 By G. Viennot

ISBN-10: 0387090908

ISBN-13: 9780387090900

ISBN-10: 3540090908

ISBN-13: 9783540090908

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4. If (X, d) is a locally compact separable metric space, then the vector space of all Lipschitz functions with compact support is dense in Cc (X) and in C0 (X) (with respect to the topology generated by the uniform norm, of course). Consequently, Lip(X) is dense in C0 (X). Proof. 3. Since the sum or the product of two Lipschitz functions is Lipschitz, it follows that all the functions in A have compact supports and are Lipschitz. Since A is dense in C0 (X), it follows that the assertions of the corollary are true.

L) For every x ∈ Γ and every Banach limit L let εx : C0 (X) → R be deﬁned (L) by εx (f ) = L ( f, T n δx )n∈N∪{0} for every f ∈ C0 (X). It is easy to see that (L) (L) εx is a positive linear functional, so εx is also continuous. Accordingly, we may (L) and do think of εx as an element of M(X). 1. If x ∈ Γ and L is a Banach limit, then εx measure. is a T -invariant Proof. Let φ : Cb (X) → R be deﬁned by φ(f ) = L ( f, T n δx )n∈N∪{0} for every f ∈ Cb (X). Clearly, φ is a positive linear functional, φ (1X ) = 1, and the restriction (L) of φ to C0 (X) is εx .

2. Invariant Probabilities 21 The next theorem summarizes various results of Chapter 4 of Revuz’s book . 6. Let (S, T ) be a Markov–Feller pair deﬁned on a locally compact separable metric space (X, d), and let µ ∈ M(X) be a T -invariant probability. e. to a µ-integrable k=0 function g, and f, µ = n∈N g dµ. Proof. It is well-known that the Radon–Nikodym theorem implies that the set of all the elements of M(X) that are absolutely continuous with respect to the measure µ is a Banach subspace of M(X) that is isometric to L1 (X, B(X), µ).