I am practicing for an exam and one of the past papers asks the following
Suppose we have a function which is complex analytic on the open unit disk $|z|<1$ satisfying the following property $$|f(z)|<1-|z| \text{ for all } |z|<1$$ The question simply asks what can be concluded from this property, though I am rather stumped.
My intuition is that we must have f being identically zero. This is obviously the case if the property extends to the boundary of the disk, $|z| = 1$.
My attempt at showing this is to break things up into two cases.
First: The radius of convergence of $f$ is greater than $1$. In this case, I think we should be able to leverage the continuity of the absolute value to extend the inequality condition to the boundary of the disk $|z|=1$ and then show $f$ is identically zero by the cauchy integral formula.
Second: The radius of convergence is exactly 1. In this case I think (though I'm not 100% sure) that we must necessarily have that $f$ diverges at some $w$ with $|w|=1$. From here, I think we should be able to claim that there is some contradiction with the inequality condition somewhere close to $w$, though I can't think of a way to do it.
Am I on the right track with this proof, or have I made a mistake? (or perhaps a far simpler line of argument I missed?)
