I am working on the following problem:
Let $\mathcal{F}$ denote the set of functions which are analytic on a neighborhood of the closed unit disk in $\mathbb{C}$. Find:
$$\sup\{|f(0)|\mid f(1/2)=0=f(1/3) \text{ and } |f(z)|\leq 1 \text{ when } |z|=1\}$$
Is the supremum attained? Hint: consider the function $\frac{z-1/2}{1-z/2}\frac{z-1/3}{1-z/3}$.
Here are my thoughts so far:
I know that by the maximum modulus principle, $|f(z)|\leq 1$ for all $|z|\leq 1$. By Montel's theorem, the new subfamily of $\mathcal{F}$, call it $\tilde{\mathcal{F}}$, is precompact.
Q: However, I am not sure if $\tilde{\mathcal{F}}$ is closed and so I cannot see if it is compact.
Q: Secondly, I understand the hint function to be a finite Blaschke product. Moreover, I see why the hint function belongs to $\tilde{\mathcal{F}}$, but I fail to see how this is helpful in computing the supremum.
Thank you for any hints/help!
We only need the maximum modulus principle, not Montel's theorem.
The Blaschke product $B(z) = \frac{z-1/2}{1-z/2}\frac{z-1/3}{1-z/3}$ has modulus one for $|z| = 1$, and zeros in the unit disk only at $z=1/2$ and $z=1/3$. It follows that $$ g(z) = \frac{f(z)}{B(z)} $$ has removable singularities, i.e. can be continued to a holomorphic function in the unit disk $\Bbb D$.
Then $|g(z)| \le 1$ for $|z|=1$, and therefore $|g(z)| \le 1$ in $\Bbb D$ because of the maximum modulus principle. In particular, $$ |f(0)| \le |B(0)| = \frac 1 6 $$ for all $f \in \mathcal{F}$. Equality holds for $f=B$, so that the supremum is attained.