6
$\begingroup$

Let $f$ be an entire analytic function which attains all but $k$ complex numbers $z_1,\ldots,z_k$. Is there any elementary proof, for some $k$, that $f$ is constant?

$\endgroup$
1
  • $\begingroup$ why do you wish to know this? $\endgroup$ Commented Jun 26, 2015 at 22:50

3 Answers 3

19
$\begingroup$

There are very many different proofs of Picard's theorem, and some of them are really "simple". (Picard's original proof occupies about 2 lines, using the things already known at that time). None of those proofs that I know simplifies if you assume that the function omits 4, 10, or any finite or countable set of points.

However, if it omits a larger set, for example an arc, then one can give a "simpler" proof, in the sense that less knowledge is required: map the complement of this arc onto a bounded region, this can be done by an elementary function, and then apply Liouville's theorem.

Remark. Let me add that the question whether there exists a "simpler" proof was asked immediately after Picard's proof was published, and it led to a very substantial progress in several areas of mathematics. People were looking not for a shorter proof but for a "more elementary" proof, using less prerequisite. Many such proofs were found and almost each of them led to a substantial generalization of Picard's theorem. However some very natural generalizations are still unproved. For example, it is conjectured that a holomorphic map from $C$ to $P^2$, the projective space of dimension $2$, omitting a generic curve of degree $5$ must be constant.

$\endgroup$
2
  • $\begingroup$ Alexandre, could you indicate a reference for the conjecture you mention at the end of the answer, and work thereon? Thanks. $\endgroup$
    – user41593
    Commented Jun 27, 2015 at 10:22
  • 3
    $\begingroup$ This is sometimes called the Kobayashi-Zaidenerg Conjecture, based on the paper of Zaidenberg MR0904640. $\endgroup$ Commented Jun 28, 2015 at 9:33
19
$\begingroup$

Suppose you knew that if $f$ is entire and nonconstant, then $f$ attains all but finitely many values. Then I claim Picard's theorem follows by a completely elementary argument. To show this, suppose $f$ is a nonconstant entire function that vanishes nowhere. Then we can write $f=e^g$ for some entire function $g$. By hypothesis, $g$ must attain all but finitely many values. It follows that $f$ must attain all values except $0$.

That is, if there were such an elementary proof that you ask for, you would get an equally elementary proof of Picard's theorem.

$\endgroup$
0
$\begingroup$

From my personal view, I recommend the paper written by Professor Zhang Guangyuan, the proof relies some fact that is called area-length equality. http://blms.oxfordjournals.org/content/34/2/205.short

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .