The [Fibonacci word](http://en.wikipedia.org/wiki/Fibonacci_word) is the limit of the sequence of words starting with $0$ and satisfying rules $0 \to 01, 1 \to 0$. Equivalently, it is obtained from the recursion $S\_n= S\_{n-1}S\_{n-2}$ under initial conditions $S\_0 = 0, S\_1 = 01$. Let us denote this Fibonacci word as $f\_k\in\{0,1\}^{\mathbb Z\_+}$.


We define the Fibonacci subshift $\Omega$ as the set of all two-sided right limits of the Fibonacci word. More precisely, we extend our Fibonacci word to $\{0,1\}^{\mathbb Z}$ by defining $\{f\_j\}\_{j=-\infty}^0$ arbitrarily, and define $(\mathcal Tf)\_k=f\_{k+1}$. The Fibonacci subshift $\Omega$ is then the set of limit points of $\{\mathcal T^n f\}\_{n=0}^\infty$ under the product topology on $\{0,1\}^{\mathbb{Z}}$.


It is easy to show that the ratio of zeroes to ones in the Fibonacci word is $\varphi$, the golden ratio. Equivalently, the density of ones in the Fibonacci word (that is, the ratio of the number of ones in a subword to the length of the subword) tends to $\frac{1}{\varphi^2}$. I would like to know if this is true uniformly. That is,



> 
> Is the following true for a.e. $\omega\in\Omega$? Given any
> $c\in\mathbb Z$, $$\lim\_{m\to\infty}
> \sum\_{j=c-m}^{c+m}\frac{\omega\_j}{2m+1}=\frac{1}{\varphi^2} $$
> uniformly in $c$.
> 
> 
> 


I also suspect this might be true for every $\omega\in\Omega$, not just a.e.



Is the density of 1's in the Fibonacci word uniform?
Yes. By Proposition 2.1.10 in [Lothaire, Algebraic Combinatorics on Words](http://tomlr.free.fr/Math%E9matiques/Fichiers%20Claude/Auteurs/aaaDivers/Lothaire%20-%20Algebraic%20Combinatorics%20On%20Words.pdf), if $u$ is any substring of the Fibonacci word then 
$$\left| \frac{\mbox{number of $1$'s in $u$}}{\mbox{length of $u$}} - \frac{1}{\phi^2} \right| \leq \frac{1}{\mbox{length of $u$}}.$$
Any length $n$ substring of any $\omega \in \Omega$ is also a length $n$ substring of the Fibonacci word, so the same bound applies to it.


