# Cotype property of measures

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.

• It is well known that $M(K)=L_1(\mu)$ for some weird $\mu$. Jul 21, 2017 at 21:06
• @ 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. Jul 22, 2017 at 6:06

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.