Let $(E, |\cdot|)$ be an infinite-dimensional Banach space. Let $T:E\to E$ be a compact (bounded linear) operator. Let $(\lambda_n)$ be 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!

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!