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Feb 4, 2021 at 22:27 comment added Mateusz Kwaśnicki Well-defined in what sense? As a distribution — sure it is! As a pure function — no, even if $u \in L^2$ (in this case $u$ is in the domain of the quadratic form associated to $(-\Delta)^s$). My guess is that in general, $u$ is in the homogeneous Sobolev space $\dot H^s$, and hence $(-\Delta)^u$ is in the negative Sobolev space $\dot H^{-s}$; but I have never worked with homogeneous Sobolev spaces (and the definitions vary, I believe), so I might be wrong here.
Feb 4, 2021 at 20:27 answer added Bazin timeline score: 2
Feb 4, 2021 at 19:13 history edited Jay CC BY-SA 4.0
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Feb 4, 2021 at 19:05 history asked Jay CC BY-SA 4.0