4
$\begingroup$

Let $X$ be a compact metric space and $C(X)=\{ f:X\rightarrow \mathbb{R} \ | \ \ f \ continuous\}$ with the uniform norm. It is a separable Banach space.

1) I'm aware of the fact that $C(X)^*$, the space of continuous linear functionals $C(X)\rightarrow \mathbb{R}$ coincide with $M(X)$, the space of signed regular borel measures on $X$.

This intuitively makes sense because $\mu\in M(X)$ can be naturally be seen as an evaluation map $f\mapsto \mathbb{R}$ by means of integration.

2) I'm also aware that $(C(X)^*)^*$, the dual of the dual, coincide with the set of bounded Borel measurable functions $F:X\rightarrow\mathbb{R}$ (source: P. Lax, Functional Analysis).

UPDATE Landscape provided a counterexample in the comments. My source was: P. Lax "Functional Analysis" 2002, Page 82, Theorem 14(ii). I guess this might be a mistake, perhaps fixed in some errata somewhere.

New question: C.f. Landscape's example, let $g_A$ be the extension (given by Hahn-Banach) of $f_A$ to $C([0,1])^{**}$. Looking at $g_A$ as a function $X\rightarrow\mathbb{R}$, $g_A$ must satisfy by construction the equation $\int_{[0,1]}g_A \ d \ \delta_x= 1$ if $x\in A$ and $0$ otherwise. This seems to imply that $g_A$ is the characteristic function of $A$. Hence not Borel if $A$ is not Borel.

Thus, $g_A$ as a function $(X\rightarrow\mathbb{R})$ is uniquely determined by the construction starting from $f_A$. But is it $g_A$ as an element of $C([0,1])^{**}$ uniquely determined? If so, it looks to me we would have a reasonably well defined notion of integration for non-measurable sets.

Thanks!


Screen shot of Lax's theorem 14 on page 82:

Lax, Theorem 14, page 82

$\endgroup$
  • 1
    $\begingroup$ Things in $C(X)^{**}$ are things that morally can be integrated with respect to every Borel measure. Does that give you intuition about Q1? $\endgroup$ – Mariano Suárez-Álvarez May 15 '13 at 8:48
  • $\begingroup$ Mariano thanks for the answer. This is helpful. But also, e.g., analytic functions can be integrated and are not borel. But beside this observation... Why is this the right way to look at C(X)**? After all, we do not look at $C(X)^*$ as things that can be "integrated" with respect to every continuous $f\in C(X)$, don't we? Please bear with my ignorance here. $\endgroup$ – IamMeeoh May 15 '13 at 8:56
  • $\begingroup$ Let me also ask, just to be sure. it correct to write that the action of $F\in C(X)^{**}$ on measures is expressed as $F(\mu)=\int F d \mu$, right? $\endgroup$ – IamMeeoh May 15 '13 at 8:58
  • $\begingroup$ The assertion in your question 1 is incorrect. Please the comment given by commenter here. $\endgroup$ – 23rd May 15 '13 at 10:20
  • $\begingroup$ Hello Landscape, thanks! this is quite surprising as I think Lex's book is a very solid reference. Regarding the counterexample proposed in the page you linked. To apply Hahn-Banach it is required for $A$ not just to be subset of $[0,1]$ but to be a sub-linear space. Does this make any difference? $\endgroup$ – IamMeeoh May 15 '13 at 11:42
2
$\begingroup$

Let $X$ be a compact metric space. Let $\mathcal B$ be the collection of Borel sets on $X$. Banach space $C(X)$ is the set of all (necessarily bounded) continuous real-valued functions $f : X \to \mathbb R$, with norm $$ \|f\|_\infty = \max\{|f(x)| : x \in X\} $$

Banach space $M(X)$ is the set of all countably-additive signed ($\mathbb R$-valued) measures on the sigma-algebra $\mathcal B$. The norm is the total variation: for $\mu \in M(X)$, write $\mu = \mu^+ - \mu^{-}$ where $\mu^+$ and $\mu^-$ are positive measures, singular to each other, and let $$ \|\mu\|_1 = \mu^+(X)+\mu^-(X) $$

The pairing $M(X) \times C(X) \to \mathbb R$ defined by $$ \langle\mu,f\rangle = \int_X f \;d\mu $$ identifies $C(X)^\ast$ isometrically with $M(X)$ in these norms.

Another description of $M(X)$ may be obtained as follows. Let $\{\tau_i : i \in I\}$ be a maximal (under inclusion) family of mutually singular probability measures on $(X,\mathcal B)$. (Such a family exists by Zorn's Lemma. But it is not unique.) Identify $M(X)$ isometrically with the $l^1$-direct sum of the family of all the $L^1(\tau_i)$ spaces: $$ M(X) \approx \left(\bigoplus_{i\in I} L^1(X,\mathcal B,\tau_i)\right)_{1} $$

Once we describe $M(X)$ in this way, we get the corresponding description of the dual as the $l^\infty$-sum of the spaces $L^\infty(\tau_i)$. $$ C(X)^{\ast\ast}\approx \left(\bigoplus_{i\in I} L^\infty(X,\mathcal B,\tau_i)\right)_{\infty} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.