Let $s\in (1/2, 1)$ and $f\in H^s(\mathbb R^n)$. By definition, $f\in L^2(\mathbb R^n)$ and $[f]_s<+\infty$, where $[\cdot]_s$ denotes the Gagliardo seminorm.
Assume that there is a constant $K>0$ such that $$[f]_s\le K.$$
Is there any way to use this information to prove that there exists $c>0$, possibly depending on $K$, such that $$\|f\|_{L^2(\mathbb R^n)}\le c?$$
I mean, if I know that the Gagliardo seminorm of $f$ is bounded by $K$, does this give any information about the bound of the $L^2$-norm of $f$?
I am looking for some hint here Hitchhiker's guide to the fractional Sobolev spaces, but I do not have success so far.
I think this is not true. The Gagliardo seminorm only measures the regularity part of the Sobolev norm and so cannot control the ``mass'' part of it. To construct a counter example I am going to use Proposition 3.4 from the paper in your question which provides a Fourier representation of the seminorm: $$ [f]_s^2 = \int_{\mathbb{R}} |\xi|^{2s}|\mathcal{F}(f)|^2(\xi) \, \mathrm{d}\xi \, , $$ where $\mathcal{F}(f)$ is the Fourier transform of $f$ which I will assume w.l.o.g is zero inside $[-1,1]$ (if not I can always subtract it away and I am still left with an $H^s$ function because the integral will still be finite). Consider a bump function $\varphi$ which is nonnegative, compactly supported in the ball of size $1$ around $0$, and has integral one. Rescale it as follows $\varphi_{\epsilon}= \epsilon^{-1}\varphi(x/\epsilon)$ for some $\epsilon>0$ so that it has support in $[-\epsilon,\epsilon]$. Now consider the function $f_\epsilon = \mathcal{F}^{-1}(\mathcal{F}(f) + \varphi_{\epsilon})$. We then have that \begin{align} [f_{\epsilon}]_{s}^2 = &\int_{\mathbb{R}}|\xi|^{2s}|\mathcal{F}(f_{\epsilon})|^2 (\xi)^2 \\ =& \int_{\mathbb{R}}|\xi|^{2s}|\mathcal{F}(f)|^2 (\xi)^2 + \int_{-\epsilon}^\epsilon|\xi|^{2s}|\varphi_\epsilon|^2 (\xi)^2 \\ \lesssim & K + \epsilon^{2s-1} \, , \end{align} where have used the fact that $|\mathcal{F}(f)|$ and $\varphi_{\epsilon}$ have disjoint supports, that $\varphi_{\epsilon} \leq \epsilon^{-1}$, and that $|\xi|^{2s} \leq \epsilon^{2s}$ on $[-\epsilon,\epsilon]$. On the other hand the $L^2$-norm is given by \begin{align} \|f_{\epsilon}\|_{L^2}^2 = &\int_{\mathbb{R}}|\mathcal{F}(f_{\epsilon})|^2 (\xi)^2 \\ =& \|f\|_{L^2}^2 + \int_{\mathbb{R}} |\varphi_\epsilon|(\xi)^2 \mathrm{d}\xi \\ \sim & \|f\|_{L^2}^2 +\epsilon^{-1} \, . \end{align} Thus, if $s>1/2$ it is clear that one bound is stable while the other blows up. There is probably a simpler construction than this to show the same result.