Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
Analytic capacity
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
===Positive length but zero analytic capacity=== Given the partial correspondence between the 1-dimensional Hausdorff measure of a compact subset of '''C''' and its analytic capacity, it might be conjectured that ''Ξ³''(''K'') = 0 implies ''H''<sup>1</sup>(''K'') = 0. However, this conjecture is false. A counterexample was first given by [[Anatoli Georgievich Vitushkin|A. G. Vitushkin]], and a much simpler one by [[John B. Garnett]] in his 1970 paper. This latter example is the '''linear four corners Cantor set''', constructed as follows: Let ''K''<sub>0</sub> := [0, 1] Γ [0, 1] be the unit square. Then, ''K''<sub>1</sub> is the union of 4 squares of side length 1/4 and these squares are located in the corners of ''K''<sub>0</sub>. In general, ''K<sub>n</sub>'' is the union of 4<sup>''n''</sup> squares (denoted by <math>Q_n^j</math>) of side length 4<sup>β''n''</sup>, each <math>Q_n^j</math> being in the corner of some <math>Q_{n-1}^k</math>. Take ''K'' to be the intersection of all ''K''<sub>''n''</sub> then <math>H^1(K)=\sqrt{2}</math> but ''Ξ³''(''K'') = 0.
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)