I asked this at math.stackexchange, but nobody answered.

Let $K$ be a (Hausdorff) compact topological space, ${\mathcal C}(K)$ the usual Banach space of continuous functions $x:K\to{\mathbb C}$, ${\mathcal C}(K)^*$ the Banach dual space of measures.

For each measure $\mu\in{\mathcal C}(K)^*$, $\mu\ge 0$, consider the natural mapping $$ \varPhi_\mu: L_1(\mu)\to {\mathcal C}(K)^*\quad\Big|\quad \varPhi_\mu(f)=f\cdot\mu,\quad f\in L_1(\mu), $$ or, in other words, $$ \varPhi_\mu(f)(x)=\int_K x(t)\cdot f(t)\cdot\mu(d t),\quad f\in L_1(\mu),\quad x\in {\mathcal C}(K). $$ Let $p:{\mathcal C}(K)^*\to{\mathbb C}$ be a linear functional, which is continuous on each subspace $L_1(\mu)$, i.e. for any $\mu$ the composition $p\circ\varPhi_\mu$ is continuous (=bounded) on the Banach space $L_1(\mu)$ (with the usual integral norm).

Is $p$ continuous on ${\mathcal C}(K)^*$? (Equivalently, is $p$ an element of ${\mathcal C}(K)^{**}$?)

  • 1
    $\begingroup$ "I asked this at math.stackexchange, but nobody answered": Well, you only waited 4 hours. The usual recommendation is only to cross-post after several days. See meta.mathoverflow.net/a/2638/4832. As a result, David Ullrich (on Math.SE) and I unnecessarily duplicated our efforts. $\endgroup$ Mar 28, 2016 at 14:32
  • $\begingroup$ Nate, excuse me, I did not know about this rule. I already apologized to David Ullrich. $\endgroup$ Mar 28, 2016 at 14:48
  • $\begingroup$ It's okay, I thought it was an interesting question, but I thought you should know this for the future. $\endgroup$ Mar 28, 2016 at 14:52
  • $\begingroup$ Of course! Actually, I think there must be a reminder for people like me, who don't know things like this. Say, a button "cross post" with an explanation when it should be used. $\endgroup$ Mar 28, 2016 at 15:01
  • $\begingroup$ If you look on meta.stackexchange.com you'll probably find some discussion of this idea. As I understand it, the general rule on Stack Exchange is never to cross post at all; the situation with Math.SE vs MO is sort of a special exception. So there might not be a lot of support for adding this to the software. $\endgroup$ Mar 28, 2016 at 16:00

1 Answer 1



Suppose $p$ were not continuous. Then we could find a sequence of signed Radon measures $\mu_n$ with norms $\|\mu_n\| \le 2^{-n}$ but $|p(\mu_n)| \ge 1$. Let $|\mu_n|$ denote the total variation measure of $\mu_n$, which is still Radon and has the same norm as $\mu_n$. Set $\mu = \sum_n |\mu_n|$; this sum converges in the Banach space $C(K)^*$, so $\mu$ is a positive finite Radon measure. Now each $\mu_n$ is absolutely continuous with respect to $\mu$, so let $f_n \in L^1(\mu)$ be its Radon–Nikodym derivative; then, in your notation, $\mu_n = \Phi_\mu(f_n)$. Moreover, we have $\|f_n\|_{L^1(\mu)} = \|\mu_n\|$, so $f_n \to 0$ in $L^1(\mu)$. Yet $|p(\mu_n)| = |p(\Phi_\mu(f_n))| \ge 1$, contradicting the continuity of $p \circ \Phi_\mu$ on $L^1(\mu)$.

Looking at it another way, this construction shows that any countable set $\{\mu_k\} \subset C(K)^*$ is contained in $L^1(\mu)$ for some $\mu$, namely $\mu = \sum_k a_k |\mu_k|$ for suitable positive coefficients $a_k$. Thus $p$ is continuous when restricted to any countable set, and by considering sequences (since $C(K)^*$ is a metric space), this is sufficient for continuity. (Indeed, since the spaces $L^1(\mu)$ are complete, this actually shows that any separable subset of $C(K)^*$ is contained in some $L^1(\mu)$.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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