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).



A generalization of Rubio de Francia's inequality
The answer is "no" for any $p<2$ (obviously the inequality holds for $p=2$), but the construction I have is rather indirect (analogous to how the Hardy-Littlewood majorant conjecture is disproved, see e.g., [this paper](https://arxiv.org/abs/1203.2378)). A shame, because restriction theory (and other related areas of harmonic analysis) would be a *lot* easier if such a powerful inequality was true!


Suppose your claim was true for some $p<2$, that is to say that
$$ \| \sum\_j f\_j \|\_{L^p({\bf T})} \lesssim \| (\sum\_j |f\_j|^2)^{1/2} \|\_{L^p({\bf T})}$$
whenever $f\_j$ are one-dimensional trigonometric polynomials with disjoint Fourier supports. This implies a higher dimensional analogue
$$ \| \sum\_j f\_j \|\_{L^p({\bf T}^d)} \lesssim \| (\sum\_j |f\_j|^2)^{1/2} \|\_{L^p({\bf T}^d)} \quad (1)$$
whenever $f\_j$ are $d$-dimensional trigonometric polynomials with disjoint Fourier supports, with constant independent of $d$. This follows from applying the equidistribution observation
$$ \int\_{{\bf T}^d} F(x\_1,\dots,x\_d)\ dx\_1 \dots dx\_d = \lim\_{N \to \infty} \int\_{{\bf T}} F( x, Nx, N^2 x, \dots, N^{d-1} x)\ dx$$
for any continuous $F$ (easily checked first for Fourier phases, then the general case follows from Stone-Weierstrass and a limiting argument) to express both sides of (1) as limiting values of one-dimensional counterparts, and applying the one-dimensional hypothesis.


From (1) and the [tensor power trick](https://terrytao.wordpress.com/2008/08/25/tricks-wiki-article-the-tensor-product-trick/) we can now eliminate the constant and conclude that
$$ \| \sum\_j f\_j \|\_{L^p({\bf T})} \leq \| (\sum\_j |f\_j|^2)^{1/2} \|\_{L^p({\bf T})}$$
whenever $f\_j$ are trigonometric polynomials with disjoint Fourier supports. In particular
$$ \| 1 + f \|\_{L^p({\bf T})} \leq \| (1+|f|^2)^{1/2} \|\_{L^p({\bf T})}$$
whenever $f$ is a trigonometric polynomial of mean zero. By a limiting argument this inequality must then hold for all bounded measurable $f$ of mean zero, thus
$$ \int\_{\bf T} |1+f|^p - pf - 1 \leq \int\_{\bf T} |1+f^2|^{p/2} - 1$$
whenever $f: {\bf T} \to {\bf R}$ is real-valued of mean zero. I've subtracted terms on both sides to make the expressions quadratic or higher order in $f$. Applying this inequality to $f = c (1\_{[0,\varepsilon]} - \varepsilon)$ for any real $c$ and small $\varepsilon$, dividing by $\varepsilon$, and then taking the limit $\varepsilon \to 0$, we conclude that
$$ |1+c|^p - pc - 1 \leq |1+c^2|^{p/2} - 1$$
for any real $c$. Setting $c = -x$ for a large $x$, we see from Taylor expansion and the hypothesis $p<2$ that
$$ |1+c|^p - pc - 1 = x^p + px + o(x)$$
and
$$ |1+c^2|^{p/2} - 1 = x^p + o(x)$$
and we obtain a contradiction for $p$ large enough.


