# Every closed subspace $A$ of $C_0(K)$ can be regarded as a subspace of continuous functions on $A^*$?

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.

• 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 $A^*$. – Christian Remling Oct 2 '18 at 20:29
• 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). – André Porto Oct 2 '18 at 22:23

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 $$(A^*,\omega^*)$$.
"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 $$(A^*,\omega^*)$$.