most recent 30 from http://mathoverflow.net 2013-05-19T11:59:56Z http://mathoverflow.net/feeds/question/37458 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/37458/converse-of-picards-big-theorem Henry Yuen 2010-09-02T05:00:20Z 2010-09-02T05:16:11Z

<p>The celebrated Big Theorem of Picard's is that, in every open set containing an essential singularity of a function $f(z)$, $f(z)$ takes on every value (except for at most one) of $\mathbb{C}$ infinitely often.</p> <p>Now - is the converse true? Is this a way to characterize the existence of an essential singularity of a function? </p> <p>For example, if you're given a non-constant function $f(z)$ that is holomorphic on some open set $\Omega$, and you know that there is an accumulation of 0's towards some point $x$ on the boundary of $\Omega$, then do you know that there must be an essential singularity at $x$? </p>

http://mathoverflow.net/questions/37458/converse-of-picards-big-theorem/37459#37459 Jonas Meyer 2010-09-02T05:16:11Z 2010-09-02T05:16:11Z

<p>If the boundary point is not an isolated singularity, then you can't say it is an essential singularity, so the answer to your final question is no. It is quite possible that $f$ cannot be extended to a larger domain that contains a punctured neighborhood of $x$. To be quasi-explicit, take a holomorphic function whose domain of holomorphy is $\Omega$ and multiply by a holomorphic function defined everywhere in the plane except with an essential singularity at $x$.</p> <p>If $f$ has an isolated singularity at $x$, then yes, this characterizes essential singularities. $f$ goes to $\infty$ at a pole, so in particular is nonzero in a neighborhood of the pole. If the singularity at $x$ is removable, then either $f$ goes to a nonzero value, in which case it is nonzero in a neighborhood of $x$, or $f$ goes to 0. In the latter case, $f$ could be extended to a holomorphic function on a domain containing $x$, so if $x$ were a limit point of the zero set of $f$, then $f$ would be identically zero. And you could easily adjust this to nonzero cases. </p> <p>You don't need anything near the strength of Picard's theorem. In fact, that is what is so amazing about Picard's theorem: Either $f$ has really nice behavior near an isolated singularity, or it maps everywhere except possibly one point in each neighborhood of the singularity.</p>