Discontinuous function in $W^{1, 1}(\mathbb{R}^{2})$

Question

What's an example of a bounded function in $W^{1, 1}(\mathbb{R}^{2})$ which is discontinuous? Can this function be discontinuous on a set of positive measure?

Answer 1

When forming Sobolev classes, we identify functions that differ on a set of measure zero. Therefore, continuity should be understood as having a continuous representative. For example, $\chi_{\mathbb Q\times \mathbb Q}$ has a continuous representative, namely $0$.

I'll answer in greater generality, in $d\ge 2$ dimensions. Let $$u(x) = \begin{cases}|x|^{-p} - 1 \quad & |x|\le 1 \\ 0 & |x|>1\end{cases} \tag{1}$$ This function is absolutely continuous on every line not passing through the origin. Therefore, it is $W^{1,1}$ provided that its pointwise gradient $\nabla u$ is integrable. Since $|\nabla u(x)| $ is a constant multiple of $|x|^{-(p+1)}$, we have $u\in W^{1,1}$ provided that $p+1<d$. That is, for any $0<p<d-1$ the function $u$ gives an example of an essentially unbounded element of $W^{1,1}(\mathbb R^d)$. (There are other ways to verify $u\in W^{1,1}$; how you do it depends on what you know about Sobolev spaces.)

But you wanted a bounded example, so take $\sin u$ instead. Since $\sin$ has derivative bounded by $1$, by the chain rule $\sin u\in W^{1,1}$ whenever $u\in W^{1,1}$. And $\sin u$ does not have a continuous representative, because in any neighborhood of the origin we have $\sin u>1/2$ and $\sin u<-1/2$ on sets of positive measure.

For your second question: no, every Sobolev function has a representative that is continuous at almost every point. One usually takes so-called precise representative, defined as $$u^*(x) = \begin{cases} {\displaystyle\lim_{r\to 0} \frac{1}{|B_r(x)|}\int_{B_r(x)} u } \quad &\text{if limit exists} \\ 0 \quad &\text{otherwise}\end{cases} \tag{2}$$ If $u\in W^{1,1}$, then the set where $u^*$ is discontinuous has zero measure. In fact more can be said: an appropriate capacity of this set is zero. This gets a bit technical, so I refer to a book: Measure theory and fine properties of functions by Evans and Gariepy, page 160. The basic idea is to look at the Lebesgue points of $\nabla u$ and apply the Poincaré inequality, which controls the oscillation of $u$ by integral average of $|\nabla u|$.

Answer 2

Examples $\ (1)\ \&\ (2)\ $ do not answer to the original question posed, since $\ p=1 \ \&\ d=2\ \ - $ as originally the question was posed $\ -\ $ is incompatible with the condition $\ p<d-1.\ $

However, I now see that this question has been answered by :

Normal Human (https://math.stackexchange.com/users/147263/normal-human), Are all functions in the Sobolev space $W_0^{1,2}(\Omega)$ continuous and bounded?, URL (version: 2014-12-31): https://math.stackexchange.com/q/1087038