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Suppose that $A_1=\{a\in A: \|a\|<1\}$ and $B=\{ b\in A: \|be\|<1\}$. Obviously $Be\subset A_1$, and $S_e(Be)=T_e(Be)$. Hence $S_e(Be)\subset T_e(A_1)$, and consequently $\overline{S_e(Be)}^w\subset \overline{T_e(A_1)}^w$. But $\overline{T_e(A_1)}^w$ is weakly compact from assumption, and so $\overline{S_e(Be)}^w$ is weakly compact. But ...
We will show that $T:L^{2}[1,\infty)\rightarrow L^{2}[1,\infty)$ is compact if and only if $-\infty<\alpha<0$. Observe that the integral kernel $K$ of $T$ is given by \begin{align*} K(x,y)=\dfrac{x^{\alpha}}{y}\chi_{[1\leq x\leq y<\infty)}, \tag{1} \end{align*} where $\chi$ is the characteristic function of the set \$\left\{(x,y)\in [1,\infty)^{2} : ...