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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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    $t=y{}{}{}{}{}$?2012-05-03
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    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.)2012-05-03
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    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 $\lambda f'' = -f$.2012-05-03

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