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This is a question from an old qualifying exam that I was trying to solve for practice:

Prove or disprove that there exist nonconstant holomorphic functions $f$ and $g$ on the open unit disk $D=\{z\in \mathbb{C}: |z|<1\}$ such that $e^{f(z)}+e^{g(z)}=1$ for all $z\in D$?

My initial thought was to show it is true on the real line and then use the uniqueness principle. But the examples I came up with on the real line use logarithms, which will not be continuous when I consider them as functions on the unit disk.

Since this is for practice, I was hoping only for hints and suggestions instead of complete answers. Thank you!

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  • $\begingroup$ "uniqueness principle" would only apply if you had existence on an open set, and the real line is not open in $\mathbb{C}$. $\endgroup$
    – John Fernley
    Jul 31, 2014 at 0:26
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    $\begingroup$ I thought it was existence on a set with a nonisolated point, for which the real line would work. $\endgroup$
    – user122916
    Jul 31, 2014 at 0:30
  • $\begingroup$ I think that's not right - to be honest uniqueness principle cannot show existence either way! It would show uniqueness $\endgroup$
    – John Fernley
    Jul 31, 2014 at 0:43
  • $\begingroup$ In practice, the idea of using the uniqueness is unlikely to be helpful. You would have to be in a scenario where (a) it was very obvious that $f,g$ are holomorphic on the whole disk, but (b) the computation to show the above formula is only "easy" to do on $\mathbb{R}$, which seems unlikely to me. $\endgroup$
    – jxnh
    Jul 31, 2014 at 0:44
  • $\begingroup$ @ John Fernley Uniqueness theorem for holomorphic functions states that if they are equal on a set with a limit point they are equal on the entire domain encyclopediaofmath.org/index.php/…. Real line is more than enough. $\endgroup$
    – Conifold
    Jul 31, 2014 at 1:23

2 Answers 2

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Just chose $g(x)$ such that $e^{g(z)}\neq 1$ for all $z$ in the unit disk. Then $f(x)=\ln (1-e^{g(x)})$ under some well defined branch of the log function.

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  • $\begingroup$ $e^{g(z)}\neq 1$ is not enough. If $1-e^{g(z)}$ maps the disk onto some annulus surrounding $0$ there won't be globally defined branch of logarithm on it even if it misses $0$ itself. $\endgroup$
    – Conifold
    Jul 31, 2014 at 2:03
  • $\begingroup$ @Conifold: If you have a nowhere-vanishing holomorphic function $h(z)$ in a simply connected region $U$ then you can define a branch of $\log h(z)$ in $U$. Here $U$ is the unit disc, which is clearly simply connected, and $h(z)$ is $1 - e^{g(z)}$. $\endgroup$
    – A. Barron
    Jul 31, 2014 at 2:11
  • $\begingroup$ I was thinking of something like $z^2$ mapping a simply connected piece of annulus onto the whole annulus, there wouldn't be a branch of $\ln$ on that. $\endgroup$
    – Conifold
    Jul 31, 2014 at 2:18
  • $\begingroup$ Fix a point $z_{0}$ in $U$. You can define a branch by letting $L(z) = \int_{\gamma} \frac{h'(w)}{h(w)} \ dw + c_{0}$, where $\gamma$ is a path in $U$ connecting $z$ to $z_{0}$ (this is well-defined by the simple connectivity assumption), and $c_{0}$ is chosen so that $e^{c_{0}} = h(z_{0})$. It's then easy to verify that $e^{L(z)} = h(z),$ e.g. see Stein and Shakarchi's 'Complex Analysis,' end of Ch. 3. This isn't a "branch" in the normal sense as when you're considering all of $\mathbb{C}$, but the log is still well-defined in $U$. $\endgroup$
    – A. Barron
    Jul 31, 2014 at 2:22
  • $\begingroup$ Look up Monodromy theorem. $\endgroup$
    – Rene Schipperus
    Jul 31, 2014 at 2:26
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Hint: it is fine to use logarithms as long as they are applied to values from a set that doesn't circle the origin. Suppose you picked some $f(z)$, than you need to solve $e^{g(z)}=1-e^{f(z)}$. One solution is obviously $g(z)=\text{Ln}(1-e^{f(z)})$, where $\text{Ln}$ is the principal branch of the logarithm with the cut along the negative half-axis. So you need to make sure that $f(z)$ maps the unit disk into a region, which $1-e^{z}$ then maps into something not crossing the negative half-axis. A judiciously chosen Möbius transformation should do the trick.

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  • $\begingroup$ Wonderful explanation, thank you. $\endgroup$
    – user122916
    Aug 9, 2014 at 21:07

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