3
$\begingroup$

Consider the operator $T_\lambda f(x)= \int_{\mathbb{R}^3}{\frac{\sin\lambda|x-y|}{|x-y|}\phi(x-y) f(y) dy}$, where $\phi\in C_0^{\infty}$, $\phi(x)=1$, when $|x|<1$ . My question is that do we have the following estimate $$ \|T_{\lambda}f T_{\mu}g\|_{L^2(\mathbb{R}^3)}\leq \frac{C}{\lambda^{\frac{1}{2}}\mu^{\frac{3}{2}}}\|f\|_{L^2(\mathbb{R}^3)}\|g\|_{L^2(\mathbb{R}^3)},\quad 1\leq \lambda\leq \mu $$ I want the deacy for "high frequency" (i.e. $\mu$ here) go faster than the "low frequency"(i.e. $\lambda$ here). Are there any references which discuss such type of bilinear oscillatory integrals?

Thanks.

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ I am not convinced that your $T_\lambda$ is a bounded operator on $L^2$. In particular, I do not see why a formular like yours should hold. $\endgroup$
    – Matthias Ludewig
    Commented May 19, 2014 at 19:36
  • $\begingroup$ @Kofi, sorry, I change the kernel now. $\endgroup$
    – Tomas
    Commented May 19, 2014 at 20:54
  • $\begingroup$ I'm sorry if this is a dumb question, but why is the new operator a bounded operator now? It isn't obvious to me. $\endgroup$
    – k3thomps
    Commented May 22, 2014 at 11:27

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .