Let bounded domain $X \subset \mathbb{R^2}$ , $$T:L^2(X) \to L^2(X) \ \ Tf(x)=\int _{X} \frac{f(t)}{|x-t|}dt $$ $X_r (x)=X \cap \{t \in \mathbb{R}; |t-x| \geq r\}$ ,and $$T_r :L^2(X) \to L^2(X) \ \ Tf(x)=\int _{X_r} \frac{f(t)}{|x-t|}dt $$ Then $\lim _{r \to 0} ||T-T_r||_{L^2 (X)} =0$ and $T$ is compact operator.
I know that I can write $Tf(x)=g *f(x)$ using $g(y)=\frac{1}{|y|}$ and convolution, so $Tf, T_r f \in L^2 (X)$ by Young's inequality (it means these operators are well-defined). But I can't use Young's inequality to $||(T-T_r)f||$, so I can't prove $\to 0$.
I know that if for any weak convergence to 0 sequence $\{f_n\}$ $\{Tf_n\}$ strong convergence to 0,then $T$ is compact operator. But in this case, I can't use it, so someone has idea?
Remark: $x,t \in \mathbb{R^2}$ not $\mathbb{R}$
If you let $g(x)= {1 \over \|x\|}$, then $g \in L^2(X)$ (since $X$ is a bounded subset of the plane).
Let $K(x,t) = g(x-t)$, then $K \in L^2(X \times X)$, and so we see that $T$ is a Hilbert Schmidt operator $(Tf)(x) = \int K(x,t ) f(t) dt$ and hence it is compact.
To show that $T_r \to T$, we can write $T f = f * g$ and $T_r f = f * (g \cdot 1_{B(0,r)^c})$. If $\|f\ \|_2\le 1$, Cauchy Schwarz shows that $|(Tf)(x)-(T_r f)(x)| \le \|f\|_2 \|g-g \cdot 1_{B(0,r)^c}\|_2 \le \|g-g \cdot 1_{B(0,r)^c}\|_2$, hence we see that $T_r \to T $.