Fourier multiplier and applications Let $\mu$ be in $\mathcal D (\mathbb R^d)$ with $\mu \geq 0$, i.e. $\mu$ is a test function. Furthermore, we assume $\mu (\xi) =1$ when $|\xi|<1$ and $\mu (\xi) =0$ when $|\xi| \geq 2$. Why is the following true? For $j=1,2,3,...,d$ and $\lambda >0$, $\|\mu (\lambda D)\partial_{j}f\|_{L^2} \leq C \lambda^{-\frac{d+2}{2}}\|f\|_{L^1}$ for some constant $C$.
I think this is related to Fourier multiplier but after I checked relevant notes, I still can't figure it out.
 A: We begin by using Parseval to compute
$$
\Vert \mu(\lambda D) \partial_j f \Vert_{L^2}^2 = \int_{\mathbb{R}^d} \vert \mu(\lambda \xi) i \xi_j \hat{f}(\xi) \vert^2 d\xi.
$$
Then we estimate 
$$
\int_{\mathbb{R}^d} \vert \mu(\lambda \xi) i \xi_j \hat{f}(\xi) \vert^2 d\xi \le \Vert \hat{f} \Vert_{L^\infty}^2 \int_{\mathbb{R}^d} |\mu(\lambda \xi)|^2 |\xi|^2 d\xi=  \Vert \hat{f} \Vert_{L^\infty}^2 \int_{\mathbb{R}^d} |\mu(\omega)|^2 \left\vert \frac{\omega}{\lambda}\right\vert^2 \frac{d\omega}{\lambda^d} \\
= \lambda^{-d-2} \Vert \hat{f} \Vert_{L^\infty}^2  \int_{\mathbb{R}^d} |\mu(\omega)|^2 |\omega|^2 d\omega = C_\mu \lambda^{-d-2} \Vert \hat{f} \Vert_{L^\infty}^2.
$$
Note here that $C_\mu$ is finite since $\mu$ is a test function, and it doesn't actually depend on $j$ due to the way we have estimated things.  On the other hand we know that 
$$
\Vert \hat{f} \Vert_{L^\infty} \le \Vert f \Vert_{L^1}.
$$
Combining all these then shows that 
$$
\Vert \mu(\lambda D) \partial_j f \Vert_{L^2}^2 \le C_\mu \lambda^{-d-2} \Vert \hat{f} \Vert_{L^1}^2,
$$
and so the estimate you want follows by taking the square root of both sides.
