# A function holomorphic on $\mathbb{D}(0,2)$ and bounded on a unit circle is a polynomial

Suppose function $$f(z)$$ is holomorphic on $$\mathbb{D}(0,2)$$ and $$N>0$$ is an integer such that: $$|f^{(N)}(0)| = N! \sup\{|f(z)|: |z|=1\}$$ show that $$f(z) = cz^N$$, $$c \in \mathbb{C}$$.

I have shown that since $$f(z)$$ is holomorphic in $$\mathbb{D}(0,2)$$, then it has a power series expansion around zero.

$$f(z) = \sum_{n=0}^{\infty} a_n z^n$$

Calculating the $$N$$-th derivative I got:

$$|f^{(N)}(0)| = N! a_N$$

from which I conclude that $$f(z)$$ is bounded by $$a_N$$ on the unit circle. Therefore by maximum principle for holomorphic function we may also conclude that it is bounded in the unit disc. But I still can't figure out how to show that $$f(z) = cz^N$$.

By Cauchy's formula, $$f^{(N)} (0) = \frac{N!}{2\pi i} \oint_{|z| = 1} \frac{f(z) \ dz}{z^{N+1}} = \frac{N!}{2\pi } \int_{0}^{2\pi} e^{-iN\theta}f(e^{i\theta}) \ d\theta.$$

Thus $$|f^{(N)}(0)| \leq N!\sup_{\theta \in [0, 2\pi)} |f(e^{i\theta})|,$$

with equality if and only if

$$e^{-iN\theta} f(e^{i\theta}) = c$$

for some constant $$c \in \mathbb C$$.

As the equality does hold by assumption, we have $$f(z) = cz^N$$ when $$|z| = 1$$.

• By the way, you wrote $f^{(N)}(0) = N! \sup_{\theta \in [0, 2\pi)} |f(e^{i\theta})|$ (no modulus sign on the $f^{(N)}(0)$). This implies that $f^{(N)}(0)$ is real and positive, so $c$ is real and positive too. (Not sure if that was a typo though...) Jan 5, 2019 at 20:37
• Yes, sure. That was a typo,it should have been a modulus. Thank you! Jan 5, 2019 at 20:45

For every $$0 \leq \theta \leq 1$$, $$f(e^{2i\pi\theta})e^{-2iN\pi\theta}$$ has an absolute value not greater than $$a_N$$, but the integral over $$[0;1]$$ is exactly $$a_N$$.

So it is constant equal to $$a_N$$, thus $$f(z)=a_Nz^N$$ on the unit circle, thus the equality holds on the whole disc.