Special density on $L^2$

Question

Let $\Omega\subset\mathbb{R}^N$ be a bounded domain, and $u\in L^2(\Omega)$ with $0\leq u(x)\leq 1$ a.e. on $\Omega$. It is well known that $C^{\infty}_c(\Omega)$ is dense in $L^2(\Omega)$. Because $C^{\infty}_c(\Omega)\subset W^{1,p}(\Omega)$ for some $1\leq p<\infty$ we have that $W^{1,p}(\Omega)$ is also a dense subset of $L^2(\Omega)$. We take $p\geq\dfrac{2N}{N+2}$ so that $W^{1,p}(\Omega)\hookrightarrow L^2(\Omega)$.

Here comes my question: Is it possible to select a sequence $(u_n)_{n\geq 1}$ with the following three properties:

1. $u_n\longrightarrow u\ \text{in}\ L^2(\Omega)$

2. $0\leq u_n\leq 1$

3. $(u_n)_{n\geq 1}$ is a bounded sequence from $W^{1,p}(\Omega)$

?

Answer 1

At least for $1 < p < \infty$ the answer is no.

By Banach-Alaoglu, after passing to a subsequence we can assume the $u_n$ converge weakly in $W^{1,p}$ to some $v \in W^{1,p}$. Now let $w \in L^2$ be arbitrary. Since $W^{1,p}$ embeds continuously in $L^2$, the linear functional $f \mapsto \int fw$ on $W^{1,p}$ is continuous, so by weak convergence, we have $\int u_n w \to \int vw$. On the other hand, since $u_n \to u$ in $L^2$, we also have $\int u_n w \to \int uw$. We conclude that $\int uw = \int vw$ for all $w \in L^2$, so that $u = v$ a.e. Thus we see that your statement cannot hold when $u \notin W^{1,p}$.

I am fairly sure the claim is also false for $p=1$ but I don't have a proof off the top of my head. This argument doesn't work for $p=1$ because $W^{1,1}$ is not reflexive.