Showing that an integral operator is bounded on a hilbert space.

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Im trying to understand how it was hsown that T is a bounded operator on H. Going through the proof I don't understand what justifies the convergence of the last integral in this proof? What is it referring to when it talks about numerators and denominators? What is r?

Isn't it sufficient to say that |x-y| is bounded because they're inside the unit ball?

part 1 of proof

part 2 of proof

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If we integrate over the ball in $\mathbb R^d$ you can write (essentially as an extension of polar or spherical coordinates to the n-dimensional case) \begin{align*} M\int_{B(0,1)} |f(y)|^2|x-y|^{-d+\alpha}dy &= M\int_{\partial B(0,1)}\int_0^1 |f(r)|^2r^{-d+\alpha}r^{d-1}drdS\\ &= M\int_{\partial B(0,1)}\int_0^1 |f(r)|^2r^{\alpha}drdS, \end{align*} where the $r^{d-1}$ term comes from your change of variables. This integral must be convergent since $f$ is bounded.

I believe saying $|x-y|$ is bounded is not sufficient because for a fixed $x \in B(0,1)$ and $d>\alpha$ we have that $|x-y|^{-d+\alpha}$ is large when $y$ is close to $x$