Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This is a question on the proof of this fact in chapter 3 of the book "Functional Analysis: Surveys and Recent Results II". There at the end the proof is outlined as follows:

Let $\mu$ be a shift invariant measure on $X=\{0,1\}^\mathbb{Z}$. For each subset $\Lambda\subset\mathbb{Z}$ define $C_\Lambda$ to be the subset of continuous functions on $X$, $C_\Lambda\subset C(X)$ which depend only on coordinates indexed by $\Lambda$. Let $\mu_\Lambda$ be the restriction of $\mu$ to $C_\Lambda$.

Now let $\Lambda_n=[1,n]$ and define $\nu_n=\otimes_{k\in\mathbb{Z}}\mu_{shift^{kn}(\Lambda_n)}$. Now define $$\mu_n=\frac{1}{n}\sum_{s=0}^{n-1}shift^{s}_*(\nu_n).$$

Then $\nu_n$ and $\mu_n$ are ergodic. Also $\mu_n=\nu_n=\mu$ on $C_{\Lambda_n}$, hence $\mu_n\to\mu$.

I do not understand precisely this last statement, that $\mu_n=\mu$ on $C_{\Lambda_n}$.

I don't know how to show it even for $n=2$: Since $C_{\Lambda_2}=C_{[1]}\otimes C_{[2]}$, any element of $C_{\Lambda_2}$ is a linear combination of functions $f(\omega_1)g(\omega_2)$. So if $\mu_2=\nu_2$ one must have $$\int f(\omega_1)g(\omega_2) d\mu_{\Lambda_2}=\int f(\omega_1)d\mu_{[1]}\int g(\omega_2)d\mu_{[2]},$$ which I don't see why should hold for any $\mu$.

share|improve this question
    
In your second paragraph, $\mu$ should be "a" shift-invariant measure, not "the" shift-invariant measure, since there are very many such measures –  Vaughn Climenhaga May 12 '12 at 16:17
    
You mention that this is from a book, and you give the title - could you also give the author(s) to make the reference more complete? –  Vaughn Climenhaga May 12 '12 at 16:18
    
It doesn't directly answer your specific question, but the fact that the simplex of invariant measures is Poulsen comes up (together with most of the proof) in the discussion on this other question: mathoverflow.net/questions/83981/… –  Vaughn Climenhaga May 12 '12 at 16:21
    
@Vaughn This is actually a collection of survey articles. Chapter 3 from which I cite the proof is written by G.H. Olsen. The full data from mathscinet is MR0565394 (81a:46004) Functional analysis: surveys and recent results. II. Proceedings of the Second Conference on Functional Analysis held at the University of Paderborn, Paderborn, January 31–February 4, 1979. Edited by Klaus-Dieter Bierstedt and Benno Fuchssteiner –  user23644 May 12 '12 at 17:37
    
@Vaughn Yes, I have seen that question; I also know some other proofs of the fact, which I do understand. However, I am interested in this particular proof. –  user23644 May 12 '12 at 17:42
show 1 more comment

1 Answer

up vote 2 down vote accepted

The language of the proof given in the book you refer to is a little different from the language I'm accustomed to, but I'll give what I believe is the exact same argument using a slightly different language, and hopefully do it in such a manner that the issue you point out doesn't arise. (Since I'm not quite sure how to explain it away using a language that's less familiar to me.)

Let $X = \{0,1\}^\mathbb{Z}$ with $\sigma\colon X\to X$ the shift map, and let $\mu$ be any $\sigma$-invariant probability measure on $X$. Given $n\in \mathbb{N}$, let $\nu_n$ be the Bernoulli measure for $\sigma^n$ that best approximates $\mu$, and let $\mu_n$ be the invariant measure generated by $\nu_n$.

More precisely, $\nu_n$ is defined as follows. Let $Y = \{0,1,\dots,2^n-1\}^\mathbb{Z}$, with $\tau\colon Y\to Y$ the shift map, and define a homeomorphism $\pi\colon Y \to X$ by identifying symbols in the alphabet of $Y$ with $n$-words in $X$: if $\phi\colon \{0,1,\dots,2^n-1\}\to \{0,1\}^n$ is a bijection, then we put $\pi(y) = \dots \phi(y_{-1}).\phi(y_0)\phi(y_1)\dots$, where juxtaposition denotes concatenation. Note that $\pi$ conjugates $\tau$ to $\sigma^n$ via $\pi\circ \tau = \sigma^n\circ \pi$.

Define a measure $\mu^\*$ on $Y$ by $\mu^\*(E) = \mu(\pi E)$. Now define a $\tau$-invariant measure $\nu$ on $Y$ by putting $\nu([y_1\dots y_k]) = \prod_{j=1}^k \mu^\*([y_j])$. This is the Bernoulli measure that best approximates $\mu^\*$. Define $\nu_n$ on $X$ by $\nu_n(E) = \nu(\pi^{-1}E)$. Then $\nu_n$ is a Bernoulli measure for $\sigma^n$ with the property that $\nu_n([x_1\dots x_n]) = \mu([x_1\dots x_n])$ for every $n$-cylinder, but $\nu_n$ is not $\sigma$-invariant.

To rectify this, let $\mu_n = \frac 1n \sum_{k=0}^{n-1} \sigma_*^k \nu_n$. Then $\mu_n$ is $\sigma$-invariant, and moreover, since for every $n$-cylinder $C$ the $\sigma$-invariance of $\mu$ gives $(\sigma_*^k \nu_n)(C) = (\sigma_*^k \mu)(C) = \mu(C)$, we have $\mu_n(C) = \mu(C)$.

I believe that this last set of equalities (the fact that $\nu_n$, $\mu_n$, and $\mu$ agree on $n$-cylinders) is the statement you wanted explained. It seems to me that agreement on $C_{\Lambda_n}$ (in the language of the book) corresponds to agreement on $n$-cylinders (in the language here). In any case, the measure $\mu_n$ is ergodic and $\sigma$-invariant, and approaches $\mu$ as $n\to\infty$, which shows that the space of $\sigma$-invariant measures is the Poulsen simplex.

share|improve this answer
    
@Vaughn Thank you! It is indeed the same construction, just interpreted directly as measures rather then functionals. The point that confused me was exactly the equality of $\nu_n$ and $\mu$ on every cylinder, but with this interpretation it is much easier to check. –  user23644 May 14 '12 at 8:16
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.