4
$\begingroup$

Let $\lVert \cdot \rVert \colon \mathbb{R}^d \to \mathbb{R}^{\geq0}$ be a norm. Prove that the area $A(S)$ of the unit-sphere $S = \{x \in \mathbb{R}^d : \lVert x \rVert = 1\}$ exists.

Integral representation

Denote by $\lVert \cdot \rVert_D, \lVert \cdot \rVert_E \colon \mathbb{R}^d \to \mathbb{R}^{\geq 0}$ an arbitrary norm and the Euclidean norm, respectively. By geometric intuition, under proper smoothness assumptions, it seems to me that the area of $S$ is given by

$$A(S) = \int_{S_D} \frac{1}{n(\Theta) \bullet n_D(\Theta)} \left(\frac{\lVert x(\Theta) \rVert_E}{\lVert x_D(\Theta) \rVert_E}\right)^{d - 1} \mathrm{d} \Theta,$$

where $n, n_D$ are the unit-normals (unit Euclidean-norm), and $x, x_D$ are the points on the corresponding unit-spheres $S$ and $S_D$.

Intuition

Intuitively, we think of mapping an infinitesimal piece of the $S_D$ sphere to $S$ by multiplication. The multiplication scales the area of the piece to the power $(d - 1)$ since the piece is stretched in every dimension. The scaled piece, and the corresponding piece on $S$ still have different normals. When the normals point to different directions, the contribution should be larger.

Examples

The formula works trivially when $\lVert \cdot \rVert = \lVert \cdot \rVert_D$.

To compute the area of the maximum norm unit-sphere in $\mathbb{R}^2$, let $\lVert \cdot \rVert_D = \lVert \cdot \rVert_E$. Then in polar coordinates $\Theta \in [-\pi / 4, \pi / 4]$ we have $n_D(\Theta) = (\cos(\Theta), \sin(\Theta))$, $n(\Theta) = (1, 0)$, $x(\Theta) = (1, \tan(\Theta))$, and $\lVert x(\Theta) \rVert_E = 1 / \cos(\Theta)$. By symmetry,

$$A(S) = 4 \int_{-\pi / 4}^{\pi / 4} \frac{1}{\cos(\Theta)^2} \mathrm{d} \Theta = 8.$$

To compute the area of the Manhattan norm unit-sphere in $\mathbb{R}^2$, let $\lVert \cdot \rVert_D = \lVert \cdot \rVert_E$. Then in polar coordinates $\Theta \in [0, \pi / 2]$ we have $n_D(\Theta) = (\cos(\Theta), \sin(\Theta))$, $n(\Theta) = (1, 1) / \sqrt{2}$, $x(\Theta) = (\cos(\Theta), \sin(\Theta)) / (\cos(\Theta) + \sin(\Theta))$, and $\lVert x(\Theta) \rVert_E = 1 / (\cos(\Theta) + \sin(\Theta))$. By symmetry,

$$A(S) = 4 \int_{0}^{\pi / 2} \frac{\sqrt{2}}{(\cos(\Theta) + \sin(\Theta))^2} \mathrm{d} \Theta = 4 \sqrt{2}.$$

To compute the area of the Euclidean norm unit-sphere in $\mathbb{R}^3$, let $\lVert \cdot \rVert_D$ be the maximum norm. Then in Euclidean coordinates $-1 \leq x, y \leq 1$ and $z = 1$ we have $n(x, y, 1) = (x, y, 1) / \sqrt{x^2 + y^2 + 1}$, $n_D(x, y, 1) = (0, 0, 1)$, $x_D(x, y, 1) = (x, y, 1)$, and $\lVert x_D(\Theta) \rVert_E = \sqrt{x^2 + y^2 + 1}$. By symmetry,

$$A(S) = 6 \int_{-1}^{1} \int_{-1}^{1} \frac{1}{\sqrt{x^2 + y^2 + 1}^3} \mathrm{d}y \mathrm{d}x = 4 \pi.$$

Compactness

It can be proved that the unit-sphere of any norm is compact. Given that the integrand is continuous, the integrand is bounded by the extreme-value theorem. The area of the Euclidean sphere $S_E$ is known and finite. Therefore, the area $A(S)$ is an integral of a bounded function on a set $S_E$ of finite measure, and so a finite number.

Measurability

The integral is well-defined when the integrand is measurable. When $n$ is continuous, the integrand is continuous and therefore measurable. The problem is, there are norm-spheres which do not have continuous normal-fields, such as the maximum norm. The solution for the maximum norm is to partition the integral along the discontinuities (i.e. the faces of the cube).

The question then seems to come down to: is it always possible to partition the norm-sphere into countable many measurable sets with continuous normal-fields?

$\endgroup$
1
$\begingroup$

I'll suggest two approaches, which one to prefer depends on what one knows about "area" (definition, basic results).

Hausdorff measure

Let's interpret the area as $(d-1)$-dimensional Hausdorff measure $\mathcal H^{d-1}$. This measure has the property that $\mathcal H^{d-1}(f(E))\le L^{d-1} \mathcal H^{d-1}(E)$ for any Lipschitz map $f$ (with $L$ being its constant constant).

  • Upper bound: pick $R$ such that the Euclidean ball $B_R$ of radius $R$ centered at the origin surrounds $S$. Let $f:\partial B_R\to S$ be the nearest-point projection. It is known that $f$ is Lipschitz with $L=1$. Hence, $\mathcal H^{d-1}(S) \le \mathcal H^{d-1}(\partial B_R) = \omega_{d-1}R^{d-1}$, with $\omega_{d-1}$ being the area of the unit Euclidean sphere.

  • Lower bound: Let $r$ be such that the Euclidean ball $B_r$ of radius $r$ centered at the origin is surrounded by $S$. Use the nearest-point projection of $S$ onto $\partial B_r$ to conclude that $\mathcal H^{d-1}(S) \ge \omega_{d-1}r^{d-1}$.

Thus, $\mathcal H^{d-1}(S)$ is finite and positive.

Coarea formula

Apply the Coarea formula $$ \int g(x)|\nabla u(x)|\,dx = \int_{\mathbb{R}} \int_{u^{-1}(t)}g(x)\,d\mathcal H^{d-1}(x)\,dt \tag{1}$$ with $g=\chi_{\{\|x\|\le 1\}}$ and $u(x)=\|x\|$. Since $|\nabla u| = 1$ at the points of differentiability of $u$ (almost everywhere), the left-hand side of (1) is just the volume of the unit ball $\{\|x\|\le 1\}$. The right hand side is the integral $\int_0^1 t^{n-1} A(S)\,dt = A(S)/n$. Note that the existence of $\int_{u^{-1}(t)}g(x)\,d\mathcal H^{d-1}(x)$ for a.e. $t$ is a part of the statement of the coarea formula: thus, almost every level set of a Lipschitz function on a set of bounded measure has finite area. Since all level sets of $u$ are similar, the statement applies to all of them.

$\endgroup$

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.