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The question I have in mind is the following: how can we prove that for any $v\in H^1(\Omega)$ and for any $u\in H^1(\Omega)$ with $\Delta u\in L^2(\Omega)$ the Gauss-Green identity takes place

$$\int_{\Omega} v(x)\Delta u(x)\ dx=\int_{\partial\Omega}\frac{\partial u}{\partial\nu}(x) v(x)\ d\sigma-\int_{\Omega}\nabla u(x)\cdot\nabla v(x)\ dx \;\;? $$

Here $\Omega\subset\mathbb{R}^N$ is a bounded Lipschitz domain: in some books I found this identity assuming that $u\in H^2(\Omega)$ (like Atkinson - Theoretical Numerical Analysis, page 325).

I tried myself to prove this, using the weak form of Gauss-Green formula, and I came across the following simple-looking fact that I do not know how to prove.
Let $h\in H^1(\Omega)$ be such that there exists $\Delta h\in L^2(\Omega)$ ($h$ may not be in $H^2(\Omega)$) and consider a sequence $(f_n)_n\subset C^{\infty}(\Omega)$ with $f_n\to h$ in $H^1(\Omega)$.
My question is: is it true that $$\lim\limits_{n\to\infty} \int_{\Omega} g(x)\Delta f_n(x)\ dx=\int_{\Omega} g(x)\Delta h(x)\ dx$$

for all $g\in H^1(\Omega)$? Knowing this is true will suffice for our aim.

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    $\begingroup$ You need $u \in H^2$ to write such a formula, in particular to give a meaning to $\nabla u$ at the boundary. $\endgroup$
    – Giorgio Metafune
    Commented Apr 22, 2023 at 23:49
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    $\begingroup$ 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. $\endgroup$
    – user378654
    Commented Apr 27, 2023 at 7:36
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    $\begingroup$ 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. $\endgroup$
    – user378654
    Commented Apr 27, 2023 at 7:40

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