Suppose that $f(z)$ is analytic on the closed unit disk $\overline{\mathbb{D}}$ and $1<|f(z)|<M$ for $|z|=1$, while $f(0)=1$. Show that $f(z)$ has a zero in $\mathbb{D}$ and that such zero $z_{0}$ satisfies $|z_{0}|>\frac{1}{M}$.
My attempt:
Suppose $f$ has no zero in the unit disk. Then, by minimum modulus principle, $f$ attains its minimum on the boundary. But maximum modulus principle suggests that $f$ attains its maximum on the boundary. This implies that $|f(z)|=1$ for all $z$, and this is a contradiction to our initial assumption that $1<|f(z)|<M$ for $|z|=1$. Hence,$f(z)$ has a zero in $\mathbb{D}$.
Now consider the mapping $$\phi_{M^{-1}}=\frac{M^{-1}-z}{1-\overline{M^{-1}}z}$$. Then $\phi_{M^{-1}}\left(\frac{f(z)}{M}\right):\mathbb{D}\rightarrow\mathbb{D}$ and maps $0$ to $0$.
Then, by Schwarz Lemma, $$\left|\phi_{M^{-1}}\left(\frac{f(z_0)}{M}\right)\right|=\left|\frac{1}{M}\right|\leq|z_0|$$
Any clarification is appreciated.
Best Answer
Your invoking minimum modulus principle is fine, but the subsequent argument using maximum principle does not produce $|f(z)|=1$ on the boundary (if there's an argument that can prove it, then you should include it in your proof.) I suggest instead to note that $|f(0)|=1<|f(z)|$ for all $|z|=1$. It says that the minimum of $|f(z)|$ cannot occur on $|z|=1$. This leads to a contradiction.
About the second part, it looks fine except that the equality sign in $$ \left|\frac{1}{M}\right|\leq|z_0| $$ can be dropped since $|f(z)|=M$ cannot occur on the boundary.
Here, I will present another way in which the Schwarz lemma can be applied. Define $$ g:\overline{\mathbb{D}}\ni w\mapsto f\left(\frac{w+z_0}{1+\overline{z_0}w}\right). $$ Then we can see that $g(0)=f(z_0)=0$ and $|g(w)|<M$ for all $|w|=1$. Schwarz's lemma implies $$ |g(w)|<M|w| $$ for all $0<|w|\le 1$. Finally the conclusion comes from noticing that $$ |g(-z_0)|=|f(0)|=1<M|-z_0|=M|z_0|. $$