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I am trying the following problem.

Let $\mu$ be a probability measure. Show that an operator $T : L_1(\mu) \rightarrow X$ is Dunford Pettis if and only if $T|_{L_2(\mu)}$ is compact.

I was trying to use the definition. But I cannot come up with anything.Could you give me some hint?

I appreciate your help!

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    $\begingroup$ The Dunford-Pettis property is usually a property of a Banach space, not of a single operator. What do you mean by "Dunford Pettis property of $T$"? $\endgroup$ – uniquesolution Apr 11 at 5:28
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    $\begingroup$ I had the same question but a search revealed that an operator is D-P if it maps weakly compact sets to norm compact sets. $\endgroup$ – Kavi Rama Murthy Apr 11 at 5:30
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    $\begingroup$ One way is easy: any $L^{2}$ bounded sequence is uniformly integrable, hence weakly relatively compact. $\endgroup$ – Kavi Rama Murthy Apr 11 at 5:33
  • $\begingroup$ @KaviRamaMurthy, I got you. How about the other way round? $\endgroup$ – Topology Apr 11 at 14:01
  • $\begingroup$ This may work: Suppose $(f_n)\subset L_1$ converges weakly to $0$. We need to show $Tf_n$ converges in norm to $0$ (see Prop 5.4.2 in Albiac/Kalton). Towards this end, fix $M>0$ and set $g_n=f_n\cdot\chi_{A_n}$ where $A_n=\{ x : |f_n (x)|<M\}$. Then $(g_n)\subset L_2$ and, being a bounded sequence, its image under $T$ tends to $0$ in norm. Since $(f_n)$ is uniformly integrable, $\sup_n \int_{A_n^c } |f_n|$ can be made small (by choosing $M$ large). It seems this will imply the result. $\endgroup$ – David Mitra Apr 11 at 17:43

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