functional analysis. Compact operator. Hilbert-Schmidt theorem. I have the following problem:
"Under which $ \alpha \in \mathbb{R}$ is the operator $ T: L_2 [1, + \infty) \to L_2 [1, + \infty) $: 
\begin{equation*}
(Tf) (x) = x^{\alpha} \int_x^{\infty} \frac{f(y)}{y}dy,~f\in L_2 [1, + \infty),~x \in [1, + \infty) 
\end{equation*}
compact?"
I tried to solve it through the first criteria of compactness
Arzela-Ascoli. That is, take a bounded set $ M \in L_2 [1, \infty] $, ie, such that $ || m ||_{L_3} \le c $, then somehow estimated $ ||Tm||_{L_3} $. I tried to pull out the conditions on $ \alpha $, but then there was a problem with the proof that the set $ T (M) $ is equicontinuous on average. Then he found a theorem on the compactness of the operators Shilberta Schmidt, that is, operators $ A: L_2 (X, \mu) \to L_2 (Y, \nu) $ with kernel $ K $, for which 
\begin{equation*}
\int_X\int_Y K ^ 2 ( x, y) d \mu (x) d \nu (y) \le + \infty , 
\end{equation*}
and all such operators are compact. However I do not understand how to use it. What is there to do?
 A: 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 $\left\{(x,y)\in [1,\infty)^{2} : x\leq y\right\}$. By the Hilbert-Schmidt result you cited, $T$ will be compact if the $L^{2}([1,\infty)\times [1,\infty))$ norm of $K$ is finite.
Observe that by Fubini/Tonelli,
\begin{align*}
\int_{1}^{\infty}\left(\int_{1}^{\infty}\left|K(x,y)\right|^{2}\mathrm{d}y\right)\mathrm{d}x&=\int_{1}^{\infty}x^{2\alpha}\left(\int_{1}^{\infty}y^{-2}\chi_{1\leq x\leq y<\infty}\mathrm{d}y\right)\mathrm{d}x\\
&=\int_{1}^{\infty}x^{2\alpha-1}\mathrm{d}x<\infty
\end{align*}
if and only if $\alpha<0$.
For $\alpha>0$, $T$ does not map $L^{2}[1,\infty)$ into $L^{2}[1,\infty)$. Indeed, consider $f=y^{-1/2-\delta}$, where $0<\delta<1/2$. Clearly, $f\in L^{2}[1,\infty)$, but
\begin{align*}
Tf(x)=x^{\alpha}\int_{x}^{\infty}y^{-3/2-\delta}\mathrm{d}y=\dfrac{x^{\alpha-1/2-\delta}}{-(1/2+\delta)},\qquad x\in [1,\infty)
\end{align*}
which is not square integrable if $\alpha-\delta\geq 0$.
It remains for us to show that $T$ is not compact when $\alpha=0$. It suffices to exhibit a sequence of functions $f_{n}\in L^{2}[1,\infty)$ such that $\left\|f_{n}\right\|_{L^{2}}=1$, $Tf_{n}(x)\rightarrow 0$ for a.e. $x$, but that there exists a $\delta>0$ such that $\left\|Tf_{n}\right\|_{L^{2}}\geq\delta>0$, for all but finitely many $n$. Indeed, if $(Tf_{n})$ had an $L^{2}$-convergent subsequence $Tf_{n_{k}}\rightarrow f$, then passing to a subsequence of this subsequence if necessary, $Tf_{n_{k}}\rightarrow f$ pointwise a.e., whence $f=0$ a.e. But $\left\|Tf_{n_{k}}\right\|_{L^{2}}\not\rightarrow 0$.
Consider the functions $f_{n}$ defined by
\begin{align*}
f_{n}(y):=\dfrac{1}{c_{n}}y\chi_{[n,n+1]}, \ \forall y\in[1,\infty), \quad c_{n}:=\left(\dfrac{(n+1)^{2}-n^{2}}{2}\right)^{1/2}=\left(\dfrac{2n+1}{2}\right)^{1/2} \tag{2}
\end{align*}
I leave it to you to check that $\left\|f_{n}\right\|_{L^{2}}=1$. Observe that
\begin{align*} Tf_{n}(x)=
\begin{cases}
c_{n}^{-1} & {1\leq x\leq n}\\
c_{n}^{-1}(n+1-x) & {n<x<n+1}\\
0 & {x>n+1}
\end{cases} \tag{3}
\end{align*}
Since $c_{n}^{-1}\rightarrow 0$, as $n\rightarrow\infty$, it follows from our computation that $Tf_{n}(x)\rightarrow 0$ pointwise a.e. However,
\begin{align*}
\int_{1}^{\infty}\left|Tf_{n}(x)\right|^{2}\mathrm{d}x\geq c_{n}^{-2}\int_{1}^{n}\mathrm{d}x=\dfrac{2(n-1)}{2n+1} \tag{4}
\end{align*}
which tends to $1$ as $n\rightarrow\infty$, hence is bounded from below for all $n$ sufficiently large.
