I just learned about the weak-* topology on $X^*$, and in this context was introduced to the 'dual of the dual of a space', and the functional $J: X \rightarrow (X^*)^*$ where $J(x)[\psi] = \psi(x)$, i.e. $J$ takes a point $x \in X$ to the functional that takes a map from $X$ to the reals, and evaluates it at the point $x$. From this, $J(X) \subset (X^*)^*$, and the weak-* topology on $X^*$ is the topology induced by this $J(X)$.
It originally took me a while to parse this definition in the right way, but it has become more intuitive after I started thinking of the elements of $(X^*)^*$ as evaluation maps $J_x: X^* \rightarrow \mathbb{R}$ i.e. $J_x(\psi) = \psi(x)$. Formally, however, for a general space we only know $J(X) (\subset (X^*)^*)$ is a set of evaluation maps. $J(X)$ might be a proper subset.
My question: am I losing intuition about the space $(X^*)^*$ if I'm thinking about its members as evaluation maps? What do members of $(X^*)^* \setminus J(X)$ (i.e. those that are not evaluation maps) look like?
There is a concept for when all elements of $X^*$ are those coming from $X$, it is called reflexive space.
The typical example of a non-reflexive space is $c_0$, the space of sequences that tend to zero, with the supremum norm.
Then $c_0^*$ is isomorphic to $\ell^1$, the space of absolutely convergent series with norm the sum of the sequence of absolute values.
The double dual $(c_0)^{**}=(\ell^1)^*$ is isomorphic to $\ell^\infty$, the space of bounded sequences, with the supremum norm. You have that $J:c_0\to \ell^{\infty}$ is the inclusion, but you can see that it misses many elements of the co-domain. For example, the sequence $(1,1,1,...)$.