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Must every holomorphic function $f:D(0,1)\longrightarrow D(0,1)$ have a fixed point?

I know that any holomorphic function with two fixed points is the identity: $f=Id$, but I can't find out an holomorphic function without a fixed point.

Appreciate any suggestion.

Old John
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Irene
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    Your comment about a holomorphic function with two fixed points having to be identity is not correct. Ex. Mobius transformations. It is correct with two replaced by three. – Raghav Nov 24 '13 at 21:57
  • @Raghav I'm pretty sure that Irene meant: any automorphism of the unit disk with two fixed points is the identity. – Daniel Fischer Nov 24 '13 at 21:59
  • Maybe you can use the fact that $D$ is homeomorphic to $\mathbb C$, and work with automorphisms of $\mathbb C$ that send $D$ to $D$. – user99680 Nov 24 '13 at 22:05
  • Maybe you can use one of the zero-counting results ; Rouche', winding number , etc. on $f(z)-z$ , to tell the number of zeros – user99680 Nov 24 '13 at 22:11

4 Answers4

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It depends on the precise definition of $D(0,1)$. If this is open disk then the answer is negative. Indeed, open disk $D$ is conformal to the upper half plane $U$, say, via the map $g: D\to U$. In the latter take the map $f(z)=z+1$ which has no fixed points in the half-plane. Thus, the holomorphic self-map $F=g^{-1}\circ f\circ g $ of $D$ has no fixed points in $D$ either.

If by a disk you mean a closed disk then every continuous self-map of the closed disk (of any dimension) has a fixed point. This is Brouwer's fixed point theorem.

Ruzayqat
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Moishe Kohan
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Need not be. Counterexamples: $f(z)=(z+1)/2$

Curious student
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If your domain $\mathbb{D}$ is the unit disk then your map $f$ is a conformal self-map. If it's not the identity map then it either has two fixed points on the boundary (counting multiplicity) or one fixed point inside the disk. This follows from the fact that You can write the equation for conformal self-map and make it equal to its fixed point. Then you will find a quadratic equation whose product of roots has modulus $1$.

Spock
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  • are you assuming $f$ is a self-homeomorphism to say $f$ is conformal? – user99680 Nov 24 '13 at 22:22
  • Considering that $f$ is analytic and domain is unit disk, yes. – Spock Nov 24 '13 at 22:28
  • I am not convinced that this argument works. We do not appear to be given that the function $f$ is onto, so it need not be a self-map, and surely a constant map would be possible for $f$, and this would not be conformal. However, Rouche's theorem should work, as shown in the answer here. – Old John Nov 24 '13 at 23:08
  • @Spock: The map $f(z)=z^2$ is a holomorphic self-map of the unit disk, which is clearly not conformal. Your argument suffices in the case of linear-fractional transformations, but not in general. – Moishe Kohan Nov 25 '13 at 10:05
  • @studiosus I thought this is a conformal map but I think I'm wrong since it's not assumed in the statement. But if the map is bijective holomorphic from the unit disk to itself then it's indeed fractional linear map. – Spock Nov 25 '13 at 12:00
  • ....composed with a rotation. – Reznick Apr 18 '18 at 06:03
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The answer is no. We know that the unit disk $\mathbb{D}$ is conformally equivalent to the upper half-plane $\mathbb{H}$. Consider any conformal map sending $\mathbb{D} \rightarrow \mathbb{H}$. Then consider the map $z \rightarrow z+1$ which certainly does not have any fixed points. Then take the inverse of the first map to get from $\mathbb{H}$ back to $\mathbb{D}$. This map does not have any fixed points.

As an example: $\displaystyle f(z) = i\frac{1+z}{1-z}$ sends $\mathbb{D}$ to $\mathbb{H}$. Then $g(z) = z+1$ is conformal and has no fixed points. Then the inverse of the above map, $\displaystyle f^{-1}(z) = \frac{z-i}{z+i}$ maps $\mathbb{H}$ back to $\mathbb{D}$. Composing these gives: $\displaystyle \frac{1-(1-2i)z}{1+2i-z}$ is a conformal automorphism of the disk with no fixed points.

SescoMath
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