# Extension of Sobolev functions

Actually I am searching if $u$ belong to $W^{1,p}(\Omega)$ then why extension out side by zero does not belong to $W^{1,p}(\mathbb{R}^n)$ in general? Can some body help?

-
Because of the "jump" on $\partial \Omega$. That destroys (weak) differentiability. – Daniel Fischer Sep 5 '13 at 9:24
Example: $\Omega=(0,1)$ and $u(x)=1$ for all $x\in \Omega$. If you extend it to be $0$ outside of $\Omega$ then its first distributional derivative will have two $\delta$s at $0$ and at $1$, and so it won't be a function. – Giuseppe Negro Sep 5 '13 at 9:33

## 1 Answer

We can resume the two comment given by @DanielFischer and @GiuseppeNegro with the following result (see Leoni page 293):

If $u\in W^{1,p}(\Omega)$ with $p\in [1,\infty)$, then for almost every segment in $\Omega$ parallel to the coordinate axes, $u$ restricted to this segment is absolutely continuous. In fact, this result is part of a classification os Sobolev spaces and I suggest you to take a look in the book of Leoni that I have cited.

Also, it is worth to note that when the dimension is $1$, because of the last result, every Sobolev function has a continuous representative.

-
Or at least has a continuous representative. – Umberto P. Sep 5 '13 at 13:28
Yes @UmbertoP., you are right, let me fix it. – Tomás Sep 5 '13 at 13:37
Sorry, but where is this answer? – Tomás Sep 5 '13 at 14:56
I think the counter example given by Giuseppe Negro is sufficient to establish a contradiction. Thank you – Acharya Sep 6 '13 at 11:36