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May 23, 2022 at 13:12 history closed Gro-Tsen
Michael Renardy
Mark Wildon
Alexandre Eremenko
Amir Sagiv
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May 17, 2022 at 8:38 comment added dohmatob Well, $\int_{\mathbb R^n}\|\nabla f(x)\|^2 dx = \int_{\mathbb R^n}\|z\|^2 |\hat{f}(z)|^2dx \asymp \|f\|_{H^1(\mathbb R^n)}^2$. On the other hand, $f \in \mathcal S(\mathbb R^n)$ implies that $\|\nabla f(x)\|^2 \lesssim (1+\|x\|)^{-m}$ for any $m$. No ?
May 17, 2022 at 8:28 comment added Dirk Your definition of the $H^1$-norm does not make sense. The first inequality is wrong and the other terms do eben contain $f$.
May 17, 2022 at 6:15 vote accept dohmatob
May 16, 2022 at 20:45 history edited dohmatob CC BY-SA 4.0
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May 16, 2022 at 20:32 comment added Michael Renardy The answer is yes. And it is the Schwartz space, not the Schwarz space.
May 16, 2022 at 19:56 review Close votes
May 23, 2022 at 13:12
May 16, 2022 at 19:20 answer added terceira timeline score: 4
May 16, 2022 at 18:55 history edited dohmatob CC BY-SA 4.0
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May 16, 2022 at 18:52 comment added Willie Wong If $H^k$ is defined by $\|f\|_{H^k}^2 = \int (1 + |\xi|^2)^k |\hat{f}(\xi)|^2 ~d\xi$, then yes.
May 16, 2022 at 18:36 comment added dohmatob Laplacian: $Lap(f)(x) := \|\nabla f(x)\|^2$.
May 16, 2022 at 18:21 comment added Daniele Tampieri Could you clarify what is the meaning of the $Lap$ notation?
May 16, 2022 at 18:09 history asked dohmatob CC BY-SA 4.0