I'm trying to prove Corollary 5.4.6. from Albiac, Kalton - Topics in Banach Space Theory:

If $ K $ is a compact Hausdorff space then $ \mathcal{M}(K) $ has (DPP).

I want to use

The Dunford-Pettis Theorem. If $ \mu $ is a $ \sigma $-finite measure then $ L_1(\mu) $ has (DPP).

Definitions from the book:

Let $ X $ and $ Y $ be Banach spaces. A bounded linear operator $ T: X \to Y $ is a Dunford-Pettis operator if whenever $ W $ is weakly compact subset of $ X $ then $ T(W) $ is a norm-compact subset of $ Y $.

A Banach space $ X $ is said to have the Dunford-Pettis property (DPP) is every weakly compact operator $ T $ from $ X $ into a Banach Space $ Y $ is Dunford-Pettis.

My progress so far:

Let $ T: \mathcal{M}(K) \to Y $ be a weakly compact operator. The space $ \mathcal{M}(K) $ can be represented as an $ l_1 $-sum of spaces $ L_1(\mu_n) $, where $ \mu_n $ are probability measures on $ K $. We define $$ T_n: L_1(\mu_n) \to Y,\quad T_n(f) = T((0, \ldots, 0, f, 0, \ldots)) $$ (where $ f $ is at the $ n $-th position in the sequence $ (0, \ldots, 0, f, 0, \ldots) $).

Then for $ \mu = (f_1, f_2, \ldots) $ we have $$ T(\mu) = \sum\limits_{n = 1}^\infty T_n(f_n). $$ I guess the proof should go like this

  1. Because $ T $ is weakly compact, each $ T_n $ is weakly compact.
  2. By the Dunford-Pettis theorem, each $ T_n $ is Dunford-Pettis.
  3. From 2. it should follow that $ T $ is Dunford-Pettis.

My question

How do I prove the corollary stated above? Specifically, how do I prove 3.? Or should I try a completely different approach?


1 Answer 1


Also in the Albiac and Kalton book (Theorem $5.4.4$) you can find the following fact: $X$ has the DPP if and only if for every weakly null sequence $(x_n)_{n=1}^\infty\subset X$ and every weakly null sequence $(x^*_n)_{n=1}^\infty\subset X^*$, $\lim_n x^*_n(x_n)=0$. This is the characterization to use to prove that $\mathcal{M}(K)$ has DPP.

First suppose that $X$ is a Banach space such that every separable subspace $Y$ of $X$ is contained in a subspace $L$ of $X$ such that $L$ has DPP. Then $X$ has DPP. This is why we use the characterization of DPP in the previous paragraph. To see this claim, fix $(x_n)_{n=1}^\infty\subset X$ weakly null and $(x^*_n)_{n=1}^\infty\subset X^*$ weakly null. Let $Y$ be the closed span of the sequence $(x_n)_{n=1}^\infty$ so $Y$ is separable. Then there exists a subspace $L$ of $X$ with DPP which contains $Y$ (by hypothesis). Let $z^*_n=x^*_n|_L$ and note that $(z^*_n)_{n=1}^\infty\subset L^*$ is weakly null. Since $L$ has DPP, $$0=\lim_n z^*_n(x_n)=\lim_n x^*_n(x_n).$$ Therefore $X$ has DPP by the characterization from the previous paragraph.

In order to show that $\mathcal{M}(K)$ has DPP, we must show that every separable subspace $Y$ of $\mathcal{M}(K)$ is contained in a subspace $L$ of $\mathcal{M}(K)$ such that $L$ has DPP. For this we will use a standard trick about separable subspace of $\mathcal{M}(K)$. Fix a separble subspace $Y$ of $\mathcal{M}(K)$ (assume it is not the zero space, of course), and let $(\mu_n)_{n=1}^\infty$ be a dense sequence in the sphere of $Y$. Let $\mu=\sum_{n=1}^\infty 2^{-n}|\mu_n|\in \mathcal{M}(K)$. Here $|\mu_n|$ is the variation measure of $\mu_n$. Now let $L$ be the subspace of all measures $\nu\in \mathcal{M}(K)$ which are absolutely continuous with respect to $\mu$ (and note that $L$ contains each $\mu_n$). For every $\nu\in L$, there exists a Radon-Nikodym derivative $f_\nu\in L_1(\mu)$. Furthermore, the map $\nu\mapsto f_\nu$ is an isometric isomorphism from $L$ to $L_1(\mu)$. Since $\mu_n\in L$ for all $n$, $Y\subset L$. Since $L$ is isomorphic to $L_1(\mu)$, $L$ has DPP. Therefore every separable subspace of $\mathcal{M}(K)$ is contained in a subspace $L$ of $\mathcal{M}(K)$ such that $L$ has DPP. By the previous paragraph, $\mathcal{M}(K)$ has DPP.

  • $\begingroup$ Very nice detailed answer, thank you! I would never think of this. $\endgroup$
    – Mikulas
    Aug 31, 2018 at 16:46
  • $\begingroup$ No one ever thinks of that trick in the last paragraph. It just gets passed along. My advisor told me that trick as a hint to solve a completely different problem many years ago. $\endgroup$
    – user520024
    Aug 31, 2018 at 16:56

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