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 16702
A Banach space is a complete normed vector space: A vector space equipped with a norm such that every Cauchy sequence converges.
4
votes
0
answers
210
views
"Cyclic vector" of sequence of operators
I recently encountered the following somewhat random-looking problem in my research. At first I thought that should not be too hard, but now, the more I think about it, the more interesting it seems.
…
- 9,853
10
votes
1
answer
359
views
Group of isometries of Banach spaces a topological group?
Let $X$ be a Banach space and let $\mathrm{Iso}(X)$ be its group of isometries, i.e., the set of surjective linear maps $T: X \to X$ with $\|Tx\| = \|x\|$.
Q: Is $\mathrm{Iso}(X)$ a topological group …
- 18.7k
1
vote
1
answer
184
views
Special kind of operators
Consider an operator $A: H \longrightarrow X$ ($H$ is a Hilbert space and $X$ is a Banach space) that has a representation
$$ A = \sum_{j=0}^\infty a_j \langle \cdot, e_j\rangle \cdot x_j,$$
where $(e …
- 25.3k
1
vote
Can a self-adjoint operator have a continuous set of eigenvalues?
The resolvent set is the set of all $\zeta \in \mathbb{C}$ for which $T-\zeta$ is invertible (which means especially that the Range is all of $H$). The spectrum $\Sigma$ is the complement of the resol …
- 9,853