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S Jan 29, 2011 at 20:27 vote accept CommunityBot moved from User.Id=5295 by developer User.Id=481663
Jan 29, 2011 at 20:27 vote accept CommunityBot moved from User.Id=5295 by developer User.Id=481663
S Jan 29, 2011 at 20:27
Sep 17, 2010 at 13:50 vote accept CommunityBot moved from User.Id=5295 by developer User.Id=481663
Jan 29, 2011 at 20:27
Sep 17, 2010 at 12:39 vote accept CommunityBot moved from User.Id=5295 by developer User.Id=481663
Sep 17, 2010 at 13:36
Sep 16, 2010 at 21:49 vote accept CommunityBot moved from User.Id=5295 by developer User.Id=481663
Sep 17, 2010 at 12:39
Sep 16, 2010 at 21:14 answer added Julián Aguirre timeline score: 2
Sep 16, 2010 at 20:53 answer added Helge timeline score: 2
Sep 16, 2010 at 20:42 comment added Jonas T @Julián Aguirre: I modified it a bit, $H_1(x)$ is the Fourier transform of the $\textrm{circ}$-function.
Sep 16, 2010 at 20:41 history edited Jonas T CC BY-SA 2.5
added 40 characters in body; deleted 1 characters in body
Sep 16, 2010 at 20:23 comment added Julián Aguirre For large $x>0$, $J_1(x)$ is asymptotic to $\sqrt{2/(\pi x)}\cos(x-3\pi/4)$, so that it is not in $L^p$ for $1\le p\le 2$. The Fourier transform of $J_1(x)$ is unbounded with support in $[-1,1]$, so that $T$ does not map $L^2$ into itself.
Sep 16, 2010 at 19:28 history asked Jonas T CC BY-SA 2.5