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Let $B\subset \mathbb{R}^2$ be a closed set.

How to prove that there is a differentiable function $f:\mathbb{R}^2\longrightarrow\mathbb{R}$ such that $$Z(f)=B$$

where $$Z(f)=\{x\in\mathbb{R}^2:f(x)=0\}$$

Any hints would be appreciated.

Matemáticos Chibchas
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felipeuni
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    Use a bump function – Avi Steiner Dec 22 '13 at 04:03
  • Take a look at the first answer here http://math.stackexchange.com/questions/40557/a-nontrivial-everywhere-continuous-function-with-uncountably-many-roots . It might help. And this http://mathoverflow.net/a/24040 . – Poppy Dec 22 '13 at 05:05
  • Here is a cute solution in one dimension (it may be standard): Given $B$ closed, start with $g(x)=\mathrm{dist}(x,B)$, so $g(x)\ge 0$ for all $x$, $g(x)=0$ iff $x\in B$, and $g$ is continuous. Now let $f(x)=\int_0^x g(t),dt$, so $f$ is increasing, and $f'(x)=0$ iff $g(x)=0$ iff $x\in B$. – Andrés E. Caicedo Dec 25 '13 at 23:40
  • @AndresCaicedo You constructed a differentiable function such that the zero set of $f' $ is $B$ ... not the zero set of $f$. – Post No Bulls Dec 31 '13 at 06:58
  • @PostNoBills Ah, how silly! Oh, well. – Andrés E. Caicedo Dec 31 '13 at 06:59

3 Answers3

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I will adapt the MO answer by Harald Hanche-Olsen, filling in some details, and taking into account that you don't ask for $C^\infty$, but only for a differentiable function.

Let $$E_0=\{x\colon\operatorname{dist}(x,B)\ge 1 \}$$ and for $k=1,2,\ldots$ let $$E_k=\{x\colon 2^{-k}\le \operatorname{dist}(x,B)\le 2^{1-k}\}$$ These sets cover the complement of $B$. Also let $$F_k=\{x\colon \operatorname{dist}(x,E_k)\le 2^{-k-2}\}$$ be an "enlargement" of $E_k$ which still stays away from $B$.

Let $\omega$ be a smooth function on $\mathbb R^2$ such that $\omega\ge 0$, $\omega(x)=0$ when $|x|\ge 1/2$, and $\omega(x)>0$ when $|x|\le 1/4$. Let $\omega_k(x) = \omega( 2^{ k}x)$.

The convolution of $\chi_{F_k}$ with $\omega_k$ has the following properties:

  • it is as smooth as $\omega$ is
  • it is nonnegative
  • it is zero on the set $ \{x\colon \operatorname{dist}(x,F_k) > 2^{-k-1}\}$, which contains the set $ \{x\colon \operatorname{dist}(x,B ) < 2^{-k-2}\}$
  • it is strictly positive on $E_k$
  • it does not exceed $4^{-k}\int_{\mathbb R^n} \omega$.

Define
$$f=\sum_{k=0}^\infty (\chi_{F_k}*\omega_k) \tag{1}$$
and observe that

  • $f$ is strictly positive on the complement of $B$, and vanishes on $B$.
  • $f$ satisfies an estimate of the form $f(x)\le C(\operatorname{dist}(x,B ))^2$, 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 $C^\infty$ smooth, e.g., $$f=\sum_{k=0}^\infty 2^{-k^2} (\chi_{F_k}*\omega_k )$$

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Post No Bulls
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Maybe you could try using the fact that every open subset of the reals can be written as a countable union of disjoint open intervals.

EDIT: Sorry, misread the question, and I believe there's a simpler solution anyway. Can you think of a solution that produces a continuous function and modify it to be differentiable? What kind of function might be zero everywhere in a set and nonzero outside it?

GaussianInteger
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Would something like this work?

Let $B \subset \mathbb{R}^2$ be a closed set. Define a function $f$ $$ f(x) = \text{dist}(x,B)$$ where $$ \text{dist}(x,B) = \inf_{b\in B} d(x,b) $$

user44541
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