parameter operator $A_a$ is compact?? I need some help in this exercise.
Let define operator on $ L^2[0,1]$:
$$ A_af(x)=\int_{0}^{1}{|x-y|}^{a-1} f(y)dy $$ for f $\in L^2[0,1] $.
Prove that A_a is compact for all $a>0$.
I see that operator is bounded  for a>1/2 using Hoelder inequalty(as first what I must show).
But for a<1/2 I don't see that operator is bounded.
Do I must use Arcela-Ascoli thm and how to use it on operator?
Thank's a lot.
 A: Let's begin with showing that $g(x,y) := |x-y|^{a-1}$ is in $L^2(\Omega)$ where $\Omega := (0,1)\times(0,1)$. For this purpose we define for $\epsilon > 0$
$$ g_{\epsilon} := g(x,y) \cdot \mathbf{1}_{(0,x-\epsilon) \cup (x+\epsilon,1)}(y)$$
We calculate
$$ \int_{0}^{1} g_{\epsilon}^{2} dy =\int_{0}^{x-\epsilon} (x-y)^{2(a-1)} dy + \int_{x+\epsilon}^{1} (y-x)^{2(a-1)} dy = - \frac{1}{2a-1} (\epsilon^{2a-1} - x^{2a-1}) + \frac{1}{2a-1} ((1-x)^{2a-1} - \epsilon^{2a-1}) = \frac{(1-x)^{2a-1}+x^{2a-1} - 2\epsilon^{2a-1}}{2a-1}  $$
Because $g_{\epsilon} \nearrow g$ as $\epsilon \to 0$ by the monotone convergence theorem we can take the limit of $\epsilon \to 0$ of this expression to  see that $g(x,\cdot) \in L^2(0,1)$ for all $x \in (0,1)$. 

EDIT: had a sign error in the integration before, this only works for $a > \frac{1}{2}$.

Integrating this over $(0,1)$ (which is not hard to do) gives us that $g(x,y) \in L^{2}(\Omega)$.
Now we know that $g$ can be approximated by step functions of the form $$ \sum_{i=1}^{n} a_i \cdot \mathbf{1}_{C_i}(x,y)$$ where $C_{i}$ is a subset of $\Omega$ that is measurable with respect to the product $\sigma$-algebra. We can in turn approximate the measure of each such $C_{i}$ by sets of the form $A_{i} \times B_{i}$ where $A_{i},B_{i}$ are measurable subsets of $(0,1)$. We conclude that the functions of the form $$\sum_{i=1}^{n} a_{i} \cdot \mathbf{1}_{A_i\times B_i}(x,y) = \sum_{i=1}^{n} a_{i} \cdot \mathbf{1}_{A_i}(x)\mathbf{1}_{B_i}(y)$$ with $n \in \mathbb{N}$ and $A_i,B_i$ defined as above are dense in $L^2(\Omega)$. Now let $t_n$ be a sequence of such step functions approximating $g$ from below. We define 
$$T_{n} : L^2(0,1) \to L^2(0,1) \quad Tf(x) := \int_{0}^{1}t_{n}(x,y)f(y) dy$$ 
It is easy to see that the range of each $T_n$ is finite dimensional and that each $T_n$ is therefore compact. An application of the Cauchy-Schwarz inequality shows you that $T_n \to A_a$ in the operator topology and therefore that $A_a$ must be compact.
