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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 3, 2013 11:03 AM

On Tue, 03 Dec 2013 01:07:45 -0500, quasi <quasi@null.set> 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.

Now let

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

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