Let $f$ be holomorphic on $\overline{N(0,1)}$. Suppose $|f(z)| \leq 1$ for every $|z| = 1$. Without using fixed point theorem, show that there exists $z \in \overline{N(0,1)}$ such that $f(z) = z$.
My attempt:
Using Rouché's Theorem, I managed to show that if $|f(z)| < 1$ for every $|z| = 1$, then there is a unique $z$ such that $f(z) = z$.
However, I am not sure how I should proceed if $|f(z)| = 1$ for some $|z| = 1$.
Best Answer
If $f(z) = z$ for some $z\in \partial \mathbb{D}$, you have your fixed point. So assume that $f(z) \neq z$ for all $z\in\partial\mathbb{D}$. For $\varepsilon \geqslant 0$ consider
$$N(\varepsilon) :=\frac{1}{2\pi i}\int_{\partial\mathbb{D}} \frac{(1+\varepsilon)-f'(\zeta)}{(1+\varepsilon)\zeta - f(\zeta)}\,d\zeta.$$
For $\varepsilon > 0$, we have $\lvert f(z)\rvert < (1+\varepsilon)\lvert z\rvert$ on $\partial\mathbb{D}$, so Rouché's theorem asserts that $(1+\varepsilon)z - f(z)$ has the same number of zeros inside the unit disk as $(1+\varepsilon)z$, namely one. So we have $N(\varepsilon) = 1$ for $\varepsilon > 0$. But since $f(z) \neq z$ on $\partial\mathbb{D}$ by assumption, $N(\varepsilon)$ depends continuously on $\varepsilon$, so we also have
$$N(0) = \frac{1}{2\pi i}\int_{\partial \mathbb{D}} \frac{1-f'(z)}{z-f(z)}\,dz = 1,$$
i.e. $z-f(z)$ has exactly one zero in the unit disk, or, put differently, $f$ has exactly one fixed point in the unit disk.