Arrrgh. Yet again I'm forced to post a correction: I omitted the last line, which should read: "And f is nonzero on int(R) as well." --Dan On Jun 2, 2014, at 6:15 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Isn't there an easy counterexample if the boundary of R is not connected?
Let R be the (compact) annulus 1 <= |(x,y)| <= 3, with f(x,y) := x^2 + y^2 - 4.
Then f is nonzero on bd(R) since f(bd(R)) = {-3,5}.
f(x,y) = 0 on the circle in R where |(x,y)| = 2.
But f has no critical point in R (i.e., no point where df/dx = df/dy = 0).
Or am I misunderstanding the meaning of "intrinsic critical point (x,y)" ?
* * *
Conversely, this time let R = the unit disk D^2, and let f(x,y) = x^2 + y^2 + 1.
Then f is clearly nonzero on bd(R), but has a critical point in int(R).
----- Theorem: Given a continuously differentiable function f(x, y) , and a compact region R of the plane ( |R^2 ) with f nonzero on the boundary of R : f has a zero within R if and only if there is some (intrinsic critical) point (x, y) in R where *** df/dx = f = 0. -----