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.

What mean that expression "closed respect a norm". For example:

$$H^{m,p}(\Omega)= \overline{\left\{{u\in C^m(\Omega); \|u\|_{W^{m,p}}<\infty}\right\}}^{\|\cdot\|}$$

share|improve this question
2  
The grammatical version of "closed respect a norm $\|\cdot\|$" is closed with respect to the norm $\|\cdot\|$" –  rschwieb Nov 7 '12 at 14:00

2 Answers 2

up vote 4 down vote accepted

It means that in the topology generated by this norm, the set is closed. Since norms are metrics, this means that all the $\|\cdot\|$-convergent sequences from this set have limits inside this set as well.

share|improve this answer
    
I will be able to write $$H^{m,p}(\Omega)= \overline{\left\{{u\in C^m(\Omega); \|u\|_{NORM1}<\infty}\right\}}^{NORM2}$$ You saw that differents norms? –  user46060 Nov 7 '12 at 14:08
    
@user46060: I just saw now that those are different norms. That doesn't matter, though. This is just a definition of a particular set, and you claim that it is closed under $NORM_2$, meaning all $NORM_2$-convergent sequences find their limit within this set. –  Asaf Karagila Nov 7 '12 at 14:10

It means that you take the closure of $\left\{{u\in C^m(\Omega); \|u\|_{W^{m,p}}<\infty}\right\}$ with respect to $\| \cdot \|$. The closure is the set itself together with all its limit points. And a limit point is a point such that every $\varepsilon$-ball around it has non-empty intersection with the set.

share|improve this answer

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.