Integral Operator in $L^2$

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.

$T$ 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}