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I'm seeking a weak differentiable function $f \in W^{1,1}(\Omega)$ which is discontinuous. I think every weak differentiable function defined on $\mathbb{R}$ is continuous by Sobolev embedding theorem, so an example should exist at least in $\mathbb{R}^2$ if it does. Is my guess correct? If so, could anyone please show me an example?

Willy Chen
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    $f:\mathbb{R}^n \to \mathbb{R}$, with $f(x) = \frac{1}{|x|^a}$ for $a<n+1$, see exmaple 3.7 here https://www.math.ucdavis.edu/~hunter/pdes/ch3.pdf – math Dec 03 '19 at 17:26
  • See here: https://math.stackexchange.com/questions/746289/discontinuous-sobolev-function – PhoemueX Dec 03 '19 at 17:27

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You require an integrable function that is absolutely continuous on almost all lines parallel to the coordinate axes, whose partial derivatives are also integrable. Try $f(x) = \dfrac 1{\sqrt{|x|}}$ and $\Omega = B(0,1)$ in $\mathbb R^2$.

Umberto P.
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Yes, you are right. On $\mathbb{R}$ you will have no such luck. An example in higher dimensions is relatively easy: Take $f(x) = |x|^s$ where $s > 1- n$. In fact you can take any $C^1$ function in higher dimensions (on bounded domain for integrability) and just make it discontinuous in one point. This function will still have a weak derivative that is bounded.

hal4math
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