HLS inequality not suitable to bound this integral

Question

I am trying to bound the following integral for $f \in L^{n/2}(\mathbb{R}^n)$: $\int_{\mathbb{R}^n}\int_{\mathbb{R}^n} f(x) \lvert x-y \lvert^{2(2-n)}f(y)dxdy$. Because of the factor 2 in the exponent, the condition $0<\lambda<n$ in the Hardy-Littlewood-Sobolev inequality fails for $\lambda=2n-4$ and $n>3$. Is there another way to bound this integral?

Answer 1

I'm probably missing something, but how about this: Suppose $0<f$ is continuous (don't even worry about integrability for now). Fix $x\in \mathbb{R}^n$ and let $r>0$ be such that $f(y)>f(x)/2>0 $ for every $y\in B:=B_\delta(x)$. Then use polar coordinates to write \begin{equation} \begin{split} \int_{\mathbb{R}^n} |x-y|^{-\alpha} f(y)\, dy & \geq \dfrac{f(x)}{2}\int_B |x-y|^{-\alpha}\, dy\\ & = \dfrac{f(x)}{2} \int_{B_\delta(0)} |y|^{-\alpha}\, dy. \end{split} \end{equation} So if $\alpha>n$ (as is the case for you if $n>3$), then the first integral is $\infty$ for any $x$, so the whole integral is infinite in this case.