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Topic: zero sets of infinitely differentiable functions
Replies: 31   Last Post: Dec 9, 2013 4:14 PM

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 David C. Ullrich Posts: 3,555 Registered: 12/13/04
Re: zero sets of infinitely differentiable functions
Posted: Dec 7, 2013 12:05 PM
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On Fri, 06 Dec 2013 16:17:29 -0500, quasi <quasi@null.set> wrote:

>quasi wrote:
>>dullrich wrote:
>>>quasi wrote:
>>>

>>>>Let's try a special case ...
>>>>
>>>>Question:
>>>>
>>>>If P,Q are closed, nowhere dense subsets of R such that between
>>>>any two distinct elements of P there is an element of Q, must
>>>>there exist a differentiable function f:R -> R such that
>>>>f^(-1)(0) = P and (f')^(-1)(0) = Q?

>>>
>>>Surely yes. In fact I'd be willing to bet that this follows
>>>easily from what's been proved already, by something
>>>sort of like the following:
>>>
>>>Start with g such that g = 0 on P and g > 0 on R\P.
>>>Now if h tends to infinity fast enough at +- infinty
>>>then the function hg satisfies hg = 0 on P, hg > 0
>>>on R\P, and hg > 1 on Q.

>>
>>I don't follow the claim "hg > 1 on Q".
>>
>>Perhaps you misread the specification for Q?

>
>For example, P,Q might have nonempty intersection,

I misread your question, sorry. I thought you were
asking about f^(-1)(0) = P and (f)^(-1)(1) = Q, in
which case you must have been assuming they were disjoint.
Oh, you actually asked about f^(-1)(0) = P and (f')^(-1)(0) = Q?
That's different...

I suspect the answer is yes, but nothing springs to
mind for a proof.

>in which case
>
> "hg > 1 on Q"
>
>will definitely fail, regardless of the choice of h.
>
>In fact, if we also require f to be continuously differentiable,
>any limit point of P _must_ be an element of Q.
>

>>>So if you choose phi appropriately then the
>>>function psi = phi(gh) satisfies psi = 0 on P,
>>>psi > 0 on R\P, 0 <= psi <= 1 everywhere,
>>>and psi = 1 on Q.
>>>
>>>Which doesn't quite answer your question,

>
>Right, it doesn't quite work since we also have to worry about
>relative minima.
>

>>>but it's an example of the sort of jiggling with the
>>>previous result I have in mind ... go for it.

>
>Yes, I get the idea.
>
>Thanks -- I'll play with it when I get a chance.
>

>>Remark:
>>
>>If the answer is "no", then the answer to my previously posted
>>more general question is also "no".

>
>quasi

Date Subject Author
12/3/13 quasi
12/3/13 Virgil
12/3/13 quasi
12/3/13 quasi
12/3/13 quasi
12/3/13 quasi
12/3/13 quasi
12/9/13 quasi
12/3/13 David C. Ullrich
12/3/13 quasi
12/4/13 quasi
12/6/13 David C. Ullrich
12/6/13 quasi
12/6/13 quasi
12/6/13 quasi
12/7/13 quasi
12/7/13 David C. Ullrich
12/7/13 quasi
12/8/13 David C. Ullrich
12/8/13 quasi
12/8/13 quasi
12/8/13 quasi
12/8/13 quasi
12/8/13 quasi
12/8/13 quasi
12/8/13 quasi
12/9/13 quasi
12/9/13 David C. Ullrich
12/6/13 quasi
12/6/13 David C. Ullrich
12/9/13 quasi
12/9/13 quasi

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