$X$ Banach space , $T \in B(X , l^ {\infty})$ be surjective , there is a bounded sequence $\{x_n\}$ in $X$ such that $T(x_n)=e_n$?

Question

Let $X$ be a Banach space , $T:X \to l^ {\infty}$ be a surjective continuous linear map . Then how to show that there exist a bounded sequence $\{x_n\}$ in $X$ such that $T(x_n)=e_n,\forall n \in \mathbb N$ ?

Answer 1

By the Banach-Schauder theorem, $T$ is open, in particular, the image $TB_X(0,1)$ of the open unit ball $B_X(0,1)$ centered at origin is open. Since $0\in TB_X(0,1)$, there exists some $r>0$ such that $B_{l^\infty}(0,r)\subset TB_X(0,1)$. By linearity, $B_{l^\infty}(0,2)\subset TB_X(0,2/r)$. Hence, every unit vector (not only $e_n$) has a preimage in $B_X(0,2/r)$.