2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $x\mapsto \lambda x$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.