I asked this at [math.stackexchange](https://math.stackexchange.com/questions/1715947/a-linear-functional-on-ck-continuous-on-each-l-1-mu?noredirect=1#comment3502382_1715947), 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 ${\mathcal C}(K)^{\*\*}$?)
> 
> 
> 



A linear functional on $C(K)^*$ continuous on each $L_1(\mu)$
Yes. 


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 $L^1(\mu)$.)


