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

To find a more direct proof of Lemma 6.2. in Brezis' Functional Analysis, I have come across this question. Let $(E, |\cdot|)$ be an infinite-dimensional Banach space. Let $(E_n)$ be a sequence of closed subspaces of $E$ such that $E_n \cap E_m = \{0\}$ for all $n \neq m$.

1. Is there a bounded sequence $(e_n)$ such that $e_n \in E_n$ and that $(e_n)$ does not have any convergent subsequence?
2. What is the answer of (1.) in the following context? 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 $\lambda_n$.

Thank you so much for your elaboration!

Answer 1

Let $\{u_n\}_{n=0}^\infty$ be an orthonormal basis of a Hilbert space $\mathcal{H}.$ Define the operator $T$ by $$Tu_0=0,\ Tu_n={1\over n}u_0+{1\over n^2}u_n$$ The operator is bounded and compact as for $w=\sum_{j=1}^\infty {1\over j}u_j$ we have $$Tx=\langle x,w\rangle u_0+\sum_{n=1}^\infty {1\over n^2}\langle x,u_n\rangle u_n$$ Thus $T$ is bounded and compact as an absolutely convergent series of one dimensional operators. Moreover for $n\ge 1$ and $v_n=u_0+{1\over n}u_n$ we have $$Tv_n={1\over n}Tu_n={1\over n^2}u_0+{1\over n^3}u_n={1\over n^2}v_n$$ The space $E_n=\ker({1\over n^2}I-T)$ is one-dimensional. Indeed, assume $Tx={1\over n^2}x$ and $x=\sum_{k=0}^\infty x_ku_k.$ Then $$\sum_{k=1}^\infty x_kTu_k={1\over n^2}\sum_{k=0}^\infty x_ku_k\quad (*)$$ Calculating the inner product with $u_l,$ for $l\ge 1,$ gives $$ {1\over l^2}x_l={1\over n^2}x_l$$ Therefore $x_l=0$ for $l\neq n.$ Then $(*)$ reduces to $$x_nTu_n={1\over n^2}[x_0u_0+x_nu_n] $$ Taking inner product with $u_0$ results in $$ {x_n\over n}={x_0\over n^2}$$ Hence $x_n={1\over n}x_0,$ which concludes the proof of $\dim\ker E_n=1.$

As $1\le \|v_n\|\le 2,$ any bounded sequence $e_n\in E_n$ is of the form $e_n=\alpha_nv_n,$ where $\alpha_n$ is a bounded sequence of complex numbers. There exists a convergent subsequence $\alpha_{n_k},$ say to $\alpha.$ As $v_n\to u_0$ we get $\alpha_{n_k}v_{n_k}\to \alpha u_0.$

Summarizing every bounded sequence $e_n\in E_n$ contains a convergent subsequence.

Remark Let $\ker T=\{0\}.$ Choose $e_n\in E_n,$ such that $\|e_n\|=1.$ Assume $e_{n_k}\to x.$ Then $\|x\|=1.$ Moreover $\lambda_{n_k}e_{n_k}=Te_{n_k}\to Tx.$ As $\lambda_{n_k}\to 0,$ we get $Tx=0,$ i.e. $x=0,$ a contradiction.