How to prove that $\frac{|u(x)-u(y)|^p}{|x-y|^{n+sp}}\in L^1(\mathbb R^{2n})$?

Question

Let $\Omega$ be an open bounded domain of $\mathbb R^n$. Let $s\in(0, 1)$, $p\in (1, \infty)$ and consider the Banach space $$X^{s, p}(\Omega)=\{u\in W^{s, p}(\mathbb R^n): u=0 \text{ in } \mathbb R^n\setminus\Omega\}$$ equipped with the usual Gagliardo seminorm defined, for any measurable $u:\mathbb R^n\to\mathbb R$ as $$[u]_{s, p}=\left(\iint_{\mathbb R^n\times\mathbb R^n}\frac{|u(x)-u(y)|^p}{|x-y|^{n+sp}} dx dy)\right)^{1/p}.$$

I am reading the notes the lecturer sent to me and to my classmates about his last two classes. I got a couple of questions (I think he should have been clearer about).

1. At some point, he wrote: let $(u_n)_n$ be a bounded sequence in $X^{s, p}(\Omega)$; hence there exists some $u$ such that $u_n\to u$ weakly in $X^{s, p}(\Omega)$, $u_n\to u$ in $L^p(\Omega)$ for $p\in [1, p^*_s)$ and $u_n\to u$ a.e. (here $p^*_s=(np)/(n-ps)$). The question is: where does that $u$ belong to? I would say that $u\in X^{s, p}(\Omega)$, but I am not completely sure about that.

2. Later, he writes that $$\frac{|u(x)-u(y)|^p}{|x-y|^{n+sp}}\in L^1(\mathbb R^{2n}).$$ The question is: why? Is that because $u\in X^{s, p}(\Omega)\hookrightarrow L^1(\Omega)$ and $$\|u\|_{L^1(\Omega)}\le [u]^p_{s, p}<+\infty?$$ (since $u=0$ in $\mathbb R^n\setminus\Omega$, hence $\|u\|_{L^1(\Omega)} = \|u\|_{L^1(\mathbb R^{n})}$). Is that the motivation?

I can not see any other way to justify the above mentioned statement. Anyone could please tell me if am I wrong? If yes, could please help me to understand why the above statements are true?

Answer 1

Ad 1: As $u_n\rightarrow u$ weakly in $X^{s,p}(\Omega)$, we have $u\in X^{s,p}(\Omega)$.
Ad 2: Considering the function $f:\mathbb{R}^{2n}\rightarrow\mathbb{R}$, $(x,y)\mapsto \frac{\vert u(x)-u(y) \vert^p}{\vert x-y \vert^{n+sp}}$, we see that $$\Vert f \Vert_{L^1(\mathbb{R}^{2n})} = \iint_{\mathbb{R}^n\times\mathbb{R}^n} \vert f(x,y) \vert dxdy = \iint_{\mathbb{R}^n\times\mathbb{R}^n}\frac{\vert u(x)-u(y) \vert^p}{\vert x-y \vert^{n+sp}} dxdy = [u]_{s,p}^p.$$ And since $u\in X^{s,p}(\Omega)$, the right-hand side is finite which entails that $f\in L^1(\mathbb{R}^{2n})$.