Let $E$ be a reflexive Banach space and $T\in L_{c}(E)$ (set of the compact operators), inyective. Show that if $(v_{n})$ is a bounded sequence of eigenvectors linearly independent of $T$, then $v_{n}$ converge weakly to $0$.
Remark: My attempts only show the existence of weakly convergent subsequences.
First, we will show the following lemma:
Let $\|v_n\|\leq M$ for all $n$. Note that if $v_n\rightharpoonup v\neq 0$, then we must have $T(v_n)\to T(v)$. As $T$ is injective, we must have $T(v)\neq 0$, so $\|T(v)\| = \epsilon > 0$. Then, there must be infinitely many $v_n$ such that $\|T(v_n)\|\geq \epsilon/2$. We let this subsequence be $\{v_{n_k}\}_{k=1}^{\infty}$, and we let $F$ be the subspace spanned by $\{v_{n_k}\}_{k=1}^{\infty}$. Then, the closed $M$-ball in $F$ is a subset of the closed $M$-ball in $E$, and the image under $T$ of the closed $M$-ball in $F$ contains the closed $\epsilon/2$-ball in $T(F) = F$ (the equality of $T(F)$ and $F$ comes from the fact that $F$ is spanned by eigenvectors of $T$). As $F$ is infinite dimensional, the closed $\epsilon/2$-ball in $F$ is not compact, so therefore the image under $T$ of the closed $M$-ball in $E$ is not relatively compact, which contradicts that $T$ is compact. This implies that $v_n\not\rightharpoonup v$ for any $v\neq 0$, so we must have $v_n\rightharpoonup 0$.
Now, given that every subsequence of the given bounded, linearly independent sequence $\{v_n\}_{n=1}^{\infty}$ has a further bounded, linearly independent subsequence that converges weakly to $0$ (existence of a weakly convergent subsequence comes from the reflexivity of $E$, weak convergence to $0$ comes from the above lemma), we must have that $v_n\rightharpoonup 0$. This uses the result that for a topological space $X$, if $\{x_n\}_{n=1}^{\infty}\subset X$ such that every subsequence of $x_n$ has a further subsequence that converges to $x$, then $x_n\to x$. Here, the space is $E$ endowed with the weak topology.