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.

  • $\begingroup$ I'm lost with your definition of $\Phi_\mu(f)$. if $K = [a,b]$, is it $\Phi_\mu(f)(g) = \int_a^b f g d\mu$ for any function $g$ continuous on $[a,b]$ ? $\endgroup$
    – reuns
    Commented Mar 27, 2016 at 17:45
  • $\begingroup$ Yes, of course! $\endgroup$ Commented Mar 27, 2016 at 17:46
  • $\begingroup$ what about $p \circ \Phi_\mu$ ? what is it ? $\endgroup$
    – reuns
    Commented Mar 27, 2016 at 17:47
  • $\begingroup$ It's the usual concatenation of mappings. $\endgroup$ Commented Mar 27, 2016 at 17:48
  • $\begingroup$ Note that $\Phi_\mu f$ is always a finite measure (I guess, by $f\mu$ you mean $f\,d\mu$?) $\endgroup$ Commented Mar 27, 2016 at 17:49

1 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 $N(\mu)\ge N(\mu_n)$.

  • $\begingroup$ David, I have to go to the university now, I'll check this later. Excuse me, I thought, after first comments people lost interest to this question, that's why I asked this also at MathOverflow: mathoverflow.net/questions/234648/… When I'll come back, I'll look at your answer, thank you anyway! $\endgroup$ Commented Mar 28, 2016 at 5:08
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    $\begingroup$ @SergeiAkbarov Well that sucks. If you'd told me that you were only going to wait three blessed hours before saying "nobody answered" I wouldn't have spent the time I did on the question! Next time you have a question please specify the time limit. I mean really, after three hours you conclude that people had lost interest? The idea that someone might be thinking about it never occurred to you? Or that someone might have something other than MSE to do? What's irritating about this is that my current version appeared an hour after Nate posted the same solution. I don't like how that looks. $\endgroup$ Commented Mar 28, 2016 at 13:12
  • $\begingroup$ @SergeiAkbarov I guess I should be satisfied that at least my first much more complicated version appeared an hour before Nate's answer. But the current version is the same as what he said, looks just like I saw his post and copied it. Three hours is much too long to wait - I see your point, my apologies for being so slow. $\endgroup$ Commented Mar 28, 2016 at 13:15
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    $\begingroup$ @DavidC.Ullrich: I just saw this (never got pinged). Certainly I didn't suspect you of plagiarism! I'll likewise plead not guilty: your original answer was posted here before mine on MO, but I didn't see it. Clearly we just independently came up with similar solutions. I was also frustrated by the rapid crossposting that resulted in us duplicating our efforts, but I recognize that the conventions of SE can be unclear, and I think Sergei now understands what to do in the future. $\endgroup$ Commented Apr 6, 2016 at 21:54
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    $\begingroup$ @NateEldredge Actually I didn't suspect you of suspecting me of plagiarism, my comment was more sort of just for the record. It does look a lot like plagiarism: The version I posted before yours was much more complicated, then I saw the much simpler second version and posted it an hour after you'd posted more or less the same thing. $\endgroup$ Commented Apr 7, 2016 at 14:22

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