Proof: Let $f$ be a Nonconstant Entire Function on the Unit Disc

[Dustinsfl]

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

 

[tarantella]

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

 

[Dustinsfl]

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

How would that be used? An open disc doesn't have a maximum modulus.

 

[tarantella]

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

 

[Dustinsfl]

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.

 

[tarantella]

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.

 

[Dustinsfl]

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.

 

[tarantella]

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 $M(r)<1$.