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 51280
Real-valued functions of real variable, analytic properties of functions and sequences, limits, continuity, smoothness of these.
0
votes
1
answer
164
views
Uniform boundedness in $L^1[0,1]$ implies finite $\limsup$ almost everywhere for a subsequence? [closed]
Given a sequence of functions $f_k \in L^1([0,1])$ such that $||f_k||_{L^1(0,1)}\leq C$.
Is there a subsequence $\{k_l\,|\,l\in \mathbb N\}\subseteq \mathbb{N}$ such that for $\mathcal{L}^1$-almost …
- 11