# Derivatives of norm of vector-valued holomorphic functions

Let $G$ be a connected domain in $\mathbb{C}^{n}$, let $H$ be a Hilbert space and let $f,g:G\to H\backslash \{0\}$ be holomorphic (in my particular situation they are also injective, but I don't think it helps).

Is it true that if $\frac{\partial^2}{\partial z_i\partial \overline{z_j}}\log \|f(z)\|=\frac{\partial^2}{\partial z_i\partial \overline{z_j}}\log \|g(z)\|$, for all $i,j\le n$, then there is a holomorphic function $h:G\to\mathbb{C}$ and an isometry $U:H\to H$, such that $g(z)=h(z)Uf(z)$?

Motivation. Consider the canonical quotient $p:H\to PH$, where the latter is the projective space over $H$. Suppose $\varphi:G\to G$ is a biholomorphism, such that $f$ and $g=f\circ\varphi$ satisfy that condition. Then $pf\varphi(pf)^{-1}$ is an isometry of $pf(G)$ with respect to the Fubini-Study metric on $PH$. I wonder, if this isometry can be lifted to an isometry $H$, i.e. if there is an isometry $U:H\to H$, such that $pUf=pf\varphi$. Finally the last condition means that there is a holomorphic function $h:G\to\mathbb{C}$, such that $h(z)Uf(z)=f(\varphi(z))$.

• The statement there says that if $\|f(z)\|=\|g(z)\|$, then $g=Uf$, which is similar, and proven kind of similarly, but does not seem to be exactly analogous. – erz May 6 '17 at 8:20
• You probably mean, $\log F$ is pluriharmonic and then $F=|h|$, right? The transition from harmonic to holomorphic usually requires that the domain is simply connected. So yes, we get the statement locally, and then using holomorphy we can "globalize" it, but this last step is not exactly trivial (at least the proof that I have), or am I missing something? – erz May 8 '17 at 15:22
• @erz: I edited my coment: In fact your derivatives condition is equivalent to $\| f\|=|F|\| g\|$, where $\log|F|$ is pluriharmonic, so take $F=\log|h|$ and you obtain your statement from the Polya-Szego statement. – Alexandre Eremenko May 8 '17 at 18:18