Dear Fred, Thank you for the clarification. Did any of my two (corrected) proofs make sense? I'm puzzled by the deafening silence. Despite my not understanding this new terminology of yours -- and I really wish you would write clearly -- my argument should be 100% relevant nonetheless, in both directions. Regards, Dan On Jun 3, 2014, at 3:41 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
"Intrinsic" ==
*********************** *** df/dx = f = 0 *** *********************** WFL
On 6/3/14, Dan Asimov <dasimov@earthlink.net> wrote:
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. ----- _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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