Let $\Omega$ be an open subset of $\mathbb R^d$. Let $u \in L^\infty (\Omega)$. I'm trying to do below exercise, i.e.,
- Prove that there is $(u_n) \subset \mathcal C_c^{\infty} (\Omega)$ such that
- $\sup_n \|u_n\|_\infty \le \|u\|_\infty$,
- $u_n \xrightarrow{n\to\infty} u$ a.e. on $\mathbb R^d$, and
- $u_n \xrightarrow{n\to\infty} u$ in $L^\infty (\Omega)$ w.r.t. the weak$^*$ topology $\sigma(L^\infty, L^1)$.
- If $u \ge 0$ a.e. on $\Omega$, we can further take $u_n \ge 0$ a.e. on $\Omega$.
- Deduce that $\mathcal C_c^{\infty} (\Omega)$ is dense in $L^\infty (\Omega)$ w.r.t. $\sigma(L^\infty, L^1)$.
Could you have a check on my below attempt for (1.)?
For $f:\Omega \to \mathbb R$, we define $\overline f : \mathbb R^d \to \mathbb R$ by $\overline f (x) := f(x)$ if $x \in \Omega$ and $0$ otherwise. Let $(\rho_n)$ be a sequence of mollifiers, i.e., $$ \rho_n \in \mathcal C_c^{\infty} (\mathbb{R}^d), \quad \operatorname{supp} \rho_n \subset \overline{B(0,1 / n)}, \quad \int \rho_n=1, \quad \rho_n \geq 0 \text { on } \mathbb{R}^d, $$ where $B(x, r)$ denotes the open ball centered at $x \in \mathbb R^d$ with radius $r>0$.
- There is a sequence $(K_n)$ of compact subsets of $\Omega$ such that $$ \bigcup_n K_n = \Omega \quad \text{and} \quad \inf_{x \in K_n}\operatorname{dist} (x, \Omega^c) \ge \frac{2}{n} \quad \forall n \in \mathbb N^*. $$
For each $n \in \mathbb N^*$, we define $\xi, \xi_n:\mathbb R^d \to \mathbb R$ by $\xi := 1_{\Omega}$ and $\xi_n := 1_{K_n}$. Then $\sup_n \|\xi_n\|_\infty = 1, \operatorname{supp} \xi_n = K_n$ and $\xi_n \xrightarrow{n\to\infty} \xi$ a.e. on $\mathbb R^d$. Let $$ v_n := (\xi_n \overline u) * \rho_n \quad \text{and} \quad v := \xi \overline u. $$
Fix $f \in L^1(\Omega)$. Then $\overline f \in L^1 (\mathbb R^d)$. I already proved that $$ \| \overline f v_n - \overline fv \|_1 \xrightarrow{n\to\infty} 0. $$
Notice that $\xi_n \overline u \in L^1 (\mathbb R^d)$, so $v_n \in \mathcal C^\infty (\mathbb R^d)$. Because $\rho_n \ge 0$ and $\int \rho_n =1$, we get $\|v_n \|_\infty \le \| \xi_n \overline u \|_\infty \le \|u\|_\infty$. We have $u = v \restriction \Omega$ and $$ \operatorname{supp} v_n \subset \overline{\operatorname{supp} (\xi_n \overline u) + \operatorname{supp} \rho_n} \subset K_n + \overline{B(0,1 / n)} \subset \Omega. $$
Then $v_n \in \mathcal C_c^\infty (\mathbb R^d)$ and $1_{\Omega} v_n \in \mathcal C_c^\infty (\Omega)$. It remains to extract a sub-sequence $\varphi$ of $\mathbb N$ such that $v_{\varphi (n)} \xrightarrow{n\to\infty} v$ a.e. on $\mathbb R^d$. This is done by diagonal method as follows.
Let $B_m := B(0, m)$. Then $\| 1_{B_m} v_n - 1_{B_m}v \|_1 \xrightarrow{n\to\infty} 0$ for every $m \in \mathbb N^*$. Recursively, there is a sequence $(\varphi_m)$ such that for each $m \in \mathbb N^*$,
- $\varphi_m$ is a sub-sequence of $\mathbb N$ with $v_{\varphi_m (n)} \xrightarrow{n\to\infty} v$ a.e. on $B_m$, and
- $\varphi_{m+1}$ is a sub-sequence of $\varphi_m$.
We define a sub-sequence $\varphi$ of $\mathbb N$ by $\varphi (m) := \varphi_m (m)$ for all $m \in \mathbb N^*$. It is easy to see that $\varphi$ is a sub-sequence of every $\varphi_m$, so $v_{\varphi (n)} \xrightarrow{n\to\infty} v$ a.e. on every $B_m$. The claim then follows by setting $u_n := v_{\varphi (n)} \restriction \Omega$.
