Fred wrote: ----- I take off my hat to a model of clarity and concision, making everything look easier than it is (as earlier discovered to my cost). However, I do not understand the necessity for C^2 --- why would not C^1 suffice? . . . ----- Andy had written: ----- nnnI 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. Since we can replace R by any smaller simply connected region that doesn't touch C, we may as well choose R to be diffeomorphic to a disc, and the theorem then becomes: For any sufficiently smooth (C2?) function f on the unit disc, if f is nonzero on the unit circle, but has zeroes inside the disc, then it has a critical point inside the disc. Proof: if f is negative on the unit circle, its maximum is a critical point; if f is positive on the circle, its minimum is a critical point. ----- Hmm, Why wouldn't C^0 (just continuous) work. In R^n, let W be a compact connected (n-1)-dimensional submanifold with connected boundary M = bd(W). Claim: ------ If a continuous function f: W -> R^1 (= the reals) for which 0 lies in f(W) but not in f(M), then f must have an absolute extremum in int(W) = W - M. Proof à la Latto: ----------------- Since f is continuous on the compact set W, it must have both an absolute minimum and an absolute maximum on W. Since f is nonzero on the connected set M, f must take M into either (0,oo) or (-oo,0). If the former, f must have a local minimum in int(W); if the latter f must have a local maximum in int(W). [ ] Corollary: ---------- If f is also differentiable on int(W), then it has a critical point there (since any local extremum x of a function differentiable in a neighborhood of x must have a critical point at x). --Dan Daniel Asimov Visiting Scholar Department of Mathematics University of California Berkeley, California