Let $V$ be a vector space over a field $\mathbb{F}$ and consider its dual, $V^*$. The natural pairing is the bilinear functional defined as follows: \begin{align*} \langle \cdot, \cdot \rangle : V^* \times V &\to \mathbb{F} \\ (f, \mathbf{v}) & \mapsto \langle f, \mathbf{v}\rangle := f(\mathbf{v}). \end{align*}
Let $U$ be a subspace of $V$. We define the orthogonal complement of $U$, $U^\perp$, as $$U^\bot := \{f \in V^* :\, \forall \mathbf{u} \in U \ ( \langle f, \mathbf{v}\rangle = 0)\}.$$
Now consider a subspace of $V^*$, $W$. Its orthogonal complement is defined as $$W^\bot := \{\mathbf{v} \in V :\, \forall f \in W \ ( \langle f, \mathbf{v}\rangle = 0)\}.$$
By definition we have that $W \subset (W^\perp)^\perp$. My question is: Is $(W^\perp)^\perp = W$?
Attempt at counterexample: Consider the vector space $V$ of all the real functions defined on $\mathbb R$. Let $W \subset V^*$ be the subspace spanned by all pointwise evaluations, $W = \text{span }\big(\{\delta_{x} :\, x \in \mathbb R\}\big)$, where $\delta_x : V \to \mathbb R$ maps $f \mapsto f(x)$. The orthogonal complement $W^\bot$ contains only the zero function. On the other hand, $(W^\perp)^\perp = V^*$ because in general there is no finite linear combination of pointwise evaluations that generates, for instance, the mean of a function, $(W^\perp)^\perp \setminus W \neq \varnothing$.