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Let $f$ be an entire function such that $\mid f(z) \mid \leq A \mid z \mid$ for all $z$, where $A$ is a fixed number. Show that $f(z)=a_1z$, where $a_1$ is a complex constant.

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Can you use Cauchy's bound? – Bombyx mori Mar 11 '13 at 11:26
Yes. But I am stuck on it. – Manjil P. Saikia Mar 11 '13 at 11:27
More generally, if $f$ is entire and if there exists a polynomial $p$ such that $|f(z)|\leq P(|z|)$ for all $z$, then $f$ is a polynomial of degree not greater than $\deg p$. This follows form Liouville. But it is actually even easier to prove it directly (and therefore prove a stronger form of Liouville) by showing that the coefficients $a_k$ of the Taylor series of $f$ at, say, $0$, are zero for all $k>n$. It suffices to use the integral formula which gives $a_k$ as an integral of $f(z)/z^{k+1}$ over a circle of radius $R$ centered at $0$, and then let $R$ tend to $+\infty$. – 1015 Mar 11 '13 at 11:44
By the above, $f(z)=az+b$. Then $b=0$ since moreover $\lim_0 f=0$. – 1015 Mar 11 '13 at 11:45

Because $f$ is entire and $f(0)=0$, $\displaystyle f(z)= \sum\limits_{n=1}^{+\infty} c_nz^n$. Therefore, $\displaystyle \frac{f(z)}{z}= \sum\limits_{n=1}^{+ \infty} c_nz^{n-1}$ is entire and bounded, so constant according to Liouville's theorem.

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Can this be done by applying only Cauchy's bounds and Liuoville? – Manjil P. Saikia Mar 11 '13 at 11:38

If you want to apply Cauchy's bounds:

The theorem says if $\left|f(z)\right| \le M$ for all $z \in D(0, R)$, then: $$ \left|f^{(n)}(0)\right| \le \frac{n!M}{R^n} \quad n \in \Bbb N, n \ge 1 $$

Here, we have $\left|f(z)\right| \le AR$ for all $z \in D(0, R)$. Thus: $$ \left|f^{(n)}(0)\right| \le \frac{n!AR}{R^n} \quad n \in \Bbb N, n \ge 1 $$

The RHS goes to $0$ as $R\to\infty$ for $n > 1$. This forces $f^{(n)}(0)$ to be $0$ for all $n > 1$. Furthermore, $f(0) = 0$ is clear from the assumptions. We conclude that $f(z) = a_1 z$ for a constant $a_1$.

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