Integral Operator in $L^2$ I was trying to do this exercise and I'm wondering if I figured it out well:
I have $\mathcal{H} := L^2(0,1)$ and $T$ the operator with integral kernel $K(x,y) = \min\{x,y\}$, $x,y \in [0,1]$. I have to show that $T$ is compact and self-adjoint.
To show that is compact I was thinking to say that because $\min\{x,y\} \in [0,1]$ then
\begin{equation}
\dim(\operatorname{Im}T) = 1
\end{equation}
(The self adjointness I think is trivial..)So T belongs to finite rank operators and so it is compact. (Is this correct?) Then it asks me to find eigenvalues and eigenvectors of $T$ and here I really don't know how to proceed...
 A: We have
$$
\int_{(0,1)^2} |k(x,y)|^2\ \mathsf d(x\times y) = \int_0^1\int_0^1 (x\wedge y)^2\ \mathsf dx\ \mathsf dy 
\leqslant \int_0^1\int_0^1\ \mathsf dx\ \mathsf dy = 1 <\infty,
$$
so T is a Hilbert-Schmidt operator and hence is compact.
A: $T$ is a Hibert-Schmidt operator because $\min\{x,y\} \in L^2([0,1]\times[0,1])$. $Tf$ may be written as
\begin{align}
    Tf & = \int_{0}^{1}K(x,y)f(y)dy  \\
   & = \int_{0}^{1}\min\{x,y\}f(y)dy \\
   & = \int_0^xyf(y)dy+x\int_x^1 f(y)dy
\end{align}
If $Tf=\lambda f$ for some $f\in L^2$ and $\lambda\in\mathbb{C}\setminus\{0\}$, then the above implies that $Tf$ is equal a.e. to a continuous function on $[0,1]$. Hence, $f$ is equal a.e. to a continuous function. So $Tf$ is continuously differentiable. So, assume without loss of generality, that $f$ is continuous. Then $(Tf)(0)=0$ and $(Tf)'$ exists with
$$
     \lambda f'=  (Tf)'=xf(x)-xf(x)+\int_x^1f(y)dty=\int_x^1 f(y)dt
$$
So $f$ is $C^2$, $f(0)=0$, $f'(1)=0$ and $\lambda f''=-f$ for every eigenfunction with eigenvalue $\lambda\ne 0$. The eigenfunctions with non-zero eigenvalues are, therefore, constant multiplies of
$$
       f_n = \sin(n\pi x/2),\;\;\; n=1,3,5,7,\cdots, \\
      \lambda_n = \frac{2}{n\pi}.
$$
The adjoint of $\int_0^x$ is $\int_x^1$. And the adjoint of $M_x$ (multiplication by $x$) is $M_x$. So $T$ is selfadjoint because
\begin{align}
              T &= \left(\int_0^x\right)M_x+M_x\left(\int_x^1\right)\\ &=\left(\int_0^x\right)M_x+M_x^*\left(\int_0^x\right)^* = T^*.\end{align}
A: Since we have the integral identity
$$ g:= Tf = \int_0^x yf(y) dy + x\int_x^1 f(y) dy $$
Then $g\in H^1$, and since $g' = \int_x^1 f(y) dy$, actually $g \in H^2$, with 
$$ -g'' = f$$
So we actually have a Poisson equation, but with boundary conditions $g(0)=0,g(1) = \int_0^1 yf(y) dy$. By setting $\tilde g = g - x \int_0^1 yf(y) dy$, we notice that we are equivalently looking for the weak solutions in $H^1_0$ to the Poisson equation with Dirichlet boundary conditions
$$ -\tilde g'' = f \text{ on } (0,1),\quad  \tilde g(0)=\tilde g(1)=0,\quad f\in L^2 $$
So if you already knew that the solution operator $f\mapsto \tilde g$ for this 1D Poisson equation was compact($H^1_0 \subset\subset L^2)$  and self-adjoint, we're done (the difference $ g-\tilde g= x\int_0^1 y f(y) dy$ is finite rank and self-adjoint).
