Compact and self-adjoint operator in $L^2$, and its eigenvalues Let $K:L^2([-\pi,\pi])\rightarrow L^2([-\pi,\pi])$ the operator defined as
\begin{align*}
(Kf)(t)=\int_\pi^\pi\vert t-s\vert f(s)\,\mathrm{d}s.
\end{align*}
I proved that $K$ is self-adjoint and how can I prove that $K$ is compact? How do I find its eigenvalues?
 A: First part: K is compact
For that, you want to show that the image of the unite ball of $L^2$ is pre-compact, that is its adherence is compact.
You can use the Arzela-Ascoli theorem. You will verify that all the function in the image are bound by the same value,which can be done by a Cauchy-Schwartz inequality. Then you have to show that function in the image are equicontinous. This can be done easily as we are on a compact and notice that $Kf$ is always a continous function.

Finding the eigenvalues
As said before $Kf$ is continuous. More specifficaly if $f$ is $\mathcal{C}^k$, $Kf$ is $\mathcal{C}^{k+1}$. We have that if $Kf =\lambda f$ then $f \in \mathcal{C}^{\infty}$.
You can now differentiate to find a differential equation. If my calculus are correct, you find:

 $ 2f = \lambda f''$

Which we know the solutions:

 if $\lambda = 0$ then $f=0$, not a eigenvalue  if $\lambda > 0$ then $f(t)=A e^{t/\tau}+B e^{-t/ \tau}$ with $\tau = \sqrt{\frac{\lambda}{2}}$  if $0 > \lambda$ then $f(t)=A cos(\omega t)+B sin(\omega t)$ with $\omega = \sqrt{\frac{-2}{\lambda}}$

Then one need to verify that this solution work. If we note $F$ a primitive of $f$ and $G$  a primitive of $F$ we can show by integrating by part that $Kf$ can be written

 $Kf(t)=2G(t) + (F(\pi)+F(\pi))(\pi-t)-(G(\pi)+G(-\pi)) $

So we need

 $(F(\pi)+F(\pi))(\pi-t)-(G(\pi)+G(-\pi))=0$ this does not happen if $\lambda > 0$ and otherwise gives $2 cos (\omega t)=0$

This give the eigenvalues

 $\lambda = \frac{-1}{2(1/2 +k )^2 } $ with $k \in \mathbb{Z}$

