Let $S^{d-1}$ be the $d$-dimensional unit sphere. Let $A\subset S^{d-1}\cap \{x_d>0\}$ be a measurable set in the Borel algebra of $S^{d-1}$. Set $C = \{ \lambda a\, | \, \lambda>0, a\in A\}$ and denote by $B_r(x)$ the ball in $\mathbb{R}^d$ with radius $r>0$ and center $x$.

We claim that \begin{align} |B_r(\xi)\cap C| \leq c \, r \,\sigma(B_r(\xi)\cap A) \end{align} holds for all $\xi \in S^{d-1}\cap \{x_d>0\}, 0<r<1$ and some constant $c>0$ independent of $r,\xi$. Here, $|\cdot|$ denotes the $d$-dimensional Lebesgue measure and $\sigma$ denotes the spherical measure.

I think one needs to do some kind of transformation here. I wanted to use $x\mapsto \frac{x}{|x|}$, but somehow I cannot recover the spherical measure if I start my integration with the Lebesgue measure. Is there another kind of transformation theorem that I am unaware of?

  • $\begingroup$ What does $B_r(\xi)$ mean here? $\endgroup$ – kimchi lover Sep 2 '19 at 12:17
  • $\begingroup$ Somehow this smells like an application of the coarea formula to me. Have a look at en.wikipedia.org/wiki/Coarea_formula. Especially the subsection for the spherical integration may be helpful. $\endgroup$ – humanStampedist Sep 2 '19 at 12:32
  • $\begingroup$ Using the Coarea formula I obtain $|B_r(\xi) \cap C| = \int_{1-r}^{1+r} H^{d-1}(B_r(\xi) \cap C \cap \partial B_y(0)) dy$, where $H^{d-1}$ is the $d-1$-dimensional Hausdorff measure and $\partial B_y(0) = \{ x \in \mathbb{R} | |x|=y \}$. But now I would need to estimate the Hausdorff measure inside the integral with the $H^{d-1}$-measure in the case $y=1$. Is that possible? Maybe only for sufficient small $r$? $\endgroup$ – user406143 Sep 3 '19 at 9:33
  • $\begingroup$ Since $H^{d-1}(B_r(\xi)\cap C\cap \partial_1B(0))=\sigma(B_r(\xi)\cap A)$ this may be useful? $\endgroup$ – humanStampedist Sep 3 '19 at 9:42
  • $\begingroup$ Yes, that is true, but I need something like $H^{d-1}(B_r(\xi)\cap C\cap \partial B_y(0)) \leq c \sigma(B_r(\xi)\cap A)$ for $c>0$ independent of $r,y$. That I cannot see. $\endgroup$ – user406143 Sep 3 '19 at 10:16

\begin{align} |B_r(\xi)\cap C| &< |(B_r(\xi)\cap A)\cdot[1-r, 1+r]| < \int_{1-r}^{1+r}\left( \int_{B_r(\xi)\cap A} r^{d-1}ds \right)dr \\ &= \sigma(B_r(\xi)\cap A)\int_{1-r}^{1+r}r^{d-1}dr = \sigma(B_r(\xi)\cap A)\frac1d[(1+r)^d - (1-r)^d] \\ &\leq \sigma(B_r(\xi)\cap A)[2r + o(r)] \end{align}

So $B_r(\xi)\cap C$ is contained in a spherical cylinder $(B_r(\xi)\cap A)\cdot[1-r, 1+r]$ and you use spherical coordinates to calculate its volume. Hope this helps!

  • $\begingroup$ Is it really true that $B_r(\xi)\cap C$ is contained in the spherical cylinder $(B_r{\xi}\cap A) \times [1-r,1+r]$? I don't think so. The thing is that if I draw a line through the origin and a boundary point of $B_r(\xi)\cap A$ then this line is not necessarily a tangent to the ball $B_r(\xi)$, right? $\endgroup$ – user406143 Sep 5 '19 at 8:29
  • $\begingroup$ The line you're talking about wouldn't be tangent to $B_r(\xi)$ indeed, but that's not relevant to the problem. I'd argue as follows: $B_r(\xi)\cap C$ is enclosed in an infinite spherical cylinder $(B_r(\xi)\cap A)\cdot (0, \infty)$ and it's maximal distance from $S^{d-1}$ is $r$, hence it is enclosed in a spherical cylinder $(B_r(\xi)\cap A)\cdot [-1-r, 1+r]$. I see you have written $B_r(\xi)\cap A \times [1-r, 1+r]$, maybe that's where confusion arises. I think of spherical cylinder $A\cdot [x, y]$ as of set $\{ar: a\in A, r\in[x, y] \}$. Maybe 'spherical cone' would be a better name. $\endgroup$ – mbartczak Sep 5 '19 at 9:47
  • $\begingroup$ Ok, maybe I just don't get it, but why is $(B_r(\xi)\cap C) \subset (0,\infty)\cdot (B_r(\xi)\cap A)$? This was actually my question in the comment above, where I used the line that is not tangent to argue that I doubt this assertion. $\endgroup$ – user406143 Sep 5 '19 at 10:34
  • $\begingroup$ That's because $B_r(\xi)\cap C = B_r(\xi)\cap (A\cdot (0, \infty)) \subseteq (B_r(\xi)\cdot(0, \infty))\cap (A\cdot (0, \infty)) = (B_r(\xi)\cap A)\cdot(0, \infty)$. Just draw it in d=2 and you will see :). $\endgroup$ – mbartczak Sep 5 '19 at 10:57

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.