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Apr 26, 2012 at 19:18 comment added Bazin Well, may I say that you should get familiar with Fourier analysis. For $u\in \mathscr S(\mathbb R^n)$, you have $$ \hat u(\xi)=\int e^{-2i\pi x\cdot \xi} u(x) dx,\quad u(x)=\int e^{2i\pi x\cdot \xi} \hat u(\xi) d\xi, $$ and both formulas can be extended to temperate distributions, i.e. to the topological dual of $\mathscr S(\mathbb R^n)$.
Apr 26, 2012 at 19:10 comment added Giuseppe Thanks. Do you know of a more explicit example or is this the only way? I am not familiar with Fourier inversion.
Apr 26, 2012 at 16:08 comment added Bazin OK, but you can modify the definition of $\hat u$ above by multiplying the numerator by $ \mathbf 1(\xi_n\ge \vert\xi\vert) $ so that $\partial_n u$ will not be bounded.
Apr 26, 2012 at 13:09 comment added Willie Wong It may not be entirely obvious that $\sqrt{(-\triangle)}u \notin L^\infty \implies \nabla u \notin L^\infty$
Apr 26, 2012 at 12:22 history answered Bazin CC BY-SA 3.0