Let $n\ge 2$, $0<s_1<s_2 <1$ and $(u_n)_n$ be a sequence such that $$u_n\to 0 \text{ in } L^r(\Omega), \ \ r\in [1, 2^*_{s_2})$$ where $2^*_{s_2} = \frac{2n}{n-2s_2}$ is the critical Sobolev exponent for $s_2$ and $$u_n\to 0 \text{ a.e.}$$
Let $[u_n]_{s_i}$, $i\in\{1, 2\}$ be the Gagliardo seminorm of $u_n$, i.e. $$ [u_n]^2_{s_i}=\iint_{\mathbb R^{2n}} \frac{|(u_n(x) -u_n(y)|^2}{|x-y|^{n+2s_i}} dx dy.$$
I am trying to prove that (but I am not sure if that's true) $$\frac{[u_n]^2_{s_1}}{[u_n]^2_{s_2}} = o(1) \quad\text{ as } n\to +\infty.$$
$\textbf{Notation:}$ we say that $f(x) = o(g(x))$ as $x\to x_0$ if $\lim_{x\to x_0} \frac{f(x)}{g(x)} =0.$
The only thing I got in mind so far is to use Proposition 2.2 (with $\Omega =\mathbb{R}^n$) of the paper Hitchhiker’s guide to the fractional Sobolev spaces to say that $[u_n]^2_{s_1} \le c [u_n]^2_{s_2}$ where $c=c(n, s_1)$ to say that $$\frac{[u_n]^2_{s_1}}{[u_n]^2_{s_2}}\le 1,$$ but it does not help me to say or not if $\frac{[u_n]^2_{s_1}}{[u_n]^2_{s_2}} = o(1)$.
I hope someone could help. Thank you.
As it is currently stated, the conjecture is false.
Take $\phi \in C^\infty_c(\Omega)$ a non-zero smooth function. Let $u_n := \frac{1}{n} \phi$. On the one hand, $u_n \to 0$ in any $L^r(\Omega)$ (and in fact, in any space). On the other hand $$ \frac{[u_n]_{s_1}^2}{[u_n]_{s_2}^2} = \frac{[\phi]_{s_1}^2}{[\phi]_{s_2}^2} \neq o(1). $$ You probably want to add another assumption preventing this situation.