Example of compact operator which has eigenvalues with 0 as limit point

Question

I was reading the Fredholm alternative which has the following consequence:

A compact linear mapping $T$ of normed linear space into itself possesses a countable set of eigenvalues having no limit point except possibly $0$. Each nonzero eigenvalue has a finite multiplicity.

I wanted to find an example of a compact operator which has eigenvalues with limit $0.$

Any Help will be appreciated

Answer 1

Let $H$ be any infinite dimensional separable Hilbert space with orthonormal basis $\{e_i; i \in \mathbb{N}\}$. Let $P_j$ be the orthogonal projection onto the span of $e_j$. Finally pick any sequence $(\lambda_j) \in \ell^1$. In particular, $\lambda_j \to 0$ as $j \to \infty$.

Let $$T = \sum_{j=1}^\infty \lambda_j P_j.$$ This sum converges absolutely in the space of bounded linear operators on $H$ since for any $n$, $$\sum_{j = 1}^n \|\lambda_j P_j\| = \sum_{j=1}^n |\lambda_j|$$ which is bounded above since we chose $(\lambda_j) \in \ell^1$.

Since $T$ is a limit of finite rank operators it is compact and it is clear that its eigenvalues are given by $\{\lambda_j: j \in \mathbb{N}\}$ which is a set with the desired properties.