Help to find all of the eigenvalues of the operator $T:L^2([0,1]) \rightarrow L^2([0,1])$ 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.
 A: I suggest that you write out explicitly what $Tf(x)$ is, and suppose that $\lambda f = Tf$.  Since functions like $x\mapsto \int_a^x g(y)\,dy$ are continuous when $g$ is integrable, the right hand side of the equation $\lambda f = Tf$ represents a continuous function.  Since $\lambda\neq 0$, this means that $f$ is continuous.
Next you can apply the Fundamental Theorem of Calculus, which says that if $g$ is continuous, then the derivative of $x\mapsto \int_a^x g(y)dy$ is $g(x)$. This allows you compute the derivatives of $Tf$.  Differentiating the equation $\lambda f=Tf$ twice yields $\lambda f'' = -f$. In particular, this makes it straightforward to see that $f$ is in $C^\infty$.  
It also allows you to solve a second order homogeneous linear differential equation to find all possibilities for $f$, namely $f(x)=A\sin(x/\sqrt\lambda)+B\cos(x/\sqrt \lambda)$ for constants $A$ and $B$. Using the boundary values of $f$ allows you to narrow the possibilities and find all candidates for $\lambda$.  
