**Definitions and some motivation:**


Let $X$ be a compact metric space, and $T$ a uniquely ergodic measure preserving transformation on $X$, with associated invariant ergodic probability measure $\mu$. Assume $\mu$ is non atomic and supp $\mu = X$.


Given a continuous function $f$ on $X$, we know by unique ergodicity that the Birkhoff averages $A\_n f := \frac{1}{n}\sum\_{k=0}^{n-1} T^k f$ converge uniformly to the constant function $Cf := \int\_X f d\mu$. But how uniform is the convergence with respect to the convergence at other points?


Given a positive real valued continuous function $f$ on $X$, and $n \in \mathbb N$, define the error function $E\_n: X \times \mathbb R^+ \to \mathbb R$ by


$E\_n(x, r) := \frac{1}{\mu(B\_r (x))} \int\_{B\_r (x)} |A\_n f - Cf| d\mu$.


Define also for each $\delta > 0$, the set $S\_\delta := \{ (x, r) | \ (x, r) \in X \times \mathbb R^+, \ \mu (B\_r (x)) \geq \delta \}$.


**Question:** For fixed positive real valued continuous $f$, is it true that for all $\delta > 0$, we have $\
\limsup\_{n \to \infty} \sup\_{(x\_1, r\_1), (x\_2, r\_2) \in S\_\delta} \frac{E\_n (x\_1, r\_1) - E\_n (x\_2, r\_2) }{E\_n (x\_1, r\_1) + E\_n (x\_2, r\_2)} = 0$?


*Note: By convention we set $\frac{0}{0} = 0$.*



Uniformity of convergence in the pointwise ergodic theorem
No. This is too much to ask for. For a counterexample, let $T$ be an irrational rotation of the circle (which I think of as $[0,1)$) and let $f(x)=\sin(2\pi x)$. Let $n\_k$ be a sequence of integers such that $d(n\_k\alpha,\mathbb Z)\to 0$. Clearly $Cf=0$ (you can add 2 to $f$ if you care about it being strictly positive, but it makes no difference whatsoever to the question). On the other hand, $(n\_k+1)A\_{n\_k+1}f(x)\to f(x)$ by summing a geometric series (the first $n\_k$ terms almost perfectly cancel just leaving the last term). In particular, considering $(0,\frac1{100})$ and $(\frac 14,\frac1{100})$, both elements of $S\_\frac{1}{100}$, then along this sequence of $n$'s, $(n\_k+1)E\_{n\_k+1}(0,\frac 1{100})\le 2\pi\cdot \frac{2}{200}$ (the $\frac 2{200}$ here is really just $\frac 1{100}+\epsilon$ coming from the error between $n\_k\alpha$ and 0), while $(n\_k+1)E\_{n\_k+1}(\frac 14,\frac1{100})\ge 1-\frac 1{100}$.


