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](https://mathoverflow.net/questions/59149/dual-space-of-ell-infty/59168#comment148669_59168), 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.)





Dual space of $\ell^\infty$
Obviously, 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 question.)


The authors refer to the book Day, M. (1973). Normed Linear Spaces. Springer, Berlin. I was not able to find the exact place in Day's book where this is shown, but I think that for this special case it is relatively easy.


For $f\in\ell\_\infty^\*$ put $a\_i=f(e^i)$. Then the sequence $a=(a\_i)$ belongs to $\ell\_1$. (Since $\sum\limits\_{i=1}^n |a\_i| = \sum\limits\_{i=1}^n |f(e^i)| = f(\sum\limits\_{i=1}^n \varepsilon\_ie^i) \le \lVert f \rVert$, where $\varepsilon\_i=\pm1$ are chosen according to the signs of $f(e^i)$.)


Now, if $x\_n\to 0$, then 
$$f(x)-a^\*(x)= \lim\limits\_{n\to\infty} f(\sum\limits\_{i=1}^n x\_ie^i)-\sum\limits\_{i=1}^n a\_ix\_i=0.$$


I hope I haven't overlooked something and that someone will provide the reference to the result (probably more general) which the authors of the above-mentioned paper had in mind.


