# Prove that: $f$ is not a constant function on the open unit disc $\Delta (0,1)$

I have come across the following questions from my text book! However, I'm not sure how to go about answering... Any help (or hint) would be greatly appreciated. Thanks.

Suppose $F=\mathbb{C}^2$ with norm $\sup$. For $f: \mathbb{C} \to F$, the function $f$ be defined by $f(z)=(1,z)$, $\forall z \in \mathbb{C}$.

Prove that:

a. $f \in \mathcal{P}(\mathbb{C},F)$

b. $\left \|f \right \|$ is constant on the open unit disc $\Delta (0,1)$

c. $f$ is not a constant function on the open unit disc $\Delta (0,1)$

d. $\left \|f \right \|$ is not a constant on $\mathbb{C}$

Hint: You have $||f(z)||= \max (1,|z|)$. So if $z \in \Delta(0,1)$, $|z| <1$ hence $||f(z)||=1$; if $z \notin \Delta(0,1)$, $|z| \geq 1$ hence $||f(z)||=|z|$.

• I'm sorry, but I still don't understand :( Can you say more details? b. $||f||$ is constant on the open unit disc $\Delta(0,1)$ and d. on $\mathbb{C}$, .I can not distinguish the difference between them. Seirios! Thanks. Jun 16, 2013 at 7:09
• I added some details. Jun 16, 2013 at 7:14
• Wow! Thanks Seirios. Now, I understand b. and d. You explained to me that is very easy to understand. Can you add your hint for c. I look forward to your help. Seirios! Jun 16, 2013 at 7:18
• You just have to find $z_1,z_2 \in \Delta(0,1)$ such that $f(z_1)=f(z_2)$, so it seems pretty easy. Jun 16, 2013 at 7:20
• In fact, I do not know the notation $\mathcal{P}(\mathbb{C},F)$. What is it? Jun 16, 2013 at 7:44

If you conpute the norm using elements of the open disc, it will always be 1, for the definition of the disc $B_0(1)$ (I use this notation) which contains every element whose norm is less of 1. Then use the definition in the "hint" and you will be ok. D) follows easily for the same reason.

the other points are immediate.

• Can you post your hint for a. Ric Ped! I have thought $(1,z)=(1,0)+(0,z)$, Is it right? Jun 16, 2013 at 7:32

Thank you very much! Seirios and Ric Ped.

With Seirios's hints, I will rewrite as follows:

We consider: $f: \mathbb{C} \to \mathbb{C}^2$, $f$ is defined by $f(z)=(1,z)$.

a. $f \in \mathcal{P}(\mathbb{C},\mathbb{C}^2)$.

it is obvious. Because, $f(z)=(f_1,f_2)=(1,z)$ is a polynomial and continuous, since $f_1=1$ and $f_2=z$ are polynomials and continuous.

b. $∥f∥$ is constant on the open unit disc $\Delta(0,1)$.

We have: $\Delta(0,1)=\{\lambda \in \mathbb{C}:|\lambda|<1\}$.

$||f||=\sup_{z}{|f(z)|}=\sup_{z} \{1,|z|\}$. Hence, if we consider $f$ on $\Delta(0,1)$ $\implies ||f||=1$.

Therefore, $||f||=1$ on $\Delta(0,1)$.

c. $f$ is not a constant function on the open unit disc $\Delta(0,1)$

If $f$ is a constant function on the open unit disc $\Delta(0,1)$ then $f(z_1)=f(z_2)$, $\forall z_1,z_2 \in \Delta(0,1)$.

I show that there exists $z_1=\dfrac{1}{2} \ne z_2=\dfrac{1}{3} \in \Delta(0,1)$ so $f(z_1) \ne f(z_2)$.

d. $∥f∥$ is not a constant on $\mathbb{C}$.

Since a. $f$ on $\Delta(0,1)$ $\implies ||f||=1$

If $z \notin \Delta(0,1)$, $|z| \ge 1$, hence $||f(z)||=|z|$. Therefore, $||f||$ is not a constant on $\mathbb{C}$. :p