Vtgyjfy

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...

We have
$$
int_{(0,1)^2} |k(x,y)|^2 mathsf d(xtimes y) = int_0^1int_0^1 (xwedge y)^2 mathsf dx mathsf dy
leqslant int_0^1int_0^1 mathsf dx mathsf dy = 1 <infty,
$$
so T is a Hilbert-Schmidt operator and hence is compact.

$T$ 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+xint_x^1 f(y)dy
end{align}
If $Tf=lambda f$ for some $fin L^2$ and $lambdainmathbb{C}setminus{0}$, 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 $lambdane 0$. The eigenfunctions with non-zero eigenvalues are, therefore, constant multiplies of
$$
f_n = sin(npi x/2),;;; n=1,3,5,7,cdots, \
lambda_n = frac{2}{npi}.
$$

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^xright)M_x+M_xleft(int_x^1right)\ &=left(int_0^xright)M_x+M_x^*left(int_0^xright)^* = T^*.end{align}

Since we have the integral identity
$$ g:= Tf = int_0^x yf(y) dy + xint_x^1 f(y) dy $$
Then $gin H^1$, 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
$$ -tilde g'' = f text{ on } (0,1),quad tilde g(0)=tilde g(1)=0,quad fin L^2 $$
So if you already knew that the solution operator $fmapsto tilde g$ for this 1D Poisson equation was compact($H^1_0 subsetsubset L^2)$ and self-adjoint, we're done (the difference $ g-tilde g= xint_0^1 y f(y) dy$ is finite rank and self-adjoint).