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$(\Omega,\Sigma,\mu)$ is a finite measure space, $X$ is a Banach space.
$B_n$ is a sequence of disjoint subsets of $\Omega$.
$f:\Omega\rightarrow X$ satisfies $\phi f\in L_1(\mu) \forall \phi\in X^*$ (in fact, we deal with Pettis integrable function $f$ ).
Let $U:X^*\rightarrow \ell^1$ be a linear bounded operator defined by $$U\phi=\left(\int_{B_n}\phi f\ d\mu\right)_n.$$ Assume $U$ is compact (in fact, we can deduce it from the Pettis integrability of f).

$i)$ Why the compactness of the operator $U$ implies $\Sigma_{n=1}^\infty \left|\int_{B_n}\phi f\ d\mu\right|$ converges uniformly in $\phi\in B_{X^*}?$
$ii)$ Why the uniformity ensures that $\Sigma_{n=1}^\infty\left|\int_{B_n}\phi_n f\ d\mu\right|$ converges for any sequence $(\phi_n)$ of $B_{X^*}?$

Thank you for help.

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  • $\begingroup$ I have a solution for $i)$, but $ii)$ is still troubling me. Do you have any thoughts? $\endgroup$ – Aweygan Aug 2 '17 at 1:44
  • $\begingroup$ @Aweygan I tried to find a subsequence $N_k$ of $\mathbb{N}$ such that $\Sigma_{n=N_k}^\infty |\int_{B_n}\phi fd\mu|<\frac{\epsilon}{2^k} \forall \phi$ so can choose $\phi_{N_k}$ such that $\Sigma_{k=1}^\infty |\int_{B_n} \phi_{N_k}fd\mu|<\epsilon$ but I can't go further $\endgroup$ – CSH Aug 2 '17 at 4:00
  • $\begingroup$ I think the second question can be reduced into this form $\endgroup$ – CSH Aug 2 '17 at 4:06
  • $\begingroup$ I believe so, I doubt that the function $f$ has much to do with it. I'll post what I have so far, and edit it if I find a solution to part $ii)$. $\endgroup$ – Aweygan Aug 2 '17 at 4:11
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This isn't a complete answer, but it is too much for a comment.

$i)$ Suppose $U$ is compact. Then $U(B_{X^*})$ is totally bounded. Given $\varepsilon>0$, there exist $\phi_1,\ldots\phi_l$ such that $$U(B_{X^*})\subset\bigcup_{k\leq l}B(U\phi_k,\varepsilon/2).$$ For each $k$, there is some $N_k\in\mathbb N$ such that $\sum_{n=m}^\infty\left|\int_{B_n}\phi_kf\ d\mu\right|<\varepsilon/2$ whenever $m\geq N_k$. If $U\phi\in B(U\phi_k,\varepsilon/2)$, then $$\sum_{n=m}^\infty\left|\int_{B_n}\phi f\ d\mu\right| \leq\sum_{n=m}^\infty\left|\int_{B_n}(\phi-\phi_k) f\ d\mu\right|+\sum_{n=m}^\infty\left|\int_{B_n}\phi_k f\ d\mu\right|<\varepsilon.$$ Put $N=\max\{N_1,\ldots,N_l\}$. Then for any $\phi\in U(B_{X^*})$ and $m\geq N$ we have $\sum_{n=m}^\infty\left|\int_{B_n}\phi f\ d\mu\right|<\varepsilon$.

$ii)$ Still working on this.

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  • $\begingroup$ In the second question, I think the structure of f and the integral must be considered together. Anyway, thank you very much. $\endgroup$ – CSH Aug 2 '17 at 4:49
  • $\begingroup$ You're welcome. I think that in your other question, it may have helped to consider that $a_n\in X^*$, not just continuous. $\endgroup$ – Aweygan Aug 2 '17 at 5:00

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