I've been reading the paper "On the Bourgain, Brezis, and Mironescu Theorem Concerning Limiting Embeddings of Fractional Sobolev Spaces" by Maz'ya and Shaposhnikova and struggling with the short style where many arguments are omitted. My question concerns with an inequality relating two integrals with complementary domains:
Let $0 < s < 1$ and $p \ge 1$. We set the fractional Sobolev norm of a function $u: \mathbb{R}^n \to \mathbb{R}$ as
$$ \| u \|_{s,p} := \left( \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} \frac{|u(x)-u(y)|^p}{|x-y|^{n+sp}} dx\ dy \right)^{\frac 1 p} $$
What I want to show is that, if $u \in C^\infty (\mathbb{R}^n)$ with compact support and $\| u \|_{s,p} < \infty$, then the following estimate holds:
$$ \int_{\mathbb{R}^n} \int_{|x-y| > |x|} \frac{|u(x)-u(y)|^p}{|x-y|^{n+sp}} dx\ dy \le \int_{\mathbb{R}^n} \int_{|x-y| \le |x|} \frac{|u(x)-u(y)|^p}{|x-y|^{n+sp}} dx\ dy $$
I tried plotting the domains of integration for $n = 1$ and using symmetries that result by interchanging the variables, but I only obtained
$$ \int_{\mathbb{R}^n} \int_{|x-y| > |x|} \frac{|u(x)-u(y)|^p}{|x-y|^{n+sp}} dx\ dy = \int_{\mathbb{R}^n} \int_{|x-y| > |y|} \frac{|u(x)-u(y)|^p}{|x-y|^{n+sp}} dx\ dy $$
I suppose there has to be a simple trick to manipulate the left hand side of the inequality I want to show, but it has not come to my mind yet. I would be thankful for any ideas tackling this seemingly easy problem!