Dual space of $\ell^\infty$

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.)

• 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 $l^\infty$ jstor.org/stable/2243023 ? – Martin Sleziak Mar 22 '11 at 11:00
• Sorry. Yeah, i mean Fremlin and Talagrand article. – Ravil Mudarisov Mar 22 '11 at 11:59
• My favorite article on this topic is sciencedirect.com/science/article/pii/S0019357798800396 sorry couldn't resist... – Abdelmalek Abdesselam Jan 24 '18 at 22:50

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.

• 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. books.google.com/… 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: thales.doa.fmph.uniba.sk/sleziak/texty/rozne/pozn/books/… – Martin Sleziak May 10 '11 at 14:01
• @YemonChoi: Can we put is a norm on the direct sum $c_0^{\perp} \oplus c_0^*?$ – Idonknow Apr 19 '18 at 2:08

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'.

• It is also a special case of a general decomposition property of Dedekind complete Riesz spaces. See 353 in Fremlin's Measure Theory. – PassingThru Jan 21 '16 at 20:16

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 $i_{X}$.

• I use this fact to give a quick and easy construction of the Borel functional calculus for a self-adjoint operator $A \in B(H)$. The continuous functional calculus is an isometric linear map from $C({\rm spec}(A))$ into $B(H)$. Double dualizing, we get an isometric linear map from $C({\rm spec}(A))^{**}$ into $B(H)^{**}$. Since $B(H)$ is a dual space, we can project that $B(H)^{**}$ onto $B(H)$; restricting the domain to ${\rm Bor}({\rm spec}(A)) \subseteq C({\rm spec}(A))^{**}$ then yields the Borel functional calculus. – Nik Weaver Jan 24 '18 at 22:55
• @NikWeaver Aren't you using Arens products somewhere in the background, then? – Yemon Choi Jan 24 '18 at 22:56
• @YemonChoi You mean, to get multiplicativity? Sure, I guess you can do it that way. Probably I would just make a continuity argument, given that the continuous functional calculus is multiplicative. – Nik Weaver Jan 24 '18 at 23:26
• @NikWeaver: Wonderful – Pietro Majer Jan 24 '18 at 23:34