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I am wondering whether the following statement is true or not.

Let $f\colon \overline{B_1(0)}\subseteq \mathbb{C} \to \mathbb{C}$ be a continuous map on $\overline{B_1(0)}$ and holomorphic on $B_1(0)$. If $f$ is not constant and $\lvert f(z) \rvert = 1$ whenever $\lvert z \rvert = 1$, then $$f\left[\overline{B_1(0)}\right]=\overline{B_1(0)} .$$

I have tried using Maximum Modulus Principle and Schwarz Lemma to no success since I do not know anything about $f(0)$.

CJ Dowd
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Jorge Gaspar
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    I'm not sure how to solve this, but if $f(0)$ is a problem, then you can compose $f$ with an automorphism of $B_1(0)$ to fix $f(0) = 0$. – Riemann Apr 17 '23 at 01:24
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    The boundary of the image is included in the image of the boundary by the open mapping property of analytic functions so if $D$ is the image of the unit disc under $f$ its boundary is included in the unit circle; now $D$ is a bounded (by the boundness of $f$) open domain and it is very easy to see that this implies that its boundary must be the whole unit circle as otherwise you could connect points in $D$ to infinity by arcs; hence $D$ is the unit disc and the function is surjective – Conrad Apr 17 '23 at 11:34
  • Use the Argument Principle. if $n\big(f\circ\ \gamma,z\big)=0$ for $\vert z\vert \lt 1$ with $\gamma(t)=\exp(2\pi i t)$ for $t\in [0,1]$ and $\gamma_\delta := (1-\delta)\cdot \gamma$ for arbitrary small $\delta \gt 0\implies n\big(f\circ \gamma_\delta,z\big)=0$ by dog on a leash lemma (first select some $z$ then select any small enough $\delta$); this implies $f$ is constant. Conclude $f\big(S^1\big)=S^1$ and for any $z$ inside the unit circle: $n\big(f\circ\ \gamma_\delta,z\big) =n\big(f\circ\gamma,z\big)=k\geq 1 $ for $\delta$ small enough i.e. $z$ is in the image of $f$. – user8675309 Apr 17 '23 at 16:14
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    maybe easier: conclude $0$ is in the image of $f$ by minimum modulus theorem. And for $w\in B(0,1)$ consider $m_w\circ f$ where $m_w$ is the univalent (moebius) map that sends $\overline B(0,1)$ to itself and sends $w\mapsto 0$; conclude $0$ is in its image so $w$ is in the image of $f$ thus $f\big(B(0,1)\big)=B(0,1)$. And $f(S^1)=S^1$ by considering the continuous function $g:\overline B(0,1)\longrightarrow \mathbb R$ given by $g(z)=\text{Re}\big(e^{-i \theta}\cdot f(z)\big)$ which must have a maximum (of 1) for any choice of $\theta \in [0, 2\pi)$ by compactness. – user8675309 Apr 18 '23 at 18:18

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So I think I have got the answer.

First, since f is holomorphic on $B_1(0)$ and not constant, then it is an open map on $B_1(0)$. Then as proved here we must have that

$$\partial f(B_1(0)) \subseteq f(\partial B_1(0)).$$

Now, since $\lvert f(z)\rvert =1$ whenever $\lvert z\rvert=1$ we have that

$$f(\partial B_1(0))\subseteq \partial B_1(0).$$

So we have that $\partial f(B_1(0)) \subseteq \partial B_1(0)$. Now, by the maximum modulus principle we have that $f(B_1(0))\subseteq B_1(0)$. Since both $f(B_1(0))$ and $B_1(0)$ are open connected sets, and $f(B_1(0))$ is not empty we must have by the first answer here that $f(B_1(0))=B_1(0)$. Since $$\partial f(B_1(0)) \subseteq f(\partial B_1(0)) \subseteq \partial B_1(0)$$ this implies that $$\partial B_1(0)=f(\partial B_1(0))$$ and we conclude that $$f\left[\overline{B_1(0)}\right]=\overline{B_1(0)}.$$

Jorge Gaspar
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