How can we prove that |f(1/2)| <= 1?

[sbashrawi]

Homework Statement

Let f ba analytic function on 0< |z| < 1 and suppose |f(z)| <= 4|z|^1.1 for all 0<|z|<1.
Prove that |f(1/2)| <= 1

Homework Equations

The Attempt at a Solution

I tried to prove it be cauchy integral formula but I got
|f(1/2)|< 8 r ^1.1 r<1

 

[Count Iblis]

Hint: Clearly f(0) = 0. The consider f(z)/z, Cearly this functon is analytic and is also zero at z = 0, so we can divide yet again by z. We comnclude that

g(z) = f(z)/z^2

is an analytic function.

What does the Maximum Modulus theorem imply for |g(z)| for
|z|<1 given the bounds on |f(z)|?

 

[sbashrawi]

Hint: Clearly f(0) = 0. The consider f(z)/z, Cearly this functon is analytic and is also zero at z = 0, so we can divide yet again by z.


the function [f(z)]/[z] is not defined at z = 0, so how it can be zero.

and if f(0 ) = 0 but 0 is not in the set. it is a boundary point

Max | f(z)| = Max |f(z)| on the boundary and it is clear that

Max | f(z)| $$\leq$$ 4 at z = 1.

then I don't know how to fugure that | f(1/2)|$$\leq$$1

 

[Count Iblis]

Yes, I see now that the point z = 0 is excluded. I would suggest you to first solve the slightly different problem in which z = 0 is in the domain. If you do that, then it is a trivialy matter to modify the proof to take into account that z = 0 is not in the domain.

 

[sbashrawi]

I think this is not related to the maximum modulus principle since it says that f cannot achieve max. on an open set or it is a constant. If it is constant, then what is the value of this constant?

 

[Count Iblis]

Just close the boundaries at first and then assume that f(z)/z^2 assumes a maximum at the boundary. Then prove the result of this modified problem and then fix the proof by taking into account that the boundary is in fact open. E.g. you can think of placing boundaries a distance of epsilon within the region and then the analogues of the above result will hold. Since epsilon is larger than zero but arbitrary, you will recover the result.