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$T$ is bounded for all $\beta > -d$. The strategy is to show that $T$ is bounded both as a function $L^1(d\mu) \rightarrow L^1(d\mu)$ and as a function $L^\infty \rightarrow L^\infty$; the Riesz-Thorin theorem then shows that it is bounded $L^p(d\mu) \rightarrow L^p(d\mu)$ for any $p$. We just need the following estimate; everything else is standard. ...

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One of the main tools on infinite matrices and Hilbert spaces operators is the so-called Schur's test. This is Exercise 45 in Halmos' A Hilbert Space Problem Book. Schur's test. Let $A=[a_{ij}]_{i,j\in\mathbb{N}}$ be an infinite matrix. Suppose that there exist positive numbers $p_i>0$, $q_j>0$ ($i,j\in\mathbb{N}$), $\beta>0$ and $\gamma>0$ ...

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If $\operatorname{range}(S)$ is not closed, then we can find a sequence $(y_n)_n\subset \operatorname{range}(S)$ such that $y_n\to y$ and $y\notin \operatorname{range}(S)$. We may write $y_n$ as $x_n-Kx_n$, where $x_n\in Z$. Claim. The sequence $(x_n)_{n\geqslant 1}$ is not bounded. Indeed, if it was, then the sequence $(Kx_n)_{n\geqslant 1}$ would ...

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