This is an exercise in my course of harmonic analysis: assume that $f:\mathbb{R}^n\rightarrow \mathbb{R}$ for $n\geq 3$ satisfies $$\lim_{r\rightarrow 0}\sup_{x\in \mathbb{R}^n} \int_{|x-y|<r} \frac{|f(y)|}{|x-y|^{n-2}}\,dy=0.$$ Prove that there exists a constant $c_0>0$ such that for all $u\in C_c^{\infty}(\mathbb{R}^n)$, we have $$\int_{\mathbb{R}^n} |f(x)|^2|u(x)|^2\,dx\leq \frac{1}{4}\|\nabla u\|_2^2+c_0(f,n)\|u\|_{2}^2.$$ Note that $c_0(f,n)$ is independent of $u$ but may depends on $f$ and $n$.
From the condition of $f$ holds, it seems reasonable to consider the potential estimate for compact support smooth function $v$: $$v(x)=C(n)\int_{\mathbb{R}^n}\frac{\nabla v(y)\cdot (x-y)}{|x-y|^n}\,dy.$$ Let $v=u^2$, we get $\nabla v(y)=2u(y)\nabla u(y)$, and $$\int_{\mathbb{R}^n} |f(x)|^2|u(x)|^2\,dx=2C(n)\int_{\mathbb{R}^n}\left( \int_{\mathbb{R}^n} \frac{|f(x)|^2 (x-y)}{|x-y|^n}\,dx\right) \cdot \nabla u(y) u(y)\,dy$$ But I am stuck in here for controlling of the term $\int_{\mathbb{R}^n} \frac{|f(x)|^2 (x-y)}{|x-y|^n}\,dx$, so do you have any idea? Any help will be appreciated.
