1
$\begingroup$

TLDR: I want to prove that never-zero holomorphic functions defined in balls in $\mathbb C^n$ have holomorphic logarithms. I have an incomplete proof which I believe is salvagable; I present my incomplete proof along with justification that it may be salvagable below.


For $p\in\mathbb C^n$, let $B_r(p)$ denote the open ball in $\mathbb C^n$ of radius $r>0$ and center $p$. Fix $n\in\mathbb N$, fix $p\in\mathbb C^n$, and fix $r>0$. It can be shown (link) that if $f:B_r(p)\to\mathbb C\setminus\{0\}$ is holomorphic, then there exists some holomorphic function $h:B_r(p)\to\mathbb C$ such that $f(z)=e^{h(z)}$ for all $z\in B_r(p)$. Suppose that one wishes to prove this, but doesn't want to make use of De Rham cohomology. One might attempt to proceed by induction, using the result for $n=1$ as a base case. Such an attempt follows.


For simplicity, we consider the case where $n=2$ and $p=(p_1,p_2)$. For $w\in B_r(p_2)\subset\mathbb C$, we shall denote by $U_w$ the (open) set $$ \{z\in\mathbb C:(z,w)\in B_r(p)\}. $$ Observe that the function $B_r(p_2)\to\mathbb C$ defined by $w\mapsto f(p_1,w)$ is holomorphic and never zero. Let $g:B_r(p_2)\to\mathbb C$ be a holomorphic logarithm of $w\mapsto f(p_1,w)$. For each $w\in B_r(p_2)$ the function $f_w:U_w\to\mathbb C\setminus\{0\}$ defined by $f_w(z)=f(z,w)$ is holomorphic, and hence has an analytic logarithm $g_w$; let $g_w$ be such. For $w\in B_r(p_2)$, observe that $e^{g(w)}=f(p_1,w)=e^{g_w(p_1)}$, and hence there exists some integer $m_w$ such that $g(w)=g_w(p_1)+2\pi im_w$. Let $m_w$ be such, and let $h_w:U_w\to\mathbb C$ be the function defined by $h_w(z)=g_w(z)+2\pi im_w$.

At this point, we would like to say that the function $h(z,w):=h_w(z)$ is an holomorphic logarithm of $f$. By construction, $h$ is a logarithm for $f$; we just have to show that it is holomorphic. One might attempt to do this by showing that if $F:B_r(p)\to\mathbb C$ is such that

  1. $w\mapsto F(p_1,w)$ is holomorphic

  2. For every $w\in B_r(p_2)$, the function $z\mapsto F(z,w)$ is holomorphic

then $F$ is holomorphic. Unfortunately, this is not the case; consider $p=(0,0)$ and $F(z,w)=z\bar w$.


One might now suspect that the line of attack outlined above is fatally flawed. After all, what if the $h$ we defined isn't holomorphic? As it turns out, this isn't the case: We can show that $h$ is holomorphic given the existence of a holomorphic logarithm of $f$:

Suppose that $G:B_r(p)\to \mathbb C$ is a holomorphic logarithm for $f$. Then $e^{G(p)}=f(p)=e^{h(p)}$. It follows that $h(p)=G(p)+2\pi im$ for some integer $m$. Let $m$ be such, and define $H:B_r(p)\to\mathbb C$ by $H(z,w)=G(z,w)+2\pi im$. Observe that $H$ is a holomorphic logarithm of $f$. It follows that $w\mapsto H(p_1,w)$ is a holomorphic logarithm for $w\mapsto f(p_1,w)$. Since $H(p_1,p_2)=g(p_2)$, we conclude that $H(p_1,w)=g(w)$ for all $w$ (their difference is a continuous $2\pi i\mathbb Z$-valued function). But then for each $w\in B_r(p_2)$, $z\mapsto H(z,w)$ is holomorphic logarithm for $f_w$. It follows just as above that for each $w\in B_r(p_2)$ and for each $z\in U_w$, we have $h_w(z)=H(z,w)$. But this means that $h=H$ on all of $B_r(p)$. In particular, $h$ is holomorphic, as desired.


Question: Is there any simple way to prove that the function $h$ defined above is holomorphic without assuming the existence of a holomorphic logarithm for $f$?

$\endgroup$
  • $\begingroup$ I'd consider the disks through $0$ instead. I vaguely recall that being continuous on a ball and and holomorphic on all such disks implies being holomorphic on the ball... $\endgroup$ – user357151 Mar 17 '18 at 2:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.