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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 $f(z) = cz^N$.

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2 Answers 2

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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 $|z| = 1$.

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  • $\begingroup$ 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...) $\endgroup$
    – Kenny Wong
    Jan 5, 2019 at 20:37
  • $\begingroup$ Yes, sure. That was a typo,it should have been a modulus. Thank you! $\endgroup$
    – s.kovalska
    Jan 5, 2019 at 20:45
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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.

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