$,$$r$$ such that the closed ball $$\overline{B(a,r)}$$ is contained in $$\Omega$$).

The function is said to be plurisubharmonic if it is subharmonic restricted to any complex line. It is straight-forward to see that plurisubharmonic implies subharmonic, see for example Demailly, "Complex Differential and Algebraic Geometry", §I.5.A. It is also not so difficult to see that plurisubharmonicity is invariant under holomorphic changes of coordinates, see for example Demailly, Theorem I.5.11.

The entry Encyclopedia of Mathematics: Pluripotential theory states that "Plurisubharmonic functions are precisely the subharmonic functions invariant under a holomorphic change of coordinates". One inclusion is clear from above, but does anyone have an idea of how to prove the other inclusion, i.e.: if $$u$$ is subharmonic with respect to any local holomorphic coordinates, then $$u$$ is plurisubharmonic?

1 Answer

It is actually enough to assume that your function $$u$$ defined on a neighborhood of $$0\in \mathbb C^n$$ remains subharmonic after composing with any linear transformation.

Indeed, let $$0\neq \lambda \in \mathbb C^n$$ and set, for $$\sigma \in \mathbb C^*$$,

$$A(\sigma) = \begin{pmatrix} \lambda_1 & \sigma a_{12} & \cdots & \sigma a_{1n}\\\ \vdots & \vdots & & \vdots \\\ \vdots & \vdots & &\vdots\\\ \lambda_n & \sigma a_{n2} & \cdots & \sigma a_{nn} \end{pmatrix}$$

with $$a_{ij}$$ chosen such that the matrix is invertible. Then one finds, at the origin

$$\begin{eqnarray} \Delta (u \circ A(\sigma))&=&\sum_{k=1}^n \frac{\partial^2(u \circ A(\sigma))}{\partial z_k \partial \bar{z_k}}(0)\\\ &=&\sum_{k=1}^n \sum_{1\leq i,j\leq n}A(\sigma)_{ik} \overline{A(\sigma)_{jk}} \frac{\partial^2 u}{\partial z_i \partial \bar {z_j}}(0)\\\ &=&Hu_{0}(\lambda) + |\sigma|^2 \sum_{\ell=2}^n Hu_{0}((a_{j \ell})_{j}) \end{eqnarray}$$

where $$H$$ denotes the complex hessian. This quantity is non-negative by assumption, and one gets the result by making $$\sigma$$ approach $$0$$.

• Are you assuming that the function is smooth? That argument can be modified to work with non-smooth functions. Commented Apr 14, 2020 at 15:27
• Yes, I was assuming that $u$ was $\mathcal C^2$. If $u$ is not $\mathcal C^2$, take a convolution $u_\varepsilon=u\ast \rho_{\varepsilon}$ of $u$ with a smoothing kernel $\rho_{\varepsilon}$. Then, $u_\varepsilon$ remains subharmonic after composing with any affine transformation and the argument above shows that $u_\varepsilon$ is psh, hence so is $u$. Commented Apr 14, 2020 at 15:37
• Hi, nice computation but I guess you want the $\lambda$'s to be on the first column of $A$, not the first row. Commented Apr 23, 2020 at 13:20
• @user111 You are perfectly right; I edited the answer accordingly. Thanks. Commented Apr 23, 2020 at 14:26