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

Question

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, more precisely, $$ \varPhi_\mu(f)(x)=\int_K x(t)\cdot f(t)\cdot\mu(d t),\quad f\in L_1(\mu),\quad x\in {\mathcal C}(K). $$ 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 norm).

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

P.S. I asked this also at MathOverflow.

Answer 1

Edit: Proof much simplified.

Yes, it's true.

I'm going to write $N(\mu)$ for what we've been calling $C_\mu$. So $N(\mu)$ is the smallest constant such that $$|p(f\mu)|\le N(\mu)\int|f|\,d|\mu|$$for $f\in L^1(\mu)$.

Lemma 0 If $c$ is a non-zero constant then $N(c\mu)=N(\mu)$.

More generally:

Lemma 1 If $f\in L^1(\mu_1)$ and $\mu_2=f\mu_1$ then $N(\mu_2)\le N(\mu_1)$.

Proof: If $g\in L^1(\mu_2)$ then $$|p(g\mu_2)|=|p((gf)\mu_1| \le N(\mu_1)\int|gf|\,d|\mu_1|=N(\mu_1)\int|g|\,d|\mu_2|.$$

Theorem $\sup_\mu N(\mu)<\infty$.

Proof. Suppose on the other hand that there exists a sequence $\mu_n$ of measures with $N(\mu_n)\to\infty$. By Lemma 0, wlog $\sum||\mu_n||<\infty$. Let $\mu=\sum|\mu_n|$. Then $\mu_n<<\mu$, so Lemma 1 shows that $N(\mu)\ge N(\mu_n)$.