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I am preparing for qualifying exams, and this is a question from the Penn State Qualifying Exam for Fall 2015. It is stated as follows

Let $\epsilon > 0$ and let $f$ be holomorphic (analytic) on the disk $S = \{z \in \mathbb{C} \ | \ |z| < 1 + \epsilon \}$. Suppose that $f(z)$ is real valued whenever $|z| = 1$. Prove that $f$ is constant.

I have tried a few things in showing this to be true, but I keep finding holes in my logic. My largest issue is that the portion of the domain in which $f$ is real valued is not open, and so I'm not able to use many of the theorems I otherwise feel would be helpful (Open Mapping Theorem, Cauchy-Riemann Equations, Maximum Modulus, etc.).

Most of my attacks towards this problem have centered around showing things for the unit disk $\mathbb{D}$ instead of $S$, because I figure if I can show that $f$ is constant on $\mathbb{D}$, then I can use the Identity Theorem to extend it to $S$. However, since I don't know anything about the specific values that $f$ obtains, I can't use Schwarz' Lemma either.

I played with the idea, also, of suggesting that, if we consider the closure of $\mathbb{D}$, then $f(\partial{\mathbb{D}}) \in \mathbb{R}$. Since $\partial\mathbb{D}$ is closed and bounded and $f$ is analytic, it's image should also be bounded. That would put, for some $M \in \mathbb{R}$, $f(\partial{\mathbb{D}}) \in [-M,M]$, which is bounded and obtains a maximum (since $f$ is continuous and real valued here). I know that this set isn't open, but there is a Corollary in my text (Complex Analysis - Freitag) that says the following:

"If $K$ is a compact subset of the domain $D$ and $f:D \rightarrow \mathbb{C}$ is analytic, then the restriction of $f|K$ being a continuous function has a maximal modulus on $K$. By the Maximum Modulus Principle, we can moreover affirm that the maximal modulus value is necessarily taken on the boundary of $\mathbb{D}$."

The proof of that statement follows from the Open Mapping Theorem. Would it be appropriate to use that in this case, then? If $K = \partial \mathbb{D}$, even though it's technically not an open set, can I say that $f$ obtains its maximum on $\mathbb{D}$, state that it's constant, and then extend this to $S$ using the Identity Theorem? My issue here is that the constant itself isn't actually in $\mathbb{D}$,

I appreciate any help for this problem (or any tips for showing that complex functions will be constant, as these types of problems show up often).

Edit: My function $f$ is not necessarily entire, so Liouville's Theorem does not apply

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    $\begingroup$ Possible duplicate of How to show that $f$ is constant by using Liouville's theorem? $\endgroup$ – Hmm. Jul 6 '16 at 16:16
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    $\begingroup$ @Hmm., I'm not sure if it is, as I don't have any information to suggest that my function is entire, just that it is analytic in this disk $S$ $\endgroup$ – cnolte Jul 6 '16 at 16:18
  • $\begingroup$ the solutions given essentially do not use entireness, only analyticity of $f$ in a region containing the unit circle. $\endgroup$ – Hmm. Jul 6 '16 at 16:43
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One can get this from the Schwarz Reflection Principle. That only works for the disk, or other special domains. It's easy to give an argument that works for any bounded domain:

The imaginary part is harmonic and vanishes on the boundary, so the imaginary part is $0$. So $f$ is real-valued. Hence $f$ is constant (Cauchy-Riemann equations, Open Mapping Theorem, who knows what else).

Or: Apply Maximum Modulus to $e^{if}$ and $e^{-if}$.

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  • $\begingroup$ I understand that the analyticity of $f$ gives rise to the statement that the imaginary part is harmonic, and clearly it vanishes on the boundary, but can you elaborate on how we draw the conclusion that it is everywhere $0$? Would this be true if we knew $f$ is be real valued on any connected subset of our domain, for example, if we had just stated that $f(z) \in \mathbb{R}$ so long as $z \in \mathbb{R}$? $\endgroup$ – cnolte Jul 6 '16 at 16:51
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    $\begingroup$ @burnmfburn Harmonic functions have their own maximum principle - if $D$ is bounded, $u$ is a real-valued function continuous on the closure, harmonic in the interior, and less than or equal to $\alpha $ on the boundary then $u\le\alpha$. See the chapter on harmonic functions in whatever text you used... $\endgroup$ – David C. Ullrich Jul 6 '16 at 16:55
  • $\begingroup$ This is exceptionally helpful. I will be sure to include this in the proof that I write up for my class. Thank you for your help $\endgroup$ – cnolte Jul 6 '16 at 17:01
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By the Schwarz reflection principle, the formula $f(z)=\overline{f(1/\overline{z})}$ for $|z|>1$ extends $f$ analytically to the entire complex plane. Then by definition of the extension, $f(\mathbb{C})\subset f(\overline{\mathbb{D}}) \cup \overline{f(\overline{\mathbb{D}})}$, which is bounded by compactness.

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  • $\begingroup$ I have a question about this - as we don't use the fact that $f$ is constant on the boundary of $\mathbb{D}$, it seems like this statement could be made about any function $f$ that is analytic on the unit disk. Am I missing something with this? $\endgroup$ – cnolte Jul 6 '16 at 17:22
  • $\begingroup$ You did not assume $f$ is constant on the boundary, only real valued. And you need $f$ to be real-valued to apply the reflection principle - otherwise what are you going to reflect across? $\endgroup$ – Noah Olander Jul 6 '16 at 17:54
  • $\begingroup$ Oh, duh. In my mind I kept telling myself it was constant (since this is what I knew I needed to show). I just looked up a proof of this outside of my text, and it makes much more sense as to how we use this. Excellent, thank you for clarifying $\endgroup$ – cnolte Jul 6 '16 at 18:22

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