Show that $f$ is constant using Maximum modulus principle

Question

Let $D=\{z:|z|<1\}$ and $ \bar{D}=\{z:|z|\leq 1\}$. Suppose that $f : \bar{D} \setminus \{0\}\to \mathbb{C}$ is a continuous function that satisfies:

(i) $f$ is analytic in $D \setminus \{0\}$;

(ii) $\limsup_{z \to 0}|f(z)|< \infty$;

(iii) $f(z)$ is equal to a finite constant on the unit circle $|z|=1$.

Prove that $ f$ is a constant function.

The first thing coming into my mind that I may use the Maximum Modulus Principle for this problem. On $|z|=1$, $f$ is bounded, I want to prove that $f$ has a maximum on $D $, but I can't see how to use the assumption (ii). Does it mean that

$$\forall \delta>0, \sup|f(z)|_{|z|\leq \delta} < \infty ?$$

Do I have to prove that $f$ is continuous at $z=0$?

Could you please help me with this problem? I really appreciate your help.

Answer 1

I am trying to solve this problem based on your advice, please let me know if my solution is correct.

We have $ \limsup_{z \to 0}|f(z)| < \infty$, so there exists a neighborhood of $0$ on which $f$ is bounded. Since $ f$ is analytic in $D \setminus \{0\}$, by Riemann's Theorem (https://en.wikipedia.org/wiki/Removable_singularity), we get $f $ is analytically extendable over $0$, then $f$ is also continuously extendable over $0$.

Put $f(z) =c$ on $|z|=1$, $c$ is a constant. Consider the function $g(z) = f(z)-c$, then $g(z)=0$ on $|z|=1$.

Since $f$ is continuous and analytic in $\bar{D}$, $g$ is continuous and analytic in $\bar{D}$ as well, by Maximum Modulus Principle, $g(z)=0$ in $\bar{D}$.

Thus, $f$ is a constant in $\bar{D}$.

Answer 2

ii) says that $f$ is bounded near $0$ and this implies that $f$ has removable singularity at $0$. So we can treat $f$ as a holomorphic function on $D$ by assigning a suitable value at $0$. See Riemann 's theorem in https://en.wikipedia.org/wiki/Removable_singularity

Let $f=c$ on $|z|=1$. By considering $f-c$ instead of $f$ we may suppose $f=0$ on the boundary. Now MMP implies that $f=0$ in $D$.