Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
4  
Because of the "jump" on $\partial \Omega$. That destroys (weak) differentiability. –  Daniel Fischer Sep 5 '13 at 9:24
1  
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
add comment

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.

share|improve this answer
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.