If $Y\subset X^\*$ is a closed subspace (where $X$ is a separable Banach space), the preannihilator of $Y$ in $X$ is $Y\_{\perp}:=\{x\in X : y^\*(x)=0, \forall y^\*\in Y \}$. If $Y$ is a proper subspace of $X^\*$, and $X$ is reflexive then it can be proved that $Y\_{\perp}\neq\{0\}$. On the other hand if $X=l\_1$ and $Y=c\_0\subset l\_{\infty}$, then $Y\_{\perp}=\{0\}$. 


* 1) If $X^\*$ is separable, is it still possible that $Y\_{\perp}=\{0\}$?
* 2) Can $c\_0$ be a subspace of a separable dual? This question is probably unrelated, and most likely well known, but I realized I do not know an answer to it.



Preannihilators of subspaces of separable duals
For (1), take $X = c\_0$ so that $X^\* = \ell^1$, which is separable. Take $Y = \{y \in \ell^1 : \sum\_i y(i) = 0\}$ which is a proper closed subspace. Since $Y$ contains all the elements of the form $e\_i - e\_{i+1}$, where $e\_i$ are the usual basis functions, you can see that any $x \in Y\_\perp$ satisfies $x(1) = x(2) = x(3) = \dots$. The only such element of $c\_0$ is $0$.


The intuitive idea is that by Hahn-Banach, $Y$ will certainly have a nontrivial annihilator living in $X^{\*\*}$. In this case it is the constants in $X^{\*\*} = \ell^\infty$ which are not in $X$.


More generally, whenever $X$ is non-reflexive, you should be able to choose any $f \in X^{\*\*} \setminus X$ and take $Y$ to be its kernel. Then if $x \in Y\_\perp$, it is also in the annihilator $Y^\perp$ when considered as an element $f\_x$ of $X^{\*\*}$. The kernel of $f\_x$ contains the kernel of $f$, so $f\_x$ is a scalar multiple of $f$. Since $f\_x \in X$, this is only possible if $f\_x = 0$, i.e. $x=0$.


