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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!

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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… – Greg Martin May 2 '13 at 17:31
up vote 18 down vote accepted

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

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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 $c_n$? – La Belle Noiseuse Jun 7 '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 7 '12 at 13:10
@learner you are right I'll fix it – Norbert May 20 '13 at 13:43

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 $a_1$.

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