2
$\begingroup$

I attempted to prove the following statement can anyone give his/her opinion about the proof? Or give some ideas or references. Thank you in advance.

Statement: Let $D\subset$ be a domain in the complex plan and $h$ be harmonic in $D$ then there is $f\in O(D)$ ( holomorphic functions) so that $h=\log|f|.$

Proof. For $m>0$ let $(D_m)_m$ be a sequence of open disc so that $D_m\cap D_{m+1}\not=\varnothing$, $\bar D_m\subset D$ and $\cup_m D_m=D.$\ For each $m>0$ we construct (by induction) a holomorphic function $f_m$ defined on $ D_1\cup\cdots\cup D_m$ so that $\Re(f_m)=h.$

For $m=1$, as $h$ is harmonic, there is $f_1\in O(D_1)$ so that $h=\Re(f_1)$ in $D_1.$ For $m=2$ there is $g\in O(D_2)$ so that $h=\Re(g)$ in $D_2.$ Since $f_1$ and $g$ are holomorphic and $\Re(g)=\Re(f)$ on $D_1\cap D_2$ we infer that there is $c_1\in R$ so that $f_1=g+ic_1$ on $D_1\cap D_2.$ Set $f_2=f_1$ in $ D_1$ and $f_2=g+ic_1$ in $ D_2$ it follows that $f_2\in O( D_1\cup D_2).$ By the same way we construct $f_m,$ $m=3,4,\dots$

For $m>0$ set $F_m=e^{f_m}$. The squence $(F_m)_m$ has a subsequence that converges locally uniformly in $D.$ Indeed let $K\subset D$ be a compact ball. Observe that $|F_m|=e^h$ in $D_m$ for all $m>0$ we infer that $F_m$ is uniformly bounded on $K.$ By Montel's theorem $(F_m)_m$ has a subsequence $(F_{m_k})_k$ that converges on $K$. In fact $(F_{m_k})_k$ converges locally uniformly in $D.$

To prove that $(F_{m_k})_k$ is convergent, it is enough to show that :

1) it does not contain divergent subsequences

2) all its convergent subsequences converge to the same limit.

First assume that there are a compact set $K'$ and a subsequence $(F_{m_{k'}})$ of $(F_{m_{k}})$ which does not converge on $K'$. Let $U$ be a relatively compact domain in $D$ containing $K'$ and $K$. By Montel theorem's theorem $(F_{m_{k}})$ has a subsequence which converges uniformly in $U$ to a holomorphic function $g$. So there is $\epsilon>0$ so that :

(1) $\sup_{K'}|F_{m_{k'}}-g|>\epsilon$ for all $k'$ big enough.

This is impossible since the sequence $(F_{m_{k'}})$ is uniformly bounded on $U$ and by Montel's theorem its has a subsequence which converges uniformly in $U$ to a function $g'$. As $g'=g$ on $K$ then they coincide in $U.$ Equality (1) can not hold in other words such a subsequence $(F_{m_{k'}})$ does not exist.

Second let $(F_{m_{k'}})$ be a subsequence of $(F_{m_{k}})$ which converges locally uniformly in $D$ to $g'$ and $(F_{m_{k''}})$ be a subsequence of $(F_{m_{k}})$ which converges locally uniformly in $D$ to $g''.$ As $g'$ and $g''$ are holomorphic in $D$ and coincide in $K$ then $g'\equiv g''$ in $D.$

For $z\in D$ set $f(z)=\lim_{k\rightarrow\infty} F_{m_k}(z)$ as the convergence is locally uniform we infer that $f$ is holomorphe and $|F_{m_k}|$ converges to $|f|$ as $k$ tends to infinite. Since $|F_{m_k}(z)|=e^{h(z)}$ for all $k$ we get $|f(z)|=e^{h(z)}$.

$\endgroup$

1 Answer 1

1
$\begingroup$

The statement is wrong (unless $D$ is simply connected). A counter-example is $$ D = \Bbb C \setminus \{ 0 \} \, , \quad h(z) = \alpha \log \lvert z \rvert \, $$ for any real, positive, and non-integer $\alpha$. $h$ is harmonic in $D$. Assume that $ h(z) = \log \lvert f(z) \rvert $ for some holomorphic function in $D$. Then $$ \lvert z \rvert^\alpha = \lvert f(z) \rvert $$ for all $z \in D$. It follows that $f$ is bounded near zero and therefore has a removable singularity at $z=0$. The continuation has a zero of some order $k $ at $z=0$. But this implies that $\alpha = k \in \Bbb N$ in contrast to the assumption.


The flaw in your proof is the definition of $f_m$ for $m \ge 3$. If $f_{m-1}$ is already defined on $D_1\cup\cdots\cup D_{m-1}$ then you choose a holomorphic function $f_m$ on $D_m$ which coincides with $f_{m-1}$ on $D_m \cap D_{m-1}$. But that does not imply that $f_m = f_{m-1}$ in $D_1 \dots D_{m-1}$, which means that $f_m$ is not well-defined on $D_1\cup\cdots\cup D_m$.

Note also that the first part of your proof would imply that a harmonic function is the real part of a holomorphic function on any domain.

$\endgroup$
6
  • $\begingroup$ Thank you Martin. I did not see clearly why the statement is wrong!!!! If $f(z)=0$ that means there are $g\in\O(D)$ and $k\in N$ so that $f(z)=z^kg(z)$. Now $|z|^{\alpha}=|z^k||g(z)|$. Why $k=\alpha$? $\endgroup$
    – Kara
    Commented Dec 21, 2016 at 19:56
  • $\begingroup$ @Kara: Because otherwise $ |g(0)| = \lim_{z \to 0} |z^{\alpha - k}|$ would be zero or infinity. $\endgroup$
    – Martin R
    Commented Dec 21, 2016 at 19:58
  • $\begingroup$ I should say wauuuuuuuu :) $\endgroup$
    – Kara
    Commented Dec 21, 2016 at 20:20
  • $\begingroup$ That is a bit strange!!! Since $D=C\setminus\{0\}$ is simply connected. There is $f\in O(D)$ so that $h(z)=\alpha\log|z|=\Re(f)$. Set $F(z)=e^{f(z)}$ we shall get $|z|^{\alpha}=|F(z)|=e^{\Re(f(z))}$. $F$ is bounded near zero so we can extend it to a holomorphic function at zero. As $F(0)=0$ there are $k\in N$ and $g\in O(D)$ so that $F(z)=z^kg(z)$ in this case we still get $k=\alpha.$ $\endgroup$
    – Kara
    Commented Dec 21, 2016 at 20:31
  • $\begingroup$ @Kara: That domain is not simply-connected. A circle around zero cannot be "shrunk" to a point without leaving the domain. – Or: its complement in the extended complex plane has two components. $\endgroup$
    – Martin R
    Commented Dec 21, 2016 at 20:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .