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bio website pcwww.liv.ac.uk/~lrempe
location Liverpool
age 37
visits member for 5 years, 3 months
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Professor of Pure Mathematics at the University of Liverpool

15h
revised Continuous extension of Riemann maps and the Caratheodory-Torhorst 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 Caratheodory-Torhorst 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 Caratheodory-Torhorst 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 Non-bijective 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 Gauss--Lucas 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 near-zeros
Feb
16
answered Palis' conjecture and Newhouse's results
Jan
30
awarded  Yearling