Volume of convex body as an integral of its radial function

Question

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?

Answer 1

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.