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. 
 A: HINT:
The Cauchy-Schwarz Inequality reveals that
$$\begin{align}
\left|Af(x)\right|^2 &= \left|\int_0^1 f(y) \frac{1}{|x-y|^\alpha} dy \right|^2\\\\
&\le \int_0^1\left|f(y)\right|^2\,dy\,\int_0^1\frac{1}{|x-y|^{2\alpha}} dy
\end{align}$$
And thus, the square of the operator norm is 
$$\begin{align}
||A||_2^2&=\sup_{f\in \mathscr{L}^2} \left(\frac{\int_0^1\left|\int_0^1 f(y) \frac{1}{|x-y|^\alpha} dy \right|^2dx}{\int_0^1\left|f(y)\right|^2\,dy}\right)\\\\
&\le \int_0^1\int_0^1\frac{1}{|x-y|^{2\alpha}}dydx
\end{align}$$
A: By the inequality $|Af| \leq A|f|$, we may assume that $f$ is non-negative. Then from the Tonelli's theorem (a.k.a. Fubini's theorem for non-negative functions),
$$\| Af \|_2^2 = \int_0^1 \int_0^1 f(y)f(z) \left( \int_0^1 \frac{dx}{|x-y|^{\alpha}|x-z|^{\alpha}} \right) \, dydz $$
and we may try to estimate the following function
$$ k(y, z) = \int_0^1 \frac{dx}{|x-y|^{\alpha}|x-z|^{\alpha}}. $$
In this regard, we make the following observation:

Claim. $k(y, z) \leq C |y - z|^{-\alpha}$ for some constant $C = C(\alpha)$ that depends only on $\alpha \in (0, 1)$.

Assuming this claim, we have
$$ \| Af \|_2^2 \leq C \int_0^1 \int_0^1 \frac{f(y)f(z)}{|y - z|^{\alpha}} \, dydz = C \langle f, Af \rangle \leq C \|f\|_2 \|Af\|_2. $$
Now assume first that $f$ is bounded. Then $\|Af\|_2 < \infty$ and we get
$$\|Af\|_2 \leq C \|f\|_2. \tag{1} $$
For general $f \in L^2$, truncation argument applied to $|f|$ shows that $Af \in L^2$ and (1) remains valid in this case. Therefore $A$ is a bounded operator on $L^2(0, 1)$ for any $\alpha \in (0, 1)$.
I am still working on the condition for compactness. At least we know that $A$ is compact for $\alpha < 1/2$ by noticing that $A$ is a Hilbert-Schmidts operator in that case.
Proof of Claim. By symmetry, we may assume $y \leq z$. Then
\begin{align*}
k(y, z)
&\leq \int_{y-1}^{z+1} \frac{dx}{|x-y|^{\alpha}|x-z|^{\alpha}} \\
&= 2\int_{0}^{1} \frac{dx}{x^{\alpha} (x + |y - z|)^{\alpha}} + \int_{y}^{z} \frac{dx}{|x - y|^{\alpha}|x - z|^{\alpha}} \\
&\leq C_1 \left( \frac{1}{|y - z|^{\alpha}} + \frac{1}{|y - z|^{2\alpha-1}} \right) \\
&\leq \frac{C_2}{|y - z|^{\alpha}}
\end{align*}
for some generic constants $C_1, C_2 > 0$ that depend only on $\alpha$. So the claim follows. ////
A: 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 $0<\alpha<\frac{1}{2}$ this operator is a Hilbert-Schmidt operator and therefore compact. Consider the case $\frac{1}{2}\leq\alpha<1$. 
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 $$2^\alpha\int\limits_2^{\frac{2}{|t-\tau|}}\dfrac{d\eta}{\eta^{2\alpha}}\leq2^\alpha\int\limits_1^{\frac{2}{|t-\tau|}}\dfrac{d\eta}{\eta^{2\alpha}}=\frac{2^\alpha}{2\alpha-1}\left(1-\left(\frac{|t-\tau|}{2}\right)^{2\alpha-1}\right)\leq\frac{2^\alpha}{2\alpha-1},\;\;\frac{1}{2}<\alpha<1.$$
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 $\beta<\dfrac{1}{2}$. 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.
