# Space of measures has the Dunford-Pettis property?

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?

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.