On Sat, May 31, 2014 at 4:25 PM, Warren D Smith <warren.wds@gmail.com> wrote:
Andy Latto: I think this is really a theorem of differential geometry, not of algebraic geometry; the only place Warren's proof uses the fact that f is polynomial is when he needs it to be sufficiently smooth.
--WDS: not so.
I gave a proof. I still think it's true.If f is a C2 function on a simply connected compact manifold with boundary R, and f is nonzero on the boundary of R, but has zeroes in R, then f has a critical point in R.
I also used the fact the zero set of a polynomial, restricted to a compact region, has finite measure, e.g. cannot be an infinite spiral.
and I have therefore proved that is such a curve is the zero set of a smooth function f, then f will have a critical point in the closure of its zero set. To see this intuitively, take a sequence of points, one from each loop of the spiral, that converge. The derivative at that point is zero in the direction tangent to the spiral (because it's zero along the spiral) and in the direction along the sequence of points (because it's zero at all of those points).
One can easily construct an INFINITELY differentiable (i.e way smooth) function f(x,y) such that f(x,y)=0 is a spiral curve that within some annulus -- or disc -- makes an infinite number of winds.
f will also be zero on the limiting closed curve (or single point) that the spiral approaches, and every point of this limit cycle will be a critical point.
The function f will be 0 on the spiral, and in between windings will have a ridge or valley, but the ridge always has a slight slope, approaching but never quite becoming horizontal. As a result there are no critical points of f within the annulus.
There will be critical points in the closed annulus.
--a counterexample to that "theorem" and "proof" is the function f(x,y)=x? Oh, by "nonzero" you meant "never zero."
Yes. The hypothesis was always "C avoids the boundary of R", which means "f is never zero on the boundary of R"
Aha. Right, this is now trivial.
I'm a little confused. My statement is false, *and* you have a counterexample, *and* it's trivial?
Well, I think the final last winding of the spiral, which never comes, is a circular ring of critical points, which lie on the boundary of the region, but not inside it, and thus technically yield a counterexample, at least to some wordings of the theorem.
If by "technically a counterexample" and "some wordings of the theorem", you mean "oops! My counterexample is nonexistent, and when I said "Not so" to Andy I was completely wrong", then I agree with you. The theorem was that if f is sufficiently smooth, R is a compact region with connected boundary (such as a simply connected region), R contains a zero of f, and f is nonzero on the boundary of R, then R contains a critical point of f. What is the "other wording" of the theorem that is not true? Andy
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