My question is on Brascamp-Lieb-inequality on the Euclidean sphere (which can be viewed as an analogue of Young's inequality on the sphere) obtained in [1].


(See also this question:
[Brascamp-Lieb inequalities on the sphere](https://mathoverflow.net/questions/325211/brascamp-lieb-inequalities-on-the-sphere) )


The inequality reads:


Let $e\_{i},i=1,\dots,N$, be orthogonal vectors in
$\mathbb{R}^{N}$. If $f\_j$ are positive functions on $[-1,1]$, $N \geq 3$, and $p\geq 2$ then
$$
\int\_{S^{N-1}} \prod\_{j=1}^N f\_j(x\cdot e\_j) d\mu(x)
\leq \prod\_{j=1}^N \left(\int\_{S^{N-1}} f\_j(x\cdot e\_j)^p d\mu(x)\right)^{1/p},
$$
where $\mu$ is the uniform Borel probability measure on the sphere $S^{N-1}$.


Fix **$(x\_{1},\dots,x\_{N})\in \mathbb{R}^{N}\setminus S^{N-1}$**. Let $0<\alpha\_{j}<1$. Suppose that
$$f\_{j}(t)=|t-x\_{j}|^{-\alpha\_{j}},\quad x\_{i}\neq x\_{j},\;\forall i\neq j.$$
Does the inequality $(1)$ hold true for these particular functions $f\_{j}$ **for some $1\leq p<2$** ?


That is, is it true that


$$
\int\_{S^{N-1}} \prod\_{j=1}^N 
|\theta\cdot e\_j-x\_{j}|^{-\alpha\_{j}}
 d\mu(\theta)
 \leq C\prod\_{j=1}^N \left(\int\_{S^{N-1}} |\theta\cdot e\_j-x\_{j}|^{-p\alpha\_{j}}
 d\mu(\theta)\right)^{\frac{1}{p}}\qquad (\*)$$
for some $1\leq p<2$ ?


[1] EA Carlen, EH Lieb, and M Loss. "A Sharp Analogue of Young's Inequality on $S^N$ and Related Entropy Inequalities." <https://arxiv.org/abs/math/0408030>


Edit Summary:


As @fedja pointed out, the estimate is immediately false when
$(x\_{1},\dots, x\_{N})$ lies on $\mathbb{S}^{N-1}$.


Latest Edit:


Instead of asking whether the inequality $(\*)$ holds with $p=1$ I tried to simplify the question to whether $(\*)$ is true for some $p\in [1,2[$.


**Most recent Edit**:
Simplified the functions $f\_{j}$ to $f\_{j}(t):=|t-x\_{j}|^{-\alpha\_{j}}$ instead of $f\_{j}(t)=|t^k-x\_{j}|^{-\alpha\_{j}}$ where $k$ is an odd integer.



(simplified further) Is Brascamp-Lieb inequality on the sphere applicable for these functions for some $1\leq p<2$
Unfortunately, the answer to the question as stated is "No" even if $N=2$. Let $k=1,\alpha\_1=\alpha\_2=\frac 12$. Choose $x=(x\_1,x\_2)$ on the unit circle (say $(3/5,4/5)$). Then the integrals on the right are finite but the integral on the left diverges (you basically integrate inverse distance to $x$). Perhaps, you meant something else?


