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Does anyone have a nice geometric example of function in $H^1(U)$ which is not continuous, where $U \subset R^2$ and has a smooth boundary. I want something that is easy to remember.

Joe
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  • Take a smooth function on the disc (even continuous to the boundary) and change it arbitrarily at a single point, so that it has a removable singularity. Is this enough, or do you want to specify some more pathological behavior? – Gyu Eun Lee Jul 17 '15 at 18:47
  • I don't want the function to be in an equivalence class with a continuous function. – Joe Jul 17 '15 at 18:49
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    Ok. See the answer at http://math.stackexchange.com/questions/569390/function-always-continuous-in-a-sobolev-space. – Gyu Eun Lee Jul 17 '15 at 18:55
  • Ok thanks alot. – Joe Jul 18 '15 at 01:50

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