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seen Feb 9 at 8:40

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