Is there a bounded sequence $(e_n)$ such that $e_n \in E_n$ and that $(e_n)$ does not have any convergent subsequence?

Question

Let $(E, |\cdot|)$ be an infinite-dimensional Banach space. Assume that

• $T:E\to E$ is a compact (bounded linear) operator, and
• $(\lambda_n)$ is a sequence of distinct eigenvalues of $T$.

Let $E_n$ be the corresponding eigenspace of $\lambda_n$. Then $(E_n)$ is a sequence of finite-dimensional subspaces of $E$ such that $E_n \cap E_m = \{0\}$ for all $n \neq m$.

Is there a bounded sequence $(e_n)$ such that $e_n \in E_n$ for all $n$ and that $(e_n)$ does not have any convergent subsequence?

Thank you so much for your elaboration!

Answer 1

The answer is negative. Take a biorthogonal sequence $(x_n,x_n^*)$ in $E$ such that $(x_n)$ converges to some unit norm vector $x_0$. Set

$T=\sum_n 2^{-n} \|x_n\|^{-1} x_n^* \otimes x_n$.

Such a sequence exists in every infinite dimensional space. For example, in $\ell_2$ let $x_n = e_0 + n^{-1} e_n$, where $(e_n)_{n=1}^\infty$ is orthonormal, and $x_n^*$ is $\langle \cdot, n e_n \rangle$.

Answer 2

Partial answer: yes for normal operators on Hilbert spaces. We can write any normal compact operator as $\sum_{n\geq 1} \lambda_n f_n g_n^*$ with $(f_n)$, $(g_n)$ orthonormal families and $\lambda_n \to 0$. A $\lambda\neq 0$ is an eigenvalue if and only if $N_\lambda = \{ n : \lambda_n = \lambda\}$ is non-empty, in which case the eigenspace is $\mathrm{Vect}(f_n : n\in N_\lambda) = \mathrm{Vect}(g_n : n\in N_\lambda)$. This means that eigenspaces associated to different eigenvalues are necessarily orthogonal. Given an infinite sequence of distinct eigenvalues, one can then easily find a sequence of eigenvectors from which no subsequence converges.