bio | website | |
---|---|---|
location | ||
age | ||
visits | member for | 1 year, 7 months |
seen | Feb 9 '14 at 8:40 | |
stats | profile views | 63 |
Feb 2 |
accepted | Riesz potential inequality |
Feb 2 |
asked | Riesz potential inequality |
Nov 9 |
accepted | Integral and conformal mappings II |
Nov 8 |
comment |
Integral and conformal mappings II
Probably it is a correct construction. |
Nov 8 |
revised |
Integral and conformal mappings II
added 109 characters in body |
Nov 8 |
comment |
Integral and conformal mappings II
If $D_n$ is smooth (for example $C^2$), then $|f'(z)|\le C_n$, so why the integral diverges? I am assuming that $D_n$ are images of $n/(n+1) D$ under a $C^2-$$K$ q.c. diffeomorphic mapping of the unit disk onto itself. |
Nov 8 |
revised |
Integral and conformal mappings II
added 7 characters in body |
Nov 8 |
asked | Integral and conformal mappings II |
Nov 8 |
accepted | Uniform convergence of conformal mappings |
Nov 8 |
comment |
Uniform convergence of conformal mappings
Yes you right, I understand the point. Thanks. |
Nov 7 |
awarded | Commentator |
Nov 7 |
revised |
Uniform convergence of conformal mappings
deleted 41 characters in body |
Nov 7 |
revised |
Uniform convergence of conformal mappings
added 46 characters in body |
Nov 7 |
asked | Uniform convergence of conformal mappings |
Oct 2 |
accepted | Holder class of analytic functions |
Oct 1 |
comment |
Holder class of analytic functions
Yes (n1) means nontangential! |
Oct 1 |
comment |
Holder class of analytic functions
&Koushik: It is related to little Bloch space. |
Oct 1 |
comment |
Holder class of analytic functions
No, when I said $|z|\to 1$ uniformly I had in mind that $z\to e^{it}$ for some $t$ and throughout the unit disk. Nontangentialy means that $z$ also tends to $e^{it}$ but inside an fixed angle. |
Oct 1 |
revised |
Holder class of analytic functions
deleted 2 characters in body |
Oct 1 |
asked | Holder class of analytic functions |