# Find necessary and sufficient conditions for $f \in S$ to exist such that $f(0) = 0$ and $f'(0)=b$

Given $b\in \mathbb{C}$, find a necessary and sufficient conditions for $f \in S$ to exist, such that $f(0) = 0$ and $f'(0) =b$, and describe the set of solutions.

$S$ denotes the set of all analytic functions $f: D \to \overline{D}$, where $D$ is the open unit disk. Note $f(z) = \frac{a+zg(z)}{1+z\overline{a}g(z)}$ describes all functions in $S$ such that $f(0)=a$, when $g$ runs though $S$.

I want to use Schwarz Lemma for this. Because $f\in S$, then $|f(z)| \leq z$ and $f(0)=0$. So, $f$ satisfies the Schwarz Lemma conditions.

Since $f'(0)=b$, then $f$ must be of the form $bz$ since we must get $b$ after taking the derivative. $f(0)=0$ would be satisfied since $b(0)=0$.

I'm not really sure what "necessary and sufficient" conditions there would be. Anyone mind giving me some insight on this?

Necessary and sufficient conditions mean we're looking for some condition(s) $P$ such that

Condition $P$ begin satisfied implies that such an $f\in S$ exists. Conversely, if such an $f\in S$ exists, then condition $P$ is satisfied.

You might know this as an "if and only if" statement. For this particular problem, we must find out what "condition $P$" is. Let's look into it.

First suppose our cherished $f$ exists. Since $f$ has the property that $f$ maps zero to itself and $f(D)=D$, the Schwarz lemma applies (as you said). However, recall there is another condition in addition to $|f(z)|\leq |z|$: that $|f'(0)|\leq 1$. Therefore it must be the case that $|b|\leq 1$.

The above discussion gives us a candidate for a "condition P". In fact, we have just shown the $(\Leftarrow)$ direction; we just need make sure we don't need to add anything to this condition in order to prove the converse direction--but this shouldn't be very difficult to show: after all, the function $h(z)=bz$ maps $D$ to itself (because $|b|\leq 1$), has the property that $h(0)=0$, and $h'\equiv b$. In particular, $h'(0)=0$, so we know such a function exists when we're given a $|b|\leq 1$. This shows the $(\Rightarrow)$ direction.

Lastly, there's the matter of describing all functions in $S$ having the properties you've stated. As you've noted, any $f: D\to\overline{D}$ with $f(0)=0$ can be written as (setting $a=0$) $$f(z)=zg(z)$$ where $g\in S$ (while this expression seems true, I'm taking your word/your teacher's word for it here). We compute $$f'(z)=g(z)+zg'(z).$$ So $$f'(0)=g(0)+0g'(0)=g(0).$$ Since we desire $f'(0)=b$, it follows that we need $g(0)=b$. Therefore $$g(z)=\frac{b+z\tilde{g}(z)}{1+z\overline{b}\tilde{g}(z)}$$ for some other $\tilde{g}\in S$. Furthermore $$f(z)=z\left(\frac{b+z\tilde{g}(z)}{1+z\overline{b}\tilde{g}(z)}\right).$$ From our discussion, the full statement should be

An $f\in S$ exists such that $f(0)=0$ and $f'(0)=b$ if and only if $|b|\leq 1$.

Hopefully this helps.

• Blake, this extremely close to what I have in what I have been working on since posting this question. Thank you! Commented Dec 10, 2015 at 11:06
• @complexLost Great to hear! I got a little carried away with my answer and was concerned that I might've ruined the fun for you to figure it out yourself! Glad to see you came to same conclusion independently of me. Commented Dec 10, 2015 at 11:08
• If you are still feeling Complex Analysis frisky, my next question is here: math.stackexchange.com/q/1569119/295773 Commented Dec 10, 2015 at 11:31