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Bogdan
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A question about Gauss-Green formula - a weaker assumption

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:

If $h\in H^1(\Omega)$ such that there exists $\Delta h\in L^2(\Omega)$ ($h$ may not be in $H^2(\Omega)$). 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 $$\lim\limits_{n\to\infty} \int_{\Omega} g(x)\Delta f_n(x)\ dx=\int_{\Omega} g(x)\Delta h(x)\ dx$$?

In other words: Is $\Delta f_n$ weakly convergent to $\Delta h$ in $L^2(\Omega)$?

Bogdan
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