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Let $\{a_k\}_{k\in \mathbb{Z}} \subset \mathbb{R}$ a real sequence and $a\in \mathbb{R}$ such that $$ \lim_{n\to +\infty} \frac{1}{n} \sum_{k=1}^n a_k = a = \lim_{n\to +\infty} \frac{1}{n+1} \sum_{k=0}^n a_{-k},$$ for all $j\in \mathbb{Z}$, the same is true for $\{a_{k+j}\}_{k\in \mathbb{Z}} \subset \mathbb{R}$. Therefore for all $\epsilon>0$ and $j\in \mathbb{Z}$ we can find a $N_{\epsilon,j}\in \mathbb{N}$ such that for all $n\geq N_{\epsilon,j}$, $$\left|\frac{1}{n} \sum_{k=1}^n a_{k+j}-a\right|< \epsilon \; \text{ and } \;\left |\frac{1}{n+1} \sum_{k=0}^n a_{-k+j}-a\right|< \epsilon.$$ If we add as an additional condition that the sequence $\{a_k\}_{k\in \mathbb{Z}}$ is bounded, I would like to show that for all $\epsilon >0$ there exists $N_\epsilon \in \mathbb{N}$ such that for all $j \in \mathbb{Z}$ and $n \geq N_\epsilon$, $$ \left|\frac{1}{n} \sum_{k=1}^n a_{k+j}-a\right| < \epsilon \; \text{ and } \; \left|\frac{1}{n+1} \sum_{k=0}^n a_{-k+j}-a\right|< \epsilon.$$ Without this additional condition, I can show that $N_{\epsilon,j}$ is proportional to $|j|$ for $|j|$ large enough. The sequence $1,-1,\sqrt{3},-\sqrt{3},\sqrt{5},-\sqrt{5},\sqrt{7},-\sqrt{7},\ldots$ gives us an example where the $N_{\epsilon,j}$ are not bounded. On the other hand, it's easy to show that if $\lim_{n \to +\infty} a_k = a = \lim_{k \to +\infty} a_{-k}$ then the $N_{\epsilon,j}$ can be bound.

My intuition is that for the $N_{\epsilon,j}$ to explode when we start the average further into the sequence we need the terms in the sequence to be larger and larger.

I can't find a counter-example with this additional condition, nor can I use this condition to prove what I'm looking for. I've already asked this question to quite a few people and nobody seems to agree whether the statement I'm trying to demonstrate seems correct or not.

If anyone has an idea of a direction to take or finds a counter-example, it will be more than welcome. Thank you very much!

Originally, my question concerned the Birkhoff average for a smooth function $h$ on a closed manifold $M$ with a diffeomorphism $\varphi$. If the Birkhoff average exists at a point $x\in M$ is there a sufficiently large $N$ so that for all $j\in \mathbb{Z}$ and $n\geq N$, $\frac{1}{n}\sum_{k=1}^n h\circ \varphi^{k+j}(x)$ is close enough to the Birkhoff average.

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2 Answers 2

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The sequence $1,0,1,1,0,0,1,1,1,0,0,0,\ldots$ is a counterexample. For each $j$ we have $\frac{1}{n}\sum_{k=1}^n a_{k+j} \to \frac{1}{2}$, but for any proposed $N_\epsilon$ we can find a value of $j$ with $a_{j+1}=a_{j+2}=\dots = a_{j+N_\epsilon} =1$, so that $\frac{1}{N_\epsilon}\sum_{k=1}^{N_\epsilon} a_{k+j} = 1$.

Not sure why people are voting to close, this wasn't obvious to me.

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    $\begingroup$ I'd tried several 0 and 1 sequences but I hadn't thought of that one, thank you very much @NikWeaver ! $\endgroup$
    – Pac's
    Feb 22 at 12:39
  • $\begingroup$ You are welcome! $\endgroup$
    – Nik Weaver
    Feb 22 at 12:54
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Regarding your original question about Birkhoff averages, the story is the following: Suppose $X$ is a compact metric space, $T\colon X\to X$ is continuous, and $f\colon X\to \mathbb{R}$ is continuous. Given $x\in X$, let $Y_x = \overline{\{T^n x : n\geq 0\}}$, and let $\mathcal{M}_x$ denote the set of $T$-invariant Borel probability measures on $Y_x$. Then $\mathcal{M}_x$ is a weak* compact convex set, and thus $I_x = \{ \int f \,d\mu : \mu \in \mathcal{M}_x \}$ is a closed interval $[a_x,b_x]$. Now exactly one of the following two cases occurs.

  1. $a_x=b_x$, in which case the answer to your question is "yes". Indeed, in this case the functions $A_nf := \frac 1n \sum_{k=0}^{n-1} f\circ T^k$ converge uniformly to $a_x = b_x$ on $Y_x$; to prove this, suppose that for some $\epsilon>0$ there are $n_k \to \infty$ and $y_k \in Y_x$ such that $| A_{n_k} f(y_k) - a_x | \geq \epsilon$ for all $k$, and consider the measures $\mu_k = \sum_{i=0}^{n_k - 1} \delta_{T^i y_k}$. Any weak*-limit point $\mu$ of this sequence satisfies $\mu \in \mathcal{M}_x$ and $|\int f \,d\mu - a_x | \geq \epsilon$, so $\int f \,d\mu \notin I_x$, a contradiction.
  2. $a_x<b_x$, in which case the answer to your question is "no". To prove this, start by observing that in this case there are ergodic measures $\mu,\nu \in \mathcal{M}_x$ and $\epsilon>0$ such that $\int f \,d\mu + 3\epsilon < \int f \,d\nu - 3\epsilon$. Choosing $y,z \in Y_x$ to be generic points for $\mu,\nu$ respectively, we see that there is $N$ such that for every $n\geq N$, we have $A_nf(y) < \int f \,d\mu + \epsilon$ and $A_nf(z) > \int f \,d\nu - \epsilon$, so in particular $A_nf(z) > A_nf(y) + 4\epsilon$. Now by the definition of $Y_x$ and the continuity of $A_n f$, there exist $j,k$ such that $A_nf(T^jx) < A_nf(y) + \epsilon$ and $A_nf(T^kx) > A_nf(z) - \epsilon$. This implies that $A_nf(T^jx) - A_nf(T^kx) > 2\epsilon$. Since such a $j,k$ can be found for every $n$, the "convergence independent of $j$" that your question asked about cannot happen.

It is worth pointing out that both cases occur. The first case occurs whenever $f$ is cohomologous to a constant ($f = c + g\circ T - g$ for some $c\in \mathbb{R}$ and $g\in C(X)$), or whenever $(Y_x, T)$ is uniquely ergodic (admits exactly one invariant probability measure). The second case occurs if $X$ is a locally maximal hyperbolic set (or a subshift of finite type) and $f$ is not cohomologous to a constant.

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