On Tue, 03 Dec 2013 01:07:45 -0500, quasi <firstname.lastname@example.org> wrote:
>I think the following is true ... > >Proposition: > >For any closed subset A of R there exists an infinitely >differentiable function f:R -> R such that A = f^(1)(0).
Modulo the obvous typo, this is certainly true. Here's the proof that springs to my mind:
Lemma. If I is an open interval there exists an infinitely differentiable f such that f > 0 on I, f = 0 on R\I, and every derivative of f is bounded.
Proof. This follows from whatever your favorite oroof of the existence of infinitely differentiable functions with compact support it. QED.
If you're wondering why the last bit about every derivative being bounded is not automatic from the rest of the conclusion, the point is I could be unbounded. If, say, I = (0, infinity) then take an f so that f(x) = 0 for x <= 0, f(x) > 0 for x > 0, and f(x) = 1 for x > 1.
Now to prove your theorem. The complement of A is the union of countably many open intervals I_n. Choose f_n corresponding to I_n as in the lemma.
Now choose a_n > 0 with the following property:
(*) If g_n = a_n f_n and k <= n then
|D_k g_n(x)| < 1/2_n everywhere.
Here D_k is a convenient notation for the k-th derivative. You can find a_n making (*) true because each D_k f_n is bounded and for a given n, (*) refers to only finitely many values of k.
f = sum_n g_n.
For every k, the series
sum_n D_k converges uniformly (because |D_k g_n| < 1/2_n for all n > k). Hence f is infinitely differentiable. QED.
Note: Of course it's natural to take the I_n to be disjoint here. But disjointness is not needed in the proof. Hence the same proof works in R^d: The complement of any closed set A is the union of countable many balls...
> >Proof: > >If A = the empty set, let f(x) = 1 for all x, and we're done. > >Thus, assume A is nonempty. > >Let B = R\A. Then B is open so can be expressed as a countable >union of pairwise disjoint nonempty open intervals, say > > B = B_1 U B_2 U ... > >where the number of intervals in the union is either finite >or countably infinite. > >Let B_n = (s_n,t_n) where s_n is either real or -oo, t_n is >either real or +oo, and s_n < t_n. > >Let g_n: R -> R be an infinitely differentiable function such >that g_n = 0 on R\B_n (via a bump function construction). > >Let d be the usual distance function on R. > >For x in B_n, let d_n(x) = min(d(s_n,x),d(x,t_n)). > >Define f:R -> R by > > f(x) = 0 if x in A > > f(x) = (d_n(x))*(g_n(x)) if x in B_n > >Then f is infinitely differentiable and A = f^(-1)(0), >as required. > >Is my proof correct? > >If not, is the claim of the proposition true? > >If so, here's a followup question ... > >For an infinitely differentiable function f:R -> R, let f^(n) >denote the n'th derivative of f if n > 0 and f if n = 0. > >Let A_0, A_1, A_2, ... be closed subsets of R such that for all >s,t in A_n with s < t, the set A_(n+1) has nonempty intersection >with the open interval (s,t). > >Question: > >Must there exist an infinitely differentiable function f:R -> R >such that, for all n, A_n = (f^(n))^(-1)(0)? > >quasi