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We say that a subset $K$ of a Banach space $X$ is relatively $p$-compact ($1\leq p<\infty$) if there exists a $p$-summable sequence $(x_n)_{n=1}^{\infty}$ in $X$ such that $$ K\subseteq \left\{\sum_{n=1}^{\infty}\alpha_{n}x_{n}:(\alpha_{n})_{n}\in B_{l_{q}}\right\}\quad\quad(\frac{1}{p}+\frac{1}{q}=1).$$ A Bananch space $X$ is said to have the $p$-approximation property if $I_{X}$ can be approximated by finite rank operators uniformly on the relatively $p$-compact subsets of $X$. An operator $T:X \rightarrow Y$ is said to be quasi $p$-nuclear if there exists a $p$-summable sequence $(x^{*}_{n})_{n=1}^{\infty}$ in $X^{*}$ such that $$\|Tx\|\leq \left(\sum_{n=1}^{\infty}|\langle x^{*}_{n},x\rangle|^{p}\right)^{1/p} \quad\quad\hbox{ for all $x\in X$.} $$

We denote the Banach ideal of quasi $p$-nuclear operators by $\mathcal{Q}\mathcal{N}_{p}$ and the Banach ideal of $p$-summing operators by $\prod_{p}$. We say that a Banach space $X$ has the right approximation property with respect to $\mathcal{Q}\mathcal{N}_{p}$ (or $\prod_{p}$)if for every Banach space $Y$ and every operator $T\in \mathcal{Q}\mathcal{N}_{p}(X,Y)$(or $T\in \prod_{p}(X,Y)$), we have $T\in \overline{\{TS:S\in \mathcal{F}(X,X)\}}^{\tau}$, where $\tau$ is the topology of uniform convergence on compact subsets of $X$. My questions are:

  1. If $X$ has the right approximation property with respect to $\mathcal{Q}\mathcal{N}_{p}$, does $X$ has the $p$-approximation property ?

  2. If $X^{*}$ has the $p$-approximation property, does $X^{*}$ have the right approximation property with respect to $\prod_{p}$ ?

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    $\begingroup$ Typographical note: Use $\langle x^*,x\rangle$ (with \langle and \rangle) rather than $<x^*,x>$. For me at least, the second formatting is difficult to read. (I made this edit.) $\endgroup$ Aug 27, 2014 at 9:16

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The answer for both questions are positive. The first is an easy consequence of a version of Grothendieck's Theorem that characterizes the p-approximation property in terms of the density of finite-rank operators w.r.t uniform convergence on p-compact sets (1). Since p-compact sets are q-compact for p

It follows the number of the results in my thesis I have mentioned above: (1) Proposition 3.1.8 (2) Proposition 2.2.4 (3) Proposition 3.3.3

Link for my master's thesis: http://www.teses.usp.br/teses/disponiveis/45/45131/tde-12112013-232741/en.php

Proposition 3.1.8

Let $X$ be a Banach sapce and $1\leq p \leq \infty$, then are equivalent:

(i) $X$ has the $p$-aproximation property.

(ii) For every Banach space $Y$, $\mathcal{F}(Y,X)$ is $\tau_p$-dense in $\mathcal{B}(Y,X)$.

Proposition 2.2.4

Let $1 \leq p< r \leq \infty$ and suppose $K$ is a relative $p$-compact set, then $K$ is relative $r$-compact.

Proposition 3.3.3

Let $X$ be a Banach space and $1 \leq p \leq \infty$. If $X^*$ has the $p$-approximation property then $X$ has the p-approximation property too.

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  • $\begingroup$ Thank you for your answer. But I can not read your master's thesis because it is not written in English. $\endgroup$ Dec 12, 2015 at 15:11
  • $\begingroup$ Sorry about that, I regret not having written in English. I'll try to translate the mentioned results and post here. $\endgroup$ Dec 18, 2015 at 12:55
  • $\begingroup$ Well, I translated (very fast, sorry for possible errors in my english) the statements, if you need the translation of some of the demonstrations, let me know. $\endgroup$ Dec 18, 2015 at 13:14

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