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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?2012-05-30
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    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#3794652013-05-02

2 Answers 2

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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

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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 $\lfloor p \rfloor$.2012-05-30
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    @Robert Israel, thanks for this observation.2012-05-30
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    Could any one tell me why he took $lim_{R\rightarrow \infty}$ to get the $c_n$?2012-06-07
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    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.2012-06-07
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    @learner you are right I'll fix it2013-05-20
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    @learner functions holomorphic on $\mathbb{C}$2013-05-20
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    @Norbert thanks a lot. got it.2013-05-20
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    @learner, You are wellcome :)2013-05-20
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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 $a_1$.