Let $\Omega \subset \mathbb{R}^N$ be a bounded open set and let $(f_n) \subset L^2(\Omega)$. Suppose there exists $f \in L^2(\Omega)$ such that $$\int_{\Omega} f_n \varphi \rightarrow \int_{\Omega} f \varphi$$ for all $\varphi \in C_c^\infty(\Omega)$.
How can we deduce that $$\int_{\Omega} f_n v \rightarrow \int_{\Omega} f v$$ for all $v \in W_0^{1,2}(\Omega)$ ?
By definition, for all $v \in W_0^{1,2}(\Omega)$ there exists a sequence $(v_m) \subset C_c^\infty(\Omega)$ such that $$\vert\vert v_m - v \vert\vert_{W^{1,2}(\mathbb{R}^N)} \rightarrow 0.$$ Therefore—using Lebesgue's dominated convergence— $$\lim_{n \rightarrow \infty} \int_{\Omega} f_n v = \lim_{n \rightarrow \infty} \lim_{m \rightarrow \infty} \int_{\Omega} f_n v_m.$$ If we can switch the two limits, then I believe the proof is complete. But I'm pretty sure that it is not that simple. Any idea?
This is not true. Let $\Omega=(0,1)$. Set $f_n(x):= n^3 \chi_{[0,1/n]}(x)$. Then $$ \int_0^1 f_n \phi \to 0 $$ for all $\phi\in C_c^\infty(\Omega)$.
Let $\psi \in C^\infty(\bar\Omega)$ be a smooth cut-off function: $\psi=1$ on $(0,1/2)$, $\psi\ge0$, $\psi(1)=0$.
Take $\phi(x)=x^{3/2-\epsilon}\psi$, $\epsilon>0$, which is in $W_0^{1,2}(\Omega)$. Then $$ \int_\Omega f_n \phi=\int_0^{1/n} n^3 x^{3/2-\epsilon} =O(n^{1/2}) \to\infty. $$
The statement is true, if one additionally assumes that $(f_n)$ is bounded in $L^2(\Omega)$. Then the conditions imply $f_n\rightharpoonup f$ in $L^2(\Omega)$ by density of smooth functions in $L^2(\Omega)$. This is equivalent to $\int_\Omega f_n\phi\to\int_\Omega f\phi$ for all $\phi\in L^2(\Omega)$. Hence, the desired estimate follows as $W_0^{1,2}(\Omega)\subset L^2(\Omega)$.