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Apr 20, 2020 at 13:46 vote accept Isaac
Apr 20, 2020 at 13:26 answer added leo monsaingeon timeline score: 5
Apr 20, 2020 at 12:59 comment added leo monsaingeon OK. I edited your question to make this clear, since this notation is not completely standard.
Apr 20, 2020 at 12:58 history edited leo monsaingeon CC BY-SA 4.0
added 53 characters in body
Apr 20, 2020 at 12:56 comment added Isaac It is full Hessian. I figured out using the orthonormal basis anyway...
Apr 20, 2020 at 12:54 comment added leo monsaingeon Is the double $\nabla\nabla u$ a typo in your statement and did you mean $(\dots)\leq C\int_U |\nabla u|^2$, or do you write $\nabla\nabla u$ the full second-order hessian $D^2 u$?
Apr 19, 2020 at 23:29 history asked Isaac CC BY-SA 4.0