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 multivariable-calculus
Search options answers only
not deleted
user 10265
3
votes
Accepted
Strong convexity inequality w.r.t. infinity norm $\lVert\cdot\rVert_{\infty}$
I think it is not possible to do much better than $\frac{1}{\sqrt n}$. More precisely, I believe the best $\alpha$ is $\frac{1}{\sqrt n}$ whenever $n$ is a power of $2$, and therefore (since $\alpha$ …
- 9,486