3
$\begingroup$

The following inequality is quite clear for $R^1$: $$\int_{B_1}1/|x-y|^\alpha dx\leq\int_{B_1}1/|x|^\alpha dx,\quad\forall y\in B_1,$$ where $B_1$ is the unit ball in $R^1$, i.e., $[-1,1]$ and $\alpha>0$ is a constant which make $1/|x-y|^\alpha$ and $1/|x|^\alpha$ are integrable. For the proof, you can use the geometric meaning of an integral. But when consider the case of $R^2$, or even $R^n$, the argument provided above is not efficient anymore, so is there anyone can show this for me?

$\endgroup$
1
$\begingroup$

The basic intuition behind this is that for any $x \in B_1$, one can choose a unique $x^{\prime} \in B_1$ so that $|x-y| \geq |x^{\prime}|$ and so the right hand integral is bigger than the left. This can be seen easily in geometric terms in the one dimensional case. So the question is how to do show this in higher dimensions? The way to do this is to rewrite $|x-y|$ to get $$ \int_{B_1} \dfrac{dx}{|x-y|^\alpha} = \int_{B_1} \dfrac{dx_1\cdots dx_n}{\big(|x_1 - y_1| + \cdots + |x_n - y_n| \big)^{\alpha/2}}. $$ Consider the $i$th coordinate. Fix $x_j$ for all $j \neq i$. Then for each $x_i$, there is a unique $x_i^{\prime}$ so that $|x_i - y_i| \geq |x_i^{\prime}|$ and so $$ |x_1 - y_1| + \cdots + |x_i + y_i| + \cdots + |x_n - y_n| \geq |x_1 - y_1| + \cdots + |x_i^{\prime}| + \cdots + |x_n - y_n|. $$ Doing this in each coordinate gives that the integral of $1/|x|^\alpha$ over $B_1$ must be bigger than that of $1/|x-y|^{\alpha}$ over $B_1$. At least, this is how I look it.

$\endgroup$

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.