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
    \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.

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