# 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 $$$$\Big\| \Big( \sum_{m} |(\hat{f}\chi_{I_m})^{\vee}|^2 \Big)^{1/2} \Big\|_{L^p}\lesssim \|f\|_{L^p}.$$$$ The constant which hides under the sign ''$$\lesssim$$'' depends only on $$p$$.

Using duality, it is not difficult to see that this inequality is equivalent to the following: $$$$\Big\| \sum_j f_j \Big\|_p\lesssim \Big\|\Big(\sum_j |f_j|^2\Big)^{1/2}\Big\|_p, \qquad 1 where the functions $$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 $$1) hold for arbitrary pairwise disjoint sets $$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).

• 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. Jul 30 at 18:52

## 1 Answer

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.

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