On Fri, 06 Dec 2013 16:17:29 -0500, quasi <firstname.lastname@example.org> 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