# If the average of a sequence converges, can I find a uniform bound that does not depend on where I start?

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.

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.

• I'd tried several 0 and 1 sequences but I hadn't thought of that one, thank you very much @NikWeaver ! Feb 22 at 12:39
• You are welcome! Feb 22 at 12:54

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. $$a_x, in which case the answer to your question is "no". To prove this, start by observing that in this case there are ergodic measures $$\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.