Bounded spherical derivative implies finite order - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T02:21:36Z http://mathoverflow.net/feeds/question/79265 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/79265/bounded-spherical-derivative-implies-finite-order Bounded spherical derivative implies finite order Malik Younsi 2011-10-27T13:48:56Z 2012-08-03T19:00:16Z <p>Hi,</p> <p>Let $f$ be an entire function. The <em>spherical derivative</em> $\rho(f)$ is defined by $$\rho(f)(z):= \frac{|f'(z)|}{1+|f(z)|^2}.$$</p> <p>A result from Clunie and Hayman states that if $\rho(f)$ is bounded, then $f$ is of exponential type. The proof uses the machinery of Nevanlinna's theory of value distribution.</p> <p>My question is the following :</p> <p>Is there an elementary proof that if $\rho(f)$ is bounded, then $f$ is of <strong>finite order</strong>?</p> <p>(Note that this is a weaker result, since I'm only asking for finite order here). Finite order means that there exists constants $K$ and $\alpha$ such that $$|f(z)| \leq Ke^{|z|^\alpha}$$ for all $z$.</p> <p><strong>Motivation :</strong> I'm interested in this because it would lead to a quick proof of Picard's little theorem. Indeed, if there exists a non-constant entire function which omits $0$ and $1$, then it is possible to obtain (using normal families techniques) a non-constant entire function $f$ which omits $0$ and $1$ and that has <em>bounded</em> spherical derivative. Write $f = e^g$ for some entire function $g$. Since $f$ is of finite order, $g$ is a polynomial. But $f$ does not take the value $1$, so $g$ must be constant, a contradiction.</p> <p>Any reference is welcome. Thank you, Malik.</p> <p>EDIT I asked the <a href="http://math.stackexchange.com/questions/77886/bounded-spherical-derivative-implies-finite-order/78182#78182" rel="nofollow">question </a> on math.stackexchange.</p> http://mathoverflow.net/questions/79265/bounded-spherical-derivative-implies-finite-order/103891#103891 Answer by Alexandre Eremenko for Bounded spherical derivative implies finite order Alexandre Eremenko 2012-08-03T19:00:16Z 2012-08-03T19:00:16Z <p>Yes, of course. A bound on spherical derivative immediately gives T(r)=O(r^2) where T is the Nevanlinna characteristic. And that finite order of T implies finite order of f is proved in the beginning pages of any book on Nevanlinna theory.</p> <p>BTW. Your idea on a simple proof of Picard theorem is not new. A proof based on this idea was published by Zalcman many years ago.</p>