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Let $D$ be a smooth domain of $\mathbb{R}^d$. Let $\partial D$ denote the boundary of $D$. We denote by $B(x,r)=\{y \in \mathbb{R}^d \mid |y-x|<r\}$ the Euclidean ball centered at $x$ with radius $r$

For $x \in \partial D$, we consider the following limit: \begin{align*} \lim_{r \to 0}\frac{1}{m(D \cap B(x,r))}\int\limits_{D \cap B(x,r)}\frac{z-x}{r}\,m(dz) \end{align*} Here, $m$ denotes the Lebesgue measure on $\mathbb{R}^d$.

Does this limit exists?

In my impression, this seems to converge to the inward (unit) normal vector at $x$.

Daniele Tampieri
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sharpe
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    If $D$ is a half-space, the integral mean is independent of $r$, and by symmetry one can replace $z$ with its normal component, getting the unit inward normal as a value. In general, the integral mean with $D$ differs by the integral mean with the “tangent inner half-space” by $o(1)$ as $r\to0$, so the answer is yes (but the question is maybe not suitable for this site). – Pietro Majer Sep 25 '21 at 20:52
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    The simplest way to get the result is likely to substitute $z = x + r y$ with $y$ in the unit ball, and then use dominated convergence. – Mateusz Kwaśnicki Sep 25 '21 at 21:09
  • @MateuszKwaśnicki Thank you for your comment. It's like a measure-theoretic definition of a unit normal vector. – sharpe Sep 25 '21 at 21:16
  • follow-up on https://mathoverflow.net/q/404848/11260 – Carlo Beenakker Sep 26 '21 at 07:13

1 Answers1

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For small $r$ the curvature of the surface $\partial D$ can be neglected, so $D \cap B(x,r)$ is half the $d$-dimensional ball with radius $r$. Choosing the origin of the coordinate system at position $x$ and orienting the $x_1$-axis along the inward normal, the integral is given by the vector $v$ with components $$v_p=\frac{2}{V_{d}(r)}\int\cdots\int_{-\infty}^\infty \theta\biggl(r-\sum_{i=1}^d x_i^2\biggr)\theta(x_1)\frac{x_p}{r}\,dx_1dx_2\ldots dx_d$$ with $V_d(r)$ the volume of the $d$-dimensional ball of radius $r$ and $\theta$ the unit step function.
Only the $p=1$ component is nonzero because of symmetry, and this component evaluates to $$v_1=\frac{\int_0^r d\rho\int_0^{\pi/2}d\theta\,\rho^{d} \cos \theta \sin^{d-2}\theta}{r\int_0^r d\rho\int_0^{\pi/2}d\theta\,\rho^{d-1} \sin^{d-2}\theta}=\frac{\Gamma \left(\frac{d}{2}+1\right)}{\sqrt{\pi }\, \Gamma \left(\frac{d+3}{2}\right)}.$$ So it's an inward normal vector, but not a unit vector.

Carlo Beenakker
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