$(\Omega,\Sigma,\mu)$ 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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.