$L^{p}$ estimates for the bilinear Hilbert transform for $1/2<p\leq2/3$: A counterexample and generalizations to non-smooth symbols

Abstract: M. Lacey and C. Thiele proved in 27 and 28 that the bilinear Hilbert transform maps $L^{p_1}\times L^{p_2}\rightarrow L^{p}$ boundedly when $\frac{1}{p_1}+\frac{1}{p_2}=\frac{1}{p}$ with $1<p_{1}, \, p_{2}\leq\infty$ and $\frac{2}{3}<p<\infty$. Whether the $L^p$ estimates hold in the range $p\in (1/2,2/3]$ has remained an open problem since then. In this paper, we prove that the bilinear Hilbert transform does not map $\mathcal{F}L^{p'_{1}}\times L^{p_{2}}\rightarrow L^{p}$ for $p_1<2$ and $L^{p_{1}}\times \mathcal{F}L^{p'_{2}}\rightarrow L^{p}$ for $p_2<2$ boundedly (Theorem 1.2). In particular, this shows that the bilinear Hilbert transform neither maps $\mathcal{F}L^{p'_{1}}\times L^{p_{2}}\rightarrow L^{p}$ nor $L^{p_{1}}\times \mathcal{F}L^{p'_{2}}\rightarrow L^{p}$ for $\frac{1}{2}<p<\frac{2}{3}$. Nevertheless, we can establish $L^p$ estimates for the bilinear Fourier multipliers whose symbols are not identical to but arbitrarily close to that of the bilinear Hilbert transform in the full range $p\in(1/2,\infty)$ (Theorem 1.3).