A scaled fractional ''Sobolev inequality'' Does a fractional interpolation inequality similar to $$
\int_{B_R(0)} |u| dx \le C R^{2} \sqrt{\log(2R)} \Big( \int_{\mathbb R^2}\int_{\mathbb R^2} \frac{|u(x)-u(y)|^2}{|x-y|^{2+2s}} dxdy + \int_{B_1(0)} |u|^2 dx \Big)^{1/2}
$$
(where $B_R(0), B_1(0) \subset \mathbb R^2$) hold? Maybe changing the powers of $R$ or of the logarithm if needed?
Is it known in the literature?
 A: Then, unless there are more typos, what you request is very cheap and no logarithmic factor is needed.
WLOG, $u\ge 0$ (just consider $|u|$ instead, if not).
Denote by $A_1$ the average of $u$ over the unit disk centered at $0$ and by $A_r$ ($r=2,4,8,16,\dots$) the average of $u$ over the annulus $\frac r2\le|x|\le r$. Then, assuming that the expression in parentheses on the $RHS$ is $1$, we get $A_0\le C$ and $|A_r-A_{\frac r2}|^2 \frac {r^4}{r^{2+2s}}\le C$ (just restrict the double integration to $x\in A_{\frac r2}, y\in A_r$ and use the trivial upper bound on the denominator and Cauchy-Schwarz). Hence $|A_r-A_{\frac r2}|\le Cr^{s-1}$ and, summing a geometric progression, we get that $A_r\le C'$ for all $r$, i.e., the integral of $u$ over any disk centered at the origin is at most constant times its area.
Edit: the details for the difference estimate
Let $Q_r$ be the region (annulus or disk) over which $A_r$ is taken.
Then
$$
|A_r-A_{r/2}|=\frac{1}{Area(Q_r)Area(Q_{r/2})}\left|\iint_{x\in Q_r, y\in Q_{r/2}}(u(x)-u(y))\,dxdy\right|\le
\\
(2r)^{1+s}\frac{1}{Area(Q_r)Area(Q_{r/2})}\iint_{x\in Q_r, y\in Q_{r/2}}\frac{|u(x)-u(y)|}{|x-y|^{1+s}}\,dxdy \le
\\
(2r)^{1+s}\left[\frac{1}{Area(Q_r)Area(Q_{r/2})}\iint_{x\in Q_r, y\in Q_{r/2}}\left(\frac{|u(x)-u(y)|}{|x-y|^{1+s}}\right)^2\,dxdy\right]^{1/2}
\\
\le\frac{(2r)^{1+s}}{\sqrt{Area(Q_r)Area(Q_{r/2})}}\left[\iint_{x\in\mathbb R^2, y\in \mathbb R^2}\frac{|u(x)-u(y)|^2}{|x-y|^{2+2s}}\,dxdy\right]^{1/2}
$$
and it remains to note that $\sqrt{Area(Q_r)Area(Q_{r/2})}\ge cr^2$.
