Suppose $f:\mathbb{D}\rightarrow\mathbb{C}$ is analytic such that $|f|<2$. Also, suppose that $f''(0)=4i$. Find the value of $f(i/2)$.

Question

Question: Suppose $f:\mathbb{D}\rightarrow\mathbb{C}$ is analytic such that $|f|<2$. Also, suppose that $f''(0)=4i$. Find the value of $f(i/2)$.

Thoughts: If this was the first derivative, I feel like I could just use Schwarz-Pick, but being given the second derivative is throwing me off a bit. I still feel like it is the "right track", because we are given the second derivative of $f$ at $0$, so it is (at least to me) suggesting something with Schwarz Lemma, but I can't quite get anything to come out nicely. Any thoughts? Thank you!

Answer 1

One possible solution is via Parseval's identity for holomorphic functions (see for example Proof of Parseval's identity):

Let $f(z) = \sum_{n=0}^\infty a_n z^n$ be the Taylor series of $f$ in $\Bbb D$, then $$ \sum_{n=0}^\infty |a_n|^2 r^{2n} = \frac{1}{2\pi}\int_0^{2\pi}|f(re^{it})|^2 \, dt \le 4 \, . $$ Taking the limit $r \to 1$ gives $$ \sum_{n=0}^\infty |a_n|^2 \le 4 \, . $$ But $|a_2|^2 = |f''(0)/2|^2 = 4$, so that all other Taylor coefficients must be zero.

It follows that $f(z) = 2i z^2$.