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Apr 27, 2023 at 7:40 comment added user378654 That said, that does not help OP with their actual question, where they would need to assume that $\Delta f_n \rightarrow \Delta h$ in $L^2$ to use this approach, in which case the conclusion is trivial anyway.
Apr 27, 2023 at 7:36 comment added user378654 It's true that you need $u \in H^2$ in order to give a ($L^2$) meaning to $\nabla u$ on the boundary, but you do not need that to discuss only the normal derivative on the boundary. Indeed, from the formula in the OP we have that $\partial_\nu u$ defines a bounded linear functional on $H^{1/2}(\partial \Omega)$ (acting by extending the $H^{1/2}$ function to $v \in H^1$), and the map $u \mapsto \partial_\nu u$ is bounded from $H^1 \cap \{\Delta u \in L^2\}$ to the dual of $H^{1/2}(\partial \Omega)$. This is common in the fluids literature, where they have vector fields with controlled div.
Apr 23, 2023 at 14:37 history edited Daniele Tampieri CC BY-SA 4.0
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Apr 22, 2023 at 23:49 comment added Giorgio Metafune You need $u \in H^2$ to write such a formula, in particular to give a meaning to $\nabla u$ at the boundary.
Apr 22, 2023 at 17:47 history edited Bogdan CC BY-SA 4.0
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Apr 22, 2023 at 17:41 history edited Bogdan
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Apr 22, 2023 at 15:45 history asked Bogdan CC BY-SA 4.0