Example 2.5.12. The function $m(\xi)=\mathrm{e}^{2\pi i\xi \cdot b}$ is an $L^p$ multiplier for all $b \in R^n$ since the corresponding operator $T_m(f)(x)=f(x+b)$ is bounded on $L^p(R^n)$. Clearly $\|m\|_{M_p}=1$.
(from the book "Classical Analysis" by Grafakos).
If $\mathrm{e}^{2\pi i\xi \cdot b}$ is not in $L^{\infty}$ (for some $b$). Why is $\mathrm{e}^{2\pi i\xi \cdot b}$ an $L^p$ multiplier?