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Topological entropy for divergence points

Published online by Cambridge University Press:  19 May 2005

CHEN ERCAI
Affiliation:
Department of Mathematics, Nanjing Normal University, Nanjing 210097, People's Republic of China (e-mail: ecchen@njnu.edu.cn, shulin003@sohu.com)
TASSILO KÜPPER
Affiliation:
Mathematical Institute, University of Cologne, D-50931 Cologne, Germany (e-mail: kuepper@mi.uni-koeln.de)
SHU LIN
Affiliation:
Department of Mathematics, Nanjing Normal University, Nanjing 210097, People's Republic of China (e-mail: ecchen@njnu.edu.cn, shulin003@sohu.com)
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Abstract

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Let X be a compact metric space, f a continuous transformation on X, and Y a vector space with linear compatible metric. Denote by M(X) the collection of all the probability measures on X. For a positive integer n, define the nth empirical measure $L_{n}:X\mapsto M(X)$ as

\[L_{n}x=\frac{1}{n}\sum _{k=0}^{n-1}\delta_{f^{k}x},\]

where $\delta_{x}$ denotes the Dirac measure at x. Suppose $\Xi:M(X)\mapsto Y$ is continuous and affine with respect to the weak topology on M(X). We think of the composite

\[\Xi\circ L_{n}:X \xrightarrow{L_n} M(X)\xrightarrow{\Xi} Y\]

as a continuous and affine deformation of the empirical measure Ln. The set of divergence points of such a deformation is defined as

\[D(f,\Xi)=\{x\in X\mid \mbox{the limit of }\Xi L_n x \mbox{ does not exist}\}.\]

In this paper we show that for a continuous transformation satisfying the specification property, if $\Xi(M(X))$ is a singleton, then set of divergence points is empty, i.e. <formula form="inline" disc="math" id="ffm008"><formtex notation="AMSTeX">$D(f,\Xi)=\emptyset$, and if $\Xi(M(X))$ is not a singleton, then the set of divergence points has full topological entropy, i.e.

\[h_{\rm top}(D(f,\Xi))=h_{\rm top}(f).\]

Information

Type
Research Article
Copyright
2005 Cambridge University Press