3
$\begingroup$

We consider a locally compact Hausdorff space $X$ and the Banach space $C_0(X)$ of continuous functions on $X$ taking values at $\mathbb K = \mathbb R$ or $\mathbb C$, equipped with the supremum norm.

I'm studying this paper: http://www.siue.edu/MATH/kj_papers/Into.pdf. At page 375, in the proof of Lemma 2, it says the following:

"If $A$ is a closed subspace of $C_0(X)$, then any functional $F\in A^\ast$ can be extended, with the same norm, to a regular Borel measure $\mu_F$ on $X$."

Until this point, there is no big deal. We use Hahn-Banach to extend $F$ to $\tilde F\in(C_0(X))^*$ with the same norm then use Riesz Representation Theorem to get a regular Borel measure $\mu_F$ at $X$ such that $$ \tilde F(f)=\int f d\mu_F,\ \forall f \in C_0(X). $$ Continuing:

"On the other hand, $A$ can be regarded as a subspace of continuous functions on $A^*$ with the weak* topology, and any such measure $\mu_F$ can be regarded as a regular Borel measure defined on a closed subset of $A^*$."

I'm having difficult to understand it.

In the first statement, I came to the conclusion that he reffers to the image of $A$ by the James map $J: A \to A^{**}$, and since the weak* topology preserves the continuity of the functionals in $J(A)$, every element of $J(A)$ is a continuous function on the topological space $(A^*,\omega^*)$. The map $J$ is linear (so $J(A)$ is a subspace of $(A^*,\omega^*)$).

I guess there is a second possibility to interpret the first statement, that is to consider the map $\delta: X \to A^*$ such that $\delta(x)$ is the evaluation map on $x$ restricted to $A$. But then I don't know where to go from here.

In the second statement, I have no idea what meant. Please help me.

$\endgroup$
  • $\begingroup$ I think all they are saying is the following: any $f\in A$ can be viewed as the function $f(F):=F(f)$ ($F\in A^*$) on its dual, and since $A^*$ separates the points of $A$, we can recover $f\in A$ from $f(F)$. Then a functional on $A$ can now be viewed as a functional on certain continuous functions on $A^*$ (continuity of $f(F)$ is of course built into the definition of the weak $*$ topology), so comes from a measure on $A^*$. $\endgroup$ – Christian Remling Oct 2 '18 at 20:29
  • $\begingroup$ Ok, but why does it come from a measure on $A^*$? I can't use Riesz representation to get a measure that represents a functional on the continuous functions on $A*$ because this space is not a $C(A^*)$ space (since $A^*$ is not compact, it does not have the supremum norm). $\endgroup$ – André Porto Oct 2 '18 at 22:23
1
$\begingroup$

I think I got it now. My interpretation of the first statement is right. Let us recall it:

"A can be regarded as a subspace of continuous functions on A∗ with the weak* topology"

Consider the map $J: A \to A^{**}$ defined by $ a\in A \mapsto \hat a\in A^{**}$. Since the weak* topology preserves the continuity of the functionals $\hat a$, it follows that every $\hat a$ is a continuous real function on $(A^*,\omega^*)$. Moreover, the map $J$ is injective so we can identify A with $J(A)$, and $J$ is linear, so $J(A)$ is a subspace of the spaces of continuous real function on $(A^*,\omega^*)$.

Now, let us interpret the secon statement:

"any such measure $\mu_F$ can be regarded as a regular Borel measure defined on a closed subset of $A^∗$."

Consider the map $\delta: X \to (A^*,\omega^*)$ such that $$ \delta(x)(f)=f(x),\ \ \forall f\in A. $$ It is simple to see that $\delta$ is a continuous map, so for every Borel set $E\subset (A^*,\omega^*)$, the set $\delta^{-1}(E)$ is a Borel set of $X$, and then we can define a measure $\tilde \mu_F$ on $(A^*,\omega^*)$ by $$ \tilde \mu_F(E) = \mu_F(\delta^{-1}(E)). $$ So we may consider that $\mu_F$ is defined on the Borel sets of $(A^*,\omega^*)$. Since $\|\delta(x)\|=1$ for any $x\in X$, we have that such measure is concentrated at the unit sphere of $A^*$, which is a closed subset of $(A^*,\omega^*)$.

$\endgroup$

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.