Estimate in Sobolev Spaces

Question

Let $u\in H^0(U)\cap H^1_0(U)$ and $v_k\in C^\infty_c$(U) such that $v_k\rightarrow u$ in $H^1_0(U)$ and $w_k\in C^\infty (U)$ such that $w_k \rightarrow u $ in $H^2(U)$.

I want to show that $ \int_U Dw_k \cdot Dv_k -|Du|^2 dx \leq \int_U |Dw_k| |Dv_k-Du|dx+\int_U |Du| |Dw_k-Du|dx$

tends to $0$. I feel like I should use that $v_k,w_k\rightarrow u$ in $H^1_0$ and $H^2$ respectively but I'm not sure how since I dont think I have the correct norms.

Answer 1

By the Cauchy-Schwarz inequality, we have $$\int_U |Dw_k| |Dv_k - Du|\,dx \le \sqrt{\int_U |Dw_k|^2} \sqrt{\int_U |Dv_k - Du|^2}.$$ Since the Sobolev norm has $\|f\|_{H^1}^2 = \int_U (|f|^2 + |Df|^2) \ge \int_U |Df|^2$, we have $$\int_U |Dw_k| |Dv_k - Du|\,dx \le \|w_k\|_{H^1} \|v_k - u\|_{H^1}.$$ Since $w_k$ converges in $H^1$ norm, $\sup_k \|w_k\|_{H^1} < \infty$. And since $v_k \to u$ in $H^1$ norm, $\|v_k - u\|_{H^1} \to 0$.

The other term is similar.

Answer 2

I am afraid that it is rather straight-forward:

We have that $$ Dw_k \cdot Dv_k -|Du|^2 =Dw_k\cdot Dv_k-Dw_k\cdot Du+Dw_k\cdot Du-Du\cdot Du= Dw_k\cdot (Dv_k- Du)+Du\cdot(Dw_k-Du), $$ and hence $$ Dw_k \cdot Dv_k -|Du|^2 \le \lvert Dw_k\cdot (Dv_k- Du)\rvert+\lvert Du\cdot(Dw_k-Du)\rvert \le \lvert Dw_k\rvert\lvert Dv_k- Du\rvert+\lvert Du\rvert \lvert Dw_k-Du\rvert, $$ and finally $$ \int_U Dw_k \cdot Dv_k -|Du|^2\,dx\le \int_U \lvert Dw_k\rvert\lvert Dv_k- Du\rvert\,dx+\int_U\lvert Du\rvert \lvert Dw_k-Du\rvert\,dx. $$