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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).

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  • $\begingroup$ Yes, the constant in the inequality is independent of the intervals (I edited the question). However, I don't see how such approximation helps (for the intervals in $\mathbb{R}$). And it is not obvious that this inequality holds for open sets $I_j$ instead of intervals. $\endgroup$ Jul 30 at 18:52
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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.

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  • $\begingroup$ Thank you, a very interesting arguement, I have never seen such tricks. And especially thank you for the useful references! $\endgroup$ Jul 31 at 11:17

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