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Nov 20, 2017 at 6:25 history edited j.c. CC BY-SA 3.0
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Nov 19, 2017 at 23:26 history edited Tim Carson CC BY-SA 3.0
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Nov 19, 2017 at 21:57 comment added Tim Carson You're right, this only shows that the functional cannot be finite on the Bryant soliton. As you point out, this is not a full argument since we can have gradient flows of functionals even if they have infinite value.
Nov 19, 2017 at 19:50 comment added Rbega Is this really the argument? I think it is generally agreed that mean curvature flow is the negative gradient flow of area, but there are also plenty of non-trivial non-compact translating solutions, e.g., the translating bowl (aka grim paraboloid) which is the MCF analog of the Bryant solitons.
Nov 19, 2017 at 14:39 vote accept L.F. Cavenaghi
Nov 19, 2017 at 13:17 history answered Tim Carson CC BY-SA 3.0