# Volume of convex body as an integral of its radial function

Let $$C$$ be a compact convex set in $$\mathbb{R}^d$$. Let the origin $$O$$ by in the internal of $$C$$. The gauge function $$\gamma_C(.) : \mathbb{R}^d \to [0, \infty]$$ of $$C$$ is defined as $$\gamma_C(x) = \inf\{t : x \in t \cdot C\}.$$ The radial function is defined as a reciprocal of gauge function or, equivalently, $$1 / \gamma_C(x) = \sup\{x : t \cdot x \in C\}$$ I'm considering the following integral: $$I(C) = \int_{S^{d - 1}} |1 / \gamma_C(u)| du.$$

I have two following notions:

1. Intuitively, it looks like a volume of $$C$$: we integrate over all directions and summarize all distances from the origin $$O$$ to the boundary of $$C$$.

2. However, if we consider the body $$a \cdot C$$ for some constant $$a > 0$$, we will have $$1 / \gamma_{a \cdot C}(u) = a \cdot 1 / \gamma_C(u)$$, and: $$I(a \cdot C) = \int_{S^{d - 1}} |a \cdot 1 / \gamma_C(u)| du = a \int_{S^{d - 1}} |1 / \gamma_C(u)| du = a \cdot I(C).$$ From this equality, we have that $$I(C)$$ is not a volume of $$C$$ because $$Vol(a \cdot C) = a^3 \cdot Vol(C)$$.

So, which of (1) and (2) is wrong?

And, if (1) is wrong, the next question: can we construct a volume of $$C$$ from the integral of its radial function?

1 is wrong. For example consider the unit disk in $$\mathbb{R}^2$$. Then the integral is $$\int_0^{2\pi} 1= 2\pi$$. The volume, however, is of course $$\pi$$. The problem is, like you point out in 2, that the volume is not actually linear with respect to scaling. The integral treats the area of a disc like a rectangle on the circle, but it’s actually much more like a triangle. (With both sides depending on the scaling factor.)
The problem is that the infinitesimal volume element in spherical coordinates is not $$d\rho du$$ where $$du = vol_{S^{d-1}}$$ is the volume form of $$S^{d-1}$$ and $$\rho$$ is the radius. You need to scale with the Jacobi determinant when you change coordinates.
• Fair enough, that was pretty unclear. Let $C$ be the disc of radius $a$. Then $\int_{S^1}| 1/\gamma_C(U)|du= 2\pi a$ I.e the area of a rectangle with sides $a$ and $2\pi$. The correct area/volume is however $\pi a^2=\frac{1}{2}(2\pi a) a$. Aug 6 '20 at 9:20
• The infinitesimal volume element is as follows, right? $dV = \rho^2 sin \theta d\rho d\phi d\theta = \rho^2 d\rho du$. Then we can say that volume is $\int_{S^{d - 1}} \int_0^\rho \rho^2 d\rho du = \frac{1}{3} \int_{S^{d - 1}} \rho^3 du = \frac{1}{3} \int_{S^{d - 1}} |1 / \gamma_C(u)|^3 du$. Is this correct? Aug 6 '20 at 9:54
• I think this is correct if $d=3$. Aug 6 '20 at 9:59
• Yes, the volume element will have $\rho^{d-1}$ as a factor. Aug 6 '20 at 10:12