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Young Women in Harmonic Analysis and PDE

December 2-4, 2016




Cristina Benea (Universit├ę de Nantes, LMJL)

On Rubio de Francia's Theorem for arbitrary Fourier projections


In [2], Rubio de Francia proved that disjointness in frequency is enough for establishing a one-dimensional orthogonality principle. That is, \begin{equation} \label{RF:thm} \big|\ \big( \sum_{k} \big| \int_{\mathbb{R}} \hat{f}(\xi) \mathbf{1}_{\left[a_k, b_k \right]}(\xi) e^{2 \pi i x \xi} d \xi \big|^\nu \big)^{1/\nu} \big\|_p \leq C \|f\|_p, \end{equation} whenever the intervals $\left[ a_k, b_k\right]$ are mutually disjoint, $\nu > 2$ and $p > \nu'$ or $\nu =2$ and $p \geq 2$. Moreover, the constant $C$ does not depend on the choice of intervals. In the bilinear setting, a similar, one-parameter question can be formulated: given an arbitrary collection $\Omega$ of mutually disjoint squares, prove that \begin{equation} \label{RF-squares} \big\| \big( \sum_{\omega \in \Omega} \big| \int_{\mathbb{R}^2} \hat{f}(\xi) \hat{g}(\eta) \Phi_\omega(\xi, \eta) e^{2 \pi i x \left(\xi+\eta\right)} d \xi d \eta\big|^r \big)^{1/r} \big\|_s \leq C \|f\|_p \|g\|_q, \end{equation} whenever $\frac{1}{p}+\frac{1}{q}=\frac{1}{s}$ and $p, q, s$ are in the ``local $r'$" range. While $r$ needs to be $\geq 2$, just like in the linear case, we were able to prove the above result only for $r>2$. This is joint work with F. Bernicot.

References

[1] C. Benea, F. Bernicot, A bilinear Rubio de Francia inequality for arbitrary squares, available at http://arxiv.org/abs/1602.01948
[2] J. Rubio de Francia A Littlewood-Paley Inequality for Arbitrary Intervals, Revista Matematica Iberoamericana, 1985