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Apr 13, 2017 at 12:19 history edited CommunityBot
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Jul 23, 2015 at 19:22 comment added md2perpe There is no problem multiplying delta with a function continuous in 0. Delta doesn't need infinitely differentiable test functions (unless you want to define derivatives of all orders).
Jul 20, 2015 at 21:02 answer added md2perpe timeline score: 2
Jul 16, 2015 at 17:39 vote accept Gateau au fromage
Jul 16, 2015 at 16:24 answer added paul garrett timeline score: 3
Jul 16, 2015 at 15:43 comment added Christian Remling There is a near infinite literature on Schrodinger operators with highly singular potentials, you might want to do a literature search.
Jul 16, 2015 at 14:18 answer added Carlo Beenakker timeline score: 10
Jul 16, 2015 at 13:41 history asked Gateau au fromage CC BY-SA 3.0