Let B⊂R2 be a closed set.
How to prove that there is a differentiable function f:R2⟶R such that Z(f)=B
where Z(f)={x∈R2:f(x)=0}
Any hints would be appreciated.
Answer
I will adapt the MO answer by Harald Hanche-Olsen, filling in some details, and taking into account that you don't ask for C∞, but only for a differentiable function.
Let
E0={x:dist(x,B)≥1}
and for k=1,2,… let
Ek={x:2−k≤dist(x,B)≤21−k}
These sets cover the complement of B. Also let
Fk={x:dist(x,Ek)≤2−k−2}
be an "enlargement" of Ek which still stays away from B.
Let ω be a smooth function on R2 such that ω≥0, ω(x)=0 when |x|≥1/2, and ω(x)>0 when |x|≤1/4. Let ωk(x)=ω(2kx).
The convolution of χFk with ωk has the following properties:
- it is as smooth as ω is
- it is nonnegative
- it is zero on the set {x:dist(x,Fk)>2−k−1}, which contains the set {x:dist(x,B)<2−k−2}
- it is strictly positive on Ek
- it does not exceed 4−k∫Rnω.
Define
f=∞∑k=0(χFk∗ωk)
and observe that
- f is strictly positive on the complement of B, and vanishes on B.
- f satisfies an estimate of the form f(x)≤C(dist(x,B))2, because at distance about 2−k from B
it takes values about 4−k - By the above, f is differentiable on B.
- f is also differentiable on the complement of B, because every point of this complement has a neighborhood in
which only finitely many terms of (1) are nonzero.
As Harald Hanche-Olsen notes, introducing a rapidly decaying weight one can make the sum C∞ smooth, e.g.,
f=∞∑k=02−k2(χFk∗ωk)
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