Is $X^{***}$ useful in banach space theory? I was wondering why i have never seen $X^{***}$ be used anywhere in functional analysis. 
I know that $X$ can be viewed as a subspace of $X^{**}$, and in this way you can identify a subset $C$ of $X^{**}$ with a subspace of $X^{***}$, but it useful in some situation?
Or there is always a way to identify $C$ with a subspace of $X^{**}$ without using the third dual?
Thanks
 A: I know a theoretical application of $X^{***}$, when you show the reflexivity of a dual space. 
We denote the canonical map from a normed space $X$ to its bidual space $X^{**}$ by $i_{X}: X \to X^{**}, x \mapsto i_{X}[x]$ where $i_{X}[x](\varphi) = \varphi(x)$ for all $\varphi \in X^*$. Further for an Operator $T:X \to Y$ we donate its dual operator by $T^*: Y^* \to X^*$.
Consider the following lemma:
Let $X$ be a normed space. For $i_{X^*}: X^* \to X^{***}$ and $(i_X)^*: X^{***} \to X^*$. Then  it holds that $$(i_X)^* \circ i_{X^*} = \operatorname{id}_{X^*}.$$
Proof. For all $\varphi \in X^*$ and $x \in X$ we have
$$\Big((i_X)^*(i_{X^*}(\varphi))\Big)(x) = \Big((i_{X^*}(\varphi)) \circ i_X\Big)(x) = ((i_{X^*}(\varphi))(i_X(x)) = (i_X(x))(\varphi) = \varphi(x).$$
So we have $(i_X)^* \circ i_{X^*} = \operatorname{id}_{X^*}$.
Now  the application of this lemma:
Let $X$ a Banach space. Then if $X$ is reflexive, $X^*$ is reflexive too. (It actually holds iff but we need the lemma just for this direction)
Proof. Since $i_X$ is bijective, $(i_X)^*$ is bijective too. Our lemma gives us $(i_X)^* \circ i_{X^*} = \operatorname{id}_{X^*}$. Hence $i_{X^*} = [(i_X)^*]^{-1}$ is surjetive and so we have that $X^*$ is reflexive.
As you see it's a pretty special case and very complicated. I don't think one should use that space since there are often better ways to show things. 
