Friday, August 23, 2019

real analysis - Find a differentiable $f$ such that $mathrm{Zeros}(f)=mathbb{any; closed ;set}$



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.,
$$f=\sum_{k=0}^\infty 2^{-k^2} (\chi_{F_k}*\omega_k )$$


No comments:

Post a Comment

analysis - Injection, making bijection

I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...