# 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?

• You don't have to bother for C because an eigenvector is only known up to a mult.constant – Jean Marie May 25 '16 at 21:58
• True, silly me! – Vorhang May 26 '16 at 5:57

## 1 Answer

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.

• Thanks! I still don't understand why $f$ must satisfy $f(1)=0$ and $f'(0)=0$. Where do these side conditions come from? – Vorhang May 26 '16 at 6:01
• @Vorhang : Examine $f(t)=\frac{1}{\lambda}\int_{t}^{1}\int_{0}^{s}f(y)dyds$. Clearly $f(1)=0$. And $f'(t)=-\int_{0}^{t}f(y)dy$ must vanish at $t=0$. – DisintegratingByParts May 26 '16 at 6:25