i got the following problem to solve.
Let $0 < \alpha < 1$, $L \in L_\infty([0,1]^2)$, $D = \{(x,y) \in \mathbb{R}^2: x = y\}$ the diaagonal of $\mathbb{R}^2$ and $k:[0,1]^2 \setminus D \to \mathbb{C}$, $k(x,y) = \frac{L(x,y)}{\vert x-y \vert^\alpha}$.
Then $T_k: L_2[0,1] \to L_2[0,1], (T_kf)(x) = \int_0^1 k(x,y) f(y) dy$ is well-definied and compact.
For the first part my Idea was to show that $k(x, \cdot) \in L_2[0,1]$, so i can use Cauchy–Schwarz inequality and Fubini thereom to show that $T_kf \in L_2[0,1]$. Hence I would like show that $\int_0^1 \vert k(x,y) \vert^2 dy < \infty$ for a fixed $x \in [0,1]$. But I'm not sure how to bound $\int_0^1 \frac{1}{\vert x - y \vert^{2 \alpha}} dy$ for a fixed $x \in [0,1]$. Am I on the right track or is this idea useless? And if I'm on the right track, how can i show that? For the second part i got no real idea, how you can show easily that $T_k$ is a compact operator. I would be grateful for some help on the topic.
Let $C=\max\{\|L\|_{\infty}^2,1\}$. Assuming that $\alpha<\frac12$, we may compute \begin{align} \int_0^1\int_0^1|k(x,y)|^2\ \mathsf dy\ \mathsf dx &= \int_0^1\int_0^1 L(x,y)^2|x-y|^{-2\alpha}\ \mathsf dy\ \mathsf dx\\\\ &\leqslant C \int_0^1\int_0^1 |x-y|^{-2\alpha}\ \mathsf dy\ \mathsf dx\\ &= 2C\int_0^1\int_0^x (x-y)^{-2\alpha}\ \mathsf dy\ \mathsf dx\\ &= 2C\int_0^1 \left(\frac{x^{1-2 \alpha }}{1-2 \alpha }\right)\ \mathsf dx\\ &= \frac C{(1-\alpha)(1-2\alpha)}<\infty, \end{align} so that $L_k$ is a Hilbert-Schmidt operator. It follows that $L_k$ is compact.