Let $\varphi(D)$ be a Fourier multiplier with symbol $\varphi(\xi) = \xi/(1+|\xi|^2)$. It's easy to prove that \begin{equation} \|\varphi(D)u^2\|_{H^s(R)}\lesssim \|u\|^2_{H^s(R)} \quad (*) \end{equation} for any $s\geq 0$. Moreover, $(*)$ fails for $s<0$. Indeed, let $$ u(x)=\frac{8}{x}\cos\frac{2N+\gamma}{2}x\sin\frac{\gamma}{2} x $$ such that the Fourier transform of $u$ $$ \widehat{u}(\xi)=1_{[-N-\gamma,-N]}+1_{[N,N+\gamma]}. $$ Now an elementary calculation yields that $(*)$ becomes $$ \gamma^{\frac{3}{2}}\lesssim N^{2s}\gamma. $$ Since $s<0$, we set $\gamma=N^{-\varepsilon}(\varepsilon$ depends on $s$) and let $N\rightarrow \infty$ leads to a contradiction.
The above argument gives us a full picture of $(*)$ in scales of $H^s$ space. Now I'm interested in the $L^p$ version of $(*)$. It's easy to show that \begin{equation} \|\varphi(D)u^2\|_{L^p(R)}\lesssim \|u\|^2_{L^p(R)} \quad (**) \end{equation} for any $p\geq 2$. Note that $p=2$ coincides the borderline case $s=0$. So my question is, does $(**)$ holds for $p<2$? If it does not, what kind of counterexamples may work. Observe that the above example fails in this problem.
