# How to “uniformly” mollify a sequence without knowing the higher derivative.

The motivation and previous discussion of some related ideas can be seen here.

My question: given a sequence $u_n\in u_0+H^1_0(\Omega)$ where $u_0\in H^1(\Omega)$ and $\Omega\subset \mathbb R^N$ is open bounded lipschitz boundary.

Assume $u_n\to u\in u_0+H^1_0(\Omega)$ weakly in $H^1$, I want to obtain a sequence of functions $u_{n,\epsilon}$ with double index $n$, $\epsilon$ such that $u_{n,\epsilon}\in C^\infty(\bar \Omega)$ and $u_{n,\epsilon}\to u_{\epsilon}$ weakly in $H^1$ and $$\lim_{\epsilon\to 0}\sup_{n\in\mathbb N}\left| \int_\Omega |\nabla u_{n,\epsilon}|^2dx- \int_\Omega |\nabla u_{n}|^2dx\right|=0,$$ i.e., the mollification is uniformly obtained.

Or this condition can be weaken as for any $\delta>0$, there exists a sequence of functions $u_{n,\epsilon}$ with double index $n$, $\epsilon$ such that $u_{n,\epsilon}\in C^\infty(\bar \Omega)$ and $u_{n,\epsilon}\to u_{\epsilon}$ weakly in $H^1$ and $$\lim_{\epsilon\to 0}\sup_{n\in\mathbb N}\left| \int_\Omega |\nabla u_{n,\epsilon}|^2dx- \int_\Omega |\nabla u_{n}|^2dx\right|<\delta.$$

If we know the Hession $\nabla^2 u$ such that $\|\nabla^2 u\|_{L^2}$ is uniformly bounded, then this would be easy by using standard mollification operator. But I am wondering can we still obtain the same without the information of Hession? Maybe some other mollification construction?