Identification of $\ell_\infty$ with $C(\beta\mathbb{N})$ and of $\ell_\infty^*$ with $C(\beta\mathbb{N})^*$

banach-spacesdual-spacesfunctional-analysisgeneral-topology

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

For example, $x=(x_n)_n$, with $x_n=1$, for exery $n\in\mathbb{N}$ I think has $\beta x=1$ (the constant function identically equal to $1$).

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

My questions are:

  1. If $\mu\in C(\beta\mathbb{N})^*$ and $x=(x_n)_n\in \ell_\infty$ can we find a closed formula for $\mu(\beta x)$?

For example, if $\delta^k\in\ell_\infty$ is given by $(\delta^k)_n=1$, if $n=k$; $(\delta^k)_n=0$, for $n\neq k$ can we say what $\beta \delta^k$ is and then what $\mu(\beta \delta^k)$ is?

When $x=(x_n)_n$, with $x_n=1$, for exery $n\in\mathbb{N}$, $\beta x=1$, so $\mu(\beta x)=\mu(\beta\mathbb{N})$ is the (finite) measure of $\beta\mathbb{N}$.

  1. Could you suggest a book where this subject is treated?

Best Answer

  1. In your special example, sure: $\beta \delta^k$ is the function on $\beta \mathbb{N}$ defined by $(\beta \delta^k)(y) = 1$ when $y=k \in \mathbb{N} \subset \beta \mathbb{N}$ and $0$ otherwise. This is easily verified by showing that the function $f : \beta \mathbb{N} \to \mathbb{R}$ defined by this formula is continuous on $\beta \mathbb{N}$ (recall that $\mathbb{N}$ is open in $\beta \mathbb{N}$, hence $\{k\}$ is open as well) and that it agrees with $\delta^k$ on $\mathbb{N}$.

    As such, $\mu(\beta \delta^k) = \mu(\{k\})$, the measure of the singleton set $\{k\}$.

    For more general $x \in \ell^\infty$, you probably can't do this in closed form, precisely because you can't express $\mu$ in closed form except in very special cases.

  2. Sorry, I don't know a reference for this off the top of my head.

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