Eigenfunctions of an integral operator Let $Tf(x):=\int_0^x f(t)dt$ be an integral Operator ($T:L_2[0,1]\rightarrow L_2[0,1]$). I am trying to find the eigenvalues and eigenfunctions of $S:=T^*T:L_2[0,1]\rightarrow L_2[0,1]$. So far I know that $T^*g(t)=\int_t^1 g(x)dx$ as well as $Sf(x)=\lambda f(x)\Leftrightarrow -f(x)=\lambda f''(x)$ and therefore $f(x)=Ce^{\frac{1}{\sqrt{\lambda}}x}$. Is this correct? How do I get $C$ and what should I do next?
 A: Suppose you have a non-trivial solution of
$$
         T^*Tf = \int_{t}^{1}\int_{0}^{s}f(y)dy ds = \lambda f(t)
$$
Then $\lambda \ne 0$ because the above would give $f=0$ after differentiating a couple of times. For $\lambda \ne 0$, any solution of the above must satisfy
$$
              \lambda f'' = -f \\
               f(1)=0,\;\; f'(0)=0.
$$
Any non-trivial solution of the above ODE cannot satisfy $f(0)=0$ because $f'(0)=0$ and $f(0)=0$ can hold for a solution of an ODE such as the above iff $f\equiv 0$. So a non-trivial solution may be normalized to satisfy $f(0)=1$ and $f'(0)=0$, which means any eigenfunction must be a non-zero scalar multiple of
$$
                 \varphi_{\lambda}(t)=\cos(t/\sqrt{\lambda}).
$$
The eigenvalue equation is then given by $\varphi_{\lambda}(1)=0$, or $\cos(1/\sqrt{\lambda})=0$, $\lambda \ne 0$. The eigenvalues are
$$
                \lambda_n = \frac{4}{(2n+1)^2\pi^2},\;\;\; n=0,1,2,3,\cdots,
$$
and the corresponding eigenfunctions are
$$
                     \cos\left(\frac{(2n+1)\pi}{2}t\right)
$$
These are the only possible eigenfunctions, and you can check that they work.
