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Jan 12, 2021 at 18:48 answer added Willie Wong timeline score: 1
Jan 5, 2021 at 18:09 vote accept BigbearZzz
Jan 5, 2021 at 17:03 answer added mlk timeline score: 4
Jan 5, 2021 at 17:03 comment added mlk @WillieWong Well, since you asked nicely, though it isn't the precise statement.
Jan 5, 2021 at 15:00 comment added BigbearZzz I agree. If he posts an answer based on this comment I'll accept it.
Jan 5, 2021 at 14:48 comment added Willie Wong @mlk since the OP tagged reference request, your comment is probably good as an answer.
Jan 5, 2021 at 13:54 comment added BigbearZzz @mlk Thank you, I'll have a look at that and perhaps just prove it myself. The formula in my question just seem so natural hence it surprised me a bit to learn that it's not already well-documented in the literatures.
Jan 5, 2021 at 13:21 comment added mlk If you look in Evans & Gariepy's "Measure theory and fine properties of functions", Sec. 5.3., you see that they define the trace on BV (which includes $W^{1,1}$) for Lipschitz domains in a similar fashion and prove a similar convergence. You should be able to adapt this proof.
Jan 5, 2021 at 13:20 comment added Giorgio Metafune This is true with some regularity on $\partial \Omega$ and follows for $u \in W^{1,1}$ from the half-space case. If $Q$ is a cube in the $x$ variable, then $\int_Q |u(x,y_2)-u(x,y_1)| dx \le \int_{Q\times (y_1,y_2)}|u_y(x,y)|dxdy$ and the function $y \to \int_Q u(x,y)dx$ is continuous.
Jan 5, 2021 at 10:37 history edited BigbearZzz CC BY-SA 4.0
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Jan 5, 2021 at 2:29 history edited BigbearZzz
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Jan 4, 2021 at 19:22 history asked BigbearZzz CC BY-SA 4.0