A generalization of Rubio de Francia's inequality 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: 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).  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 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.
