5

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 $|z| = 1$.

  • if you could explain how you say "if $|f(z)| < 1$ for every $|z| = 1$, then there is a unique $z$ such that $f(z) = z$." – Myshkin Nov 07 '13 at 13:01
  • Let $|f(z)| < 1$ for every $|z| = 1$. Inside the unit disk, $|f(z)| < 1 = |z| \leq |z| + |f(z)-z|$. By Rouche's Theorem, $z$ and $f(z)-z$ have the same number of zeros inside the unit disk. That is, there is a unique $z$ with $|z|<1$ such that $f(z)-z = 0$. –  Nov 08 '13 at 04:42
  • I know this: if $|f(z)|>|g(z)|\forall |z|=a$ then $f,f+g$ has same no. of roots inside $|z|<a$, now tell me what is your $f, g$ here – Myshkin Nov 08 '13 at 05:13
  • In this case, I used $F(z) = -z$ and $G(z) = f(z)$. Then for $|z|=1$, $|F(z)| = 1 > |G(z)|$. So $F(z)=-z$ and $F(z)+G(z)=f(z)-z$ both have exactly one root in the unit circle. –  Nov 08 '13 at 09:50

1 Answers1

1

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.

Daniel Fischer
  • 206,697
  • May I find out how did you come up with that integral? –  Nov 08 '13 at 04:47
  • $$\frac{1}{2\pi i}\int_\gamma \frac{g'(\zeta)}{g(\zeta)},d\zeta$$ is the standard way to count the zeros of $g$ that $\gamma$ winds around. So that counts the zeros of $(1+\varepsilon)z-f(z)$. By Rouché's theorem, we know the value. Now if we let $\varepsilon \to 0$, what happens? – Daniel Fischer Nov 08 '13 at 09:30
  • @user60736 Daniel uses the argument principle that shows the number of roots of $f(z)-z$ strickly inside the unit disk http://en.wikipedia.org/wiki/Argument_principle – Haha Nov 09 '13 at 12:07