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 $f(0)$.
Best Answer
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 $$f\left[\overline{B_1(0)}\right]=\overline{B_1(0)}.$$