I would like to find an example of a non-reflexive normed vector space $X$ and a proper closed subspace $W \subset X^*$ such that $ ^\perp W = \{0\}$.
I was thinking maybe taking $X = C^0([0,1])$ the space of all continuous functions on $[0,1]$ and $W = \{ f \in C^0([0,1]) : f \text{ is constant} \}$.
I know that $X$ is separable but its dual is not, which implies that its double dual is also not separable and hence $X$ cannot be reflexive. Also $W$ is closed.
However, I am not completely sure if $ ^\perp W = \{0\}$. Would that be true ? Otherwise what other example could I take ?
For $X=c_0$ we have $X^*=\ell^1.$ Let $$W=\left \{y\in \ell^1\,:\,\sum_{n=1}^\infty y_n=0\right \}$$ Then $$^\perp W=\left \{ x\in c_0\,:\, \sum_{n=1}^\infty x_ny_n=0, \ y\in W\right \}=\{0\}$$ Indeed, let $\{e_n\}_{n=1}^\infty$ denote the standard basis in $c_0.$ The elements $e_n-e_{n+1}$ belong to $W.$ If $c_0\ni x\perp e_n-e_{n+1}$ then $x_n=x_{n+1}.$ Hence $\{x_n\}_{n=1}^\infty$ is a constant sequence. Thus $x=0.$
We can construct a similar example starting with $$X=\{f\in C[0,1]\,:\, f(0)=0\}$$ The dual space consists of measures $\mu$ such that $\mu(\{0\})=0.$ Let $$W=\left \{\mu\in M(0,1)\,:\,\mu(\{0\})=0,\ \int\limits_0^1d\mu(x)=0\right \}$$ Then $^\perp W=\{0\}.$ Indeed the measures of the form $\mu=\delta_{x_1}-\delta_{x_2}$ for $0<x_1<x_2\le 1$ belong to $W.$ If $f\in ^\perp\hspace{-3pt} W,$ then $f(x_1)=f(x_2).$ Hence $f$ is constant on $(0,1].$ As $f(0)=0$ we get $f=0.$
Remark It seems that the problem is connected with the Goldstine theorem.