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Jun 24, 2018 at 16:17 history edited Michael Hardy CC BY-SA 4.0
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Jun 21, 2018 at 16:50 comment added ABIM I don;t know this book. What is it called?
Jun 21, 2018 at 1:38 comment added Piotr Hajlasz Did you check the book by Fonseca and Leoni?
Jun 20, 2018 at 23:15 comment added ABIM True, in that case the most I can assume is that $f$ is convex in its second variable for almost-every $x$ (I made the update, as I was trying for a weaker result)
Jun 20, 2018 at 23:14 history edited ABIM CC BY-SA 4.0
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Jun 20, 2018 at 23:07 comment added Nate Eldredge Oh yeah, I missed "weak". But Borel is certainly not enough; take $f(x,u) = 1_{[0,1]}(u)$ and $u_n = 1+1/n$.
Jun 20, 2018 at 23:05 history edited ABIM CC BY-SA 4.0
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Jun 20, 2018 at 23:05 comment added ABIM Even under those assumptions, whouldn't that only give strong lower semi continuity which is less than weak lower semi continuity?
Jun 20, 2018 at 23:02 comment added Nate Eldredge What are you assuming about $f$? If it's continuous in the second variable and nonnegative then this is just Fatou's lemma. And if it's not continuous then I don't see why you would expect any semicontinuity at all.
Jun 20, 2018 at 22:50 comment added ABIM Aren't sequential weak lower-semi-continuity and weak lower-semi-continuity the same in a Hilbert space?
Jun 20, 2018 at 22:42 comment added Piotr Hajlasz You should clarify what statement you have in mind. For example $\int_{\mathbb{R}^n}|u|^2$ is (sequentially) weakly lower semicontinuous since the norm is (sequentially) weakly lower semicontinuous.
Jun 20, 2018 at 22:32 history asked ABIM CC BY-SA 4.0