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!



Is there a bounded sequence $(e_n)$ such that $e_n \in E_n$ and that $(e_n)$ does not have any convergent subsequence?
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 $\langle \cdot, n e\_n \rangle$.


