Find the spectrum of compact operator "min"

Question

Consider the (compact) operator $T:C([0,1],\mathbb{R})\rightarrow C([0,1],\mathbb{R})$ s.t. \begin{equation} T(f)(x)=\int_0^1\min\{x,y\} f(y)dy \; . \end{equation} How could one find its spectrum? I have tried to impose $Tf-\lambda f \equiv 0$ but then I couldn't solve anything...

Answer 1

A fundamental fact about this operator is that its inverse is the opposite of the second derivative. I show it in (Looking for examples of Discrete / Continuous complementary approaches).

What I give here is not an answer for the continuous operator "min" but for its discrete equivalent as the $n \times n$ matrix $M_n$ (or $M$) defined by

$$M_{i,j}=min(i,j) \ \ \ \ 1 \leq i,j \leq n.$$

Studying this issue is worthwile for at least two reasons:

• it helps to understand the behaviour of the continuous operator "min" .

• it is essential in the implementation of this continuous operator on a digital computer.

In the same reference, I show also that the inverse of the discrete operator "min" as a $n \times n$ matrix still is the opposite $D_n$ of the discrete analog of the second derivative.

In my answer to a question (https://math.stackexchange.com/q/1880704), I show that the spectrum of $D_n$ is the set of

$$\lambda_p=4 \left(\sin\dfrac{p \pi}{n+1}\right)^2 \ \ (p=1 \cdots n)$$

Therefore the spectrum of $M$ is the set of all $\dfrac{1}{\lambda_p}$.

Remark : $M_n$ and $D_n$, having positive eigenvalues, are symmetric positive definite matrices.