I am not sure how to approch this question.
$c_0$ is the space of sequences converging to zero, with the endowed norm being supreme of $|y_n|$ for $y_n$ in the sequence y. $j$ is an isomorphism between $c_0$ and $X_0$, a closed subspace of a seperable Banach space $X$.
Prove that there exists a bounded sequence $x^*_n$ in $X_0$ such that for all $y$ in $c_0$ $<j(y),x^*_n>=y_n$ for all natural number n.
My idea: I kind of find this question trivial.
I consider the linear map from $c_0$ to $X_0$, say, $T_n$ where $T_n(y)=y_n$.
Then $|T_n(j(y))|<=|y_n|<=||j(y)||$, since $|y_n|$ is less than or euqal to the norm of y (supreme of y_n in the sequence), and unitary mapping preserves norm so we have |T(j((y))|<= ||j(y)||.
That makes T a bounded linear transformation for all elements in $X_0$. Then I just let $<x^*_n,u>=T_n(u)$ for u in $X_0$ and zero otherwise. $x^*_n$ is an element of $X^*$ obviously. Then the sequence is found.
However I think this is unlikely the solution. Can anyone please help?
