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Let $f$ be an entire function such that : $$ |f(z)| \leqslant 1 + |z|^{\frac{3} {2}} \forall z $$

What we can conclude about $f$ . Sorry for asking this , but I want to see some examples of the contents of the chapter that I'm reading, this problem it's from the chapter of maximum modulus principle.

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    I can't close this question but it is exact duplicate of http://math.stackexchange.com/questions/151700/if-f-is-entire-and-fz-leq-1z1-2-why-must-f-be-constant/151703#1517032012-10-07
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    @Norbert I'm not too sure. Robert Israel's comment certainly answers the problem beautifully but I would technically consider this a separate question.2012-10-07
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    Ok here is another one question with general solution for this kind of problems http://math.stackexchange.com/questions/171610/an-entire-function-is-identically-zero/1716152012-10-07

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Edit: oop... misread... revised: The given inequality and Cauchy's formula for the second derivative $f''$, letting the large circle go to infinity, show that $f''(z)=0$, so $f$ is (not constant, but) linear. This is just a little extension of the argument for Liouville's theorem, so not really so much about maximum modulus, perhaps.

Edit-edit: explicitly, by the Cauchy integral formula for the derivatives, $f''(z)={2!\over 2\pi i}\int_\gamma {f(\zeta)\,d\zeta\over (\zeta-z)^3}$, where $\gamma$ is a large circle of radius $R$. The numerator is bounded by $R^{3/2}$, and the denominator is essentially $R^3$. The length of the curve is $2\pi R$, so the integral expressing the second derivative is bounded by a constant multiple of $1/R^{1/2}$, which goes to $0$ as $R$ goes to $+\infty$.

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    I don't think that can be the full argument. If we only use the asymptotic behavior then $f(z) = z$ satisfies the inequality but the derivative isn't $0$.2012-10-07
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    @EuYu ... thanks! I was too hasty in reading. The same sort of application of Cauchy's estimate proves not that it's _constant_, but _linear_, because the _second_ derivative is 0. Sorry not to have read more carefully, and thanks for the correction! :)2012-10-07
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    @paulgarrett could you write the estimates and inequality for me please? I am having problem to get $f''(0)=0$2013-05-07