From: "David C. Ullrich" <ullrich@math.okstate.edu>
Subject: Re: Analytic Bijections
Date: 11 Jun 2000 14:00:02 GMT
Newsgroups: sci.math
Summary: [missing]

	Finally recalled the thing I knew about this years ago -
it makes a lot more sense now that I've figured it out for 
myself.

	We know that if f' vanishes at a point then f is not 1-1.
Ie, if 0 is in the range of f' then f is not 1-1. The converse
is false, but:

	Thm. If f is analytic in a convex set and 0 is not in 
the _convex hull_ of the range of f' then f is 1-1.

	Pf. Suppose z, w are in the domain of f, z <> w, but 
f(z) =  f(w). Then

(f(z) - f(w)) / (z - w) = 0 .

But (f(z) - f(w)) / (z - w) is precisely the _average_ of f'
on the segment [w,z]. This says that 0 is in the closed
convex hull of f'([w,z]); since the convex hull of a 
compact subset of R^n is compact it follows that 0 is
in the convex hull of f'([w,z]), hence in the convex hull
of the range of f. QED

	This has some easy-to-use corollaries:

Cor 1. If f is defined in a convex set and Re(f') > 0 
      then f is 1-1.

(This is actually more or less equivalent to the 
theorem, since a convex set not containing the
origin is contained in some half-plane.)

Cor 2. If f is defined in the unit disk and c_n is the
coefficient of z^n in the power series, then

|c_1| > sum(n=2..infinity; n*|c_n|)

implies that f is 1-1.

Pf. WLOG c_1 > 0. Now the result follows from
Cor 1.

       Of course Cor 2 is "obvious", and it's probably
what you wanted. But the theorem seems interesting
per se (I saw a proof of Cor 1 years ago by a more
direct route - the "direct" proof made me wonder
what stronger results could be proved by more or
less the same method, while the theorem about
the convex hull of the range of the derivative makes
me think that the hypotheses may actually be "right".)

Dave

Stephen Montgomery-Smith <stephen@math.missouri.edu> wrote in article
<39406007.4021CB25@math.missouri.edu>...
> What are easy to use SUFFICIENT conditions on a polynomial
> p(z) that gurantees that it is a bijection from the unit
> disk to its image?  Something in terms of the coefficients
> of the polynomial (like some norm of the coefficients is
> bounded by the absolute value of the z coefficient) would
> be great.
> 
> Stephen
> 
> -- 
> Stephen Montgomery-Smith
> Department of Mathematics, University of Missouri, Columbia, MO 65211
> Phone 573-882-4540, fax 573-882-1869
> http://www.math.missouri.edu/~stephen  stephen@math.missouri.edu
>