Let $K$ be a compact Hausdorff space and $M(K)=C(K)^{*}$ the set of all bounded complex Radon measures on $K.$ Is it true that $M(K)$ is of cotype 2?

I think the answer is true, and to prove this it is enough to check that $$\Bigg(\sum\|\mu_k\|^2\Bigg)^{\frac{1}{2}}\leq C\Bigg(\int\|\sum_k\epsilon_k(\omega)\mu_k\|^2\Bigg)^{\frac{1}{2}},$$for any measure $(\mu_k),$ which are Dirac mass.

$(\epsilon_k)$ is the sequence of Rademacher functions.

  • $\begingroup$ It is well known that $M(K)=L_1(\mu)$ for some weird $\mu$. $\endgroup$
    – Norbert
    Jul 21, 2017 at 21:06
  • $\begingroup$ @ Norbert. Yes and that is already hidden in the question. Consider the set of Dirac masses $\{\delta_{x}:x\in K\}$ Then as the Dirac masses are extreme points of the set $M(K)_1,$ it is not hard to see that the map defined as $$T:span\{\delta_{x}:x\in K\}\to \ell_1(K)$$ $T(\sum_{x}\alpha_x\delta_{x}):=(\alpha_x)_{x\in K}$ extends to an isometric isomorphism. Sadly. when I posed the question I did not notice that I have already solved it. $\endgroup$
    – Mathbuff
    Jul 22, 2017 at 6:06

1 Answer 1


I wanted to comment that I am concerned by @Mathbuff 's conclusion made in the comment, and I hope that this suffices for a complete answer.

We do not have that $$ \text{cl}\big(\text{span}\{ \delta_x : x \in K \}\big) = \mathscr{M}(K),$$ I believe. This can be seen by recognizing that the Lebesgue measure lies distance 2 from any finite combination of point measures in $\mathscr{M}([0,1])$. So I'm uncertain about their argument that such a map extends to an isometric isomorphism, and I found the conclusion particularly concerning. One point against it is that $\mathscr{M}(K)$ is not separable, as any two point masses lie distance 2 apart whilst $l_1(K)$ is separable for $K = [0,1].$

I think that we can show that $\mathscr{M}(K)$ has cotype 2 by recognising that any separable subspace can be isometrically embedded into a $L_1(\mu)$ space. Then from here, we can recall that a $L_1(\mu)$ space has cotype 2 and then as each separable space has cotype 2 with a UNIFORM bound, we have that $\mathscr{M}(K)$ has cotype 2.

We can embed any separable subspace $X$ into an $L_1(\mu)$ space, by the fact that if $(\mu_i)_{i=1}$ is a dense sequence in $X$ of norm-one measures, then any measure $\nu$ is absolutely continuous with respect to $\mu = \sum_{i=1}^\infty 2^{-i} \mu_i$, a finite measure.

This is as: If $\mu(A)= 0$, then $\mu_i(A) = 0 \quad \forall i \in \mathbb{N}.$

Then given $\varepsilon > 0$, there exists $(\alpha_i)_{i=1}^\infty$ a finite sequence with $||\nu - \sum_{i=1}^\infty \alpha_i \mu_i || < \varepsilon$ and so $|\nu(A)| < \varepsilon$. Hence $\nu(A) = 0 \implies \nu << \mu.$

Thus by the Radon-Nikodym Theorem, we have that X embeds into $L_1(\mu)$. Thus the cotype of $\mathscr{M}(K)$ is 2 as required.


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