A variant of Hardy's inequality for "convolutions"?

Question

Consider Hardy's inequality on $L^{2}(\mathbb{R}^3)$. This inequality states that: $$\int_{\mathbb{R}^3} dx \, \frac{|\psi(x)|^2}{|x|^2} \le K \int_{\mathbb{R}^3} dx \, |\nabla \psi(x)|^2.$$ I want to understand how/if I can use Hardy's inequality to bound integrals of the form: $$\int_{\mathbb{R}^3 \times \mathbb{R}^{3}} \, dx \, dy \, \frac{|\psi(x)|^2|\varphi(y)|^2}{|x-y|^2}.$$ Do we have some formula as well? For instance, I would expect something like: $$\int_{\mathbb{R}^3 \times \mathbb{R}^3} dx\, dy \frac{|\psi(x)|^2|\varphi(y)|^2}{|x-y|^2} \le K \int_{\mathbb{R}^{3}\times \mathbb{R}^3} dx \, dy \, |\nabla \psi(x)|^2|\varphi(y)|^2$$ or something like this? The point is that I can't deal with this translation $|x-y|$ in the denominator of the second expression.

This question reduces to the following equivalent question. Does Hardy's inequality hold for translations? $$\int_{\mathbb{R}^3} dy \, \frac{|\psi(x+y)|^2}{|y|^2} \le \int_{\mathbb{R}^3} dy \, |\nabla \psi(x+y)|^2 $$ for each fixed $x$? If the answer is yes, then the first inquality follows from translation invariance of the integral.

Answer 1

I believe that for $n=3$ the optimal Hardy inequality is $$ \int_{\mathbb R^3}\vert(\nabla w)(y)\vert^2 dy\ge \frac14 \int_{\mathbb R^3}\frac{\vert w(y)\vert^2}{\vert y\vert^2} dy, $$ say for $w$ in the Schwartz space. Then applying this to $w(y)=v(y+x_0)$ where $v$ is in the Schwartz space you get $$ \int_{\mathbb R^3}\vert(\nabla v)(y+x_0)\vert^2 dy\ge \frac14 \int_{\mathbb R^3}\frac{\vert v(y+x_0)\vert^2}{\vert y\vert^2} dy. $$ Indeed, derivatives are commuting with translations.