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How can we prove that if $f:\mathbb{C}\rightarrow\mathbb{C}$ is holomorphic (analytic) and $|f(z)| \leq 1+|z|^{1/2} \forall z$, then $f$ is constant?

Liouville's theorem springs to mind, but I can't see how to use it since $1+|z|^{1/2}$ is not holomorphic. The maximum modulus principle doesn't seem easily usable either. And the principle of isolated zeroes can't really be applied since all we know is an inequality, not an equation.

Many thanks for any help with this!

Harry Macpherson
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    Maybe consider what kind of singularity is at infinity? – GEdgar May 30 '12 at 19:05
  • This is a special case of a more general theorem, which I address in my answer to http://math.stackexchange.com/questions/213491/show-that-an-entire-function-bounded-by-z10-3-is-cubic/379465#379465 – Greg Martin May 02 '13 at 17:31

2 Answers2

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Since $f\in\mathcal{O}(\mathbb{C})$, then $$ f(z)=\sum\limits_{n=0}^\infty c_n z^n $$ for all $z\in \mathbb{C}$. Moreover, for all $R>0$ we have integral representation for coefficients $$ c_n=\frac{1}{2\pi i}\int\limits_{\partial B(0,R)}\frac{f(z)}{z^{n+1}}dz $$ Then, we get an estiamtion $$ |c_n|\leq\frac{1}{2\pi} \oint\limits_{\partial B(0,R)}\frac{|f(z)|}{|z|^{n+1}}|dz|\leq \frac{1}{2\pi}\oint\limits_{\partial B(0,R)}\frac{1+|z|^{1/2}}{|z|^{n+1}}|dz|= \frac{1}{2\pi}\frac{2\pi R(1+R^{1/2})}{R^{n+1}} $$ Hence for $n\in\mathbb{N}$ we obtain $$ |c_n|\leq\frac{1}{2\pi}\lim\limits_{R\to+\infty}\frac{2\pi R(1+R^{1/2})}{R^{n+1}}=0 $$ which implies $c_n=0$. Finally we get $$ f(z)=c_0+\sum\limits_{n=1}^\infty c_n z^n=c_0=\mathrm{const} $$ Here you can find generalized version of this answer

Community
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Norbert
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    And more generally, if the entire function $f$ satisfies $|f(z)| = O(|z|^p)$ as $|z| \to \infty$ for some real number $p$, this argument shows that $c_n = 0$ for all $n > p$, so $f$ is a polynomial of degree at most $\lfloor p \rfloor$. – Robert Israel May 30 '12 at 19:22
  • @Robert Israel, thanks for this observation. – Norbert May 30 '12 at 19:27
  • Could any one tell me why he took $lim_{R\rightarrow \infty}$ to get the $c_n$? – Myshkin Jun 07 '12 at 12:15
  • We take this limit because it is zero, and this gives us that $c_n=0$. If this limit will be greater than zero, the only thing I will get is an estimation on the absolute value of $c_n$. In fact my approach is a slick trick. – Norbert Jun 07 '12 at 13:10
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    @learner you are right I'll fix it – Norbert May 20 '13 at 13:43
  • @learner functions holomorphic on $\mathbb{C}$ – Norbert May 20 '13 at 14:26
  • @Norbert thanks a lot. got it. – learner May 20 '13 at 14:28
  • @learner, You are wellcome :) – Norbert May 20 '13 at 14:29
  • @Norbert I know this was a long time ago but hopefully you're still around... How did you go from $\oint \frac{1+|z|^{½}}{|z|^{n+1}}$ to the next statement with the $R$'s? – BalsamicVinegar Oct 19 '20 at 00:32
  • @BalsamicVinegar, since you are integrating over the circle $\partial B(0,R)$ you can use parametrisation $z=R e^{it}$ with $t\in[0,2\pi]$ to reduce this integral to the ordinary Rienmann integral. – Norbert Oct 19 '20 at 12:25
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A slightly different way: $|z f(1/z)| \leq |z| + |z|^{1/2}$ for $z \neq 0$ so $z f(1/z)$ extends to an entire function $\sum_{k \geq 1} a_k z^k$ by Riemann's extension theorem. Then $f(z) = \sum_{k \geq 1} a_kz^{1-k}$. This implies that all coefficients $a_k$ vanish except possibly $a_1$.

WimC
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