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Define the linear operator $T : L_2[0, 1] \rightarrow L_2[0, 1]$ by $$Tf(x) = \int_0^x \int_y^1 f(z)\,dz\,dy$$ I want to prove the operator is self-adjoint and compact.

Here's my attempt. For the self-adjoint part, we have for $f,g\in L_2[0, 1]$ $$\langle f,Tg \rangle = \int_0^1 f(x) Tg(x)dx$$ $$ = \int_0^1 f(x)(\int_0^x\int_y^1g(z)dzdy)dx$$ $$ = \int_0^1 \int_0^x\int_y^1 f(x) g(z)dzdydx $$ $$ = $$

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    $\begingroup$ The following link has some general results in this direction: en.wikipedia.org/wiki/Hilbert%E2%80%93Schmidt_integral_operator $\endgroup$ – Kavi Rama Murthy Dec 6 '19 at 7:23
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    $\begingroup$ this is like a double application of the Volterra operator defined by $V(f):=\int_0^x f(t)dt$ for $f\in L^2([0,1])$. If you can show that this operator is compact you are almost done (the self-adjointness of $T$ follows from it definition) $\endgroup$ – Masacroso Dec 6 '19 at 7:43
  • $\begingroup$ @Masacroso This is the answer to the question! Actually $T(f)=V(V^*(f))$ and by Schauder's theorem ( cf.here) $V^*$ is compact if and only if $V$ is compact, though boundedness of one of them and compactness of the other would suffice. $\endgroup$ – Peter Melech Dec 6 '19 at 12:58
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I will let you show that $T$ is self-adjoint.

Observe \begin{align} T(f)(x) =&\ \int^1_x\int^{x}_0 f(z)\ dydz+\int^x_0 \int_0^z f(z)\ dydz\\ =&\ \int^1_x xf(z)\ dz+\int^x_0 zf(z)\ dz\\ =&\ \int^1_0[x\cdot\chi_{[x, 1]}(z)+z\cdot \chi_{[0, x]}(z)]f(z)\ dz\\ =&\int^1_0 k(x, z) f(z)\ dz \end{align} where $k(x, z) = x\cdot\chi_{[x, 1]}(z)+z\cdot \chi_{[0, x]}(z)$. To show that $T$ is compact, it suffices to show that $T$ is a Hilbert-Schmidt operator. In particular, it suffices to show that \begin{align} \int^1_0\int^1_0|k(x, z)|^2\ dxdy<\infty. \end{align} But that's clear since $k(x,z)$ is a bounded function on $[0, 1]\times [0, 1]$.

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  • $\begingroup$ Don't You mean $k(x,z)=x\cdot\chi_{[x,1]}(z)+z\cdot\chi_{[0,x]}(z)$? I don't understand the first equation, it looks like the variable $y$ in the bound of the integral is lost somewhere...? $\endgroup$ – Peter Melech Dec 6 '19 at 12:39
  • $\begingroup$ @PeterMelech Yes. It was a typo. The first line is just Fubini. It should be clear once you draw the picture of the domain of integration. $\endgroup$ – Jacky Chong Dec 7 '19 at 6:54
  • $\begingroup$ Could you explain the first line "T(f)(x) =&\ \int^1_x\int^{x}_0 f(z)\ dydz+\int^x_0 \int_0^z f(z)\ dydz" and what does $\chi_{[x,1]}$ mean? $\endgroup$ – Teodorism Dec 7 '19 at 18:44

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