Let $G$ be a connected simply connected domain in $\mathbb{C}^{n}$, let $H$ be a Hilbert space.

Q1. Which functions $F:G\to(0,+\infty)$ are such that there is a holomorphic $f:G\to H\backslash \{0\}$, such that $F(z)=\|f(z)\|$?

I guess that plurisubharmonicity is necessary, but is it sufficient?

Note that $f$ is "almost" uniquely determined by $F$ in the sense that if $\|f(z)\|=\|g(z)\|$, for every $z\in G$, for another holomorphic $g:G\to H\backslash \{0\}$, then there is an isometry $U:H\to H$, such that $g(z)=Uf(z)$ (see also my previous question). In particular $K(x,y)=\left<f(x),f(y)\right>$ is completely determined by $F$. This leads to the second question.

Q2. Is there a formula that expresses $K$ in terms of $F$?


1 Answer 1


Q1: Plurisubharmonicity is not sufficient. $\log F$ is plurisubharmonic, which is a stronger property, but still far from sufficient. Plurisubharmonic functions may not be continuous. In fact $F$ is continuous and $\log|F|$ is pluriHARMONIC on the set where $F\neq 0$. But this is also not sufficient (consider the case of dimension $1$).

In dimension 1, the characterization will be: $F$ is continuous, $\log|F|$ is subharmonic and $(1/2\pi)\Delta\log|F|$ is a discrete measure with integer atoms.


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