quasi
Posts:
11,740
Registered:
7/15/05


zero sets of infinitely differentiable functions
Posted:
Dec 3, 2013 1:07 AM


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).
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

