Show that $T$ is compact

Question

Let $H=L^2(B)$, where $B$ is the closed unitary ball in $\mathbb{R^d}$ and consider $K(x,y)$ a measurable functions in $B\times B$ such that $$|K(x,y)|\leq C|x-y|^{-d+\alpha}, \ \ x,y\in B\ \ (\alpha>0).$$

Now define $T:H\to H$ by $$Tf(x)=\int_{B}K(x,y)f(y)dy$$

How can I show that T is compact? I know that if $\alpha>d/2$ then $K(x,y)\in L^2(B\times B)$, so I can conclude that T is a Hilbert-Schmidt operator (and ergo compact). But what can I do in the case $\alpha\leq d/2$?

Answer 1

Define the operator $T_n$ given by $$T_nf(x)=\int_{B}K_n(x,y)f(y)dy$$ where $K_n(x,y)=K(x,y)$ if $|x-y|\geq 1/n$ and 0 otherwise.

Notice that $K_n\in L^2(B\times B)$ and so $T_n$ is a Hilbert-Schmidt operator (and therefore Compact).

Since $$\int_{B}|K_n(x,y)-K_(x,y)|dy\leq C\int_{|x-y|\leq 1/n}|x-y|^{-d+\alpha}dy\to 0 $$ we can conclude that $T$ is compact.