On $C^1$ convex domain Let $D$ be a $C^1$ domain of $\mathbb{R}^d$. Then we know that there exists a $C^1$ function $\rho:\mathbb R^d\rightarrow \mathbb R$ such that
$$
D=\{x\in \mathbb R^d, \rho(x)<0\}, \quad \partial D=\{x\in \mathbb R^d, \rho(x)=0\},
$$
and
$
x\in \partial D\Longrightarrow d\rho(x)\not=0.
$
Assume now that $D$ is convex. Can we choose $\rho$ to be convex with the same properties?
I know that the first claim follows by the local definition and use of partition of unity, but I can't manage to prove existence of such convex function.
 A: (Original answer deleted)
$\newcommand{\R}{\mathbb{R}}$
Let $D \subset \R^d$ be a compact $C^1$ convex domain that contains the origin in its interior. Its gauge function is defined to be  $g: \R^d \rightarrow [0,\infty)$, where
$$
  g(x) = \inf \{ t > 0 : x/t \in D\}.
$$
Observe that $g$ is homogeneous of degree $1$. Therefore, to prove that $g$ is convex, it suffices to show that it satisfies the triangle inequality:
$$
g(x+y) \le g(x) + g(y).
$$
A proof can be found here: Show that a closed convex ball implies the triangle inequality
Now observe that the boundary of $D$ is parameterized by the map
\begin{align*}
  G: S^{d-1} &\rightarrow \partial D\\
  u &\mapsto \frac{u}{g(u)}
\end{align*}
The boundary $\partial D$ is $C^1$ if and only if $G$ is a $C^1$ map. And $G$ is $C^1$ if and only if $g$ is $C^1$ away from the origin. Moreover, since $\nabla g$ is homogeneous of degree $0$ and continuous, it is uniformly bounded.
Let $\rho = g^2-1$. $D = \{ \rho \le 0\}$ and $\partial D = \{ \rho = 0\}$. Since $g$ is $C^1$ on $\R^d\backslash\{0\}$, so is $\rho$. On other hand, the derivative of $\rho$ is equal to
$$
  \nabla\rho = 2g\nabla g,
$$
which converges to $0$ at the origin and therefore is continuous on all of $\R^d$.
