# Show that $f$ is constant using Maximum modulus principle

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 $$z=0$$?

• (ii) means that $f$ is bounded near $0$. Since it is analytic on $D \setminus \{0\}$, (ii) implies that $f$ extend to an analytic function on the whole $D$
– user598294
Commented Sep 27, 2020 at 0:23
• Then you can use the Maximum modulus principal on each $\{ z \, : \, |z| < r \}$ for every $0<r<1$ and see what happens when $r$ tends to $1$ using (iii)
– user598294
Commented Sep 27, 2020 at 0:26
• @AlexL Thank you for your help. Do you mean if $f$ is constant on each $\{z: |z|<r\}$ for every $0<r<1$, then taking $r \to 1$, since $f$ is continuous and $f=c$ on $|z|=1$, this means that $f=c$ throughout $\bar{D}$?
– anna
Commented Sep 27, 2020 at 5:29
• $f$ is a priori not constant on each $\{ z \; : \; |z|<r \}$, But it hits its maximum M_r and minimum m_r on $\{ z \; : \; |z|=r \}$. $M_r$ must be non decreasing as $r \to 1$ and $m_r$ is non increasing as $r \to 1$. Then use (iii) and the continuity of $f$ on $\bar{D}$ to see that $|c| \leqslant m_r \leqslant M_r \leqslant |c|$ ($c$ is the value of $f$ on $\{ |z|=1\}$).
– user598294
Commented Sep 27, 2020 at 10:52

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 $$D$$.

– anna
Commented Sep 27, 2020 at 4:42
• What do you think about my solution in the Answer Part below? I applied Riemann's Theorem and then Max.Mod.Principle. Thank you for your help.
– anna
Commented Sep 27, 2020 at 4:51
• @anna The part where you wrote ??? is not clear. I think MMP applied dierectly to $f$ does not yield the proof. You have to apply it to $f-c$. Commented Sep 27, 2020 at 4:55
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 $$\bar{D}$$.