# Mappings

#### dwsmith

##### Well-known member
Let $f$ be a nonconstant entire function that maps the unit circle, $\{z: |z| = 1\}$, into itself. Prove that $f$ maps the open unit disc, $\{z: |z| < 1\}$, into itself.

I am having a little trouble starting this one. z in C

#### girdav

##### Member
Did you try by contradiction, using the maximum modulus principle?

#### dwsmith

##### Well-known member
Did you try by contradiction, using the maximum modulus principle?
How would that be used? An open disc doesn't have a maximum modulus.

#### girdav

##### Member
Use the fact that the maximum of the modulus is reached at the boundary.

#### dwsmith

##### Well-known member
Use the fact that the maximum of the modulus is reached at the boundary.
That still doesn't make sense. Every time we get close to the boundary, we can get a little bit closer. Moreover, we can get a little bit closer and infinite amount of times.

#### girdav

##### Member
In fact, we have to work with the map $M(r):=\sup_{|z|=r}|f(z)|$. We can show thanks to maximum modulus principle that this map is strictly increasing.

#### dwsmith

##### Well-known member
In fact, we have to work with the map $M(r):=\sup_{|z|=r}|f(z)|$. We can show thanks to maximum modulus principle that this map is strictly increasing.
I don't understand what you are getting at.

#### girdav

##### Member
If $M(r_1)\geq M(r_2)$ for some $r_1<r_2$, the maximum modulus principle shows that $f$ is constant, so $M$ is a strcily increasing map. Now, we have that $M(1)=1$, so if $r<1$ then $M(r)<1$.