The question is as follows:
Show that $(Ax)(t) = \int_{-\pi}^{\pi} | t - s | x(s) ds$ for $A : L_2[-\pi , \pi ] \to L_2[-\pi , \pi ] $, is compact, self adjoint and find its norm.
$\textbf{Some efort:}$
We know that for operator $(Ax)(t) = \int_{a}^{b} k(t , s) x(s) ds$ we have that $(A^*x)(t) = \int_{a}^{b} \bar{k(s , t)} x(s) ds$ is its adjoint operator. Here we have $ k(t , s) = | t - s | = | s - t | = \bar{k(s , t)}$. So it is self adjoint.
For to show it is compact, we show that it is Hilbert–Schmidt integral operator, meaning that $\left( \int_{-\pi}^{\pi} \mid \int_{-\pi}^{\pi} t - s ds \mid^2 dt \right)^{\frac{1}{2}} < + \infty.$
We have \begin{align}\int_{-\pi}^{\pi} \mid \int_{-\pi}^{\pi} |t - s | ds \mid^2 dt &\le \int_{-\pi}^{\pi} \int_{-\pi}^{\pi} |t - s |^2 ds dt \\&= \int_{-\pi}^{\pi} \left( \int_{-\pi}^{t} (t - s )^2 ds + \int_{t}^{\pi} ( s - t )^2 ds \right) dt \\& = \int_{-\pi}^{\pi} \left( -\frac{(t+s)^3}{3} \mid_{-\pi}^{t} + \frac{(s-t)^3}{3} \mid_{t}^{\pi} \right) dt \\& = 0 \end{align}
So it is Hilbert–Schmidt integral operator and so is compact.
But I do not know how to find its spectrum.
Can you please let me know if my calculation is correct so far and can you please let me know what will be its spectrum?
Thanks!