Koebe's 1/4 Theorem

[wpolly]

This theorem claims that the range of a 1-1 analytical mapping f on the unit disk with f(0)=0,f'(0)=1 contains the closed disk B(0,1/4).

Can the 1/4 in the theorem be improved if we assume that |f(z)| is bounded by some constant M?

[R.D. Silverman]

Originally Posted by wpolly

This theorem claims that the range of a 1-1 analytical mapping f on the unit disk with f(0)=0,f'(0)=1 contains the closed disk B(0,1/4).

Can the 1/4 in the theorem be improved if we assume that |f(z)| is bounded by some constant M?

I'm not sure that I understand the question. If f is almost everywhere analytic and bounded, it can not satisfy f'(0) = 1, because it must be the
constant function.

[R.D. Silverman]

Originally Posted by R.D. Silverman

I'm not sure that I understand the question. If f is almost everywhere analytic and bounded, it can not satisfy f'(0) = 1, because it must be the
constant function.

A slight clarification. A more exact statement of the theorem would be that
the map contains a disk centered at f(0) whose radius is f'(0)/4. So
for a bounded analytic function, the constant 1/4 can be replaced by
any constant (so yes, it can be improved). However, since the radius of
the disk becomes 0, any question involving the value of the constant
becomes more or less moot.

[wpolly]

Clarification: we assume |f(z)| is bounded on the unit disk.