Cantor lebesgue as an $L^1$ limit

Question

Is it possible to view the Cantor Lebesgue function (CLF) as an $L^1$ limit of bounded sequence in $W^{1,1}$?

My motivation is to proof that this cannot be:

Let $I=(a,b)$ be an interval with $−∞≤a<b≤∞$, let $u∈L^1(I)$ and let $(u_k)_{k∈\mathbb N} ⊆ W^{1,1}(I)$ be a bounded sequence with $\lim_{k→∞} u_k = u$ in $L^1(I)$.

Is $u∈W^{1,1}(I)$ ?

And I already that the CLF has no weak derivative in $L^1$.

So I'd need to show that there is a $W^{1,1}$-bounded sequence of functions converging to the CLF in the $L^1$-norm.

Any thoughts on that?

Answer 1

$\newcommand{\dd}{\text{d}}$ $\newcommand{\norm}[1]{\left \lVert #1 \right \rVert}$

I was thinking maybe you can argue directly.

Let's define the CLF approaching sequence:

$f_0(x) = x \\ f_1(x) = \begin{cases} \frac{1}{2}, & x \in \big[\frac{1}{3},\frac{2}{3}\big] \\ \frac{3 \, x}{2}, & x \in \big[0,\frac{1}{3}\big] \cup \big[\frac{2}{3},1 \big] \end{cases} \\ f_2(x) = \begin{cases} \frac{1}{4}, & x \in \big[\frac{1}{9},\frac{2}{9}\big] \\ \frac{1}{2}, & x \in \big[\frac{1}{3},\frac{2}{3}\big] \\ \frac{3}{4}, & x \in \big[\frac{7}{9},\frac{8}{9}\big] \\ \frac{9 \, x}{4}, & x \in \big[0,\frac{1}{9}\big] \cup \big[\frac{2}{9},\frac{3}{9} \big] \cup \big[\frac{6}{9},\frac{7}{9} \big] \cup \big[\frac{8}{9},1 \big] \end{cases} \\ \qquad \vdots$

For each $n \in \mathbb N$ we have $\norm{f_n}_{L^1([0,1])} = \int_0^1 |f'_n(x)| \, \dd x = 1$ thus $(f_n)_{n \in \mathbb N} \subset W^{1,1}([0,1])$ and $(f_n)_{n\in \mathbb N}$ is bounded in $W^{1,1}([0,1])$.

Answer 2

Yes. Extend the Cantor function to be $u(x)=0$ for $x<0$ and $u(x)=1$ for $x\ge 1$. Then use mollifiers, $u_\varepsilon=u\star \varphi_\varepsilon$. Since $u$ is continuous you have that $u_\varepsilon\to u$ uniformly on compact sets, and so in $[0,1]$. In particular, $u_\varepsilon\to u$ in $L^1([0,1])$. The functions $u_\varepsilon$ are in $W^{1,1}$ since they are in $C^\infty(\mathbb{R})$.