Let $V = ( \mathbb{R}^2,\ l^p)$. Then dual space is $V^{*}$ and it consists of functions $v^* : V \rightarrow \mathbb{R} $. Let $A \subset V $.
Consider two sets. Set $Z $ consists of functions which annihilate all elements of $A$: $$ Z:= \lbrace v^* \in V^* : \forall \ v \in A \quad v^*(v) = 0 \rbrace \text{,} $$ Set $W$ consists of functions which annihilates all elements of $Z$: $$ W:= \lbrace v^{**} \in V^{**} : \forall \ v^* \in Z \quad v^{**}(v^*) = 0 \rbrace \text{.} $$
I have to proof that $A \subset W$. However, since $V^{**} $ consists of functions $ v^{**} : V^* \rightarrow \mathbb{R} $, then $W$ is set of functions as well, while $A$ is set of vectors in $\mathbb{R} $ with norm $l^p$.
I can't grasp the idea of double dual space, especially what means that $V^{**} = V $ which is the case of $l^p $ and $L^p$ spaces, since $V$ can be set of any objects, while $V^{**} $ is set of functions defined of functions.
Technically, the double dual $V^{**}$ is only isomorphic to $V$. However, the isomorphism is canonical and does not depend on the choice of basis, hence $V$ and $V^{**}$ are often identified. The isomorphism is given by mapping $v\in V$ to the Map that sends a dual vector $\phi$ to $\phi(v)$. The set $W$ is thus identified with the set of all vectors on which all elements of $Z$ vanish.