My question reads:

Let M be a locally compact Hausdorff space. A continuous real valued function $f$ : $M →R$ is said to vanish at infinity if, for every $\epsilon > 0$, there exists a compact set $K \subset M$ such that $$sup_{x\in M-K} |f(x)| < \epsilon$$ Denote by $C_0(M)$ the space of all continuous functions $f : M →R$ that vanish at infinity (see Exercise 3.2.10).

(a) Prove that $C_0(M)$ is a Banach space with the supremum norm.

no problem here

''(b)The dual space $C_0(M)^∗$ can be identified with the space $\mathcal{M}(M)$ of signed Radon measures on M with the norm (1.1.4) [They refer to the total variation to be the norm], by the Riesz Representation Theorem (see [75, Thm. 3.15 & Ex. 3.35]). Here a signed Radon measure on $M$ is a signed Borel measure μ with the property that, for each Borel set $B \subset M$ and each $\epsilon$ >0, there exists a compact set $K \subset B$ such that |$\mu(A)−\mu(A \cap K)| <\epsilon$ for every Borel set $A \subset B$.''

What exactly am I being asked to do here, other than quote the result of Reisz? I'm very confused here. What exactly does it mean to prove spaces ''can be identified'' with another?

(c) Prove that the map $δ : M → C_0(M)^∗$, which assigns to each x ∈ ${M}$ the bounded linear functional $δ_x : C_0({M}) →R$ given by $δ_x(f) := f(x)$ for $f ∈ C_0({M})$, is a homeomorphism onto its image $δ(M) \subset C_0({M})^∗$, equipped with the weak* topology. Under the identification in (b) this image is contained in the set $P(M)$ := {$\mu \in \mathcal{M}(M)\ $ s.t.: $\mu ≥ 0, ||\mu|| = \mu(M) = 1$} of Radon probability measures. Determine the weak* closure of the set $δ(M)$ = {$δ_x | x \in M$} $\subset$ $P(M)$.

  • $\begingroup$ The Riesz Representation theorem usually refers only to positive linear functionals. I guess the question is to extend this to a statement about all bounded linear functionals. $\endgroup$ – Prahlad Vaidyanathan Mar 10 '20 at 2:26
  • $\begingroup$ There's no question that I can see. $\endgroup$ – Martin Argerami Mar 10 '20 at 4:40
  • $\begingroup$ You should put in the minimal effort to make your question readable by using MathJax. $\endgroup$ – s.harp Mar 16 '20 at 23:07
  • $\begingroup$ @s.harp Edited as per request. $\endgroup$ – Muselive Mar 20 '20 at 23:01

I think part (b) is not really meant as a question, just as information that you will need to solve part (c). However, you might take it as an opportunity to solve Exercise 5.35 (not 3.35, it's a typo) of the book [75], which is Measure and Integration by Dietmar A. Salamon. Note that the statement at hand is not exactly what is shown by [75] Theorem 3.15, which is only about positive measures and positive linear functionals.

Formally, "can be identified with" would be to say that there is an isometric isomorphism between these two normed spaces: a bijective linear map $T : \mathcal{M}(M) \to C_0(M)^*$ which is an isometry (i.e. $\|T\mu\|_{C_0(M)^*} = \|\mu\|_{TV}$). The map should be understood to be $(T\mu)(f) = \int f\,d\mu$. Proving that this map is a bijection and an isometry are parts (iii) and (i) of [75] Exercise 5.35. So once this has been shown, you can think of the measure $\mu$ and the functional $T\mu$ as "the same object", and whenever you are asked to prove something about a measure, you could instead prove the corresponding fact about the corresponding functional, or vice versa.

In particular, the sentence "under the identification in (b)" in part (c) should be understood as saying that $T^{-1}(\delta(M)) \subset P(M)$.

  • $\begingroup$ I think the traditional Reiz Rep. Thrm is only on compactly supported functions (identified with Borel measures), so I have to extend this notion to functions vanishing at infinity and identify them with Radon measures? I suspect I have to use Hahn-Banach extension in some way but cannot parse the details. $\endgroup$ – Muselive Mar 21 '20 at 1:42
  • $\begingroup$ @Muselive: I think for this proof, you only need the easy converse of that: every linear functional on $C_0(M)$ is automatically a linear functional on $C_c(M)$. But what you said is also true: in general, if $X$ is a Banach space and $Y \subset X$ is a dense subspace, then the restriction map from $X^*$ to $Y^*$ is a bijection (and an isometry). In particular, every continuous linear functional on $Y$ extends to a unique continuous linear functional on $X$. You don't need Hahn-Banach to prove this. $\endgroup$ – Nate Eldredge Mar 21 '20 at 12:51
  • $\begingroup$ I'm confused. I have, via Reiz, that all elements of $C_C^*$ have the form: $T_\mu(f)=\int f d\mu$ I want this extend the map $T_\mu(f)$ to a functional on $C_0$. If not Hahn Banach, do I make some kind of compactification argument? That is, add the point $\infty$ to $M$ then use Reiz? This feels complicated. $\endgroup$ – Muselive Mar 21 '20 at 20:34
  • 1
    $\begingroup$ @Muselive: Or at least, we have $\Lambda(f) = \int f\,d\mu$ for all $f \in C_c$. But every $f \in C_0$ is a uniform limit of a sequence $f_n \in C_c$. Pass to the limit, using the continuity of $\Lambda$ on the left and the triangle inequality on the right. $\endgroup$ – Nate Eldredge Mar 21 '20 at 20:42
  • 1
    $\begingroup$ @Muselive: Since $\Lambda$ is a continuous linear functional on $C_0$, in which the norm is the uniform norm, we have $\Lambda(f_n) \to \Lambda(f)$. Next, the triangle inequality tells us that $\left| \int f_n \,d\mu - \int f\,d\mu \right| \le \int |f_n - f| \,d|\mu| \le \|f_n-f\|_{\sup} |\mu(M)|$. Since $f_n \to f$ uniformly, this goes to 0, telling us that $\int f_n\,d\mu \to \int f\,d\mu$. And since $\Lambda(f_n) = \int f_n\,d\mu$ for all $n$, we conclude that $\Lambda(f) = \int f\,d\mu$. $\endgroup$ – Nate Eldredge Mar 21 '20 at 23:57

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.