Show $g(0)=\sup\{ Re(f(0)) : f\in\mathfrak{F}\}$

Define a family of functions by $$\mathfrak{F}:=\left\{ f\in Hol(\mathbb{D}) \ : \ \sum_{n=0}^\infty \left( \frac{|f^{(n)}(0)|}{n!}\right)^2\leq 1 \textrm{ and } f\left(\frac{1}{2}\right)=0\right\}$$.

I am trying to show that there exists a $g\in\mathfrak{F}$ s.t. $$g(0) = \sup\{Re(f(0)) : f\in\mathfrak F\}.$$

Now since f is holomorphic on the disc we know it has a power expansion and that $$a_n = \frac{f^{(n)}(0)}{n!},$$ which means that we know $$\sum_{n=0}^\infty |a_n|^2\leq 1$$ and so $a_n\in\mathbb{D}$ for $n=0,1,2,\dots.$ The part I am having a issue with is using the assumption that $f\left(\frac{1}{2}\right)=0.$ I've worked similar problems where we had that the functions mapped to the disc, which meant using some conformal maps we could use Schwarz's Lemma. I want to do the same thing here, but I can't see how the $\sum |a_n|^2\leq 1$ means $f:\mathbb{D}\rightarrow\mathbb{D}.$

I appreciate any guidance or hints you can provide. Thanks.

For $f \in \mathfrak{F}$ and $z \in \Bbb D$ the AM-GM inequality gives $$|f(z)| \le \sum_{n=0}^\infty |a_n z^n| \le \sum_{n=0}^\infty \frac 12 \bigl( |a_n|^2 + |z|^{2n} \bigr) = \frac 12 \bigl( \sum_{n=0}^\infty |a_n|^2 + \sum_{n=0}^\infty |z|^{2n} \bigr) \le \frac 12 \bigl( 1 + \frac{1}{1-|z|^2} \bigr)$$ so that $\mathfrak{F}$ is uniformly bounded on compact subsets of $\Bbb D$. Alternatively, one can use the Cauchy-Schwarz inequality for infinite series to conclude that $$|f(z)| \le \frac{1}{\sqrt{1-|z|^2}} \, .$$

It follows from Montel's theorem that $\mathfrak{F}$ is normal, i.e. every sequence in $\mathfrak{F}$ has a subsequence which converges uniformly on compact subsets of $\Bbb D$.

Now let $(f_k), f_k \in \mathfrak{F}$ be a sequence such that $$\lim_{k \to \infty} \text{Re} \, f_k(0) = \sup\{\text{Re} \, f(0) : f\in\mathfrak F\}$$ $(f_k)$ has a subsequence $(f_{k_j})$ which converges uniformly on compact subsets to a function $g$ which is holomorphic in $\Bbb D$.

It is clear that $g(\frac 12) = 0$ and $$\text{Re} \, g(0) = \sup\{\text{Re} \, f(0) : f\in\mathfrak F\} \, .$$

Finally, since $\lim_{j \to \infty} f_{k_j}^{(n)}(0) = g^{(n)}(0)$ for all $n$, $$\sum_{n=0}^\infty \left( \frac{|g^{(n)}(0)|}{n!}\right)^2\leq 1$$ so that $g \in \mathfrak F$. (Hint: consider a finite sum $\sum_{n=0}^N$ first.)

(In response to your comment:) The extremal function $g$ satisfies $g(0) > 0$. To see this, consider the function $\tilde g$ defined by $$\tilde g(z) = \frac{|g(0)|}{g(0)} g(z) \, .$$ Then $\tilde g \in \mathfrak F$, so that $$\text{Re} \, g(0) \ge \text{Re} \, \tilde g(0) = \text{Re} \, |g(0)| = |g(0)| \ge \text{Re} \, g(0) .$$ So equality must host in this chain: $$\text{Re} \, g(0) = |g(0)|$$ which implies that $g(0)$ is a positive real number.

Remark: $\sum |a_n|^2\leq 1$ does not imply $f:\mathbb{D}\rightarrow\mathbb{D}$. For example, $$f(z) = - \frac{\sqrt 6}{\pi} \log(1-z) = \sum_{n=0}^\infty \frac{\sqrt 6}{\pi n} z^n$$ is not bounded in $\Bbb D$, but has $$\sum_{n=0}^\infty |a_n|^2 = \frac {6}{\pi^2 }\sum_{n=0}^\infty \frac{1}{n^2} = 1 \, .$$

• That makes sense. I wouldn't have thought to use the AM-GM inequality there. I also appreciate you taking the time to add the remark. Is it possible to guarantee that $g(0)= Re [g(0)]$? Possibly by altering the members of the sequence which converges to it? – Scott Jan 7 '16 at 2:48
• @Scot: That is "automatically" satisfied, see updated answer. – Martin R Jan 7 '16 at 6:18
• Sorry, I hadn't logged on lately and I forgot to accept your answer. This was perfect thanks so much. – Scott Jan 24 '16 at 5:15