# Hardy-Littlewood-Sobolev inequality using fractional sobolev norm on the RHS

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?

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 $$H^{-1/4}((0,1))= \{f\in \dot{H}^{-1/4}(\mathbb R)\mid \operatorname{supp}f\subseteq [0,1]\}.$$
• Thanks. This answer is great! What reference would you suggest for fractional Laplacian and homogeneous Sobolev spaces? As a subsidiary question, I was wondering if the opposite inequality could also be proved, maybe with an additional hypothesis on $f$? That is, my goal would be to prove that we also have: $$\left| \int_0^1\int_0^1 |x-y|^{-\frac{1}{2}} f(x)f(y) \mathrm{d}x \mathrm{d}y \right| \geq C \left\| f \right\|_{H^{-1/4}(0,1)}^2.$$ – F.M. Apr 14 '15 at 19:15
• Indeed, $H^{-1/4}((0,1))$ is a closed subspace of $\dot{H}^{-1/4}(\mathbb R)$, so $\|(-\Delta)^{-1/8}f \|_{L^2}$ is an equivalent norm on $H^{-1/4}((0,1))$. Furthermore, the operator $(-\Delta)^{-1/8}$ is selfadjoint. Hence, $\left((-\Delta)^{-1/4}f,f\right) =\left((-\Delta)^{-1/8}f,(-\Delta)^{-1/8}f\right) = \left\|(-\Delta)^{-1/8}f\right\|_{L^2}^2\eqsim \|f\|_{H^{-1/4}((0,1))}^2$. – ifw Apr 14 '15 at 22:40