Proving operator is surjective

Question

Let $T\colon \ell_1\to c_0^*$ be a operator, such that $$(Ta)(x)=\sum_{j=1}^\infty \alpha_j\xi_j$$ and $a=(\alpha_j)_{j=1}^\infty \in \ell_1$, $x=(\xi_j)_{j=1}^\infty\in c_0$. I want to prove that $T$ is surjective. Any ideas on how to approach this proof?

Answer 1

Let $\varphi \in c_0^{*}$ be a linear functional on $c_0$. You want to find $a \in \ell_1$ such that $T(a) = \varphi$. If we denote by $e_i$ the sequence that has $1$ at the $i$th place and zero everywhere else, then we should have

$$ \varphi(e_i) = T(a)(e_i) = \alpha_i. $$

Reversing the reasoning above, given $\varphi$, define the sequence $a = (\alpha_j)_{j=1}^{\infty}$ by $\alpha_j = \varphi(e_j)$ and show that $a \in \ell_1$. Since both $T(a)$ and $\varphi$ are continuous and they agree by construction on the set $\{e_i\}_{i=1}^{\infty}$ whose span is dense in $c_0$, they must agree on $c_0$ and so $T(a) = \varphi$.