# Show the function doesnt exist

1. Mar 19, 2012

### symbol0

Show that there is no holomorphic function f in the unit disc D that extends continuously to |z|=1 such that f(z) =1/z for |z|=1

Some thoughts that might not be relevant:
If such f existed then, I can see that f maps the unit circle to the unit circle and the unit disc onto the unit disc.
On |z|=1 f would be equal to the conjugate function which is not differentiable anywhere.

I'm kind of stuck. I appreciate any suggestions.

2. Mar 19, 2012

### Office_Shredder

Staff Emeritus
Edit: wow it's way too late. Think integration

3. Mar 20, 2012

### symbol0

Do you mean that if I integrate f on circles with radii approaching 1, I get
0 = int f approaches int 1/z =2ipi ?

but does the continuity of f imply the continuity of the integral like that?

4. Mar 20, 2012

### jgens

Does integration work with this problem? I assume you had an argument like the following in mind:
But since the extension of $f$ to $\partial D$ is only continuous, none of the integration theorems for analytic functions will work here. The notion of what it means for $f$ to be analytic on $D \cup \partial D$ is ill-defined as well, since $D \cup \partial D$ is not open. Maybe I am just missing something here and you found a way to do this with integration (which would be pretty neat).

I am pretty sure that you can do this without integration though. Notice that the maximum/minimum modulus principle apply since $f$ is holomorphic on $D$ and continuous on $\partial D$. This means $|f|$ assumes its maximum and minimum values on $\partial D$; in particular, it follows that $|f|=1$ on $D$. From here you should be able to derive a contradiction.

5. Mar 20, 2012

### symbol0

Thank you jgens.
About your approach, you say f doesnt have a min in D.
How do we know 0 is not a min ?

6. Mar 20, 2012

### jgens

I said $|f|=1$ on $D$ which means that every point of $D$ is a minimum for $|f|$.

7. Mar 20, 2012

### symbol0

I want to understand how do you conclude |f| = 1 on D.
I know f cannot have a maximum in D. So |f|<1 on D.
....what else?

8. Mar 20, 2012

### jgens

The maximum modulus principle says that $|f| \leq 1$ on $D$. The minimum modulus principle says that $1 \leq |f|$. This means that $|f|=1$. There is not much to it.

9. Mar 20, 2012

### Office_Shredder

Staff Emeritus
To get the minimum you can take 1/f as long as f us non-zero.

If you integrate around circles of radius r and let r approach one you can get a contradiction can't you?

10. Mar 20, 2012

### symbol0

To jgens: you say 1 leq |f|. How do we know f is not a function like f(z)=z which has a minimum in D ?

To office shredder: to get that contradiction don't we need to know that integrals are continuous ? is that a fact?

11. Mar 21, 2012

### mathwonk

wouldn't the winding number around the origin be positive? whereas it seems to be -1.

12. Mar 21, 2012