A condition for surjectivity of a linear map Let $V,W$ be vector spaces (not necessarily finite dimensional!), and let $W^*$ the dual of $W$. Let
$$A:V\longrightarrow W^*$$
be a linear map. What conditions do I have to put on $V$ and especially $W$ for the following to hold?

$A$ is surjective $\Leftrightarrow$ $\forall w\in W\backslash\{0\}$ $\exists v\in V$ such that $\langle Av,w\rangle\neq 0$.

(Here $\langle\cdot,\cdot\rangle$ is the natural pairing of $W^*$ and $W$.)
I know that this condition is true for Hilbert spaces and not true in general (consider as a counterexample the inclusion of $\ell^1$ in $\ell^\infty\cong(\ell^1)^*$).  Would it be true for example when $W$ is reflexive?
 A: Note that your statement can be written as

$A$ surjective $\Leftarrow$ $R(A)_\perp =\{ w\in W: \ \langle Av,w\rangle = 0 \ \forall v\in V\}= \{0\}$

The desired implication fails if $A$ has dense range but is not surjective (provided $V,W$ are normed spaces, $A$ continuous, $W^*$ continuous dual). If the range of $A$ is dense then 
$$
\langle Av,w\rangle = 0 \ \forall v\in V \ \Rightarrow 
\langle w^*,w\rangle = 0 \ \forall w^*\in W^* \ \Rightarrow w=0.
$$
So $A$ is not surjective but $R(A)_\perp=\{0\}$.
In the 'topology-free' setting, i.e., solely taking algebraic duals, the following implication is true:

$A$ surjective $\Leftarrow$ $R(A)^\perp =\{ f\in W^{**}: \ \langle f,Av\rangle = 0 \ \forall v\in V\}= \{0\}$

Assume $R(A)^\perp = \{0\}$ but $A$ is not surjective. Then there is $w^*\in W^*$ with $w^*\not\in R(A)$. By Hahn-Banach there is a linear functional $f\in W^{**}$ such that $f(w^*)=1$ but $f(Av)=0$ for all $v$. A contradiction.
(If $W$ and $W^{**}$ are isomorphic then also the original claim can be recovered. I am not sure - as I am not used to work with Hamel basis)
