Re: SMOOTH URYSOHN'S LEMMA
- From: quasi <quasi@xxxxxxxx>
- Date: Tue, 31 Jul 2007 22:24:10 -0400
On Tue, 31 Jul 2007 22:01:17 -0400, quasi <quasi@xxxxxxxx> wrote:
On Tue, 31 Jul 2007 21:02:16 -0400, quasi <quasi@xxxxxxxx> wrote:
On Tue, 31 Jul 2007 20:02:22 EDT, craigt <ctcowan1972@xxxxxxxxx>
wrote:
On Mon, 30 Jul 2007 21:08:05 -0700, bigli
<nimbigli@xxxxxxxxx> wrote:
Let K and F be disjoint closed subsets of theEuclidean space R^n,
then is there a smooth (C^infinite) function f: R^n---> [0,1] such that
f(K) = {1} and f(F) ={0} ?
Yes, in fact, here's an explicit function ...
Define f:R^n->[0,1] by
f(p) = d(p,F) / (d(p,K) + d(p,F))
This ain't smooth.
Agreed.
But the fix is easy (I think).
Just square the distances.
In other words,
define f:R^n->[0,1] by
f(p) = d(p,F)^2 / (d(p,K)^2 + d(p,F)^2)
where d(p,F) and d(p,K) denote the Euclidean distances from p to the
sets F and K, respectively.
Doesn't this work?
To answer my own question, no it doesn't always work.
I think f will always be differentiable, although I'm not positive
about even that.
However f is not necessarily infinitely differentiable.
For a counterexample in R^2, let F={(-1,0)} and K={(1,0)}.
Then the function f, as defined above, can be expressed as a piecewise
function as follows ...
f(x,y) =
(x+1)^2 + y^2, if x<1
y^2, if -1<=x<=1
(x-1)^2 + y^2, if x>1
Whoops -- I don't know what I was thinking.
That formula is nowhere close to correct -- sorry.
There's no need for piecewise -- that was an hallucination on my part.
The correct f is simply
f(x,y) = ((x+1)^2 + y^2) / ((x+1)^2 + (x-1)^2 + 2*y^2)
which is clearly infinitely differentiable.
So maybe the functions f, based on squared distances, will always
work?
No, I doubt it.
It was the fact that F and K were singletons that made it so easy. If,
for example, we let F be a V shape, then f will probably fail to be
infinitely differentiable (for any nonempty closed set K disjoint from
F).
quasi
.
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