Let $a\in L^{\infty}(\mathbb{R}^N)$ with $a^{+}\neq 0$, here $a^{+}=\max\{a(x),0\}$. Given $$ f(u)=\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{a(x)u^{2}(x)a(y)u^{2}(y)}{|x-y|^{N-2}}dxdy,~~\forall u\in H^{1}(\mathbb{R}^N).$$ Show that there exist a function $\varphi\in H^{1}(\mathbb{R}^N)$ such that $f(\varphi)>0$.
I tried to assume that for all $\varphi\in H^{1}(\mathbb{R}^N)$ such that $f(\varphi)\leq 0$, but I can't find a contradiction.
It is not true. Take $$a(x) = \begin{cases} 1,\ x=0 \ \\ 0, \ x\neq 0\end{cases}$$ Then $a^+ = 1$, but $f(\phi) = 0$ for any $\phi \in H^1(\mathbb{R}^n)$. However if you assume that $a > C > 0$ on some set of positive measure $A \subset \mathbb{R}^N$. Then let $\phi = \chi_A$, where $\chi_A$ is the characteristic function of $A$. Then $$f(\phi) = \int_A\int_A \frac{a(x)a(y)}{|x-y|^{N-2}}dydx > C \int_A \int_{A}\frac{1}{|x-y|^{N-2}}dydx >0$$ Note that the Hilbert transform is well defined on $C^1_c$, so the above integral converges.