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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.

Migalobe
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1 Answers1

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(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 $\R^d$.

Martin R
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Deane
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  • Nice proof. I can't see why $\nabla g$ is bounded and also why you consider a compact set. – S. Euler Mar 28 '21 at 18:35
  • @SEuler, here's why $\nabla g$ is bounded. Let $u$ be a unit vector. If $g$ is $C^1$, then the directional derivative of $g$ is $$ u\cdot \nabla g = \lim_{t\rightarrow 0} \frac{g(x+tu)-g(x)}{t} \le \frac{g(tu)}{t} = g(u) < C$. – Deane Mar 28 '21 at 21:16
  • If $D$ is not compact, the proof I provided (it's not mine) does not work, because the domain of the function $g$, as defined, is not all of $\mathbb{R}^d$. – Deane Mar 28 '21 at 21:18
  • I don't think that boundedness of directional derivatives yields so for gradient. 2) For $g$ to be defined on $R^d$, isn't [$D\ni 0$ open bounded and convex] enough?
    • – S. Euler Mar 29 '21 at 22:53
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    Suppose $\nabla g(x) \ne 0$ and $u\cdot \nabla g(x) \le C$ for all unit vectors $u$. Let $u = \nabla g(x)/|\nabla g(x)|$, Then $|\nabla g| = u\cdot \nabla g(x) < C$. – Deane Mar 30 '21 at 00:34
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    Yes, $D$ bounded and containing the origin is enough. – Deane Mar 30 '21 at 00:35