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Sep 26, 2020 at 17:33 answer added Martin Väth timeline score: 1
Jul 21, 2020 at 8:58 vote accept ABIM
Jul 21, 2020 at 8:52 answer added Jochen Wengenroth timeline score: 2
Jul 21, 2020 at 8:51 comment added ABIM What about when $L^1$ is replaced by $C(\mathbb{R})$? I don't this works then... we probably need some integrability contstraint also?
Jul 21, 2020 at 8:07 comment added Jochen Wengenroth A simple sufficient condition for the continuity is that $f$ is Lipschitz with respect to the second argument, i.e., $|f(x,y)-f(x,z)|\le c|y-z|$ for some constant independent of $x,y,z$.
S Jul 21, 2020 at 6:20 history suggested ABIM CC BY-SA 4.0
formatted and fixed some typos... Moved bottom bit of intuition to top for continuity of read.
Jul 21, 2020 at 6:20 review Suggested edits
S Jul 21, 2020 at 6:20
Jul 21, 2020 at 6:17 history asked ABIM CC BY-SA 4.0