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Apr 16, 2017 at 14:33 comment added Henry.L @Bazin Could you add a reference for your claim on "iff" part in your answer? Thanks!
Feb 1, 2013 at 16:27 comment added Bazin @Matt Jacobs I would recommend the Bahouri-Chemin-Danchin book Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], 343. Springer, Heidelberg, 2011.
Feb 1, 2013 at 0:30 comment added Matt Jacobs Thanks! Is there a text or paper where I can find this result?
Feb 1, 2013 at 0:30 vote accept Matt Jacobs
Jan 31, 2013 at 9:58 comment added Bazin @Matt Jacobs As said in the previous comment, $f(D)u$ is the function whose Fourier transform is $f(\xi)\hat u(\xi)$. The operator $f(D)$ is called a Fourier multiplier for this reason. An integral representation is $$ (f(D)u)(x)=\int e^{2i\pi x\cdot \xi} f(\xi) \hat u(\xi) d\xi, $$ with $$ (\hat u)(\xi)=\int e^{-2i\pi x\cdot \xi} u(x) dx. $$
Jan 31, 2013 at 9:45 comment added Gian Maria Dall'Ara $D_x$ should stand for $i\partial_x$ (the $i$ is to make it self-adjoint, i.e. to make $e^{i\xi x}$ an eigenfunction of $D_x$ with real eigenvalue). If $\phi$ is a measurable function on $\mathbb{R}$, $\widehat{\phi(D_x)f}(\xi):=\phi(\xi)\widehat{f}(\xi)$.
Jan 30, 2013 at 21:52 comment added Matt Jacobs I'm not sure I understand the notation $\phi_{\nu}(D_x)u$
Jan 30, 2013 at 20:29 history answered Bazin CC BY-SA 3.0