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 hausdorff-dimension
Search options answers only
not deleted
user 44463
Questions about dimensions of possibly highly irregular or "rough" sets, Hausdorff–Besicovitch dimension and related concepts such as box-counting or Minkowski–Bouligand dimension.
14
votes
Accepted
Existence of subset with given Hausdorff dimension
First of all, $\dim_{H} (A) = \alpha$ iff $ H^k(A)=\infty$ for all $k<\beta$ and $H^k(A) = 0$ for all $k>\beta$. Then $H^\alpha(A) = \infty$ for all $\alpha \in (0,\beta)$.
If $A$ is closed then by …
- 1,277
2
votes
Accepted
Hausdorff dimension and $W^{1,1}$ functions
If $f$ is Lipschitz then $\dim_H f(A) \le \dim_H A$, since $H^s_\delta(f(A)) \le Lip(f) \cdot H^s_\delta(A)$ for any $\delta>0$.
However there exists an absolutely continuous function $f\in W^{1,1}$ …
- 1,277