

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer

I'm trying to understand a claim I heard in class. To be concrete, suppose $X$ is a compact, hausdorff topological space, and let $C(X)$ be the space of continuous functions on $X$ with the supremum norm.

Now let $M(X)$ be the space of finite signed borel measures on $X$; $M(X)$ is isomorphic to $C(X)^*$; let's give $M(X)$ the topology of weak* convergence.

My question: what is the dual of $M(X)$ in the weak* topology? To be precise (or just to be redundant) I'm asking what the topological dual of $M(X)$ is, given that it has the topology of weak* convergence.

Clearly $M(X)^*$ in the total-variation-norm-topology is not just $C(X)$ (under the embedding $C(X) \rightarrow C(X)^{**}$), but the weak* topology on $M(X)$ is different; shouldn't the dual be different as well? Is it, in fact, merely $C(X)$ where the action is given by integration?

share|cite|improve this question
Yes, that dual is $C(X)$ again. Maybe you need to think about weak topologies in general a bit. – GEdgar Feb 6 '13 at 20:00
@GEdgar Ok, that's what I wanted to hear. Having thought about it, it seems like given such a continuous functional $l$ on $M(X)$ that I should define a continuous function $\phi(x) = l(\delta_x)$, where $\delta_x$ is the unit mass at $x \in X$. This is continuous by the continuity of $l$. The missing piece for me now is whether every $\mu \in M(X)$ is the weak* limit of finite linear combinations of $\delta$-masses. This is probably the result of some separability argument, however. – A Blumenthal Feb 6 '13 at 20:21
up vote 6 down vote accepted

Let $E$ be a locally convex space with topological dual $E'$. Equip $E'$ with the weak*-topology.

Every continuous linear functional $\varphi \colon E' \to \mathbb C$ is of the form $f \mapsto f(e)$ for some $e \in E$, so $(E',\text{weak*})' = E$. In particular, the dual space of $M(X)$ with the weak*-topology is $C(X)$ again.

Since $\varphi$ is continuous, the set $U = \left\{f \in E' \mid \lvert \varphi(f) \rvert \lt 1\right\}$ is an open neighborhood of $0$. Thus, recalling how the basis of neighborhoods at a point in $(E',\text{weak*})$ is defined, there exist $e_1,\dots,e_n \in E$ and $\varepsilon \gt 0$ such that $V = \left\{f \in E' \mid \max\left\{\lvert f(e_i)\rvert \mid i = 1,\dots,n\right\} \lt \varepsilon\right\}$ is contained in $U$. Let $\varphi_i(f) = f(e_i)$.

From $V \subseteq U$ it follows that $\ker{\varphi} \supseteq \bigcap_{i=1}^n \ker{\varphi_i}$ and from linear algebra (see here) we deduce that $\varphi = \sum_{i=1}^n \lambda_i \varphi_i$ for some $\lambda_i \in \mathbb{C}$. In other words, $$ \varphi(f) = \sum_{i=1}^n \lambda_i \varphi_i(f) = \sum_{i=1}^n \lambda_i f(e_i)= f\left(\sum_{i=1}^n \lambda_i e_i \right)$$ and we've shown that $\varphi$ is evaluation at $e = \lambda_1 e_1 + \cdots + \lambda_n e_n \in E$.

share|cite|improve this answer
That's more general than I suspected. Thanks! – A Blumenthal Feb 6 '13 at 22:02
You're welcome. You might be interested in this thread as well. – Martin Feb 13 '13 at 12:28

Your Answer


