1
$\begingroup$

I have a question concerning to the integral on sphere. It's maybe true and simple but I don't know how to prove it. Could anyone have some suggestions? Thanks.

Denote $S^{n-1}$ the unit sphere in $R^n$. Let $\xi \in S^{n-1}$, $r\in [0,1)$ and let $1\leq k < \frac n2$ be an integer. We define $$F(r) = \int_{S^{n-1}} (1 -2r \langle \xi, \omega\rangle +r^2)^{k-\frac n2} d\omega,$$ where $d\omega$ denote the surface area measure on $S^{n-1}$. Is it true that $$F(r) \leq F(0),$$ for any $0\leq r < 1$?

Improve this question
$\endgroup$
2
  • $\begingroup$ for $k=1$ we have equality (since $|x|^{2-n}$ is Green function for Laplacian), for $k>1$ use power mean inequality $\endgroup$
    – Fedor Petrov
    Commented Mar 9, 2016 at 7:23
  • $\begingroup$ @FedorPetrov: Thank you very much. Your answer is great. In fact, I can show that $F$ is an even polynomial of order $2(k-1)$ in $r$, and can compute explicitly the coefficients, but it is not easy to use this form of $F$ to prove the inequality above. I didn't think about using Holder inequality here. A further question arises is that is $F$ decreasing in $r$? $\endgroup$
    – nguyen0610
    Commented Mar 9, 2016 at 13:32

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .