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.


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., Math. Ann. 97, 159-209 (1926). ZBL52.0267.01.

  • $\begingroup$ Looks good to me - thanks! I'll check the details more closely in the next day or two. $\endgroup$ – Yemon Choi Dec 3 '19 at 3:40
  • $\begingroup$ (My original question specified real scalars but, to be honest, that was only because I thought a counter-example might occur more naturally in the real setting. So your answer is good enough for my purposes) $\endgroup$ – Yemon Choi Dec 3 '19 at 16:57
  • $\begingroup$ @YemonChoi just replace the exponentials by the sine function. $\endgroup$ – Aleksei Kulikov Dec 3 '19 at 17:40
  • $\begingroup$ Thank you Aleksei - yes, of course, that will do $\endgroup$ – Yemon Choi Dec 3 '19 at 17:50

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