Simple proporties

Let X be a compact Hausdorff space, \mathcal{B}(X) be Borel \sigma-algebra, and \mathcal{M}(X) be the collection of Borel Probability measures on X. This again is a compact Hausdorff space.

Fo r each B\in\mathcal{B}(X) there exists a G_\delta set \tilde{B}\supset B such that \mu(\tilde{B}\backslash B)=0 for every \mu\in\mathcal{M}(X).

If \mu_k,\mu\in\mathcal{M}(X), then \mu_k\rightarrow\mu in the weak*-topolgy is equivalent to either one of the following:

For each closed subset F\subseteq X, \limsup\mu_k(F)\le\mu(F).

For each open subset U\subseteq X, \liminf\mu_k(U)\ge\mu(U).

For every A\subseteq E with \mu(\partial A)=0, \mu_k(A)\rightarrow\mu(A).

If \{x\in X:f(x)>c\} is an open set in X for all c\in\mathbb{R}, then f is said to be lower semicontinuous. If \{x\in X:f(x)<c\} is an open set in X for all c\in\mathbb{R}, then f is said to be upper semicontinuous. If f_\lambda are a family of l.s.c. (or g_\lambda u.s.c.), then so is \sup_{\lambda}f_\lambda (or \inf_{\lambda}g_\lambda) since

\{x\in X:\sup_{\lambda}f_\lambda(x)>c\}=\bigcup_{i\ge1}\bigcup_{\lambda}\{x:f_\lambda(x)>c-1/i\}.
\{x\in X:\inf_{\lambda}g_\lambda(x)<c\}=\bigcup_{i\ge1}\bigcup_{\lambda}\{x:g_\lambda(x)<c-1/i\}.

The Lyapunov exponent of a smooth diffeo f:M\rightarrow M is defined as

\chi(x,v)=\limsup\frac{1}{n}\log\|D_xf^n(v)\| for each v\in T_x M and x\in M. It behaves quite good in the measure sense. But in general there is no more information about its topological properties. For examle it may not be semicontinuous.

Let f:M\to M be a diffeo and \Lambda be a compact f-invariant set. Denote
D^s(x) =\{v\in T_xM|\,\|Df^nv\|\to0\text{ as }n\to+\infty\},
D^u(x) =\{v\in T_xM|\,\|Df^{-n}v\|\to0\text{ as }n\to+\infty\}.

These are Df-invariant linear subspaces of T_xM, and vectors of D^s and D^u are asymptotic to zero, but not necessarily exponentially fast. However, if the two subspaces form a direct sum at every point of \Lambda, exponential
convergence will follow:
Proposition (Mane, Liao). \Lambda is hyperbolic if and only if D^s(x) \oplus D^u(x) = T_xM for all x\in\Lambda.

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  • Pengfei  On February 13, 2010 at 10:38 pm

    I was confused every time when came up with things like this.

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