# 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.

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$t=y{}{}{}{}{}$? – Jonas Meyer May 3 '12 at 0:06
Suggestions for part of the problem: Write out explicitly what $Tf(x)$ is. Observe that $x\mapsto \int_a^x g(y)dy$ is continuous when $g$ is integrable. Conclude that $f$ is continuous. Then proceed by induction, using the Fundamental Theorem of Calculus to show that if $f$ is in $C^k$, then it is in $C^{(k+1)}$. (Evaluating $f(0)$ and $f(1)$ is trivial.) (It could probably be made easier by showing that $\lambda f'' = f$ in 3 steps, and then the fact that $f\in C^{\infty}$ would follow easily.) – Jonas Meyer May 3 '12 at 0:19
Yeah, after showing that $f$ is continuous, the Fund. Thm. of Calculus allows $\lambda f''$ to be computed directly. But maybe I slipped up, because I got $\lambda f'' = -f$. – Jonas Meyer May 3 '12 at 0:34

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 $\lambda$.