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Let $\Omega$ be a smooth domain in $\mathbb R^n$ and let $x\in\partial\Omega$. Let $n(x):\partial\Omega\to\mathbb S^{n-1}$ be the normal vector field. Then there exist a vector field $\phi\in C_c^\infty(\Omega,\mathbb R^n)$ such that $\phi(x)|_{\partial\Omega}=n(x)$ and $|\phi(x)|\leq 1$

How to prove this? In particular, if $\Omega$ is a ball then what is the explicit expressions of such $\phi$? I couldn't do this.

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  • $\begingroup$ $\phi \in C^\infty (\Omega,\mathbb{R}^n)$ and $\Omega$ is relatively compact ? $\endgroup$
    – HK Lee
    May 21, 2019 at 11:00
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    $\begingroup$ If $\Omega$ is not relatively compact, there is no chance that $\phi$ is compactly supported. If $\Omega$ is relatively compact, just use a tubular neighbourhood of $\partial\Omega$ and a smooth cut-off function to construct $\phi$. $\endgroup$
    – user10354138
    May 21, 2019 at 11:05
  • $\begingroup$ @user10354138 how to use tubular neighborhood of $\partial\Omega$ and smooth cut off? can you tell more or give some references? $\endgroup$
    – MathRock
    May 21, 2019 at 11:12
  • $\begingroup$ Let's assume $\Omega$ is relatively compact. $\endgroup$
    – MathRock
    May 21, 2019 at 11:43

1 Answer 1

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Assume $\Omega$ is relatively compact, so $\partial\Omega=M$ is a smooth compact hypersurface of $\mathbb{R}^n$.

By the Tubular Neighbourhood Theorem, and the fact $M$ is closed, the exponential map defines a diffeomorphism $B_\varepsilon(\nu_M\cong M\times\mathbb{R})\to V$, where $\nu_M$ is the normal bundle of $M$ in $\mathbb{R}^n$, $V$ is a neighbourhood of $M$ in $\mathbb{R}^n$. Thus we can define $\phi\colon\mathbb{R}^n\to\mathbb{R}$ by $$ \phi(x)= \begin{cases} \eta(d(x,M))\cdot n(\operatorname{proj}_M(x)) & x\in V\\ 0 & x\notin V \end{cases} $$ where $\operatorname{proj}_M\colon V\to M$ is the closest point projection to $M$, $\eta\colon\mathbb{R}\to[0,1]$ is a bump function that is $1$ on $B_{\varepsilon/3}(0)$ and $0$ outside $B_{2\varepsilon/3}(0)$. You can check that this $\phi$ (restricted to $\bar\Omega$) satisfies all the desired conditions.

For $\Omega$ the unit ball, it is easy to describe what happens. We pick the unit radial vector $x/\lvert x\rvert$, multiply it by $\xi(\lvert x\rvert^2)$ where $\xi$ smoothly interpolates say between $0$ on $[0,\frac13]$ and $1$ on $[\frac23,1]$. If this is not explicit enough for your taste, we can have a very explicit $$ \phi(x)= \begin{cases} \dfrac{x}{\lvert x\rvert}\cdot \exp\left(1-\dfrac1{\lvert x\rvert^2}\right) & x\neq 0\\ 0 & x=0 \end{cases} $$

A counterexample when $\Omega$ is not relatively compact (actually $\partial\Omega$ is not compact is what is needed): Consider $\Omega\subset\mathbb{R}^2$ be the region $xy<1$. Then $\partial\Omega=\{xy=1\}$ is not compact (since it is not bounded), so there are no compactly supported function on $\overline{\Omega}$ that restricts to a unit vector field on $\partial\Omega$.

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  • $\begingroup$ Thank you. But, how to define the exponential map and how its help to define $\phi$? Actually, I don't have much knowledge about the manifold. $\endgroup$
    – MathRock
    May 21, 2019 at 12:59
  • $\begingroup$ So its enough to take $\partial\Omega$ is bounded instead of taking the set $\Omega$ is relatively compact. $\endgroup$
    – MathRock
    May 21, 2019 at 13:03
  • $\begingroup$ For $\mathbb{R}^n$, the exponential map is just $\exp_p(v)=p+v$. $\endgroup$
    – user10354138
    May 21, 2019 at 13:04
  • $\begingroup$ Sorry, yes, $\partial\Omega$ is bounded is enough. $\endgroup$
    – user10354138
    May 21, 2019 at 13:07
  • $\begingroup$ How can ensure that $\phi$ is smooth because distance function $d(x,M)$ may not smooth enough right? $\endgroup$
    – MathRock
    May 21, 2019 at 20:26

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