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.

  • $\begingroup$ I can't close this question but it is exact duplicate of math.stackexchange.com/questions/151700/… $\endgroup$ – Norbert Oct 7 '12 at 20:53
  • $\begingroup$ @Norbert I'm not too sure. Robert Israel's comment certainly answers the problem beautifully but I would technically consider this a separate question. $\endgroup$ – EuYu Oct 7 '12 at 20:58
  • $\begingroup$ Ok here is another one question with general solution for this kind of problems math.stackexchange.com/questions/171610/… $\endgroup$ – Norbert Oct 7 '12 at 22:14

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 $+\infty$.

  • $\begingroup$ 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 $0$. $\endgroup$ – EuYu Oct 7 '12 at 20:45
  • $\begingroup$ @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! :) $\endgroup$ – paul garrett Oct 7 '12 at 20:55
  • $\begingroup$ @paulgarrett could you write the estimates and inequality for me please? I am having problem to get $f''(0)=0$ $\endgroup$ – Marso May 7 '13 at 3:12

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