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Jul 9, 2018 at 16:33 comment added Bart Michels Also see Borel, Automorphic forms on SL2(R), Lemma 8.3, where it is stated that this result is due to Godement.
Jul 10, 2017 at 19:56 comment added Pietro Majer Also check problem 349 of Bernard Gelbaum's Problems in Analysis (Springer 1982) , chapter on Hilbert spaces, where it is also shown by elementary inequalities: "If $\|f\|_\infty\le C\|f\|_2$ for all $f\in M$, $M$ a subspace of $L^2([0,1])$, then $\rm{dim}(M)\le C^2$".
Jan 20, 2011 at 1:46 comment added Yemon Choi Since it seems that using the open mapping theorem is allowed, perhaps the reference to Rudin might still be of interest? It's Theorem 5.2 in Functional Analysis (2nd ed)
Jan 19, 2011 at 19:12 comment added Yemon Choi Sorry, I left a reference (to Rudin's book) as an answer without properly reading your question, and in particular the fact that you wanted a proof avoiding functional-analytic tricks
Jan 19, 2011 at 18:44 vote accept Rostyslav Kravchenko
Jan 19, 2011 at 17:50 vote accept Rostyslav Kravchenko
Jan 19, 2011 at 17:57
Jan 19, 2011 at 16:24 vote accept Rostyslav Kravchenko
Jan 19, 2011 at 16:28
Jan 19, 2011 at 16:24 vote accept Rostyslav Kravchenko
Jan 19, 2011 at 16:24
Jan 19, 2011 at 16:24 vote accept Rostyslav Kravchenko
Jan 19, 2011 at 16:24
Jan 19, 2011 at 16:08 answer added Homology timeline score: 30
Jan 19, 2011 at 15:43 answer added Nate Eldredge timeline score: 7
Jan 19, 2011 at 15:39 vote accept Rostyslav Kravchenko
Jan 19, 2011 at 15:39
Jan 19, 2011 at 15:35 vote accept Rostyslav Kravchenko
Jan 19, 2011 at 15:35
Jan 19, 2011 at 14:45 answer added Mikael de la Salle timeline score: 7
Jan 19, 2011 at 12:59 history edited Rostyslav Kravchenko
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Jan 19, 2011 at 12:50 history asked Rostyslav Kravchenko CC BY-SA 2.5