TITLE: Self-adjoint and compact operator on $L_2$
QUESTION [1 upvotes]: 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 $f,g\in L_2[0, 1]$
$$\langle f,Tg \rangle = \int_0^1 f(x) Tg(x)dx$$
$$ = \int_0^1 f(x)(\int_0^x\int_y^1g(z)dzdy)dx$$
$$ = \int_0^1 \int_0^x\int_y^1  f(x)  g(z)dzdydx $$
$$ = $$

REPLY [1 votes]: 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 $[0, 1]\times [0, 1]$.
{"set_name": "stack_exchange", "score": 1, "question_id": 3465286}