Fourier multiplier and Navier Stokes equations

Question

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.

Answer 1

Boundedness for $\lambda=1$ follows from the fact that $\phi(\xi)=\mu(\xi)\xi_j$ is also a test function, $\mu(D)\partial_j f=\phi(D)f$ can be written as the convolution $\widehat \phi * f$, and one can apply Young's inequality for convolutions since $\widehat \phi$ is in the Schwartz class. For different values of $\lambda>0$, the result follows by scaling $x\mapsto \lambda x$.