# 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 $C$.

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 $$C$$.

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?

• Use Cauchy's representation formula for $f^{(n)}(0)$, considering large or small $r$ according to how $n$ compares with $k$. – user10354138 Jun 18 '19 at 16:41
• I don't see the problem: $k$ can be $0$. – Chris Custer Jun 18 '19 at 16:42
• A similar question here. – rtybase Jun 18 '19 at 22:16

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 $$(1)$$ $$f(z)=a_kz^k$$

• @Bach, re your update, $2\pi$ dissapears because $\int\limits_{0}^{2\pi}|dt|=2\pi$. – rtybase Jun 18 '19 at 23:30
• Yes, but that's what the condition gives us. – Bach Jun 18 '19 at 23:43
• $2\pi$ does not cancl out. – Bach Jun 18 '19 at 23:44