I'm trying to prove this that but I can't . Any help/reference ?



Is it true that $c_0(X)^* = \ell_1(X^*)$ ?
True. For any $n\in \mathbb{N}$ consider the inclusion to the $n$-th coordinate $j \_ n : X\to c \_ 0(X)$ which is right inverse to the evaluation at $n$, so that $(j \_ n x)(n)= x$, for any $x\in X$. Let $j \_ n ^ T : c \_ 0(X) ^ \* \to X^\*$ its transpose operator. Any $\eta \in c \_ 0(X)^ \* $ defines a sequence $y:\mathbb{N}\to X ^ \*$ such that $y(n) := j \_ n ^T \eta $. 
The $\ell \_ 1(X^\*)$- norm of $y$ is 
$$\|y\|\_ { \ell \_ 1 (X^\*)}=\sum \_{n\in\mathbb{N}}\, \|y(n)\| \_ {X ^ \*} = \sum \_{n\in\mathbb{N}}\, \, \sup \_ {\|x\| \_ X \le 1} \langle y(n), x \rangle=$$$$ = \sup \_ {m\in\mathbb{N}}\, \, \sup \_ {\|\xi\| \_ { c \_ {0} (X)} \le 1} \, \sum \_{n\le m}\, \langle y(n), \xi(n) \rangle =$$$$= \sup \_ {\|\xi\| \_ { c \_ {0} (X)} \le 1} \, \langle \eta, \xi \rangle = \|\eta\| \_ {c \_ 0(X)^\*}\, \, .$$
This shows that the inclusion $\ell \_ 1(X^\*)\to c \_ 0(X)^ \* $ is actually a linear (surjective) isometry.


