Uniformity of convergence in the pointwise ergodic theorem

Question

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 $\frac{0}{0} = 0$.

Answer 1

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 $(n_k+1)E_{n_k+1}(\frac 14,\frac1{100})\ge 1-\frac 1{100}$.