Eigenvalues of the operator $(Tu)(x)=\int_0^x (\int_t^1 u(s)ds)dt.$ Consider the linear operator $T$ in $L^2(0,1)$ defined by:
$$(Tu)(x)=\int_0^x \left(\int_t^1 u(s)ds\right)dt.$$
I have managed to prove that it's continuous,self adjoint,compact but now I have to find the eigenvalues of T. 
I know that $\sigma(T)={0}\cup EV(T)$ and $\sigma(t)\subset [-1/\sqrt 2,1/\sqrt 2]$ because $||T||\leq 1/\sqrt 2 $. Can anyone help me? Thank you in advance.
 A: Assume $\lambda$ is a non-zero eigenvalue. Since $T$ is seladjoint we may assume that $\lambda$ is real. We have some non-zero eigenvector $u\in L_2[0,1]$, such that
$$
\int_0^x \left(\int_t^1 u(s)ds\right)dt=\lambda u(x)\tag{1}
$$
Note that $u(0)=0$. Since left hand side of the equation is absolutely continuous as any Lebesgue integral, then $u\in AC[0,1]$ and we can  differentiate $(1)$ and get
$$
\int_x^1 u(s)ds=\lambda u'(x)
$$
Note, $u'(1)=0$. Again, $u'\in AC[0,1]$ and we differentiate one more time
$$
\lambda u''(x)=-u(x)\tag{2}
$$
Since $u''=-\lambda^{-1}u\in AC[0,1]\subset C[0,1]$ we are in the range of standard differential equations theory with solutions in the class of continuous functions. Hence $(2)$ can be easily solved
$$
u(x)=
\begin{cases}
c_1 \sin\frac{x}{\sqrt{\lambda}}+c_2 \cos\frac{x}{\sqrt{\lambda}}&\quad\mbox{if}\quad \lambda>0\\
c_1 \exp\frac{x}{\sqrt{\lambda}}+c_2 \exp\frac{-x}{\sqrt{\lambda}}&\quad\mbox{if}\quad\lambda<0\\
\end{cases}
$$
You can check that in both cases boundary conditions $u(0)=0$, $u'(1)=0$ imply that non zero solution exists iff $\lambda=(\pi(2n+1))^{-2}$ and $n\in\mathbb{Z}_+$. Therefore 
$$
\sigma_p(T)=\{(\pi(2n+1))^{-2}:n\in\mathbb{Z}_+\}
$$
Thanks to Daryl for pointing out the mstake in the original proof.
