Suppose $H = L_2(B)$ where $B$ is the unit ball in $\mathbb{R}^d$. Let $K(x,y)$ be a measurable function on $B \times B$ that satisfies $|K(x,y)| \leq A|x-y|^{-d+\alpha}$ for some $\alpha > 0$ whenever $x,y \in B$. Define $$Tf(x)=\int_B K(x,y)f(y) \, dy$$
$\bullet$ Prove that $T$ is a bounded operator on $H$.
$\bullet$ Prove that $T$ is compact.
$\bullet$ Prove that $T$ is a Hilbert-Schmidt operator if and only if $\alpha > d/2$.
The above exercise is from Exercise $28$ of Chapter $4$ of Stein & Shakarchi (pg. $199$). I'm attempting to show the last bullet point, i.e. that $T$ is a HS operator $\iff \alpha > \frac{d}{2}$. I found a solution online, but I'm somewhat confused by one of the steps in the proof. See their proof below:
$\begin{align} T \, \, \mathrm{is} \, \, \mathrm{HS} &\iff A|x-y|^{-d+\alpha} \in L_2(B \times B)\\ & \iff \int_B \int_B A^2|x-y|^{-2d+2\alpha} < \infty \\ & \iff -2d+2\alpha > -d \\ & \iff \alpha > \frac{d}{2} \hspace{1cm} \blacksquare \\ \end{align}$
I have a problem with the $\iff$ in the first line, working left to right. If $T$ is HS, then by definition $K(x,y) \in L_2(B \times B)$, i.e. $\int_B \int_B |K(x,y)| \,dx \, dy < \infty$. But we're only given that $|K(x,y)| \leq A|x-y|^{-d+\alpha}$. Squaring both sides of this and applying monotonicity of the integral doesn't necessarily give that $A|x-y|^{-d+\alpha} \in L_2(B \times B)$ (certainly this isn't guaranteed over $\mathbb{R}^d$).
I would think that this would actually only hold if the inequality was reversed. Then the fact that $K(x,y) \in L_2(B \times B)$ would force $A|x-y|^{-d+\alpha}$ by monotonicity of $\int$.
What am I missing here? Is it the fact that in $B \times B$, $|x-y| \leq 1$?