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Prove that a function which is analytic in the whole plane and satisfies the inequality $|f(z)|<|z|^n$ for some $n$ and all sufficiently large $|z|$ reduces to a polynomial.

The function is analytic, so $f^{n}(z)$ exists for all $n$, all $z$. We have the Cauchy's integral formula for higher derivatives

$$f^{(n)}(a)=\frac{n!}{2\pi i}\int_C\frac{f(z)}{(z-a)^{n+1}}dz.$$

So $$\int_C\frac{f(z)}{(z-a)^{n+1}}dz<|z|^n$$ for all large $|z|$. How does that help?

PJ Miller
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2 Answers2

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Use the integral formula for the $n+1^{\text{st}}$ derivative,

$$f^{(n+1)}(a) = \frac{(n+1)!}{2\pi i}\int_{\lvert z\rvert = R} \frac{f(z)}{(z-a)^{n+2}}\,dz.$$

Consider $\lvert a\rvert < 1$, and let $R \to \infty$. The standard estimate shows

$$\left\lvert f^{(n+1)}(a)\right\rvert \leqslant \frac{(n+1)!}{2\pi} \frac{R^n}{(R-1)^{n+2}}\cdot 2\pi R \leqslant \frac{C}{R},$$

so $f^{(n+1)} \equiv 0.$

Daniel Fischer
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  • Since you have assumed that $|a|<1$, is your final step a tacit appeal to the Identity Theorem? (https://en.wikipedia.org/wiki/Identity_theorem) – 311411 Jul 24 '21 at 17:49
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Take $C$ to be a circle of radius $R$ around the origin, large enough to contain $a$. Parameterize via $z = Re^{it}$. Then

\begin{align} |f^{(n)}(a)| &= \frac{n!}{2\pi} \left|\int_C \frac{f(z)}{(z - a)^{n + 1}} dz \right| \\ &= \frac{n!}{2\pi} \left|\int_0^{2\pi} \frac{f(Re^{it})}{(Re^{it} - a)^{n + 1}} Rie^{it} dt\right| \\ &\le \frac{n!}{2\pi} \int_0^{2\pi} \frac{|f(Re^{it})|}{|Re^{it} - a|^{n + 1}} R dt \\ &\le \frac{n!}{2\pi} \int_0^{2\pi} \frac{R^n}{|Re^{it} - a|^{n + 1}} R dt \end{align}

The denominator can be bounded below by, say, $\frac{1}{2} R^{n + 1}$ by choosing $R$ sufficiently large, and so we see that

$$|f^{(n)}(a)| \le n! \frac{R^{n + 1}}{\frac{1}{2} R^{n + 1}} = 2n!$$

Hence, $f^{(n)}$ is a bounded entire function, and so is constant.

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    You say "the denominator can be bounded below by, say, ${1 \over 2} R^{n+!}$ by choosing $R$ sufficiently large". How does choosing $R$ sufficiently large guarantee that? – user1770201 Apr 21 '14 at 13:16
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    Why can you bound the denominator in that way? – BingerBiter Mar 31 '21 at 04:33
  • I am not sure about bounded below by $\frac{1}{2}R^{n+1}$, but we can bound the denominator below by $\frac{\alpha}{2}R^{n+1}$ for a positive constant $\alpha$. To do so, pick $R$ so large that $a$ is within the circle of radius $\frac{R}{\sqrt[n+1]{2}}$. – 311411 Jul 24 '21 at 17:27
  • The bound on the denominator is clear geometrically because $|Re^{it} -a|$ is the distance from $a$ to the point on the circle. This is always at least $R - |a|$, so by making $R$ bigger we can assume $$|Re^{it}-a| \geq R - |a| > R/2$$ – Square Feb 24 '22 at 17:35