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Edited because I need more help than the linked "duplicate" provides.

ORIGINAL QUESTION:

This is a problem from a past complex analysis qualifying exam and I'm studying for my own qual next month. I'm not quite sure how to approach this one.

Problem:

Suppose $f(z)$ is an analytic function that maps the unit disc into itself and has zeros at $\frac{i}{2}$ and $-\frac{i}{2}$. Determine the largest possible value of $|f(1/2)|$ and give an example of a function $f$ that attains your upper bound.

What I've tried/considered:

We know by the maximum modulus principle that the function attains its maximum on the boundary, thus since it maps from the unit disc to the unit disc, the max on the boundary is 1. But what we need is the maximum at $|z| = 1/2$.

I thought about using the Schwarz Lemma, but it doesn't seem like we have $f(0)=0$ here. I considered using the argument principle, but there are zeros on the circle of radius ½.

I think I can use the Schwarz-Pick Theorem (though I’m not sure how I’d remember this specific one on a qual?) to get a bound such that $|f(1/2)|<= \sqrt{8/17}$ but I’m not sure if that is right, and if it is right, I haven’t been able to figure out a function that attains this value?

EDIT TO REQUEST ADDITIONAL HELP:

The question was closed, stating that itt is simply a special case of this. However, I do not understand how to apply it. Perhaps the suggestion is that the bound I'm looking for can be found using the zeros at $z_1 = \frac{i}{2}$ and $z_2 = -\frac{i}{2}$ by

$$ |f(z)| \leq \prod_{k=1}^n \left\vert \frac{z-z_k}{1-\bar{z_k}z} \right\vert = \left\vert \frac{z-i/2}{1+(i/2)z} \right\vert \left\vert \frac{z+i/2}{1-(i/2)z} \right\vert = \left\vert \frac{4z^2+1}{z^2+4} \right\vert $$

and thus $$ \left\vert f\left(\frac{1}{2}\right)\right\vert \leq \frac{8}{17}. $$

Is this correct? If this is the case, two questions remain for me:

(1) Is this inequality found in some lemma or theorem that I should know to use? As stated in my question, I was struggling to find which theorem applies and why.

(2) If this is the correct maximum, how do I find an example of a function $f(z)$ which attains this value at $f(1/2)$? I am new to this kind of problem, so would really appreciate a clear and complete answer if possible.

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    $\begingroup$ I'm not sure you can use Schwarz-Pick since $f$ isn't guaranteed to be a bijection. $\endgroup$ Commented Jul 10, 2022 at 16:24
  • $\begingroup$ @CharlesHudgins Good point. Can a function map the unit disk into itself and not be a bijection? $\endgroup$
    – Serafina
    Commented Jul 10, 2022 at 16:44
  • $\begingroup$ This is a special case of the more general estimate math.stackexchange.com/q/94122/42969, therefore I closed the question as a duplicate. $\endgroup$
    – Martin R
    Commented Jul 10, 2022 at 17:49
  • $\begingroup$ @MartinR I would have found it more helpful if you could have linked to this other answer and helped me see how to apply it here in my specific case, while perhaps leaving the opportunity for others to answer as well. Sometimes one person’s way of explaining things is clearer than another’s. $\endgroup$
    – Serafina
    Commented Jul 10, 2022 at 20:55
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    $\begingroup$ @Serafina Sure, $f(z) = z^2$. Observe that $|z| \leq 1$ implies $|f(z)| \leq 1$. To see that $f$ is a surjection, suppose $w = re^{i\theta}$. Take $z = \sqrt{r} e^{i\theta / 2}$, then clearly $f(z) = w$. $\endgroup$ Commented Jul 10, 2022 at 21:54

1 Answer 1

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The topic here is finite Blaschke products which are the analog for the unit disc of polynomials in the plane; these are holomorphic on the closed unit disc (meaning on a slightly bigger open disc depending on the particular Blaschke product of course), have modulus $1$ on the unit circle and you can specify their finitely many zeroes in the unit disc, so if the zeroes are $z_1,...z_n, |z_k| <1$ (with repeats for multiplicities), one can define for example $$B(z)=\Pi_{k=1}^n\frac{z-z_k}{1-\bar z_k z}$$ and then if some $f$ holomorphic in the disc has finitely many zeroes, there is a finite Blaschke product $B$ (unique up to multiplication by $|\alpha|=1$ and one can take $B=1$ if there are no zeroes) st $f=Bg$ with $g$ holomorphic in the unit disc, and having no zeroes, while $|f|, |g|$ behave the same on the unit circle.

In this case, since $|f| \le 1$ we have $|g| \le 1$ so $|f(z)| \le |B(z)|$ for any $z$ in the unit disc; however since now $g$ has no zeroes, one, of course, can take $g=1$ to attain the maximum, so we solved the extremal problem in the OP, or more generally, the extremal problem where $f$ is a holomorphic self-map of the unit disc to itself, taking zeroes at prescribed finitely many points $z_1,...z_k$ (including repeats for multiplicity here) and one asks for the maximum of $|f(y)|$ for some $y$ in the unit disc, the answer being precisely $|B(y)|$ where $B$ is the Blaschke product above, with the extremal functions being any $\alpha B, |\alpha|=1$

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  • $\begingroup$ THANK YOU. I was really having trouble with this because I had never even heard of Blaschke products before. $\endgroup$
    – Serafina
    Commented Jul 12, 2022 at 17:46
  • $\begingroup$ Happy to be of help $\endgroup$
    – Conrad
    Commented Jul 12, 2022 at 17:52

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