Weak convergence - restriction

Question

Let $\Omega$ be a open subset of $\mathbb{R}^{d}$ and consider $(f_n) \subset L^2(\Omega)$. Let $K$ be a compact subset of $\Omega$. Suppose that $f_n \rightharpoonup f$ in $L^2(\Omega)$ and $f_n|_{\Omega \setminus K} \rightharpoonup g$ in $L^2(\Omega \setminus K)$. Can I conclude that $f|_{\Omega\setminus K}=g$?

Answer 1

Yes. If $f_n \rightharpoonup f$ in $L^2(\Omega)$ then $f_n|_{\Omega \setminus K} \rightharpoonup f|_{\Omega \setminus K}$ in $L^2(\Omega \setminus K)$. This is because if $\phi \in L^2(\Omega \setminus K)$ then $$\Phi(x) = \begin{cases} \phi(x) \quad x \in \Omega \setminus K \\ 0 \quad\text{otherwise}\end{cases}$$ is an element of $L^2(\Omega)$ and so $\int_{\Omega \setminus K} (f_n|_{\Omega \setminus K} - f|_{\Omega \setminus K}) \phi = \int_{\Omega} (f_n - f) \Phi \to 0$ as $n \to \infty$. The result then follows because the weak topology is Hausdorff and so sequences have at most one limit which implies that $g = f|_{\Omega \setminus K}$.