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The Stone-Cech compactification of $\mathbb{N}$, denoted by $\beta\mathbb{N}$ has the property that every $x\in\ell_\infty$ is identified with (extended uniquely) to $\beta x\in C(\beta\mathbb{N})$.

The dual of $C(\beta\mathbb{N})$ is identified with the space of finite Borel measures on $\beta\mathbb{N}$, that is, by the Riesz's Representation Theorem: - for every $F\in C(\beta\mathbb{N})^*$ there is a unique finite Borel measure $\mu$ on $\beta\mathbb{N}$ such that $$ \forall f\in C(\beta\mathbb{N}),\ F(f)=\int_{\beta\mathbb{N}} f d\mu $$

My question is the following conjecture:

Let $\mu\in \ell_\infty^*$. Then $\mu\in\ell_1$ iff $\mu(\beta\mathbb{N}\smallsetminus\mathbb{N})=0$.

Neutral Element
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  • It seems to me that $\ell^_\infty$ contains elements that are not restrictions of elements in $C^(\beta\mathbb{N})$. – Mittens Mar 10 '23 at 15:46
  • Let $$\mu(x)=\lim_n\frac1n\sum^n_{k=1}x(k)$$for $x\in\mathcal{c}0\oplus\mathbb{R}$ equipped with the norm in $\ell\infty$. $\mu$ is a bounded positive functional and by Hahn-Banach-Kantorovich can be extended to a bounded positive functional on $\ell_\infty$. Let $\mu'$ one such extension. $\mu'$ is not a measure in $(\mathbb{N},2^{\mathbb{N}})$ for it is not $\sigma$-additive: $\mu'(\mathbb{1}_{{k}})=0$, but $\mu'(\mathbb{N})=1\neq\sum_k\mu'({k})$. Hence $\mu'\notin\ell_1(\mathbb{N})$ and is not the restriction to $\mathbb{N}$ of any measure on $\mathcal{C}(\beta\mathbb{N})$. – Mittens Mar 10 '23 at 15:57
  • @OliverDíaz $\ell_\infty$ is isomorphic to $C(\beta\mathbb{N})$ and so are their duals. – Neutral Element Mar 10 '23 at 18:58
  • As you can see, the element $\mu'$ that I gave you is not a measure in $(\mathbb{N}, 2^{\mathbb{N}})$, it is a charge though, and yet $\mu'\in \ell^*_\infty$. – Mittens Mar 10 '23 at 19:11
  • @OliverDíaz The Hahn-Banach theorem suffices. If $\mu'$ is such an extension of of $\mu$ then $|\mu'|=1$ and $\mu'({\mathbf 1}_{\mathbb{N}})=1.$ Therefore $\mu'$ is a positive functional on $\ell^\infty.$ I do not quite get you first comment. I thought that $\ell^\infty$ was isomorphically isometric to $C(\beta\mathbb{N})$ and so were their dual spaces. – Ryszard Szwarc Mar 10 '23 at 19:18
  • @RyszardSzwarc: The dual of $C(\beta\mathbb{N})$ is the space of Berel measures on $\beta\mathbb{N}$ a thus each element there is $\sigma$-additive. The charge $\mu’$ is not sigma additive – Mittens Mar 10 '23 at 19:45
  • @OliverDíaz How do you know that $\mu'(\mathbb{N})=1$ ? The indicator function of natural numbers does not belong to $C(\beta \mathbb{N}).$ We clearly have $\mu'(\beta \mathbb{N})=1.$ – Ryszard Szwarc Mar 10 '23 at 20:23
  • @RyszardSzwarc: I guess my point is that $\mu’$ is in the dual of $\ell_\infty$ but it is not the restriction to $\mathbb{N}$ of any measure (in view of not being countably additive). – Mittens Mar 10 '23 at 20:41
  • If $\mu$ is a nonnegative functional the assumption $\mu(\beta\mathbb{N}\setminus \mathbb{N})=0$ should suffice. For general $\mu$, I think, you should assume that $\mu(A)=0$ for any Borel subset $A$ disjoint with $\mathbb{N}.$ – Ryszard Szwarc Mar 10 '23 at 20:43
  • @OliverDíaz I think $\mu'(\mathbb{N})=0$ and the countable additivity holds. – Ryszard Szwarc Mar 10 '23 at 21:15
  • @RyszardSzwarc: no it is $1$ since $\mu(\mathbb{1}{\mathbb{N}})=\lim_n\frac{1}{n}\sum^n{k=1}\mathbb{1}_{\mathbb{N}}(k)=1$. $\mu'$ is an extension of $\mu$. – Mittens Mar 10 '23 at 21:50
  • Let us continue this discussion in chat. – Mittens Mar 10 '23 at 21:57

