modulation-translation operator continuous in $L^{p}$ norm? We put,
$T_{y}f(x):=f(x-y), \ (x, y\in \mathbb R^{n}).$
It is well-known that $\|T_yf-f\|_{L^{p}} \to 0$ as $y\to 0$ for $1\leq p <\infty.$
Next we put, $M_tT_yf(x):= f(x-ty) e^{i t (x\cdot y)}, (x, y \in \mathbb R^{n}, t\in \mathbb R).$

My Question is: Can we expect $\|M_tT_y f-f\|_{L^{p}} \to 0$ as $t\to 0$ for $1\leq p <\infty.$
  (In other words, roughly speaking, modulation-translation operator continuous in L^{p} norm)

(we note that $t\to 0$ implies $ty\to 0$)

The answer is yes. 
Since $y$ is fixed we can assume without loss of generality that $\|y\| = 1$. 
Fix $\epsilon > 0$. 
Since $f$ is in $L^p$ with $p < \infty$, there exists $R \gg 1$ such that the $L^p$ norm of $f$ outside of the ball of radius $R$ is at most $\epsilon / 4$. 
Choose $t \ll 1 / R$ such that $|\exp(i t x\cdot y) - 1| < \epsilon / (4 \|f\|_{L^p})$ for $\|x\| \leq 2R$.
Then splitting the integral up we have that
$$\| M_t T_y f - f\| \leq \|M_t T_y f - f\|_{L^p(B(3R/2))} + \|M_t T_y f\|_{L^p(B(3R/2)^C)} + \|f\|_{L^p(B(3R/2)^C)} $$
The last two terms together are bounded by $\epsilon / 2$. The first term is bounded by 
$$ \|M_t T_y f - T_{ty} f\| + \|T_{ty} f - f\| $$
with the norm taken in $L^p(B(3R/2))$. The first of the above is 
$$ \| [\exp(it x\cdot y) - 1] T_{ty} f \|_{L^p(B(3R/2))} $$
and so by construction is bounded by $\epsilon / 4$. The final term is bounded by $T_{ty}f - f$ measured in $L^p$ of the whole space, and by choosing $t$ even smaller we can make it bounded by $\epsilon / 4$ also. 
