We define for some fixed $L$ $$\Omega:=\{(x_1,x_2) \in [-L,L]^2 \times [-L,L]^2 \setminus \{x_1=x_2\}\}.$$ Let $f \in C_c^{\infty}(\Omega)$, then I am wondering if we have the bound for some universal $C$ depending only on $\Omega$ $$ \int_{\Omega} \frac{\vert f(x)\vert^2}{\vert x_1-x_2 \vert^2} \ dx_1 \ dx_2 \le C (\Vert f \Vert^2_{L^2} + \Vert \nabla f \Vert^2_{L^2}).$$

At first there seems to be an issue with the integral on the left-hand side namely that $1/|x|^2$ is not integrable in two dimensions, so any $f(x)=g(x_1-x_2)$ with $g(0)\neq 0$ would yield infinity on the left-hand side. So one might try to approximate such $f$ by $C_c^{\infty}(\Omega).$

However, the expression on the right-hand side is the Dirichlet form and thus anything I can expect to get in the limit should vanish somehow on $\{x_1=x_2\},$ i.e. has to be in $H_0^1(\Omega).$ I am however not even sure if the left-hand side makes sense for objects in $H_0^1(\Omega).$

We define for some fixed $L$ $$\Omega:=\{(x_1,x_2) \in ([-L,L]^2 \times [-L,L]^2) \setminus \{x_1=x_2\}\},$$ in particular $x_1,x_2 \in \mathbb R^2.$ Let $f \in C_c^{\infty}(\Omega)$, then I am wondering if we have the bound for some universal $C$ depending only on $\Omega$ $$ \int_{\Omega} \frac{\vert f(x)\vert^2}{\vert x_1-x_2 \vert^2} \ dx_1 \ dx_2 \le C (\Vert f \Vert^2_{L^2} + \Vert \nabla f \Vert^2_{L^2}).$$

At first there seems to be an issue with the integral on the left-hand side namely that $1/|x|^2$ is not integrable in two dimensions, so any $f(x)=g(x_1-x_2)$ with $g(0)\neq 0$ would yield infinity on the left-hand side. So one might try to approximate such $f$ by $C_c^{\infty}(\Omega).$

However, the expression on the right-hand side is the Dirichlet form and thus anything I can expect to get in the limit should vanish somehow on $\{x_1=x_2\},$ i.e. has to be in $H_0^1(\Omega).$ I am however not even sure if the left-hand side makes sense for objects in $H_0^1(\Omega).$