# Can I pass to the limit?

Let be a sequence $$u_n$$ of $$C_0^\infty (\mathbb{R}^N)$$-functions converging to $$u$$ in $$H^1(\mathbb{R}^N)$$, which implies that $$u_n\rightarrow u$$ in $$L^2(\mathbb{R}^N)$$ and $$\nabla u_n\rightarrow \nabla u$$ in $$L^2(\mathbb{R}^N)$$, such that $$\int_{\mathbb{R}^N} div\ (\vert u_n(x)\vert ^2 x)\ dx \rightarrow 0$$ and I wonder if I can say that $$\int_{\mathbb{R}^N} div\ (\vert u _n(x)\vert ^2 x)\ dx \rightarrow \int_{\mathbb{R}^N} div\ (\vert u (x)\vert ^2 x)\ dx=0$$ Note that $$div (|u(x)|^2x) = \nabla|u(x)|^2x+|u(x)|^2div(x)=(\overline{u}(x)\nabla u(x)+u(x)\nabla \overline{u}(x) )\cdot x +N |u(x)|^2$$ I need some help or hint here, please. I have never seen a rigurous proof by density. Clearly, by the fact that $$u_n\rightarrow u$$ in $$L^2(\mathbb{R}^N)$$, one has $$N\int_{\mathbb{R}^N } |u_n(x)|^2 \rightarrow N\int_{\mathbb{R}^N } |u(x)|^2$$ but I need help por the other part $$(\overline{u}(x)\nabla u(x)+u(x)\nabla \overline{u}(x))\cdot x$$

• Is this not just continuity of the inner product in $L^2$? – Rhys Steele Mar 30 at 19:19
• For example, $<u_n,<\nabla u_n, x>>$ tends to $<u,<\nabla u, x>>$ by continuity of inner product in L^2? – Senna Mar 30 at 19:26