How to prove $\int_{B_1} \frac{1}{|x-y|^\alpha} \le \int_{B_1} \frac{1}{|x|^\alpha}$ when $y$ is also in $B_1 \in \mathbb{R}^2$ with $\alpha<2$

Asked Aug 22 '23 at 06:13

Active Aug 22 '23 at 06:35

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How to prove that $\int_{B_1} \frac{1}{|x-y|^\alpha} \le \int_{B_1} \frac{1}{|x|^{\alpha}}$ when $y$ is also in the unit ball $B_1 \in \mathbb{R}^2$, and $\alpha < 2$.

Because $\alpha<2$, so the RHS is integrable, for the LHS, at first, I try to transfer the coordinate to polar coordinate and calculate, and I saw this answer Here, but I think there must be some intuitive proofs.

edited Aug 22 '23 at 06:35

asked Aug 22 '23 at 06:13

Elio Li

1

This is a consequence of the [Rearrangement inequality][https://en.wikipedia.org/wiki/Hardy%E2%80%93Littlewood_inequality]. Use the inequality with $f(x) = |x - y|^{-\alpha}$ and $g(x) = \chi_{B_1}$. – r9m Aug 22 '23 at 06:19
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Thanks for your answer! This is the first time I heard of this, following the answer in https://math.stackexchange.com/questions/2178548/calculation-of-symmetric-decreasing-rearrangement , I have figured this out. – Elio Li Aug 22 '23 at 07:29

How to prove $\int_{B_1} \frac{1}{|x-y|^\alpha} \le \int_{B_1} \frac{1}{|x|^\alpha}$ when $y$ is also in $B_1 \in \mathbb{R}^2$ with $\alpha<2$

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