# maybe maximum modulus principle $|f(z)| \leqslant 1 + |z|^{\frac{3} {2}} \forall z$

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.

• I can't close this question but it is exact duplicate of math.stackexchange.com/questions/151700/… – Norbert Oct 7 '12 at 20:53
• @Norbert I'm not too sure. Robert Israel's comment certainly answers the problem beautifully but I would technically consider this a separate question. – EuYu Oct 7 '12 at 20:58
• Ok here is another one question with general solution for this kind of problems math.stackexchange.com/questions/171610/… – 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 $+\infty$.
• 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 $0$. – EuYu Oct 7 '12 at 20:45
• @paulgarrett could you write the estimates and inequality for me please? I am having problem to get $f''(0)=0$ – Marso May 7 '13 at 3:12