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Let $\mathbb{D} = \{ z\in \mathbb{C}: |z| < 1\}$ be the unit disk.

I want to show that for any $k\in\mathbb{N}$, there is no holomorphic function $f$ which extends continuously to $\partial \mathbb{D}$, such that $$f(z) = \dfrac{1}{z^k}\quad \forall z\in\partial\mathbb{D}$$

There have been attempts to solve this question for $k=1$, see e.g. here. However, I cannot use the mean value theorem for holomorphic functions, because in our lecture, it is defined as follows:

Let $f$ be holomorphic on a region $G$ and $B_r(z)$ (the closed ball around $z$ with radius $r$) be a proper subset of $G$. Then $$f(z) = \dfrac{1}{2\pi} \int_0^{2\pi} f(z+re^{it}) dt$$ The proof from here uses $r=1$ and $z=0$, but clearly, $B_1(0)$ is not a proper subset of the region $\mathbb{D}$.

I also tried Schwarz' Lemma using $g(z) = z^k f(z)$ (to ensure $g(0) = 0$), but I couldn't conclude what I want.

Aericura
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1 Answers1

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First proof:

Let $h(z)=z^kf(z)-1$. Then $h$ is holomorphic in $\mathbb{D}$, continuous on $\mathbb{D}\cup\partial\mathbb{D}$ and $h(\partial\Bbb{D})=0$. Applying the maximum modulus principle, we obtain that $h\equiv 0$, which implies that $\lim_{z\to 0}f(z)=\infty$, contradicting the fact that $f$ is holomorphic near $0$.

Second proof:

First note that the condition implies that, on the boundary, $$|f(z)|=1\ \ (*)$$

I claim that $(*)$ implies that $f$ is a polynomial; this in turn clearly implies (by the identity principle) that we cannot have $f(z)= \frac{1}{z^k}$ on all $\partial\mathbb{D}$.

First note that $(*)$ implies that $f$ has a finite number of zeros in $\Bbb{D}$. Let us call those different from $0$ $\{z_1,\dots,z_k\}$ (with the respective multiplicities), and let $m$ be the multiplicity of $0$ as a zero of $f$. Consider now $$g(z)=\frac{f(z)}{z^m}\prod_{j=1}^k\frac{1-\bar{z}_jz}{z-z_j}$$ This function does not have any zeros in $\mathbb{D}$ and satisfies $(*)$ again (since every factor of the product has modulus one on $\partial\Bbb{D}$). The fact that $g(z)$ is constant follows from the maximum modulus principle applied to $\frac{1}{g}$ and this in turn implies that $f(z)$ is a polynomial.

Note:

The factors I included in the product are the inverse of the Blaschke factors, an incredibly useful tools for bounded holomorphic functions on $\mathbb{D}$.

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  • Thank you, I tried something similar to Proof 1, but I wasnt aware of taking $z\to 0$. However, how can one conclude from (*) that $f$ has a finite number of zeros? Why can't $f$ have countably many? – Aericura Jun 11 '20 at 10:01
  • Ah, I think one uses a compactness argument, because $\bar{\mathbb{D}}$ is compact. So assume $f$ has countably many zeros $(z_n)_{n\in\mathbb{N}}$, we can find a convergent subsequence. Clearly, the limit point cannot be in $\partial \mathbb{D}$. Hence it lies in $\mathbb{D}$. By identity principle, we conclude $f\equiv 0$, contradiction. Is that correct? – Aericura Jun 11 '20 at 10:10
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    @Aericura Exactly –  Jun 11 '20 at 10:19
  • Why does $g$ have no zeros in $\mathbb{D}$? Consider $f(z_1) = 0$, hence $g(z_1) = f(z_1)\cdot \prod = 0$. What am I missing? – Aericura Jun 11 '20 at 10:28
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    Because the factors in the product have a pole of the same order of the zero of $f$, so the product is not zero. To be precise, $g$ is defined by that expression in $\mathbb{D}-{z_1,\dots,z_k}-{0}$, and is defined in $\mathbb{D}$ by continuation (thanks to the Riemann theorem on singularities) –  Jun 11 '20 at 10:33
  • Yes, thank you. – Aericura Jun 11 '20 at 10:40