Operator on $L^2 (0,1)$ defined by convolution with $|x-y|^{-\alpha}$ Define $A: L^2 (0,1) \to L^2(0,1)$
$$Af(x) = \int_0^1 f(y) \frac{1}{|x-y|^\alpha} dy \quad , \quad \alpha \in (0,1)$$

For what values of $\alpha$ is it well defined? Bounded? Compact?

I tried doing the appropriate integrals but I really don't understand what am I supposed to do in this specific case.
HINT:

The Cauchy-Schwarz Inequality reveals that

$$\begin{align}
\left|Af(x)\right|^2 &amp;= \left|\int_0^1 f(y) \frac{1}{|x-y|^\alpha} dy \right|^2\\\\
&amp;\le \int_0^1\left|f(y)\right|^2\,dy\,\int_0^1\frac{1}{|x-y|^{2\alpha}} dy
\end{align}$$

And thus, the square of the operator norm is 

$$\begin{align}
||A||_2^2&amp;=\sup_{f\in \mathscr{L}^2} \left(\frac{\int_0^1\left|\int_0^1 f(y) \frac{1}{|x-y|^\alpha} dy \right|^2dx}{\int_0^1\left|f(y)\right|^2\,dy}\right)\\\\
&amp;\le \int_0^1\int_0^1\frac{1}{|x-y|^{2\alpha}}dydx
\end{align}$$
Operator is well-defined and bounded for all $\alpha\in(0,1)$: $$\|Af\|^2=\int\limits_0^1\left|\int\limits_0^1\frac{f(y)}{|x-y|^\alpha}dy\right|^2dx=\int\limits_0^1\left|\int\limits_0^1\frac{1}{|x-y|^\frac{\alpha}{2}}\frac{f(y)}{|x-y|^\frac{\alpha}{2}}dy\right|^2dx\leq$$$$\leq\int\limits_0^1\int\limits_0^1\frac{dy}{|x-y|^\alpha}\int\limits_0^1\frac{|f(y)|^2dy}{|x-y|^\alpha}dx.$$$$\int\limits_0^1\frac{dy}{|x-y|^\alpha}=\int\limits_0^x\frac{dy}{(x-y)^\alpha}+\int\limits_x^1\frac{dy}{(y-x)^\alpha}=\frac{x^{1-\alpha}}{1-\alpha}+\frac{(1-x)^{1-\alpha}}{1-\alpha}\leq\frac{2}{1-\alpha}.$$$$\|Af\|^2\leq\frac{2}{1-\alpha}\int\limits_0^1\int\limits_0^1\frac{|f(y)|^2dy}{|x-y|^\alpha}dx=\frac{2}{1-\alpha}\int\limits_0^1|f(y)|^2\int\limits_0^1\frac{dx}{|x-y|^\alpha}dy\leq\frac{4}{(1-\alpha)^2}\|f\|^2.$$

Operator is compact for all $\alpha\in(0,1)$. If $0&lt;\alpha&lt;\frac{1}{2}$ this operator is a Hilbert-Schmidt operator and therefore compact. Consider the case $\frac{1}{2}\leq\alpha&lt;1$. 

Consider the operator $A^2f(t)=\displaystyle\int\limits_0^1\left(\int\limits_0^1\dfrac{ds}{|t-s|^\alpha|s-\tau|^\alpha}\right)f(\tau)d\tau$ and let $k(t,\tau)=\displaystyle\int\limits_0^1\dfrac{ds}{|t-s|^\alpha|s-\tau|^\alpha}$. Performing first a replacement $t-s=\xi$ and then a replacement $\xi=|t-\tau|\eta$, we obtain: $$k(t,\tau)=\displaystyle\int\limits_{t-1}^t\dfrac{d\xi}{|\xi|^\alpha|t-\tau-\xi|^\alpha}\leq\int\limits_{-1}^1\dfrac{d\xi}{|\xi|^\alpha|t-\tau-\xi|^\alpha}=|t-\tau|^{1-2\alpha}\int\limits_{-\frac{1}{|t-\tau|}}^{\frac{1}{|t-\tau|}}\dfrac{d\eta}{|\eta|^\alpha\left|\frac{t-\tau}{|t-\tau|}-\eta\right|^\alpha}\leq$$$$\leq|t-\tau|^{1-2\alpha}\int\limits_{-\frac{1}{|t-\tau|}}^{\frac{1}{|t-\tau|}}\dfrac{d\eta}{|\eta|^\alpha|1-|\eta||^\alpha}=\frac{2}{|t-\tau|^{2\alpha-1}}\int\limits_0^{\frac{1}{|t-\tau|}}\dfrac{d\eta}{\eta^\alpha|1-\eta|^\alpha}\leq\frac{2}{|t-\tau|^{2\alpha-1}}\int\limits_0^{\frac{2}{|t-\tau|}}\dfrac{d\eta}{\eta^\alpha|1-\eta|^\alpha}=$$$$=\frac{2}{|t-\tau|^{2\alpha-1}}\left(\int\limits_0^2\dfrac{d\eta}{\eta^\alpha|1-\eta|^\alpha}+\int\limits_2^{\frac{2}{|t-\tau|}}\dfrac{d\eta}{\eta^\alpha|1-\eta|^\alpha}\right).$$

The first integral in brackets converges to some positive number $c(\alpha)$. In second integral $\eta-1\geq\dfrac{\eta}{2}$, therefore, the second integral does not exceed $$2^\alpha\int\limits_2^{\frac{2}{|t-\tau|}}\dfrac{d\eta}{\eta^{2\alpha}}\leq2^\alpha\int\limits_1^{\frac{2}{|t-\tau|}}\dfrac{d\eta}{\eta^{2\alpha}}=\frac{2^\alpha}{2\alpha-1}\left(1-\left(\frac{|t-\tau|}{2}\right)^{2\alpha-1}\right)\leq\frac{2^\alpha}{2\alpha-1},\;\;\frac{1}{2}&lt;\alpha&lt;1.$$

Finally, $k(t,\tau)\leq\dfrac{d(\alpha)}{|t-\tau|^{2\alpha-1}}$. For $A^4$ similarly $k_1(t,\tau)\leq\dfrac{d_1(\alpha)}{|t-\tau|^{2(2\alpha-1)-1}}$. Then for sufficiently large $n$ for operator $A^{2^n}$ $k_n(t,\tau)\leq\dfrac{d_n(\alpha)}{|t-\tau|^{\beta}}$, where $\beta&lt;\dfrac{1}{2}$. So $A^{2^n}$ is Hilbert-Schmidt operator, and it's compact. The compactness of the operator $A$ follows from the statement: if $A$ -- self-adjoint and $A^2$ -- compact in hilbert space, then $A$ also compact.

The case $\alpha=\dfrac{1}{2}$ remains. But since $\dfrac{1}{|t-s|^\frac{1}{2}}\leq\dfrac{1}{|t-s|^\frac{5}{8}}$, then this case reduces to the previous one.