## A problem in complex analysis

Let $f:\mathbb{C} \rightarrow \mathbb{R}^{+}$ be smooth. Under what condition, f can be sepearate to two part, that is, $f(z)=g(|z|)|h(z)|^2$, where $g:\mathbb{C} \rightarrow \mathbb{R}^{+}$ be smooth such that g depends only in $|z|$ and h is holomorphic. (I do not know if it is useful or not:$\partial \bar{\partial} log(f) <0$.

There is one example: $exp(-|z-c|^2)=(exp(-c^2)exp(-|z|^2)) (|exp(\bar{c}z)|^2)$ , where c is a constant in $\mathbb{C}$.

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### 3 Responses to A problem in complex analysis

1. KKK says:

I can give a necessary and sufficient condition as follows.
First of all let us assume $f$ is of the form $f(z)=g(z)|h(z)|^2$ for some radially symmetric $g$ and holomorphic $h$. By absorbing a (positive) factor into $g$, we can further assume that $|h(0)|=1$. Taking logarithm on both sides, we have
$\log f(z)=\log g(z) + 2 \log |h(z)| .$
As $h$ is holomorphic, $2 \log |h(z)|$ is harmonic, so it gives 0 when averaging over circle centered at $0$ (as we have normalized to have $|h(0)|=1$):
$\frac{1}{2\pi r}\int_{|w|=r}\log f(w)ds=\frac{1}{2\pi r} \int_{|w|=r}\log g(w)ds=\log g(z)$ for all $|z|=r>0$.
This implies that the function
$\displaystyle H(z):=\left\{\begin{matrix}\log f(z)-\frac{1}{2\pi |z|}\int_{|w|=|z|}\log f(w)ds &\text{ if }z\neq 0\\ 0&\text{ if }z=0.\end{matrix}\right.$
is harmonic on the complex plane (as it is exactly $2\log|h|$!). We have proved half of the following
Zorn Leung’s theorem, 2011
A smooth function $f: \mathbb{C}\rightarrow \mathbb{R}^+$ is of the form
$f(z)=g(z)|h(z)|^2$ for some smooth, radially symmetric $g: \mathbb{C}\rightarrow \mathbb{R}^+$ and holomorphic $h$ if and only if the following function $H$ is harmonic:

$\displaystyle H(z):=\left\{\begin{matrix}\log f(z)-\frac{1}{2\pi |z|}\int_{|w|=|z|}\log f(w)ds &\text{ if }z\neq 0\\ 0&\text{ if }z=0.\end{matrix}\right.$

2. KKK says:

To show the converse, suppose the function $H$ defined above is harmonic. We then take a function $\Theta(z)$ to be the harmonic conjugate of $\frac{1}{2}H(z)$. We can choose such $\Theta$ because $\mathbb{C}$ is simply connected. (Also, the choice of $\Theta$ is unique up to an additive constant.) We then define the function $h(z)=\exp (\frac{H(z)}{2}+i \Theta(z))$ which is clearly holomorphic (as $\frac{H}{2}+i \Theta$ is holomorphic). We also take
$g(z)=\exp(G(z))$
where
$\displaystyle G(z):=\left\{\begin{matrix}\frac{1}{2\pi |z|} \int_{|w|=|z|}\log f(w)ds &\text{ if }z\neq 0\\ \log f(0)&\text{ if }z=0.\end{matrix}\right.$

Clearly $g$ is radially symmetric and it is not hard to see that $f(z)=g(z)|h(z)|^2$. $g$ is also smooth as $g=f/|h|^2$. Also, the choice of the additive constant mentioned above does not affect the result as it will only change $h(z)$ to $h(z)e^{i\theta_0}$ where $\theta_0$ is constant.

3. KKK says:

One last word: the last condition you’ve mentioned: $\partial \bar \partial \log f<0$ is just equivalent to $\log g$ being (strictly) superharmonic i.e. $\Delta \log g<0$. I don't think it is essential in the argument, but that implies $g$ cannot be "trivial", e.g. it can't be constant.