Any bounded operator $T:E\to\ell_\infty$ can be expressed as $Te = (e_n^*(e))_{n=1}\infty$ for some sequence $(e_n^*)$ in $E^*$?

Question

Fact: Any bounded operator $T:E\to\ell_\infty$ can be expressed as $Te = (e_n^*(e))_{n=1}^\infty$ for some sequence $(e_n^*)$ in $E^*$?

The fact above is written in the book Topics in Banach Space Theory, page $45,$ Chapter $2,$ section $2.5,$ complementability of $c_0,$ first line of the proof of Proposition $2.5.2.$

I have no idea how to prove the fact. Any hint is appreciated.

Answer 1

Let $f_n^*\in (l^\infty)^*$ be the evaluation functional, i.e., $f_n(x)=x_n$. Then define $$ e_n^*(e):= f_n(T(e)). $$ The phrasing suggests that there is some magic involved, which is not the case.