# 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?

• Is the * a modified convolution? – DisintegratingByParts May 20 '15 at 19:58
• No, of course not. It's just multiplication – user241965 May 20 '15 at 20:58
• Using * is never customary notation in Mathematics for multiplication, unless you're using C++. You should probably remove the *. – DisintegratingByParts May 20 '15 at 21:01
• tell me how to edit a post? thanks for previously – user241965 May 20 '15 at 21:32
• I changed it for you. – DisintegratingByParts May 20 '15 at 21:41

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.