Dual space of $\ell^\infty$

Question

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$(c_0)$?

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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 $l^\infty$ https://www.jstor.org/stable/2243023 (Which is different claim from what was in the original version of the question.)

Answer 1

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 $l^\infty$ https://www.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.

Answer 2

The fact stated above by Martin is a special case of the general property of a bounded functional on a von Neumann algebra - it can be always decomposed into a sum of a normal functional (in other words an image of a functional in the predual, in this case a functional represented by a sequence in $l^1$) and a singular functional (a `highly non-normal' functional, in the special case a functional vanishing on $c_0$). One can even achieve the decomposition respecting the functional norms in a suitable sense

The general result together with some discussion can be found in the first volume of Takesaki's `Theory of Operator Algebras'.

Answer 3

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. So $\ell_1$ is a particular case, for $X=c_0$.

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 $X^{***}=X^*\oplus X^\perp$.

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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 $i_{X}$.