Spectrum of an integral operator. For any $f\in C([0,1],\mathbb{R})$ set
$$
Tf(x) = \int_0^1 [\min\{x,y\}\cdot f(y)]dy.
$$
I have just proved that $T$ is a compact operator from  $C([0,1],\mathbb{R})$ into itself. I would like to know how to calculate his spectrum. (Since $T$ is compact I know that his spectrum is made of only eigenvalues.) Thank you for your help.

here's my solution:
The function $\min \{ x, y \}$ can be written as follows:
$$
\min\{x, y \}=
\begin{cases}
 & y, \mbox{ if } 0 \le y \le x \\
               & x, \mbox{ if } x \le y \le 1
\end{cases}
$$
so we found the form for $T$:
$$
Tf(x) = \int_0^x yf(y) \, dy + x \int_x^1 f(y) \, dy. 
$$
Now let $Tf = \lambda f, \lambda \ne 0$. So we have:
$$
\lambda f(x) = \int_0^x yf(y) \, dy + x \int_x^1 f(y) \, dy. 
$$
 Observe that $f(0)=0$. Now derive once and obtain:
$$
\lambda f'(x) = \int_x^1 f(y) \, dy,
$$
and here we found $ \lambda f'(0) = \int_0^1 f(y) \, dy $. Now derive another time and obtain the condition: 
$$
\lambda f''(x) = - f(x),
$$
so the eigenvector solves the differential equations:
\begin{cases}
& \lambda f'' + f = 0, \\
& f(0) = 0, \\
& \lambda f'(0) =  \int_0^1 f(y) \, dy.
\end{cases}
Now we only had to solve the system; I don't write the computations right now.
However if my calcula are corrects, putting for every $k \in \mathbb{N}$, $k \ne 0$ 
$$
f_k (x) = \sin \left ( \left (\frac{3 \pi}{2} + 2k \pi \right )x \right ) 
$$
We find that $f_k$ if an eigenvector for the eigenvalue 
$$
\lambda_k = \frac{4}{\pi^2 (3+ 4k)^2}.
$$
I hope they're correct!
Saluti da Pisa, il DM
