Vanishing eigenvalues and compactness of the operator

Question

Suppose we have an infinite-rank operator that generates a complete set of eigenfunctions and eigenvalues with finite multiplicity. The eigenvalues are accumulating around zero and can be ordered such that: $\lambda_i \to 0$. I am wondering whether such a operator is compact or not?

Answer 1

Let $(u_n)_{n=1}^{\infty}$ be an orthonormal basis of $H$ such that $Tu_n= \lambda_n u_n$ for all $n$.

For $x \in H $ we have

$Tx= \sum \lambda_n <x,u_n>u_n$

For $ n \in \mathbb N$ define $T_n:H \to H$ by

$T_nx= \sum_{k=1}^n \lambda_k <x,u_k>u_k$.

Then $T_n$ is a bounded linear operator on $H$ with $ \dim T_n(H) < \infty$.

Hence $T_n$ is compact.

For $x \in H$ we get

$||(T-T_k)x||^2= \sum_{n>k}| \lambda_n|^2|<x,u_n>|^2 \le \sup_{n>k}| \lambda_n|^2||x||^2$,

hence

$||T-T_k||^2 \le \sup_{n>k}| \lambda_n|^2 \to 0$ for $k \to \infty$

Since the ideal of compact operators on $H$ is closed, $T$ is compact.

Answer 2

Yes. If $(v_i)_{i=1}^{\infty}$ is a complete orthonormal basis of $H$ that consists of eigenvalues of $T \colon H \rightarrow H$, then by identifying $H$ with $\ell^2(\mathbb{N})$ isometrically using the basis $(v_i)_{i=1}^{\infty}$, the operator $T$ is identified with a multiplication operator $\hat{T} \colon \ell^2(\mathbb{N}) \rightarrow \ell^2(\mathbb{N})$ which acts by

$$ \hat{T}((x_n)_{n=1}^{\infty}) = (\lambda_n x_n)_{n=1}^{\infty}. $$

Since $\lambda_n \to 0$, the operator $\hat{T}$ is compact and so is $T$.