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For $x\in \ell^\infty,$ let $f_x$ denote the corresponding function on $\beta\mathbb{N}.$ Thus $f_x(n)=x(n)$ for $n\in \mathbb{N}\subset \beta\mathbb{N}.$ For $\mu\in C(\beta \mathbb{N})^*$ let $\varphi_\mu$ denote the corresponding linear functional on $\ell^\infty.$ Assume $\mu$ restricted to $\beta\mathbb{N}\setminus \mathbb{N}$ is zero, i.e. $\mu(A)=0$ for all Borel sets $A \subset \beta\mathbb{N}\setminus \mathbb{N}.$ As $\mu$ is a signed measure with finite variation we get $$\sum_{n=1}^\infty |\mu(n)|<\infty $$ For any $x\in \ell^\infty$ we have $$\varphi_\mu(x)=\mu(f_x)=\int\limits_{\beta \mathbb{N}}f_x(t)\,d\mu(t)=\int\limits_{\mathbb{N}}f_x(t)\,d\mu(t)=\sum_{n=1}^\infty x(n)\mu(n)$$ Hence $\varphi_\mu$ corresponds to $\ell^1$ sequence $\{\mu(n)\}_{n=1}^\infty.$

Ryszard Szwarc
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  • You have suggested in one of your previous comments that for any Borel subset disjoint with ℕ, ()=0 because we want $\int_{\beta\mathbb{N}}=\int_{\mathbb{N}}$ and I agree. Here $\mu$ is a signed measure. Could you explain why the equality $\int_{\beta\mathbb{N}\smallsetminus\mathbb{N}}=0$ holds for a signed measure only when $\mu(\beta\mathbb{N}\smallsetminus\mathbb{N})=0$? – Neutral Element Mar 11 '23 at 11:13
  • I will re-edit. I took for granted that $\mu$ was nonnegative, basing on the discussuon with Oliver Diaz. – Ryszard Szwarc Mar 11 '23 at 11:36
  • @RyszardSzwarc: I added my comments and description of $\ell^_\infty$ as measures plus pure charges below. The issue is how to reconcile the isometry between $\ell^_\infty$ and $C^*(\beta\mathbb{N})$, and particular how such an isometry carries pure charges on $(\mathbb{N},2^{\mathbb{N}})$ into finite measures on $(\beta\mathbb{N},\mathscr{B}(\beta\mathbb{N}))$. There is an old posting (without a good answer) related to this problem here – Mittens Mar 12 '23 at 13:10
  • @OliverDíaz Thanks for the comments. – Ryszard Szwarc Mar 12 '23 at 13:30
  • @RyszardSzwarc: (+1) for your answer and for taking the time for my follow-up questions. – Mittens Mar 12 '23 at 14:23
  • @RyszardSzwarc About one of your previous comments: " If is a nonnegative functional the assumption (ℕ∖ℕ)=0 should suffice. For general , I think, you should assume that ()=0 for any Borel subset disjoint with ℕ." Can't we use the fact that $\mu$ is outer and inner regular to derive from (ℕ∖ℕ)=0 that ()=0 for any Borel subset disjoint with ℕ? – Neutral Element Mar 21 '23 at 12:52
  • I don't think so. For example choose two points $x_1\neq x_2$ in $\beta\mathbb{N}\setminus \mathbb{N}.$ The signed measure $\mu=\delta_{x_1}-\delta_{x_2}$ satisfies $\mu(\beta\mathbb{N}\setminus \mathbb{N})=0.$ – Ryszard Szwarc Mar 21 '23 at 15:38