Let $h$ be a strictly convex function such that $h(0) = h'(0)=0$. Let $\Phi: \mathbb{R}^2 \to \mathbb{R}$ be a $C^{\infty}$-function with compact support (say, $\Phi$ is supported on $[-1,1]\times[-1,1]$). Let $\delta$ be a complex number, $\Re \delta \in (0,2)$. Consider a distribution \begin{equation*} v.p. \frac{\Phi(\xi,\eta)}{sign(\eta - h(\xi))|\eta - h(\xi)|^{\delta}}, \end{equation*} which acts on a test function $\varphi \in \mathfrak{D}$ by the following rule: \begin{multline*} \Big\langle vp \frac{\Phi(\xi,\eta)}{sign(\eta - h(\xi))|\eta - h(\xi)|^{\delta}},\varphi(\xi,\eta)\Big\rangle = \\ \lim\limits_{\varepsilon \to 0} \int\limits_{\{|\eta - h(\xi)| \geq \varepsilon\}}\frac{\Phi(\xi,\eta)\varphi(\xi,\eta)\,d\xi d\eta}{sign(\eta - h(\xi))|\eta - h(\xi)|^{\delta}}. \end{multline*} The Fourier multiplier $M_{\delta}$ is defined by the following formla: \begin{equation*} M_{\delta}[f] = \mathcal{F}^{-1}\Big[vp \frac{\Phi(\xi,\eta)}{sign(\eta - h(\xi))|\eta - h(\xi)|^{\delta}} \mathcal{F}[f](\xi,\eta)\Big], \end{equation*} so it is a Fourier multiplier with the symbol $vp \frac{\Phi(\xi,\eta)}{sign(\eta - h(\xi))|\eta - h(\xi)|^{\delta}}$.
The question is: for what pairs $(p,q)$ does $M_{\delta}$ act from $L^p$ to $L^q$? Or, if the answer is known, is it written anywhere?
If $\delta = 1$, then I know the answer (except some "endpoints"): $M_{\delta}$ acts from $L^p$ to $L^q$ if $1 \leq p < \frac43$, $4 < q \leq \infty$, and $\frac{1}{p} - \frac{1}{q} > \frac32$, and does not act otherwise, except the segment $\frac{1}{p} - \frac{1}{q} = \frac23$ (I don't know what is the answer for such pairs of $p$ and $q$). The idea is to use the restriction estimate for the curves $\Gamma_{t} = \{\xi = h(\eta) + t\}$, integrate it, and then multiply the achieved operator by its adjoint. This works for the case $\Re\delta < 1$. To pass to the case $\delta = 1$, one gets some weak estimate for $M_{\delta}$ with $\Re \delta > 1$ and interpolates using Stein's lemma on analytic family of operators. Unfortunately, such method is not sharp enough to cover the case $\frac{1}{p} - \frac{1}{q} = \frac23$. What is more, it does not give a complete answer for the cases $1 < \Re \delta \leq \frac32$.
