The following operator is defined on $L_2(0,1)$: $$Kf(t)=\int_0^1|s-t|f(s)ds$$
I am wondering how I can calculate the eigenvalues and eigenfunctions of such an operator. I start with $\int_0^1|s-t|f(s)ds=\lambda f(t)$ and I took the derivative of both side twice then I got: $$2f(t)=\lambda f''(t)$$ However, I don't know what boundary condition I should apply to $f(t)$. I would appreciate your help with this.
Hint.
You have two cases.
The first one is $\lambda = 0$ which implies $f(t) = 0$ for all $t \in (0,1)$. Hence $\lambda = 0$ is not a eigenvalue.
The second one is $\lambda \neq 0$. In that case, $f$ has to satisfy the differential equation $$f^{\prime \prime}(t)-\frac{2}{\lambda}f(t)=0 \tag{2}$$ If $\lambda < 0$, you can write $\omega=\sqrt{\frac{2}{-\lambda}}$ and the solutions of $(2)$ are $$f(t)=A \cos (\omega t +\varphi), (A,\varphi) \in \mathbb R^2$$ Now you can plug those solutions into the definition of the operator $K$ to find the values of $\omega, \varphi$. If my intermediate computations are correct, you'll find $$\begin{cases} \varphi = - \omega\\ \omega = \sqrt{\frac{2}{-\lambda}} \in \{2 k \pi + \pi : k \ge 0\} \end{cases}$$ Which provides the eigenvalues and eigenvectors for $\lambda < 0$.
Finally, you have to consider the case $\lambda >0$ which can be solved in a similar way using hyperbolic trigonometry.