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Is there an analytic function $f\colon\Bbb{C}\longrightarrow \Bbb{C}$ such that for any $z$ on the unit circle $\lvert f(z) - \overline{z}\rvert < 1 $?

Rasmus
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Nemes
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2 Answers2

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No, there is no such function, because if there were, we would have

$$1 = \left\lvert\frac{1}{2\pi i}\int_{\lvert z\rvert = 1} \overline{z}\, dz\right\rvert = \left\lvert\frac{1}{2\pi i} \int_{\lvert z\rvert = 1} \bigl(\overline{z} - f(z)\bigr)\, dz \right\rvert \leqslant \frac{1}{2\pi} \int_0^{2\pi} \lvert \overline{z} - f(z)\rvert \, dt < 1$$

by Cauchy's integral theorem and the standard estimate.

We find $\frac{1}{2\pi i}\int_{\lvert z\rvert = 1} \overline{z}\,dz = \frac{1}{2\pi i}\int_0^{2\pi} e^{-it} ie^{it}\,dt = 1$ by direct evaluation, or by observing that $\overline{z} = 1/z$ on the unit circle. Cauchy's integral theorem asserts that $\int_{\lvert z\rvert = 1} f(z)\, dz = 0$, from which we get the second equality.

A more geometric argument:

The condition $\lvert \overline{z} - f(z)\rvert < 1$ on the unit circle implies that, as mappings of the unit circle to $\mathbb{C}^\ast$, $f$ is homotopic to $\overline{z}$, in particular, both have the same winding number around $0$, namely $-1$. But the winding number of $f$ (restricted to the unit circle) around $0$ is

$$\frac{1}{2\pi i} \int_{\lvert z\rvert = 1} \frac{f'(z)}{f(z)}\, dz$$

which is the number of zeros of $f$ in the unit disk (counted with multiplicity): a non-negative integer.

Daniel Fischer
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  • Would you please explain why the first equality above is correct ? – Nemes Aug 07 '13 at 09:09
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    I'll ad a fuller explanation to the answer, have a few minutes patience. – Daniel Fischer Aug 07 '13 at 09:12
  • what do you mean by $f$ is homotopic to $\bar{z}$ and why it is true? – Nemes Aug 10 '13 at 11:09
  • There is a homotopy between $f\lvert_{S^1}$ and $\overline{z}\lvert_{S^1}$, that is, a continuous map $H\colon[0,,1]\times S^1\to\mathbb{C}^\ast$ with $H(0,z)=f(z)$ and $H(1,z)=\overline{z}$. We can choose $H(t,z)=(1-t)\cdot f(z)+t\cdot\overline{z}$. The condition $\lvert f(z)-\overline{z}\rvert<1$ on $S^1$ implies that the line segment connecting $f(z)$ and $\overline{z}$ does not contain the origin, hence indeed $H\colon[0,,1]\times S^1\to\mathbb{C}^\ast$. Then $n(t)=\frac{1}{2\pi}\int_0^{2\pi} \frac{\partial H(t,e^{is})/\partial s}{H(t,e^{is})},ds$ is continuous. – Daniel Fischer Aug 10 '13 at 11:28
  • Since $n(t) \in \mathbb{Z}$ for all $t$ (it is the winding number around $0$ of the closed path $H(t,,\cdot,)$), $n(t)$ is constant, in particular $n(0) = n(1)$. – Daniel Fischer Aug 10 '13 at 11:29
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Let me give an alternative proof. Suppose $f$ is the desired function. Then for $z\bar{z} = 1$:

$$ |z f(z) - 1| = |z| |f(z) - \bar{z}| < 1 $$

so $g(z) = zf(z) - 1$ is an analytic function satisfying $|g(z)| < 1$ whenever $|z| = 1$. By the maximum modulus principle

$$ 1 = |0 \cdot f(0) - 1| = |g(0)| < 1$$

which gives a contradiction.

Willie Wong
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