Let $( \mathbb{R}^n, \| \cdot \|\_P)$ be the $n$-dimensional Euclidean space equipped with $\ell\_p$-norm $\| \cdot \|\_p$ for some $p\in [1, + \infty]$. Let $A$ be a convex set in $\mathbb{R}^n$ and define 
\begin{align}
A^{\epsilon} = \{ y \in \mathbb{R}^n \colon \exists x \in A~\text{such that}~\| x -y \| \_{p} \leq \epsilon \},
\end{align}
where $\epsilon >0$ is a real number. In general, what is the condition that there exists a $3$-order differentiable function $f \colon \mathbb{R}^n \rightarrow [0,1]$ such that $f$ is a good approximation of $\mathbb{1}\_{A}$, which is the indicator function of set $A$. 


In specific, it is ideal that $f(x) = 1$ when $x\in A$ and $f(x) = 0$ when $x\in \mathbb{R}^n \setminus A^{\epsilon}$. Moreover, the $\ell\_q$-norm of the $i$-th order gradient of $f$ is proportional to $\epsilon^{-i}$ for $i = 1,2,3$. That is
\begin{align}
\| D^{(i)} f (x) \|\_{q} \leq C\_i \cdot \epsilon^{-i}, ~\text{for any}~~ i = 1,2,3.
\end{align}
Here the high-order gradients are taken as vectors, $1/q + 1/p = 1$, and $\| \cdot \|\_q$ is the dual-norm of $\| \cdot \|\_p$.


An inspiration of this problem is that for $p = q= 2$, the problem is solved for any convex sets. In addition, for rectangles of the form $\{ x \in \mathbb{R}^n \colon a\_i \leq x\_i\leq b\_i, i =1, \ldots, n\}$, approximation under $\ell\_{\infty}$ is also established. This approximation depends on the function $g(x） = 1/\rho \cdot \log [\sum\_{i=1}^n \exp( \rho\cdot x\_j)]$ which approximates the function that returns the maximum element of a vector in $\mathbb{R}%d$. See <https://arxiv.org/abs/1412.3661v4>. But general convex sets in $(\mathbb{R}^n, \| \cdot \|\_{\infty})$ is not covered.


Moreover, for Banach spaces, similar results are also established for $\| \cdot \|\_{B}$, which is the norm in the Banach space. However, such result depends on the differentiability of $\| \cdot \|\_B$, which does not handle convex sets in $(\mathbb{R}^n, \| \cdot \|\_{\infty})$.



Smooth Approximation of Indicator Function of Convex Sets in $\mathbb{R}^n$
Let's pursue Jochen's idea. We assume $A \ne \emptyset.$ 


Let 
$$ \varphi(t) = \begin{cases} 
 e^{-\frac{1}{t}} &\text{if $ t>0$}\\
 0 &\text{otherwise.}
\end{cases}$$ 


This function is $\mathcal C^{\infty}$, and $0 < \varphi(t)$ iff $0<t.$ 


Define $\rho$ as 
$$\rho(x) := k \varphi(1- \|x\|\_2^2)$$ 
where $k$ is chosen so that $\int \rho(x)\space dx =1.$
$\rho$ is $\mathcal C^{\infty}$, $0 \le \rho$, and $0< \rho(x)$ iff $\|x\|\_2^2 < 1.$ 


Let $\delta > 0$. Set $\rho\_{\delta}(x) = \frac{1}{\delta^n} \rho(\frac{x}{\delta})$. Note that $0<\rho\_{\delta}(x)$ iff $ \|x\|\_p < \delta$, and $\int \rho\_\delta = 1.$ 


Let $\mathscr O = A+B\_2(0,\delta)$, and consider the function  

$$ f = I\_{\mathscr O} \ast \rho\_{\delta} .$$ 


Clearly $0 \le f \le 1$ and $f$ is $\mathcal C^{\infty}.$


If $a \in A$, then $B\_2(a,\delta) \subset A+B\_2(0,\delta) = \mathscr O $, so
$ I\_{\mathscr O } \ast \rho\_{\delta}(a) = 1.$ 


If $x \notin A+B\_2(0,\delta)$ then 
$B\_2(x,\delta) \cap (A + B\_2(x,\delta)) = emptyset$, 
so 
$ I\_{\mathscr O } \ast \rho\_{\delta}(x) = 0.$


For derivatives of any order:  

$$ D^\nu f(x) = 
D^\nu \int\_{\mathscr O} \frac{1}{\delta^n} \rho(\frac{x-y}{\delta}) dy = \frac{1}{\delta ^{|\nu|}} \int\_{\mathscr O} \frac{1}{\delta^n} (D^\nu\rho)(\frac{x-y}{\delta}) dy
$$
so
$$ |D^\nu f(x)| \le 
( \int |(D^\nu\rho)(y)| dy)\frac{1}{\delta ^{|\nu|}}.
$$


Finally, for any $p$, there is a constant C (depending on $p$ and $n$) such that $B\_2(0,1) \subset B\_p(0,C\_p)$. Therefore:  

$$ A \subset A+B\_2(0,\delta) \subset A+B\_2(0,2\delta) \subset A+B\_p(0,2C\_p\delta) = A + B\_p(0,\epsilon) = A^{\epsilon} $$
if we take $\delta = \epsilon /(2C\_p) .$ 


With this choice of $\delta$, the function $f$ satisfies all the requirements.


$$\dots\dots\dots$$
In order to apply the result without distinguishing the dimention $n$ of the underlying space, it is of interest to consider the following lemma.


$\mathbf{Lemma}$: Given $ p \in \{1,2\} \cup (3,+\infty]$, there is $\rho:\mathbb R^n \rightarrow \mathbb R$ such that:  

i) $\rho$ is $\mathscr C^3$  

ii) $0 \le \rho$  

iii) $\rho(x) > 0$ iff $\|x\|\_p < 1$  

(iv) and, by re-scaling, $\int \rho(x) \space dx = 1$) 


With this choice of $\rho$ the original argument can be carried without resorting to the 2-norm.


$\mathbf{Proof}$: With $\varphi$ as above, if $p \in \{2\} \cup (3,+\infty)$, define $\rho$ as
$$\rho(x) := \varphi(1- \|x\|\_p^p) .$$
With this choice, $\rho$ satisfies i,ii,iii. The condition $p>3$ is needed to justify the claimed smoothness of $\rho.$ 


If $p=1$ or $p=\infty$, note that the open unit ball in $(\mathbb R^n , \|.\|\_p)$ is a polytope of the form $P=\{x| v\_i^{'}x<1\}.$ When $p=\infty$, $\{v\_i\} = \{\pm e\_j\}$ where $\{e\_j\}$ is the canonical (vector space) bases of $\mathbb R^n$. When $p=1$, $\{v\_i\} = \{-1,1\}^n$. In these cases, take $\rho$ as
$$ \rho(x) = \prod\_i \space \varphi(1-v\_i^{'}x) . \blacksquare$$


