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.
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 $C^\infty$, 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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