# 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