Re: -- strict local mimima and level curves
- From: quasi <quasi@xxxxxxxx>
- Date: Sun, 01 Feb 2009 20:10:08 -0500
On Sun, 01 Feb 2009 19:42:31 -0500, quasi <quasi@xxxxxxxx> wrote:
On Sun, 01 Feb 2009 19:33:20 -0500, quasi <quasi@xxxxxxxx> wrote:
On Sun, 01 Feb 2009 19:14:32 -0500, quasi <quasi@xxxxxxxx> wrote:
On Sun, 01 Feb 2009 15:42:58 -0800, W^3 <aderamey.addw@xxxxxxxxxxx>
wrote:
In article <ldnbo4hpjmnt1teruh5keajlm4g9hsletg@xxxxxxx>,
quasi <quasi@xxxxxxxx> wrote:
On Sun, 01 Feb 2009 11:44:35 -0500, quasi <quasi@xxxxxxxx> wrote:
Prove or disprove:
If f : R^2 --> R is continuous and if f has a strict local minimum at
(0,0) then there exists a simple closed curve S containing (0,0) such
that f is constant on S.
Of course, when I said
"a simple closed curve S containing (0,0)"
I meant
a simple closed curve S such that (0,0)
is in the region interior to S
quasi
For a > 0 let P_a denote the parabola {(x, x^2/a - a) : x in R}. As a
-> 0+, these parabolas (which are pairwise disjoint) move up, rise
faster and fold towards the nonnegative y-axis. We have U_{a>0} P_a =
R^2 minus the nonnegative y-axis. So define f : R^2 -> R by setting f
= a on P_a, f(0,y) = y for y >= 0. Then f is continuous and none of
the level sets of f is a simple closed curve.
Is f continuous on the positive y-axis?
I can see it's continuous everywhere else.
Assuming my objection is valid, I think the following minor
modification will fix it.
Define f : R^2 -> R by
f(x,y) = (a, a + |y|) for (x,y) on P_a
f(0,y) = y if y >= 0
I think it works now.
No, it doesn't work -- tthere now are simple closed curves
containing the origin
on which f is constant.
In fact, for every positive constant c, the level curve f = c is a
simple closed curve.
In trying to fix it, I made it worse -- sorry.
quasi
.
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