A holomorphic function which has $|f(z)|>1$ for every $|z|=1$ and $|f(0)|<1$ has a fixed point inside the unit circle

Solution 1:

The condition that $|f(0)|<1$ while $|f(z)|>1$ on the unit circle allows you to conclude by continuity on the closed unit disk that $|f(z)|$ must have a local minimum in the open unit disk. This minimum must be $0$, or else you would derive a contradiction from the maximum modulus theorem applied to $\dfrac1f$. Thus, you know that $f$ has a zero inside the disk.

You can combine this with hjr's suggestion to use Rouché's theorem.

Solution 2:

Here is my writeup of Stein-Shakarchi 3.17 (b) which is essentially the same question for future users stuck on this problem.

(b) If $|f(z)| \geq 1$ whenever $|z|=1$ and there exists a point $z_{0} \in \mathbb{D}$ such $\left|f\left(z_{0}\right)\right|<1$, then the image of $f$ contains the unit disc.

proof. Since $|f|$ is continuous on the compact closed unit disk, the modulus of $f$ attains a minimum on the closed unit disk. By the maximum modulus principle, the minimum must be attained inside the open unit disk since $f(z_0)<1$ but $f(z)\ge 1$ when $|z|=1$.

Now assume that the minimum is attained at $z'\in \mathbb{D}$ and $|f(z')|>0$. Then $f(z)$ is never zero inside $\mathbb{D}$ so $\frac{1}{f(z)}$ is holomorphic in $\mathbb{D}$. Now we have that $$\frac{1}{|f(z)|} \geq 1 \qquad z\in \mathbb{D}$$ by the maximum modulus principle but this contradicts the existance of the point $1>|f(z')|>0$.

Therefore, there is a point $z\in \mathbb{D}$ at which $f(z)=0$.

Now consider $f(z)-w_0$ for $w_0\in \mathbb{D}$ Since $|f(z)|\geq |-w_0|$ when $z\in \partial \mathbb{D}$ by rouche's theorem, $f(z)-w$ has the same number of zeros as $f$ which has at least one.