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Why the element of dual space of l-infinity can be represented as sum of l1 and c0 elements?

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But that is not true, as is well known. (Hint: Ultrafilters on $\mathbb{N}$.) – Harald Hanche-Olsen Mar 22 '11 at 9:46
And Fremlin and Talgats paper it is. And i didnt understand it – Ravil Mudarisov Mar 22 '11 at 10:18
Perhaps you could be more specific in your question - you could at least provide info about the article you're studying. Did you mean D. H. Fremlin and M. Talagrand: A Gaussian Measure on $l^\infty$ ? – Martin Sleziak Mar 22 '11 at 11:00
Sorry. Yeah, i mean Fremlin and Talagrand article. – Ravil Mudarisov Mar 22 '11 at 11:59

2 Answers 2

up vote 7 down vote accepted

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 $l^\infty$ (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.

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This answers my question. Thank you. – Ravil Mudarisov Mar 22 '11 at 12:05
+1 for finding the question that was supposed to be asked. (The proof is tantamount to showing that the dual of $c_0$ is $\ell_1$ and then saying that $\ell_\infty^* = c_0^\perp \oplus c_0^*$.) – Yemon Choi Mar 22 '11 at 19:27
thanks. sometimes i think that the main aim of a students is to find that kinfd of questions and defects.) Good comment. – Ravil Mudarisov Mar 22 '11 at 23:03
Now I accidentaly stumbled upon the Hewitt-Yosida decomposition of a finitely additive measure into purely additive and $\sigma$-additive part. See e.g.… If I understand it correctly, after representing the functionals as finitely additive measures it is basically the same thing. It is sumarized nicely in Theorem 6.31 of Aliprantis-Border - the page is not viewable at google books, but you can find the claim here:… – Martin Sleziak May 10 '11 at 14:01

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'.

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