If $f:\mathbb{C}\to\mathbb{C}$ is an entire function and it holds that:
"For every $z\in \mathbb{C}$, either $|f'(z)|\leq1$ or $|f''(z)|\leq 1$."
Then there exist $a,b,c \in \mathbb{C}$ such that $2|a|\leq 1$ and $f(z)=az^2+bz+c$ .
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Possible duplicate of Suppose that $ f $ is entire and that for each $ z $, either $ |f(z)| \leq 1 $ or $ |f^\prime (z) |\leq 1 $. Prove that $ f $ is a linear polynomial. – Martin R Sep 02 '18 at 17:47
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Fix $z$ for a minute, and suppose $|f'(z)|>1$. Let $L$ be the line from $0$ to $z$. If $|f'|>1$ on all of $L$ then $|f''|\le 1$ on $L$, hence $$|f'(z)|\le|f'(0)|+|z|.$$If there exists a point $p$ on $L$ where $|f'|\le 1$, then let $q$ be the closest such point to $z$; now $|f''|\le1$ on the segment from $q$ to $z$, hence $$|f'(z)|\le|f'(q)|+|z-q|\le 1+|z|.$$
So we have $$|f'(z)|\le c+|z|$$ for all $z$ (we proved this assuming $|f'(z)|>1$, but it's certainly also true when $|f'(z)|\le1$). Hence $$|f'(z)|\le 2|z|\quad(|z|>c),$$so$$g(z)=\frac{f'(z)-f'(0)}z$$is a bounded entire function. (Because if $|z|>c$ then $$|g(z)|\le 2+\frac{|f'(0)|}{|z|}\le2+\frac{|f'(0)|}c,$$while $g(z)$ is bounded for $|z|\le c$ by compactness.)
David C. Ullrich
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1Hence, $g'(z) = 0$ for all $z$ which implies (quotient rule) $f''(z) = g(z) = \operatorname{const}$. – amsmath Sep 02 '18 at 17:23
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1@amsmath Right, but there's no need to get so fancy: $g=a$ says that $f'(z)=f'(0)+az$. – David C. Ullrich Sep 02 '18 at 17:27
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1Uh ah. Ouch. I think this is fancier. ;o) Sometimes I don't see the obvious. – amsmath Sep 02 '18 at 17:32