# The convegrence of $L^2$ norm of the gradient of a mollified sequence of functions

Let $u\in W^{1,2}(\Omega)$ where $\Omega\subset R^N$ is open bounded, smooth boundary.

Given a sequence of function $u_n\in L^2(\Omega)$ such that $\nabla u_n$ is defined a.e. and $\nabla u_n\to\nabla u$ a.e. We also know that $u_n\to u$ in $L^2$. Notice that $u_n$ may not in $W^{1,2}$. Here we can think $u_n$ is in $BV$ and $\nabla u_n$ is the absolutely continuous part of $Du_n$, i.e., the weak derivative represented by measure.

Now consider $\Omega'\subset\subset \Omega$, i.e., compact subset. Then I can define the convolution $(u_n)_{\epsilon_n}$ for $\epsilon_n$ small enough and $\epsilon_n\to0$ as $n\to\infty$.

The paper I am reading now conclude that $$\limsup_{n\to\infty}\int_{\Omega'}|\nabla (u_n)_{\epsilon_n}|^2dx\leq \int_{\Omega'}|\nabla u|^2dx$$ and it stated that it could be proved by Jensen's inequality and Fubini's theorem. I can't find out how.

## 1 Answer

This is not true as stated. For example, in one dimension let $\Omega=(-1,1)$ and $$u_n (x) = \begin{cases} 1/n,\quad &x>0\\ 0,\quad &x\le 0 \end{cases}$$ This converges to $0$ in $L^2$. The convolution $(u_n)_{\epsilon_n}$ goes from $0$ to $1/n$ on the interval $[-\epsilon_n,\epsilon_n]$. Therefore (using FTC and Cauchy-Schwarz), $$\int |\nabla (u_n)_{\epsilon_n}|^2 \ge \frac{1}{n^2 \epsilon_n}$$ and this need not go to zero in general.