Does the following estimate hold for any value of $p \in [1,+\infty)$?
$$\int\int_{\mathbb{R}^N \times \mathbb{R}^N} \dfrac{|u(x) - u(y)|^p}{|x - y|^{N + sp}}dxdy \leq 2\left[\left(\int\int_{\mathbb{R}^N \times \{|y| \geq 2|x|\}} \dfrac{|u(x)|^p}{|x - y|^{N + sp}}dxdy\right)^{1/p} + \left(\int\int_{\mathbb{R}^N \times \{|y| \geq 2|x|\}} \dfrac{|u(y)|^p}{|x - y|^{N + sp}}dxdy\right)^{1/p}\right]^{p} + 2\int\int_{\mathbb{R}^N \times \{|y| < 2|x|\}} \dfrac{|u(x) - u(y)|^p}{|x - y|^{N + sp}}dxdy$$
Based on some calculations performed using the traditional inequality $|a + b|^p \leq 2^p(|a|^p + |b|^p)$, I noticed that the constant $2^p$ appears in the first part of the second inequality expression. This presence of the constant $2^p$ represents a problem in my specific case.
Yes, it followsfrom the triangle inequality. Let me write $K = 1/|x-y|^{s+p/N}$, $\Omega = \Bbb R^N\times\Bbb R^N$, $\Omega_1 = \{(x,y)\in \Omega, |y|\geq 2|x|\}$ and $\Omega_2 = \Omega \setminus\Omega_1$. Then cutting the integral in two parts gives $$ \iint_\Omega K^p |u(x)-u(y)|^p \\ = \iint_{\Omega_1} K^p |u(x)-u(y)|^p + \iint_{\Omega_2} K^p |u(x)-u(y)|^p $$ Now the first integral of the right hand side can be written $$ \|Ku(x)-Ku(y)\|_{L^p(\Omega_1)}^p. $$ And so by the triangle inequality is bounded by $$ (\|Ku(x)\|_{L^p(\Omega_1)} + \|Ku(y)\|_{L^p(\Omega_1)})^p. $$ This gives even better than what you want since there is no $2$ in front.