# On a characterization of inward unit normal vector

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

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

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.