Define $K: [0,1] \times [0,1]\rightarrow \mathbb{R}$ by
$K(x,y) =\begin{cases} (1-x)y &\text{if } 0 \le t \le x\\ (1-y)x& \text{if }x \le y \le 1\end{cases}$
Also, Consider the operator $T: L^2([0,1]) \rightarrow L^2([0,1])$ defined by $Tf(x) = \int_0^1 K(x,y)f(y) \, dy$
How to prove that if $Tf = \lambda f$ for some $\lambda \neq 0$, then $f \in C^\infty([0,1])$, $f(0)=f(1)=0$ and $\lambda f'' = f$? Also, what are the eigenvalues of $T$?
Thank you for your help.