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.
Best Answer
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.