bio  website  pcwww.liv.ac.uk/~lrempe 

location  Liverpool  
age  37  
visits  member for  5 years, 3 months 
seen  1 hour ago  
stats  profile views  982 
Professor of Pure Mathematics at the University of Liverpool
15h

revised 
Continuous extension of Riemann maps and the CaratheodoryTorhorst Theorem
clarified what is and is not statedin Torhorst's paper. Also added link for Wilder's paper. 
2d

comment 
Continuous extension of Riemann maps and the CaratheodoryTorhorst Theorem
Not sure if Torhorst explicitly mentions continuous extension but this is clearly understood. Will try to clarify in question when I get the chance. 
2d

comment 
Continuous extension of Riemann maps and the CaratheodoryTorhorst Theorem
Thanks for the interest. Posting from mobile so apologies for typos. 1. Carathéodory considered SIMPLE closed curves, not general curves. 2. Carathéodory proves continuous extension iff all prime ends of first kind. Torhorst proves and states all prime ends of first kind iff lc. 
2d

awarded  Promoter 
2d

answered  Nonbijective conformal maps between annuli 
2d

revised 
Conformal map and Jordan curve
Provided further details for the argument and adjusted notation. 
2d

comment 
Conformal map and Jordan curve
No. If you look carefully, the curve $\gamma$ itself is analytic. I shall see whether I can clarify the answer. 
2d

comment 
A Generalization of growth exponents
@Catman You might wish to post another question, giving all the details. (You could then post a link here if you wish.) 
2d

answered  Conformal map and Jordan curve 
May 20 
answered  A Generalization of growth exponents 
May 19 
comment 
A Generalization of growth exponents
This is clearly not continuous, unless I am missing something. E.g. consider $f(s,a):= a\cdot s^2$, at $a=0$. 
May 5 
comment 
GaussLucas type theorem for tracts and higher derivatives of a polynomial
Nice example  how did you go about finding it? 
May 5 
comment 
is there a diffeomorphism with only finite orbits but of infinite order?
@asafshachar  it seems to me that if the set of periodic points has interior, then this interior is invariant under the map, and every connected component is obviously periodic. Hence it follows from the result cited by Igor that the map has finite order on each such component. 
Apr 23 
answered  Extension of conformal map and annulus 
Apr 23 
comment 
Extension of conformal map and annulus
Hi Neil, the answer is correct when the question is taken literally, i.e. "circle" actually means "round circle". 
Apr 20 
awarded  Nice Answer 
Apr 13 
answered  Which way for reading the proofs? 
Apr 7 
answered  Generalized Schwarz Lemma for nearzeros 
Feb 16 
answered  Palis' conjecture and Newhouse's results 
Jan 30 
awarded  Yearling 