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Edit: the answer suggested by @ShakedBader posted here works for this question as well. But I'm curious about the extra assumption here about $|f(0)|$. Does it allow to solve the question using Rouche's lemma?

Let $f$ be a holomorphic function on a domain which contains the unit circle.

We know that $|f(z)|>1$ for every $|z|=1$, and that $|f(0)|<1$. Show that there exists a point $a\in \mathbb C$, such that $|a|<1$ and $f(a)=a$.

I see that due to continuity, we can find a path in the unit circle which circles zero, and where $|f(z)|=1$ hold for each $z$. But I'm not sure how to proceed. Any clues?

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mbrg
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2 Answers2

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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.

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Jonas Meyer
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    how is applying mmp on $\frac{1}{f(z)}$ deriving a contradiction? – R_Squared Oct 06 '23 at 10:33
  • I can't see really how to reach the contradiction from applying the Maximum Module Theorem to $\frac{1}{f}$? Would it consist of proving $\frac{1}{f}$ can't be constant? . Could you @Jonas Meyer please detail it a bit more? Thanks for your answer by the way – CharlesJA Oct 22 '23 at 20:21
  • @CharlesJA: We know the function is not constant because it has values with modulus <1 and >1. If the minimum is not 0, then the function is nonzero everywhere so 1/f is analytic on the disk (including boundary) and |1/f| has a max at a point interior to the disk, not on the boundary, where |f| has its min. – Jonas Meyer Oct 23 '23 at 21:13
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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.

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