Show that $|f(z)| \leq |z|$ on annulus

Question

Let $D = \{ z \in \mathbb{C} : 2 < |z| < 3 \}$. Suppose that $f$ is holomorphic on $D$ and $f$ is continuous on $\overline{D}$. Suppose that $\max \{ |f(z)| : |z| = 2\} \leq 2$ and $\max \{ |f(z)| : |z| = 3 \} \leq 3$. Show that $\forall z \in D, |f(z)| \leq |z|$. The solution I am thinking of I believe is wrong, but here is what it is. Assume $f$ is non-constant. By the maximum modulus principle $|f|$ assumes its maximum on $\partial D$, Hence, $|f(z)| \leq 3, \forall z \in D$. Define $g(z) = \frac{f(z)}{z}$, which is analytic. Consider $r \in (2,3)$. For $|z| = r$, $|g(z)| \leq \frac{3}{r}$. Since for $|z| = 2$, $|g(z)| \leq 1$, by the maximum modulus principle, $\forall |z| \in (2,r), |g(z)| \leq \frac{3}{r}$. Let $r \rightarrow 3$, then $\forall |z| \in (2,3), |g(z)| \leq 1 \implies |f(z)| \leq |z|$. I essentially followed the same technique as in Schwarz's Lemma, but I am not sure if this is right.

Answer 1

Your proof is correct, but you can simplify it. The function $\lvert g(z) \rvert$ assumes its maximum on $\partial D$. For $\lvert z \rvert = 2$ we have $\lvert f(z) \rvert \le 2 = \lvert z \rvert$, thus $\lvert g(z) \rvert \le 1$. Similarly $\lvert g(z) \rvert \le 1$ for $\lvert z \rvert = 3$. Thus necessarily $\lvert g(z) \rvert \le 1$ for all $z \in D$.