Suppose the function is f(x, y) = x^2 + y^2 ; then the maximum over the origin projects into the unit disc, so the 3-space method correctly rejects it as a zero-free region. WFL On 5/30/14, Allan Wechsler <acwacw@gmail.com> wrote:
My vague memory from algebraic geometry in college is that theorems of this sort start being true more reliably in *complex, projective* space. R being compact and simply-connected is still skewered by Warren's example, replacing the annulus with a solid disk. But in complex space, the circle has parts that run out to infinity, so it cuts the disk boundary somewhere.
On Thu, May 29, 2014 at 7:15 PM, Andy Latto <andy.latto@pobox.com> wrote:
Suppose f(x,y) = y, so C is the x-axis, there are no critical points, and R is given by
(x,y) is in R when |y| < 1.
Maybe if R is compact and simply connected?
On Thu, May 29, 2014 at 4:46 PM, David Makin <makinmagic@tiscali.co.uk> wrote:
Was the region R meant to be restricted to a single continuous area without holes (i.e. topologically a solid circle but maybe extending to infinity) ?
On 29 May 2014, at 20:17, Warren D Smith wrote:
Lunnon: Given a plane curve C defined by f = 0 with f(x, y) polynomial, and a region R of the plane (possibly extending to infinity), I assert that C avoids R provided C avoids the boundary of R ; and C has no critical points within R .
There must be a well-known theorem to this effect (unless, of course, it's actually false...
WDS: a counterexample: C=unit circle, R = thickened version of unit circle (annulus).
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