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 $x$.

  • 2
    $\begingroup$ 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). $\endgroup$ Sep 25, 2021 at 20:52
  • 2
    $\begingroup$ 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. $\endgroup$ Sep 25, 2021 at 21:09
  • $\begingroup$ @MateuszKwaśnicki Thank you for your comment. It's like a measure-theoretic definition of a unit normal vector. $\endgroup$
    – sharpe
    Sep 25, 2021 at 21:16
  • $\begingroup$ follow-up on mathoverflow.net/q/404848/11260 $\endgroup$ Sep 26, 2021 at 7:13

1 Answer 1


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.

  • 1
    $\begingroup$ Thank you for your answer. It seems natural to use half-space. $\endgroup$
    – sharpe
    Sep 25, 2021 at 21:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.