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 $

Thanks in advance.

  • $\begingroup$ Is this not just continuity of the inner product in $L^2$? $\endgroup$ – Rhys Steele Mar 30 at 19:19
  • $\begingroup$ I dont see it clearly.. can you explain a little bit? $\endgroup$ – Senna Mar 30 at 19:23
  • $\begingroup$ For example, $<u_n,<\nabla u_n, x>>$ tends to $<u,<\nabla u, x>>$ by continuity of inner product in L^2? $\endgroup$ – Senna Mar 30 at 19:26

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