Let $\{x_n\}$ be a basis for a Banach space $X$ and let $\{f_n\}$ be the associated sequence of coefficient functionals. Prove or disprove: if $X^\ast$ is separable, then ${f_n}$ is complete (and hence a basis) in $X^\ast$.
I thought taking a space which is not Reflexive like $\ell^1(\mathbb{R})$ (sequences from $\mathbb{R}$).A dense countable subspace in ${\ell^1}^{\ast}(\mathbb{R})$ can be ${\ell^1}^\ast(\mathbb{Q})$ since rationals are countable dense subspace in the reals. I take the netural basis, where the $m$-th component of the $n$-th element is given by $e_n^m=\cases{1\quad m=n\\0\quad m\neq n}$. But then I cannot procceed proving that the $\{f_n\}$ are not complete. How can I do so?