Closed respect a norm $\|\cdot\|$

Question

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

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

The grammatical version of "closed respect a norm $|\cdot|$" is closed with respect to the norm $|\cdot|$" — rschwieb, Nov 07 '12 at 14:00

Asaf Karagila · Accepted Answer · 2012-11-07T14:10:41.257

6

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.

edited Nov 07 '12 at 14:10

answered Nov 07 '12 at 13:56

Asaf Karagila

393,674

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?
– Nov 07 '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 07 '12 at 14:10

score 2 · Answer 2 · answered Nov 07 '12 at 14:01

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.

Closed respect a norm $\|\cdot\|$

2 Answers2