Using *Hardy-Littlewood-Sobolev* inequality, we can prove that:
$$\left| \int\_0^1\int\_0^1 |x-y|^{-\frac{1}{2}} f(x)f(y) \mathrm{d}x \mathrm{d}y \right| \leq C \left\| f \right\|\_{L^{4/3}(0,1)}^2 \leq C \left\| f \right\|\_{L^{2}(0,1)}^2.$$


However, the left-hand side looks very similar to the singular integrals used to define fractional Sobolev norms (actually, *Gagliardo* seminorms), maybe for a negative fractional index. Thus, I would expect that we could actually prove something like:
$$\left| \int\_0^1\int\_0^1 |x-y|^{-\frac{1}{2}} f(x)f(y) \mathrm{d}x \mathrm{d}y \right| \leq C \left\| f \right\|\_{H^{-1/4}(0,1)}^2,$$
where $H^{-\frac{1}{4}}(0,1)$ is defined either as the dual of the usual fractional Hilbert space $H^{\frac{1}{4}}(0,1)$ or using *Fourier* series expansions for functions on $(0,1)$.


**Is this second inequality true? If so, where should I be looking for its proof?** 



Hardy-Littlewood-Sobolev inequality using fractional sobolev norm on the RHS
One has, for $f,\,g\in \dot{H}^{-1/4}(\mathbb R)$,
$$
\begin{aligned}
 \left|\int\_{-\infty}^\infty\int\_{-\infty}^\infty |x-y|^{-1/2}f(x)
 \,\overline{g(y)}\,dx dy\,\right| &= C\_0 \left|\left((-\Delta)^{-1/4}f,g\right)
 \right| \\ 
 & \leq C\_0\|(-\Delta)^{-1/4}f\|\_{\dot{H}^{1/4}}\|g\|\_{\dot{H}^{-1/4}} 
 = C\_0\|f\|\_{\dot{H}^{-1/4}} \|g\|\_{\dot{H}^{-1/4}},
\end{aligned}
$$
where $(\;,\,)$ denotes the $L^2$ scalar product and $C\_0>0$ is some constant. Now use the fact that
$$
 H^{-1/4}((0,1))= \{f\in \dot{H}^{-1/4}(\mathbb R)\mid
 \operatorname{supp}f\subseteq [0,1]\}.
$$


