Find a discontinuous function from $\mathbb R^n \to \mathbb R$ which is weakly differentiable.

156 Views Asked by At

For $k \in \mathbb N$, I use the notation $$H^k(\mathbb R^n) = \{ u \in L^2(\mathbb R^n) : D^\alpha u \in L^2(\mathbb R^n) \text{ for all multi-indices } \alpha \text{ with } \lvert \alpha \rvert \le k\}$$

I've seen the following version of the Sobolev Embedding theorem: For $k > \tfrac n 2$, $H^k(\mathbb R^n) \hookrightarrow C_0(\mathbb R^n).$

My question is this: are there functions $u: \mathbb R^n \to \mathbb R$ where $n \ge 2$ such that $f \in H^1(\mathbb R^n)$ but there exists no continuous $g: \mathbb R^n \to \mathbb R$ such that $f = g$ almost everywhere?

This of course would not violate the theorem, but I have thought about it for a bit and I can't immediately think of an example of such an $f$.