# Fattened volume of a curve

convex-geometrydifferential-geometrymeasure-theory

Let $$\gamma:[0,1] \rightarrow \mathbb{R}^3$$ be a smooth curve with nonvanishing velocity and let $$C_\gamma = \gamma([0,1])$$ be the image. Denote by $$B_r = \{ x : \|x\| < r \}$$ the open ball of radius $$r$$ centered at the origin. Then I expect the following to hold:
$$\lim_{r \rightarrow 0} \frac1{r^2} \mathcal{H}^3 (C_\gamma + B_r) = \pi \cdot \mathcal{H}^1(C_\gamma)$$
where $$A + B = \{ x+y : x \in A, y \in B \}$$ is the Minkowski sum of two sets $$A, B$$ and $$\mathcal{H}^d(A)$$ is the $$d$$-dimensional Hausdorff measure of $$d$$.

The above setup can be easily generalised by replacing a curve with a $$d$$-dimensional embedded submanifold $$M$$ with boundary and by replacing $$\mathbb R^3$$ by $$\mathbb R^D$$:
$$\lim_{r \rightarrow 0} \frac1{r^{D-d}} \mathcal{H}^D (M + B_r) = \omega_{D-d} \cdot \mathcal{H}^d(M)$$
with $$\omega_k = \pi^{k/2}/\Gamma(\frac k2 + 1)$$ being the volume of the $$k$$-dimensional unit ball.

Does anyone know if this type of result was proven elsewhere before?

In the beginning of this paper, we see the Weyl's formula: if $$M$$ is a $$d$$-dimensional compact, connected submanifold of $$\mathbb{R}^D$$ with boundary and if $$M^\perp _r = \{ x + y | x \in M, y \in T_x^\perp M, \|x\| \le r \}$$ denotes the normal thickening of $$M$$, then there exist constants $$k_l(M)$$ such that $$\text{Vol}(M_r^\perp) = \omega_{D-d} r^{D-d} \sum_{\substack{0 \le l \le D \\ l \text{ even} }} \frac{k_{l}(M) r^{l} }{(D-d+2) \cdots (D-d+l) }$$ where $$\text{Vol}(M_r)$$ is the Euclidean volume (which I believe is the Lebesgue measure). In particular, $$k_0(M)$$ is the Riemannian volume of $$M$$. Therefore, if $$M_r = \{ x+y | x \in M, \|y\| \le r \}$$ is the full thickening of $$M$$, then we can sandwich $$\text{Vol}(M_r)$$ as follows: $$\text{Vol}(M_r^\perp) \le \text{Vol}(M_r) \le \text{Vol}(M_r^\perp) + \text{Vol}((\partial M)_r^\perp)$$ Here, to get the second inequality, we assume that for each point $$z \in \mathbb{R}^D$$ there is a point $$z' \in M$$ such that either (1) $$z' \in M$$ and $$z - z' \in T_{z'}^\perp M$$ or (2) $$z' \in \partial M$$ and $$z - z' \in T_{z'}^\perp (\partial M)$$.
Seeing that the leading term of $$\text{Vol}((\partial M)_r^\perp)$$ in the Weyl's formula is of order $$r^{D-d+1}$$, we then see by a sandwiching argument that: $$\lim_{r \rightarrow 0} \frac{\text{Vol}(M_r)}{\omega_{D-d} r^{D-d}} = k_0(M)$$