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.


1 Answer 1


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
    Commented Dec 3, 2019 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
    Commented Dec 3, 2019 at 16:57
  • $\begingroup$ @YemonChoi just replace the exponentials by the sine function. $\endgroup$ Commented Dec 3, 2019 at 17:40
  • $\begingroup$ Thank you Aleksei - yes, of course, that will do $\endgroup$
    – Yemon Choi
    Commented Dec 3, 2019 at 17:50

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