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.
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.