For this question, all Banach spaces are over the reals.


Let $1\leq p<\infty$. Recall that a sequence $(x\_n)$ in a Banach space $E$ is weakly $p$-summable if
$$ \Vert (x\_n) \Vert\_{p,w} := \sup\_{\gamma\in E^\* \colon \Vert\gamma\Vert\leq 1} \left( \sum\_{n=1}^\infty \vert\gamma(x\_n) |^p \right)^{1/p} < \infty .$$


Another way to think about this: for a Banach space $X$, bounded linear maps $X\to\ell\_p$ correspond (isometrically) to weakly $p$-summable sequences in $X^\*$.


Now suppose $1<p<2$ and let $q$ be the conjugate index of $p$.



> 
> **Question.** Let $(x\_n)$ be a weakly $p$-summable sequence in $L\_q[0,1]$. Is the sequence $(|x\_n|)$ also weakly $p$-summable?
> 
> 
> 


I suspect the answer is negative, but only because I've not had any luck finding a "soft" proof of a positive answer. On the other hand, the question seems natural enough that it must be in the literature one way or the other.



Does taking the modulus preserve weak $p$-summability of sequences in $L_q$?
It seems to me that the answer is no for all $1 < p < 2$: consider $x\_n = \frac{e^{2\pi i nx}}{n^{r}}, n = 1, 2, \ldots$ for $r = \frac{1}{p}$. It is easy to see that the sequence $\{ |x\_n|\}$ is not weakly $p$-summable by testing against $\gamma = 1$. Let us show that $\{ x\_n\}$ is weakly p-summable:


We want to show for any $\gamma = \sum\_{n = -\infty}^\infty a\_n e^{2\pi inx}$ with $||\gamma||\_p = 1$ that $\sum\_{n = 1}^\infty \frac{|a\_n|^p}{n} \le C$ for some absolute $C$. It is a classical result of Hardy and Littlewood that $\sum\_{n = 1}^\infty \frac{|a\_n|^p}{n^{2 - p}} \le C\_1$. Since $2 - p < 1$ we are done. 


*Hardy, G. H.; Littlewood, J. E.*, [**Some new properties of Fourier constants.**](http://dx.doi.org/10.1007/BF01447865), Math. Ann. 97, 159-209 (1926). [ZBL52.0267.01](https://zbmath.org/?q=an:52.0267.01).


