Suppose that $\{I_m\}$ is a sequence of pairwise disjoint intervals in $\mathbb{Z}$. The well known Rubio de Francia's inequality says that for any function $f\in L^p(\mathbb{T})$, $2\le p<\infty$, we have \begin{equation} \Big\| \Big( \sum_{m} |(\hat{f}\chi_{I_m})^{\vee}|^2 \Big)^{1/2} \Big\|_{L^p}\lesssim \|f\|_{L^p}. \end{equation}

Using duality, it is not difficult to see that this inequality is equivalent to the following: \begin{equation} \Big\| \sum_j f_j \Big\|_p\lesssim \Big\|\Big(\sum_j |f_j|^2\Big)^{1/2}\Big\|_p, \qquad 1<p\le 2, \label{RdFclass} \end{equation} where the functions $f_j$ are such that $\mathrm{supp} \hat{f}_j\subset I_j$ and $\{I_j\}$ are pairwise disjoint intervals in $\mathbb{Z}$. The inequality in such form also holds for $p=1$ as it was shown by Bourgain and for $p<1$ (this is the result of Kislyakov and Parilov).

My question is the following: can the second inequality (for $1<p\le 2$) hold for arbitrary pairwise disjoint sets $I_j$ instead of the intervals? The first one can't for obvious reasons but I couldn't find a simple counterexample for the second one (probably there should be a simple counterexample and I just don't see something).

Suppose that $\{I_m\}$ is a sequence of pairwise disjoint intervals in $\mathbb{Z}$. The well known Rubio de Francia's inequality says that for any function $f\in L^p(\mathbb{T})$, $2\le p<\infty$, we have \begin{equation} \Big\| \Big( \sum_{m} |(\hat{f}\chi_{I_m})^{\vee}|^2 \Big)^{1/2} \Big\|_{L^p}\lesssim \|f\|_{L^p}. \end{equation} The constant which hides under the sign ''$\lesssim$'' depends only on $p$.

Using duality, it is not difficult to see that this inequality is equivalent to the following: \begin{equation} \Big\| \sum_j f_j \Big\|_p\lesssim \Big\|\Big(\sum_j |f_j|^2\Big)^{1/2}\Big\|_p, \qquad 1<p\le 2, \label{RdFclass} \end{equation} where the functions $f_j$ are such that $\mathrm{supp} \hat{f}_j\subset I_j$ and $\{I_j\}$ are pairwise disjoint intervals in $\mathbb{Z}$. The inequality in such form also holds for $p=1$ as it was shown by Bourgain and for $p<1$ (this is the result of Kislyakov and Parilov).

My question is the following: can the second inequality (for $1<p\le 2$) hold for arbitrary pairwise disjoint sets $I_j$ instead of the intervals? The first one can't for obvious reasons but I couldn't find a simple counterexample for the second one (probably there should be a simple counterexample and I just don't see something).