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This is a question from an exam and we've had it before in our homework. After trying the come up with a solution I was dissapointed to get my homework back with a big red cross. I hope my new solution is better.

Let $f:\mathbb{C}\to\mathbb{C}$ be an entire analytic function and let there be $C\geq 0,~p\in\mathbb{R}$ such that $|f(z)|\leq C(\log|z|)^p$ for all $z\in\mathbb{C}$ with $|z|\geq 2$. Show that $f$ is constant.

Because $f$ is entire, we can write: $f(z)=\sum_{n=0}^\infty a_nz^n$. Let $R>0$. Then because of Cauchy's inequality we have $$|f'(0)|\leq \frac{M_R}{R}\leq\frac{C(\log|R|)^p}{R}\to0$$ as $R\to\infty$. So $f'(0)=0$, so $f(z)=a_0z^0+0+0+0...=a_0$ is constant.

Is that right or am I missing something here?

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  • $\begingroup$ You need to show that all derivatives (except maybe the $0$-th) vanish at $0$, not only the first. Alternatively, you can show that the first derivative vanishes in a neighbourhood of $0$. $\endgroup$ – Daniel Fischer Jul 9 '13 at 19:21
  • $\begingroup$ Or what you can do is to show that $f(z)$ is bounded and then, use Cauchy Estimate and Liouville's Theorem to show that $f(z)$ is a constant (by making $R$ arbitrarily as large as possible). Remember that as Daniel said, $f$ needs to be constant for all $z$. $\endgroup$ – NasuSama Jul 9 '13 at 19:23
  • $\begingroup$ If $p \le 0$, then you automatically know that $f$ is bounded (since $(\log|z|)^p$ is bounded if $p \le 0$ and any entire function is bounded on a compact set). All you then need to consider is the case of $p > 0$. For this case, your proof just needs to be adjusted as per Daniel's statement. $\endgroup$ – Cameron Williams Jul 9 '13 at 19:26
  • $\begingroup$ Well I think that Cauchy's inequality states that $|f^{(m)}(0)|\leq \frac{m!C(\log|R|)^p}{R^m}$. So that every derivative vanish at $0$ for $m>0$ but goes to infinity for $m=0$. Is that right? $\endgroup$ – Ginger Jul 9 '13 at 19:28
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That's the idea, but you need more than $f'(0)=0$ to conclude that $f$ is constant. (I have just seen your last comment. So yes, you are right about this one. But there is no going to infinity for $m=0$).

By Cauchy's integral formula $$ f^{(n)}(0)=\frac{n!}{2i\pi}\int_{\gamma_R}\frac{f(z)}{z^{n+1}}dz\qquad \forall n\geq 0\quad\forall R>0 $$ where $\gamma_R$ denotes the circle of radius $R$ centred at $0$, counterclockwise, once around $0$.

With the assumption, this yields $$ |f^{(n)}(0)|\leq \frac{n!}{2\pi}\int_0^{2\pi}\frac{|f(Re^{it})|}{|R^{n+1}e^{i(n+1)t}|}|iRe^{it}|dt\leq \frac{n!}{2\pi}\int_0^{2\pi}\frac{C(\log R)^p}{R^n}dt=\frac{n!C(\log R)^p}{R^n}. $$ For every $n\geq 1$, this upper bound tends to $0$ as $R$ tends to $+\infty$. So $f^{(n)}(0)=0$ for every $n\geq 1$. Which means that the power series representation of $f$ is simply $$ f(z)=\sum_{n\geq 0}\frac{f^{(n)}(0)}{n!}z^n=f(0)\quad \forall z\in \mathbb{C}. $$

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  • $\begingroup$ Thanks a lot. I'm glad I understand it! $\endgroup$ – Ginger Jul 9 '13 at 19:34
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Here's another way using only the regular Liouville's Theorem.

For large $z$, $f(z) \le C(\log|z|^p) < z$. Look at $g(z) = \displaystyle \frac{f(z) - f(0)}{z}$. You see that $g(z)$ is also entire. But $g(z)$ is also bounded! So $g(z)$ is constant (Liouville's Theorem), say $g(z) = K$ (which might be zero...). Then $f(z) = Kz + f(0)$ is either a constant or a degree 1 polynomial. But it is not true that $Kz + f(0) \le C(\log|z|^p)$ for large $z$, unless $K$ is a constant.

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