Let $f(z)$ be analytic in $\Bbb C$ with a bounded second derivative. I would like to prove that $f(z)=az^2+bz+c$ with constants $a,b$ and $c$. I know that if $f''$ is entire and bounded then $f''$ is constant. Because $f(z)$ is analytic in $\Bbb C$, I think $f''(z)$ should also be analytic in $\Bbb C$ [I am not entirely sure though]. So $f''$ is entire and bounded and $f''$ is constant, implying that $f(z)$ must take the form of $f(z)=az^2+bz+c$ with constants $a,b$ and $c$. Is this correct or am I missing something? Any clarification will be greatly appreciated.
Proving the form of an analytic function
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complex-analysis
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0You are perfectly correct. – 2017-01-22
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1You're correct. Regarding your uncertainty, if $f(z)$ is analytic then all of it's derivatives are also analytic, that is, it is infinitely differentiable. – 2017-01-22
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0More specifically, if $f$ is analytic then it is by definition representable by a power series in an neighborhood of any point. Its derivative can be computed by differentiating that power series term by term. The radius of convergence is the same for the differentiated power series (you can see this using the root test), so $f'$ is also analytic at any point. By induction this is true for all derivatives of $f$. – 2017-01-24