TITLE: Dual space of $\ell^\infty$
QUESTION [8 upvotes]: Why can the elements of the dual space of $\ell^\infty(\mathbb N)$ be represented as sums of elements of $\ell^1(\mathbb N)$ and Null$(c_0)$?

EDIT: As confirmed in the comments, the OP intended to ask about this sentence
"$f\in\ell_\infty^*$ is the sum of an element of $\ell_1$ and an element null on $c_0$" from the paper D. H. Fremlin and M. Talagrand: A Gaussian Measure on $l^\infty$ http://jstor.org/stable/2243023 (Which is different claim from what was in the original version of the question.)

REPLY [6 votes]: Let's recall a simple, elementary, and general fact that hasn't been explicitly mentioned: a dual Banach space   is always   a splitting subspace in the isometric embedding into its double dual.
Let $i_X:X\to X^{**}$ denote the natural isometric embedding of $X$ in $X^{**}$. If we dualize, we have a transpose operator,  $i_X^*:X^{***}\to X^*$ (that we may identify as the restriction map, which takes a linear form on $X^{**}$ to its restriction on $X$ as a subspace of $X^{**}$). On the other hand we also have the isometric embedding $i_{X^*}:X^*\to X^{***}$. It is a straightforward (though a bit formal) computation checking that $i_{X}^*$ is left-inverse to $i_{X^*}$, that is $i_{X}^*i_{X^*}=1_{X^*}.$ As a  consequence of this,  $P:=i_{X^*}i_{X}^*$ is a linear projector with $\operatorname{ker}P=\operatorname{ker}i_X^*=X^\perp$ corresponding to the splitting $X^{***}=X^*\oplus X^\perp$.
$$*$$
Checking the identity $i_{X}^*i_{X^*}=1_{X^*}.$ This means $i_{X}^*i_{X^*}f=f$ for all $f\in X^*$, which also means $\langle i_{X}^*i_{X^*}f, x\rangle=\langle f, x\rangle$ for all $x\in X$ and $f\in X^*$. Indeed
$$\langle i_{X}^*i_{X^*}f, x\rangle_{X^*\times X}=\langle i_{X^*}f, i_{X} x\rangle_{X^{***}\times X^{**}}=\langle i_{X} x,f\rangle_{X^{**}\times X^*}=\langle f, x\rangle_{X^*\times X},$$
by the definition of the transpose operator $i_{X}^*$, respectively by the  definition of the embeddings $i_{X^*}$ and $i_{X}$.