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Jun 1, 2022 at 6:52 comment added Titouan Vayer Yes indeed. It is because I did not catch the "rotationally invariant" argument of this post (which implies that w.l.o.g you can choose a standard basis vector). So I preferred to prove it with a arbitrary $\theta$.
May 31, 2022 at 11:31 comment added dohmatob @TitouanVayer Hum, why do you have a dimension-dependent multiplicative constant $C=d^{k+1}$ in the RHS of your upper bound in lemma 7 ? The computations in this post show that one can take $C=1$.
May 31, 2022 at 9:41 comment added Titouan Vayer Hi @dohmatob I just saw your comment. I used kind of the same result here arxiv.org/pdf/2112.00423.pdf in Lemma 7. Maybe this will be useful for you
May 6, 2022 at 6:52 comment added dohmatob Also, why do you require $f$ be compactly supported in the calculations ?
May 5, 2022 at 7:10 comment added dohmatob This result can be extended to hold for any $p \in [1,\infty)$, right ?
Jul 8, 2021 at 17:12 comment added Titouan Vayer Thank you ! I do not really catch why we can consider the standard basis vector, could you elaborate ? (it is not clear what you mean by "Sobolev norms are all rotationally invariant")
Jun 16, 2021 at 11:18 history answered Johannes Hahn CC BY-SA 4.0