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

M. Lacey and C. Thiele proved in [27] (Annals of Math. (1997)) and [28] (Annals of Math. (1999)) that the bilinear Hilbert transform maps $L^{p_1}\times L^{p_2}\to L^p$ boundedly when $\frac{1}{p_1}+\frac{1}{p_2}=\frac{1}{p}$ with $1<p_1,p_2\le \mathrm{\infty }$ and $\frac{2}{3}<p<\mathrm{\infty }$. 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 $\mathcal{F}{L}^{p_1^{\mathrm{\prime }}}\times L^{p_2}\to L^p$ 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 $p\in (1\mathrm{/}2,\mathrm{\infty })$ (Theorem 1.3).