Friday, August 23, 2019

real analysis - Find a differentiable f such that mathrmZeros(f)=mathbbany;closed;set



Let BR2 be a closed set.



How to prove that there is a differentiable function f:R2R such that Z(f)=B



where Z(f)={xR2: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:2kdist(x,B)21k}
These sets cover the complement of B. Also let
Fk={x:dist(x,Ek)2k2}

be an "enlargement" of Ek which still stays away from B.



Let ω be a smooth function on R2 such that ω0, 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)>2k1}, which contains the set {x:dist(x,B)<2k2}


  • it is strictly positive on Ek

  • it does not exceed 4kRnω.



Define
f=k=0(χFkωk)
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 2k from B
    it takes values about 4k


  • 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=02k2(χFkωk)


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