Singular integral bounded by Dirichlet form?

Question

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 $H_0^1(\Omega).$

Answer 1

$\newcommand{\Om}{\Omega}$The answer is no.

Indeed, by rescaling, without loss of generality $L=1$. Suppose that \begin{equation} f(x_1,x_2)=g_a(x_1-x_2) \end{equation} for $(x_1,x_2)\in\Om$, where $g_a\in W^{1,2}((-2,2)^2)$ with support in $[-1,1]^2$ such that $g_a(u):=|u|^a$ for a real $a>0$ and all $u\in(-1/2,1/2)^2$.

So, if your inequality were true, then we would have \begin{equation*} L(a):=\int_{(-1,1)^2} \frac{g_a(u)^2}{|u|^2}\,du \\ \le C_1 \int_{(-1,1)^2} (g_a(u)^2+|\nabla g_a(u)|^2)\,du=:R(a) \tag{1}\label{1} \end{equation*} for some real $C_1>0$ and all real $a>0$.

However, for $a\downarrow0$ we have $L(a)\asymp1/a$ whereas $R(a)\asymp1$, and hence \eqref{1} fails to hold.