Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with banach-spaces
Search options not deleted
user 54609
A Banach space is a complete normed vector space: A vector space equipped with a norm such that every Cauchy sequence converges.
4
votes
Accepted
Corson-Lindenstrauss : Weakly compact sets as intersection of finite unions of cells
Let $x_n=-\frac{\sum_{k=1}^n e_k}{n}$. Then $\|e_i-x_n\|^2=\frac{n+3}{n}$ if $i \leq n$ and $\|e_i-x_n\|^2=\frac{n+1}{n}$ if $i>n$. So your set is the intersection of the sets $B(x_n,\sqrt{\frac{n+1}{ …
- 716
4
votes
relation between of uniformly rotund in every direction and uniformly rotund and locally uni...
First, the fact that UR implies URED is obvious (the definition of UR requires less from the sequences $(x_n)$, $(y_n)$ in order to conclude the convergence $\|x_n-y_n\|\to 0$.)
Second, the notions L …
- 716
1
vote
Subspaces of $l_{1}$ are not Lipschitz complemented in $l_{1}$
This is not an answer, just a related question: In the particular case when $U$ is complemented in its bidual, you have that $U$ is Lipschitz complemented in $\ell_1$ if and only if $U$ is complemente …
- 716