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 real-analysis
Search options not deleted
user 75500
Real-valued functions of real variable, analytic properties of functions and sequences, limits, continuity, smoothness of these.
2
votes
1
answer
232
views
Can a weighted $\ell^p$ norm be bounded by an unweighted $\ell^q$ norm?
For any sequence $\omega\in[1, \infty)^{\mathbb{N}}$, define the weighted $\ell^p_\omega$-norm of the sequence $v$ by
$$\Vert v\Vert_{\ell^p_\omega} := \left(\sum_{k=1}^\infty \omega_k^p |v|_k^p\right …
2
votes
Accepted
Can a weighted $\ell^p$ norm be bounded by an unweighted $\ell^q$ norm?
The answer to both questions is no.
Taking $q=1$, $p=2$, $v_k = k^{-(1+\varepsilon)}$ and $\omega_k = k^{(1+3/2\varepsilon)/2}$ for some $\varepsilon>0$ provides a counter-example to the first assert …