Let $0<s<1$, $p>1$ and $n>sp$. For $u\in C_0^{\infty}(\mathbb R^n))$ let $$ [u]_{s, p} =\left(\iint_{\mathbb R^n} \frac{|u(x)-u(y)|^p}{|x-y|^{n+sp}}\right)^{\frac1p}$$ be the ussual Gagliardo seminorm of $u$. Let us define $$D^{s, p}(\mathbb R^n)=\{ f\in L^{p^*_s}(\mathbb R^n) : [u]_{s, p}<+\infty\},$$ where $p^*_s$ is the fractional critical exponent $p^*_s=np/(n-sp)$.
For $p\in (1, \infty)$, the space $D^{s, p}(\mathbb R^n)$ is a separable and reflexive Banach spaces with respect to the Gagliardo seminorm. Moreover, it can also be understood as the respective completions of the space $C_0^\infty(\mathbb R^n)$ with respect to $[\cdot]_{s, p}$.
The question is: if $u\in D^{s, p}(\mathbb R^n)$, hence one can say that $u\in C(\mathbb R^n)$. There exists any embedding theorem or something like that?
When $sp>n$, if the Gagliardo seminorm is finite, you can prove that $u$ has a Holder continuous representative. But in the range $sp<n$ you can have discontinuous functions as ToGle wrote in the comments. Take $n=1$ and $u=\chi_{[0,1]}$. You should get two double integrals. One is $$2\int^1_0\int^\infty_1\frac1{(x-y)^{1+sp}}dxdy=\frac2{sp}\int^1_0\frac{1}{(1-y)^{sp}}dy<\infty$$ and the other is $$2\int^1_0\int_{-\infty}^0\frac1{(y-x)^{1+sp}}dxdy=\frac2{sp}\int^1_0\frac{1}{y^{sp}}dy<\infty$$ When $n>1$ take the characteristic function of a cube although the calculation is more complicated.