Operator on $L^2 (0,1)$ defined by convolution with $|x-y|^{-\alpha}$

Question

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.

Answer 1

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. ////

Answer 2

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

$$\begin{align} ||A||_2^2&=\sup_{f\in \mathscr{L}^2} \left(\frac{\int_0^1\left|\int_0^1 f(y) \frac{1}{|x-y|^\alpha} dy \right|^2dx}{\int_0^1\left|f(y)\right|^2\,dy}\right)\\\\ &\le \int_0^1\int_0^1\frac{1}{|x-y|^{2\alpha}}dydx \end{align}$$