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Dec 1, 2021 at 18:58 history edited SolubleFish CC BY-SA 4.0
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May 24, 2021 at 13:31 vote accept smallest quanta
May 24, 2021 at 13:06 comment added Charlie Ah sure, makes sense ty.
May 24, 2021 at 13:01 comment added SolubleFish Sure, $\int \nabla \phi f = - \int \phi \nabla f$ for any test function $f$. The result is the same
May 24, 2021 at 12:52 comment added Charlie Can the derivative of operator valued distributions actually be defined rigorously?
May 24, 2021 at 12:40 history answered SolubleFish CC BY-SA 4.0