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Feb 25, 2022 at 17:41 comment added Denis Serre @coudy. Of course you're right. And this is a way to prove that $H^1({\mathbb R})\subset C^0$.
Feb 25, 2022 at 15:00 comment added coudy @DenisSerre Are you sure about your comment? I am pretty sure that square integrable functions with square integrable generalised derivative have integrable Fourier transforms, as a consequence of the Cauchy-Schwarz inequality.
Feb 25, 2022 at 10:35 comment added coudy @Suvrit Link broken.
Feb 19, 2019 at 0:59 comment added Hans @Suvrit: The link is broke. Could you please write out the author and title of the paper?
Apr 21, 2011 at 9:07 comment added jessica No, but page 6 does :), thanks a lot,... I'm trying to see now if this result can be upgraded to $\mathbb{R}^{n}$
Apr 21, 2011 at 7:35 comment added Suvrit Does page 4 of the following link help? math.unc.edu/Faculty/met/s14.pdf
Apr 20, 2011 at 19:57 answer added Anatoly Kochubei timeline score: 3
Apr 20, 2011 at 19:37 history asked jessica CC BY-SA 3.0