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This is a past prelim and homework problem for which my proof is missing a detail or two:

Suppose $f$ and $g$ are non-vanishing holomorphic functions which extend continuously to $\bar{\mathbb D}$. If $|f(z)| = |g(z)|$ on $\partial\mathbb D$, then $|f(z)| = |g(z)|$ for all $z \in \bar{\mathbb D}$.

My proof is as follows: Since $f$ and $g$ are non-vanishing and holomorphic on $\mathbb D$, then $f/g$ and $g/f$ are well defined and holomorphic on $\mathbb D$. Also, $f/g$ and $g/f$ are continuous at all points such that $f(z) \neq 0$ and $g(z) \neq 0$, respectively. I go on to use the Maximum Modulus Theorem to show that $|f/g| = 1$ on $\mathbb D$, under the assumption that $f/g$ and $g/f$ are actually continuous on the boundary of $\mathbb D$.

My question: Is it possible to extend $f/g$ to a continuous function on the boundary? I can also prove the statement if I can show that $|f/g|$ or $|g/f|$ is constant on the boundary (in which case, $f/g$ is constant or has a zero on the interior; the latter can't happen). I tried to show that $|f/g|$ is constant by showing that all of the singularities of $f/g$ are removable, and thus extending the function via continuity. We know that if $g(z) = 0$ for some $z$ on the boundary, then $f(z) = 0$ by assumption as well, but do these zeros have the same multiplicity?

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  • $\begingroup$ I don't know the answer but if you take $f(z) = \sqrt{z+1},$ this clearly has a zero at a boundary point. $\endgroup$
    – Mark Joshi
    Mar 24, 2015 at 0:28
  • $\begingroup$ I agree, thanks. Even $x + 1$ would work I believe. $\endgroup$
    – dannum
    Mar 24, 2015 at 0:30
  • $\begingroup$ yes, i wanted to give an example where you couldn't factor out the zero. $\endgroup$
    – Mark Joshi
    Mar 24, 2015 at 0:34

2 Answers 2

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The statement is false.

Let $h(z)=e^{-\frac{1+z}{1-z}}$. The linear fraction $\frac{1+z}{1-z}$ maps the unit circle to the imaginary axis, and it maps the unit disk into the right half-plane. So we have $0<|h|<1$ inside the disk and $|h|=1$ along the boundary circle, except for the point $1$ where $h$ is undefined.

Let $$ f(z)=1-z \quad\text{and}\quad g(z)=\begin{cases}(1-z)\cdot h(z) & z\ne1 \\ 0 & z=1.\end{cases} $$ Both functions are nonzero inside the disk.

Since $h$ is bounded, $g$ is continuous at $1$ as well as on the whole closed disk. At point $1$ we have $f(1)=g(1)=0$; at other boundary points we have $|h|=1$ so $|f|=|g|$ is satisfied along the complete boundary.

But inside the disk we have $|h|<1$, so $|g|<|f|$.

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  • $\begingroup$ Thanks for your counterexample. I found a similar exercise in Conway's book, but instead he says it's analytic on the whole disk. I'm pretty sure that's what was intended, and I was able to prove that. $\endgroup$
    – dannum
    Aug 25, 2015 at 13:15
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After we have supplied a proof of the theorem, we'll have shown that $f/g$ is a holomorphic function with constant modulus 1 on the open disks of radius $\epsilon < 1$, which by the open mapping theorem means that $f/g = e^{i\theta}$ is a constant. Clearly, this constant won't depend on the $\epsilon$ chosen, and so will work all the way up to the boundary. So $g$ is obtained from $f$ by rotation by a fixed angle.

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  • $\begingroup$ I also think you could make use of a similar idea to actually prove the original statement. $\endgroup$
    – Zach L.
    Mar 24, 2015 at 2:44
  • $\begingroup$ Could you explain your answer a little more clearly please? Are you saying that we assume $f$ is nonzero on the disk and that $|f| = |g|$ on $\partial B(0,\varepsilon)$ for $0 < \varepsilon < 1$? Thanks. $\endgroup$
    – dannum
    Aug 22, 2015 at 18:19

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