Let be smooth. Under what condition, f can be sepearate to two part, that is, , where be smooth such that g depends only in and h is holomorphic. (I do not know if it is useful or not:.

There is one example: , where c is a constant in .

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I can give a necessary and sufficient condition as follows.

First of all let us assume is of the form for some radially symmetric and holomorphic . By absorbing a (positive) factor into , we can further assume that . Taking logarithm on both sides, we have

As is holomorphic, is harmonic, so it gives 0 when averaging over circle centered at (as we have normalized to have ):

for all .

This implies that the function

is harmonic on the complex plane (as it is exactly !). We have proved half of the following

Zorn Leung’s theorem, 2011A smooth function is of the form

for some smooth, radially symmetric and holomorphic if and only if the following function is harmonic:

To show the converse, suppose the function defined above is harmonic. We then take a function to be the harmonic conjugate of . We can choose such because is simply connected. (Also, the choice of is unique up to an additive constant.) We then define the function which is clearly holomorphic (as is holomorphic). We also take

where

Clearly is radially symmetric and it is not hard to see that . is also smooth as . Also, the choice of the additive constant mentioned above does not affect the result as it will only change to where is constant.

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