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I have to classify entire functions satisfying $|f(z)|\leq (1+|z|)^2$ for all $z\in \mathbb{C}$.

Using Cauchy integral's formula, I've shown that $f^{(3)}=0$. Thus $f(z)=a+bz+cz^2$ for some $a,b,c \in \mathbb{C}$.

Now, I want to show that $|a+bz+cz^2|\leq 1+2|z|+|z|^2$ iff $|a|,|c|\leq 1$ and $|b|\leq 2$. This would end the "classification".

$\Leftarrow$ is easy.

For $\Rightarrow$, evaluating at $z=0$ yields $|a|\leq 1$ and dividing by $z^2$ and evaluating at $|z|\rightarrow \infty$ yields $|c|\leq 1$.

But for $|b|\leq 2$, I have no idea how to start. Is it even true ?

Nitrogen
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1 Answers1

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This is false. For example, you can take $f(z) = 3z$.

Notice that $ |a| \leq 1, |b| > 2$ and $|c| \leq 1$ but $|f(z)| \leq (1 + |z|)^2$.

Indeed, $|f(z)| \leq (1 + |z|)^2 \Leftrightarrow 0 \leq 1 - |z| + |z|^2$. Why is this true ? Simply because the real function $g(t) = 1 - t + t^2$ is always positive, since it take its minimum at $t_0 = \frac{1}{2}$ and $g(\frac{1}{2}) = 0.75$, showing $g(t)$ is positive everywhere.