If $ \int_0^{2\pi}|f(re^{i\theta})|d\theta\le Ar^k $ for $f$ holomorphic and every $r>0$, then $f(z)=Cz^k$ for some constant $C$.

Question

Suppose that $f(z)$ is a holomorphic function on all of $\mathbb C$. Assume that there are constants $A>0$ and a non-negative integer $k$ so that $$ \int_0^{2\pi}|f(re^{i\theta})|d\theta\le Ar^k $$ for all $r>0$. Prove that $f(z)=Cz^k$ for some constant $C$.

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My attempt:

Note that $$2\pi|f(0)|\le\int_0^{2\pi}|f(re^{i\theta})|d\theta\le Ar^k$$ for all $r>0$, we have $f(0)=0$. If we can show that there exists $C$ such that $$g(z):=\begin{cases}\frac{f(z)}{z^k},&z\ne 0 \\ C,& z=0\end{cases}$$ is bounded and holomorphic on $\mathbb C$, then we are done. I was going to imitate the proof of Schwarz Lemma, but then I got stuck since $k$ can be less than $1$ which probably makes the limit $\lim_{z\to 0}\frac{f(z)}{z^k}$ go to infinity. So how to move on?

Answer 1

A holomorphic function on entire $\mathbb{C}$ is, technically, entire. It has a Taylor expansion $$f(z)=a_0+\sum\limits_{n=1}a_nz^n \tag{1}$$

also, using $$\int\limits_{\gamma}f(z)dz=\int\limits_{a}^{b}f(\gamma(t))\gamma'(t)dt \tag{2}$$

and applying Cauchy's estimate to $(1)$ $$a_n=\frac{f^{(n)}(0)}{n!}=\frac{1}{2\pi i}\int\limits_{C_R}\frac{f(z)}{z^{n+1}}dz$$ we have

$$|a_n|=\left|\frac{1}{2\pi i}\int\limits_{|z|=r}\frac{f(z)}{z^{n+1}}dz\right|= \left|\frac{1}{2\pi i}\int\limits_{0}^{2\pi}\frac{f(re^{it})}{(re^{it})^{n+1}}ire^{it}dt\right| \leq \\ \frac{1}{2\pi}\int\limits_{0}^{2\pi}\left|\frac{f(re^{it})}{r^{n}}\right||dt| \leq \frac{1}{2\pi}\frac{Ar^k}{r^n}=\frac{Ar^k}{2\pi r^n}$$

Taking the $\lim\limits_{r\rightarrow\infty}$ we have $a_n=0,\forall n\geq k+1$. Similarly, $\lim\limits_{r\rightarrow 0}$ we have $a_n=0,\forall n\leq k-1$. The remaining option is $n=k$ and from $(1)$ $$f(z)=a_kz^k$$