Bound involving an area regular partition of $\mathbb{S}^k$

Let $$\mathbb{S}^k \subset \mathbb{R}^{k+1}$$, $$k \geq 1$$, be the unit sphere.

Leopardi proved that, given a natural number $$n \geq 2$$, there exists a partition of $$\mathbb{S}^k$$ into pairwise disjoint sets $$\{S_j\}_{j=1}^n\subset \mathbb{S}^k$$, such that $$\sigma(S_j)=\frac{\sigma(\mathbb{S}^k)}{n}\quad \text{and} \quad \text{diam}(S_j)\leq \frac{c_1(k)}{n^{\frac{1}{k}}},$$ for all $$j=1,\ldots,n$$, where $$c_1(k)>0$$ is a positive constant only depending on the dimension $$k$$, diam$$(\cdot)$$ denotes the diameter and $$\sigma(\cdot)$$ the surface area. These partitions are called area regular partitions.

Given a point $$x\in\mathbb{S}^k$$ and $$-1\leq t\leq1$$, we define the spherical cap $$C=C(x,t)=\{y\in\mathbb{S}^k;\,\langle x,y\rangle\leq t\},$$ where $$\langle\cdot,\cdot\rangle$$ is the standard scalar product.

Fixed a spherical cap $$C=C(x,t)$$, it can be proved that the cardinality of the set of indices $$\mathcal{J}(C) \subset \{1,\ldots,n\}$$ such that, if $$j\in\mathcal{J}(C)$$ then $$C\cap S_j \neq \emptyset$$ and $$(S^k\setminus C)\cap S_j \neq \emptyset$$, is less or equal than a constant depending only on $$k$$, $$c_2(k)$$, times $$n^{1-\frac{1}{k}}$$, i.e., $$\text{card}(\mathcal{J}(C)) \leq c_2(k)n^{1-\frac{1}{k}}.$$

The proof I am following gives no detail: it only says that we have to use the properties of the partition of $$\mathbb{S}^k$$ (which it is quite obvious), but I do not know how to apply them. Any ideas or possible proofs?

• Hint: Show that the union of the $S_j$'s in contained in a strip around the boundary of $C$. – Nate May 13 at 14:51

For each such index $$j$$, the set $$S_j$$ intersect the boundary of the spherical cap, so it is contained in the $$c_1(k)n^{-1/k}$$-neighborhood of this boundary. The spherical area of this neighborhood is bounded above by $$c(k) n^{-1/k}$$ (maximized when it is a hemisphere), and since the $$S_j$$ are disjoint and all have the same spherical area, you get $$\textrm{card}(\mathcal{J}) \frac{\sigma(S^k)}{n} \le \frac{c(k)}{n^{1/k}},$$ which implies what you want to show.
• Which is the formula for the spherical area of this neighbourhood in $\mathbb{R}^{k+1}$? If we work in $\mathbb{R}^3$ I know it is $2\pi rh$ and since $r=1$ and $h\leq c_1(k)n^{-\frac{1}{k}}$ I obtain what you say. – user614222 May 14 at 9:15
• Somehow I was only thinking of the 2-dimensional unit sphere when I wrote it, which is why I have the $4\pi$ in there. In the general case you don't really need a formula, just need to know that the $r$-neighborhood of a $k-1$-dimensional sphere sitting inside $S^k$ has a $k$-dimensional volume bounded by the $r$-neighborhood of the equator, which itself is bounded by $cr$ for some constant $c>0$. – Lukas Geyer May 14 at 14:47